+
In the case where you have two sets of data that you want to show on the same plot, you can also do that easily with the scatter function. To help make it understandable, you’ll want to add a labels and a legend. An example of how to do that is shown below.
@@ -659,7 +659,7 @@Visualizing Data
-
+
If you wanted to compare the distribution of cats and dogs along the x-axis, a histogram is probably the easiest way to do that…
@@ -679,7 +679,7 @@ Visualizing Data
-
+
If the overlap in the samples is too confusing, you can change the format of the histogram (histtype) so that it’s not filled in. Different line styles (ls) will make sure the plot is readable even if it’s printed in black and white.
@@ -698,7 +698,7 @@ Visualizing Data
-
+
Normalizing the histograms (using the keyword: density=‘True’) makes it easier to compare the shapes of distributions even when one sample is much more numerous than the other. This is calculated by dividing each bin by the total area of the histogram.
@@ -717,7 +717,7 @@ Visualizing Data
-
+
diff --git a/notebooks/chem/CO_Fundamental.html b/notebooks/chem/CO_Fundamental.html
index 5a5babd..e200fdf 100644
--- a/notebooks/chem/CO_Fundamental.html
+++ b/notebooks/chem/CO_Fundamental.html
@@ -527,63 +527,63 @@ Computing Fundamental Transition of CO under different levels of approximati
We will use the Morse potential as a model for the “exact” interatomic potential, and we will approximate this potential by different orders of a Taylor expansion: including up to quadratic (which is the harmonic oscillator approximation), cubic, and quartic terms. The harmonic and Morse potentials are exactly solvable, and the eigenfunctions and eigenvalues of the vibrational Hamiltonian with cubic and quartic potentials can be approximated using perturbation theory. Therefore, we will
compare the fundamental transition computed exactly for harmonic and Morse potentials, and approximately at 2nd order of perturbation theory for cubic and quartic potentials to see the impact of various levels of potential truncation and approximation.
Within the Morse model, the vibrational Hamiltonian can be written as
-
-(21)#\[\begin{equation}
+
+(21)#\[\begin{equation}
\hat{H}_{vib} = -\frac{\hbar^2}{2\mu} \frac{d^2}{dr^2} + V_{Morse}(r), \tag{1}
\end{equation}\]
where
-
-(22)#\[\begin{equation}
+
+(22)#\[\begin{equation}
V_{Morse}(r) = D_e \left(1 - e^{-\beta(r-r_{eq})} \right)^2. \tag{2}
\end{equation}\]
The Morse parameters for \({\rm CO}\) are as follows: \(D_e = 11.225 \: {\rm eV}\), \(r_{eq} = 1.1283 \: {\rm Ang.}\), \(\beta = 2.5944 \: {\rm Ang.}^{-1}\),
and \(\mu = 6.8606 \: {\rm amu}\).
The exact energy eigenvalues for Equation (1) can be written as
-
-(23)#\[\begin{equation}
+
+(23)#\[\begin{equation}
E_n = \hbar \omega \left( \left(n+ \frac{1}{2} \right) - \chi_e \left(n+ \frac{1}{2} \right)^2 \right) \tag{3}
\end{equation}\]
where
-
-(24)#\[\begin{equation}
+
+(24)#\[\begin{equation}
\omega = \sqrt{\frac{2D_e \beta^2}{\mu}} \tag{4}
\end{equation}\]
and
-
-(25)#\[\begin{equation}
+
+(25)#\[\begin{equation}
\chi_e = \frac{\hbar \omega}{4 D_e}. \tag{5}
\end{equation}\]
The Morse potential can be approximated by a Taylor expansion as follows:
-
-(26)#\[\begin{equation}
+
+(26)#\[\begin{equation}
V_T(r) = \sum_{n=0}^{\infty} \frac{ f^{(n)}(r_{eq})}{n!} \left(r-r_{eq} \right)^n, \tag{6}
\end{equation}\]
where \(f^{(n)}(r_{eq})\) is the \(n^{th}\)-order derivative of the Morse potential evaluated at the equilibrium bondlength, e.g. \(f^{(1)}(r_{eq}) = \frac{d}{dr}V_{Morse}(r_{eq}).\)
We will define the Harmonic approximation to the potential as
-
-(27)#\[\begin{equation}
+
+(27)#\[\begin{equation}
V_H(r) = \frac{ f^{''}(r_{eq})}{2} \left(r-r_{eq} \right)^2 = \frac{1}{2} k \left(r-r_{eq} \right)^2 \tag{7}
\end{equation}\]
the cubic approximation to the potential as
-
-(28)#\[\begin{equation}
+
+(28)#\[\begin{equation}
V_C(r) = V_H(r) + \frac{ f^{'''}(r_{eq})}{6} \left(r-r_{eq} \right)^3 = V_H(r) + \frac{1}{6} g \left(r-r_{eq} \right)^3, \tag{8}
\end{equation}\]
and the quartic approximation as
-
-(29)#\[\begin{equation}
+
+(29)#\[\begin{equation}
V_Q(r) = V_C(r) + \frac{ f^{''''}(r_{eq})}{24} \left(r-r_{eq} \right)^4 = V_C(r) + \frac{1}{24}h(r-r_{eq})^4. \tag{9}
\end{equation}\]
Because we are using the Morse model as the “exact” interatomic potential in this notebook, we can compute these derivatives at \(r_{eq}\) analytically:
-
-(30)#\[\begin{align}
+
+(30)#\[\begin{align}
k = 2 D_e \beta^2 \\
g = -6 D_e \beta^3 \tag{10} \\
h = 14 D_e \beta^4.
\end{align}\]
However, in general we do not have an analytical form for the interatomic potential, so we must rely on numerical derivatives of the potential evaluated at the \(r_{eq}\). In the context of interatomic potentials computed by quantum chemistry methods (e.g. CCSD(T)), one must first identify the equilibrium geometry, and then compute derivatives by taking a number of single point calculations along all displacement coordinates to compute differences among. We will write the explicit expression for the second derivative using centered finite differences along the one displacement coordinate relevant for our \({\rm CO}\) molecule:
-
-(31)#\[\begin{equation}
+
+(31)#\[\begin{equation}
k=\frac{V_{Morse}(r_{eq}+\Delta r)-2V_{Morse}(r_{eq})+V_{Morse}(r_{eq}-\Delta r)}{\Delta r^2}+\mathcal O (\Delta r^2) \tag{11}
\end{equation}\]
where \(\Delta r\) represents a small displacement along the coordinate \(r\). Higher-order derivatives can also be computed, but will require larger numbers of displacements and therefore more energy evaluations by your quantum chemistry method. Expressions for higher-order derivatives along a single coordinate can be found here. Note that the number of displacement coordinates \(N\) grows linearly with the number of atoms, and that the number of displacements required to form the \(n^{{\rm th}}\)-order approximation to the potential grows as \(N^n\).
@@ -591,20 +591,20 @@ Computing Fundamental Transition of CO under different levels of approximati
Perturbation Theory#
We can compute the exact vibrational transition energies for the Morse oscillator and the Harmonic oscillator using Equation (3), where the Harmonic oscillator transition energies come from Equation (3) with \(\chi_e = 0\). However, the transition energies when the potential is approximated as \(V_C(r)\) or \(V_Q(r)\) must be approximated. We will illustrate the use of Perturbation Theory approximate these transition energies.
Here we will consider the Hamiltonian
-
-(32)#\[\begin{equation}
+
+(32)#\[\begin{equation}
\hat{H}_{vib} = -\frac{\hbar^2}{2\mu} \frac{d^2}{dr^2} + V_{H}(r) + V'(r) = \hat{H}_0 + V'(r) \tag{12},
\end{equation}\]
where \(\hat{H}_0\) is exactly solved by the Harmonic oscillator energy eigenfunctions and eigenvalues (\(\psi^{(0)}_n(r)\), \(E^{(0)}_n\)), and \(V'(r)\) is the perturbation which will take the form of either
\(V'(r) = \frac{1}{6}g(r-r_{eq})^3\) or \(V'(r) = \frac{1}{6}g(r-r_{eq})^3 + \frac{1}{24}h(r-r_{eq})^4\) in the cubic and quartic approximations, respectively.
We can calculate the energy of state \(n\) at 2nd order of perturbation theory as follows:
-
-(33)#\[\begin{equation}
+
+(33)#\[\begin{equation}
E_n = E_n^{(0)} + \langle \psi_n^{(0)} | V'(r) | \psi_n^{(0)} \rangle + \sum_{k \neq n} \frac{|\langle \psi_k^{(0)} | V'(r) | \psi_n^{(0)}|^2}{E_n^{(0)}-E_k^{(0)}}. \tag{13}
\end{equation}\]
Recall that for the zeroth-order functions have the form
-
-(34)#\[\begin{align}
+
+(34)#\[\begin{align}
\psi_n^{(0)}(r) &= \sqrt{\frac{1}{2^n n!}} \cdot \left(\frac{\alpha}{\pi} \right)^{1/4} \cdot H_n \left(\sqrt{\alpha} r \right) \cdot {\rm exp}\left(\frac{-\alpha }{2} r^2 \right) \\
\alpha &= \frac{\mu \omega}{\hbar} \\
\omega &= \sqrt{\frac{k}{\mu}}
diff --git a/notebooks/chem/Spin_one_half_solutions.html b/notebooks/chem/Spin_one_half_solutions.html
index 3a052dc..c1c5e75 100644
--- a/notebooks/chem/Spin_one_half_solutions.html
+++ b/notebooks/chem/Spin_one_half_solutions.html
@@ -574,16 +574,16 @@ Numpy arrays: VectorsNumpy arrays are special types of variables that can make use of different mathematical operation in the numpy library. We will see that a lot of linear algebra operations can be performed with numpy arrays using very simple syntax. Numpy arrays can have an arbitrary number of dimensions, but we will use 2-dimensional numpy arrays with
a single column and multiple rows to denote a column vector. We can take the transpose of these numpy arrays to represent a row vector.
Here we will introduce the vector representation of special spin states that have precise value of z-spin, that is, they are the eigenstates of the \(\hat{S}_z\) operator:
-
-(35)#\[\begin{equation}
+
+(35)#\[\begin{equation}
|\chi_{\alpha}^{(z)} \rangle=
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix}
\end{equation}\]
-
-(36)#\[\begin{equation}
+
+(36)#\[\begin{equation}
|\chi_{\beta}^{(z)}\rangle =
\begin{bmatrix}
0 \\
@@ -715,8 +715,8 @@ Numpy arrays: MatricesWe will use 2-dimensional numpy arrays with
a an equal number of rows and columns to denote square matrices.
Let’s use as an example matrix representation of the \(\hat{S}_z\) operator.
-
-(37)#\[\begin{equation}
+
+(37)#\[\begin{equation}
\mathbb{S}_z = \frac{\hbar}{2}
\begin{bmatrix}
1 & 0 \\
@@ -835,8 +835,8 @@ Eigenvalues and eigenvectors
Build matrix form of \(\mathbb{S}_x\) and \(\mathbb{S}_y\)#
The operator \(\hat{S}_x\) has the matrix form
-
-(38)#\[\begin{equation}
+
+(38)#\[\begin{equation}
\mathbb{S}_x = \frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
@@ -844,8 +844,8 @@ Build matrix form of \(\mathbb{S}
\end{bmatrix}
\end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-
-(39)#\[\begin{equation}
+
+(39)#\[\begin{equation}
\mathbb{S}_y = \frac{\hbar}{2}
\begin{bmatrix}
0 & -i \\
diff --git a/notebooks/chem/carbonyl_binary_classification.html b/notebooks/chem/carbonyl_binary_classification.html
index 6c98474..79c1c44 100644
--- a/notebooks/chem/carbonyl_binary_classification.html
+++ b/notebooks/chem/carbonyl_binary_classification.html
@@ -655,52 +655,36 @@ Installing RDKit Module0.0/29.4 MB ? eta -:--:--
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
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+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
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+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
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+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
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- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
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+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
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-
-
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-
-
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-
-
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+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
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?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions
+
If the overlap in the samples is too confusing, you can change the format of the histogram (histtype) so that it’s not filled in. Different line styles (ls) will make sure the plot is readable even if it’s printed in black and white.
@@ -698,7 +698,7 @@Visualizing Data
-
+
Normalizing the histograms (using the keyword: density=‘True’) makes it easier to compare the shapes of distributions even when one sample is much more numerous than the other. This is calculated by dividing each bin by the total area of the histogram.
@@ -717,7 +717,7 @@ Visualizing Data
-
+
diff --git a/notebooks/chem/CO_Fundamental.html b/notebooks/chem/CO_Fundamental.html
index 5a5babd..e200fdf 100644
--- a/notebooks/chem/CO_Fundamental.html
+++ b/notebooks/chem/CO_Fundamental.html
@@ -527,63 +527,63 @@ Computing Fundamental Transition of CO under different levels of approximati
We will use the Morse potential as a model for the “exact” interatomic potential, and we will approximate this potential by different orders of a Taylor expansion: including up to quadratic (which is the harmonic oscillator approximation), cubic, and quartic terms. The harmonic and Morse potentials are exactly solvable, and the eigenfunctions and eigenvalues of the vibrational Hamiltonian with cubic and quartic potentials can be approximated using perturbation theory. Therefore, we will
compare the fundamental transition computed exactly for harmonic and Morse potentials, and approximately at 2nd order of perturbation theory for cubic and quartic potentials to see the impact of various levels of potential truncation and approximation.
Within the Morse model, the vibrational Hamiltonian can be written as
-
-(21)#\[\begin{equation}
+
+(21)#\[\begin{equation}
\hat{H}_{vib} = -\frac{\hbar^2}{2\mu} \frac{d^2}{dr^2} + V_{Morse}(r), \tag{1}
\end{equation}\]
where
-
-(22)#\[\begin{equation}
+
+(22)#\[\begin{equation}
V_{Morse}(r) = D_e \left(1 - e^{-\beta(r-r_{eq})} \right)^2. \tag{2}
\end{equation}\]
The Morse parameters for \({\rm CO}\) are as follows: \(D_e = 11.225 \: {\rm eV}\), \(r_{eq} = 1.1283 \: {\rm Ang.}\), \(\beta = 2.5944 \: {\rm Ang.}^{-1}\),
and \(\mu = 6.8606 \: {\rm amu}\).
The exact energy eigenvalues for Equation (1) can be written as
-
-(23)#\[\begin{equation}
+
+(23)#\[\begin{equation}
E_n = \hbar \omega \left( \left(n+ \frac{1}{2} \right) - \chi_e \left(n+ \frac{1}{2} \right)^2 \right) \tag{3}
\end{equation}\]
where
-
-(24)#\[\begin{equation}
+
+(24)#\[\begin{equation}
\omega = \sqrt{\frac{2D_e \beta^2}{\mu}} \tag{4}
\end{equation}\]
and
-
-(25)#\[\begin{equation}
+
+(25)#\[\begin{equation}
\chi_e = \frac{\hbar \omega}{4 D_e}. \tag{5}
\end{equation}\]
The Morse potential can be approximated by a Taylor expansion as follows:
-
-(26)#\[\begin{equation}
+
+(26)#\[\begin{equation}
V_T(r) = \sum_{n=0}^{\infty} \frac{ f^{(n)}(r_{eq})}{n!} \left(r-r_{eq} \right)^n, \tag{6}
\end{equation}\]
where \(f^{(n)}(r_{eq})\) is the \(n^{th}\)-order derivative of the Morse potential evaluated at the equilibrium bondlength, e.g. \(f^{(1)}(r_{eq}) = \frac{d}{dr}V_{Morse}(r_{eq}).\)
We will define the Harmonic approximation to the potential as
-
-(27)#\[\begin{equation}
+
+(27)#\[\begin{equation}
V_H(r) = \frac{ f^{''}(r_{eq})}{2} \left(r-r_{eq} \right)^2 = \frac{1}{2} k \left(r-r_{eq} \right)^2 \tag{7}
\end{equation}\]
the cubic approximation to the potential as
-
-(28)#\[\begin{equation}
+
+(28)#\[\begin{equation}
V_C(r) = V_H(r) + \frac{ f^{'''}(r_{eq})}{6} \left(r-r_{eq} \right)^3 = V_H(r) + \frac{1}{6} g \left(r-r_{eq} \right)^3, \tag{8}
\end{equation}\]
and the quartic approximation as
-
-(29)#\[\begin{equation}
+
+(29)#\[\begin{equation}
V_Q(r) = V_C(r) + \frac{ f^{''''}(r_{eq})}{24} \left(r-r_{eq} \right)^4 = V_C(r) + \frac{1}{24}h(r-r_{eq})^4. \tag{9}
\end{equation}\]
Because we are using the Morse model as the “exact” interatomic potential in this notebook, we can compute these derivatives at \(r_{eq}\) analytically:
-
-(30)#\[\begin{align}
+
+(30)#\[\begin{align}
k = 2 D_e \beta^2 \\
g = -6 D_e \beta^3 \tag{10} \\
h = 14 D_e \beta^4.
\end{align}\]
However, in general we do not have an analytical form for the interatomic potential, so we must rely on numerical derivatives of the potential evaluated at the \(r_{eq}\). In the context of interatomic potentials computed by quantum chemistry methods (e.g. CCSD(T)), one must first identify the equilibrium geometry, and then compute derivatives by taking a number of single point calculations along all displacement coordinates to compute differences among. We will write the explicit expression for the second derivative using centered finite differences along the one displacement coordinate relevant for our \({\rm CO}\) molecule:
-
-(31)#\[\begin{equation}
+
+(31)#\[\begin{equation}
k=\frac{V_{Morse}(r_{eq}+\Delta r)-2V_{Morse}(r_{eq})+V_{Morse}(r_{eq}-\Delta r)}{\Delta r^2}+\mathcal O (\Delta r^2) \tag{11}
\end{equation}\]
where \(\Delta r\) represents a small displacement along the coordinate \(r\). Higher-order derivatives can also be computed, but will require larger numbers of displacements and therefore more energy evaluations by your quantum chemistry method. Expressions for higher-order derivatives along a single coordinate can be found here. Note that the number of displacement coordinates \(N\) grows linearly with the number of atoms, and that the number of displacements required to form the \(n^{{\rm th}}\)-order approximation to the potential grows as \(N^n\).
@@ -591,20 +591,20 @@ Computing Fundamental Transition of CO under different levels of approximati
Perturbation Theory#
We can compute the exact vibrational transition energies for the Morse oscillator and the Harmonic oscillator using Equation (3), where the Harmonic oscillator transition energies come from Equation (3) with \(\chi_e = 0\). However, the transition energies when the potential is approximated as \(V_C(r)\) or \(V_Q(r)\) must be approximated. We will illustrate the use of Perturbation Theory approximate these transition energies.
Here we will consider the Hamiltonian
-
-(32)#\[\begin{equation}
+
+(32)#\[\begin{equation}
\hat{H}_{vib} = -\frac{\hbar^2}{2\mu} \frac{d^2}{dr^2} + V_{H}(r) + V'(r) = \hat{H}_0 + V'(r) \tag{12},
\end{equation}\]
where \(\hat{H}_0\) is exactly solved by the Harmonic oscillator energy eigenfunctions and eigenvalues (\(\psi^{(0)}_n(r)\), \(E^{(0)}_n\)), and \(V'(r)\) is the perturbation which will take the form of either
\(V'(r) = \frac{1}{6}g(r-r_{eq})^3\) or \(V'(r) = \frac{1}{6}g(r-r_{eq})^3 + \frac{1}{24}h(r-r_{eq})^4\) in the cubic and quartic approximations, respectively.
We can calculate the energy of state \(n\) at 2nd order of perturbation theory as follows:
-
-(33)#\[\begin{equation}
+
+(33)#\[\begin{equation}
E_n = E_n^{(0)} + \langle \psi_n^{(0)} | V'(r) | \psi_n^{(0)} \rangle + \sum_{k \neq n} \frac{|\langle \psi_k^{(0)} | V'(r) | \psi_n^{(0)}|^2}{E_n^{(0)}-E_k^{(0)}}. \tag{13}
\end{equation}\]
Recall that for the zeroth-order functions have the form
-
-(34)#\[\begin{align}
+
+(34)#\[\begin{align}
\psi_n^{(0)}(r) &= \sqrt{\frac{1}{2^n n!}} \cdot \left(\frac{\alpha}{\pi} \right)^{1/4} \cdot H_n \left(\sqrt{\alpha} r \right) \cdot {\rm exp}\left(\frac{-\alpha }{2} r^2 \right) \\
\alpha &= \frac{\mu \omega}{\hbar} \\
\omega &= \sqrt{\frac{k}{\mu}}
diff --git a/notebooks/chem/Spin_one_half_solutions.html b/notebooks/chem/Spin_one_half_solutions.html
index 3a052dc..c1c5e75 100644
--- a/notebooks/chem/Spin_one_half_solutions.html
+++ b/notebooks/chem/Spin_one_half_solutions.html
@@ -574,16 +574,16 @@ Numpy arrays: VectorsNumpy arrays are special types of variables that can make use of different mathematical operation in the numpy library. We will see that a lot of linear algebra operations can be performed with numpy arrays using very simple syntax. Numpy arrays can have an arbitrary number of dimensions, but we will use 2-dimensional numpy arrays with
a single column and multiple rows to denote a column vector. We can take the transpose of these numpy arrays to represent a row vector.
Here we will introduce the vector representation of special spin states that have precise value of z-spin, that is, they are the eigenstates of the \(\hat{S}_z\) operator:
-
-(35)#\[\begin{equation}
+
+(35)#\[\begin{equation}
|\chi_{\alpha}^{(z)} \rangle=
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix}
\end{equation}\]
-
-(36)#\[\begin{equation}
+
+(36)#\[\begin{equation}
|\chi_{\beta}^{(z)}\rangle =
\begin{bmatrix}
0 \\
@@ -715,8 +715,8 @@ Numpy arrays: MatricesWe will use 2-dimensional numpy arrays with
a an equal number of rows and columns to denote square matrices.
Let’s use as an example matrix representation of the \(\hat{S}_z\) operator.
-
-(37)#\[\begin{equation}
+
+(37)#\[\begin{equation}
\mathbb{S}_z = \frac{\hbar}{2}
\begin{bmatrix}
1 & 0 \\
@@ -835,8 +835,8 @@ Eigenvalues and eigenvectors
Build matrix form of \(\mathbb{S}_x\) and \(\mathbb{S}_y\)#
The operator \(\hat{S}_x\) has the matrix form
-
-(38)#\[\begin{equation}
+
+(38)#\[\begin{equation}
\mathbb{S}_x = \frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
@@ -844,8 +844,8 @@ Build matrix form of \(\mathbb{S}
\end{bmatrix}
\end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-
-(39)#\[\begin{equation}
+
+(39)#\[\begin{equation}
\mathbb{S}_y = \frac{\hbar}{2}
\begin{bmatrix}
0 & -i \\
diff --git a/notebooks/chem/carbonyl_binary_classification.html b/notebooks/chem/carbonyl_binary_classification.html
index 6c98474..79c1c44 100644
--- a/notebooks/chem/carbonyl_binary_classification.html
+++ b/notebooks/chem/carbonyl_binary_classification.html
@@ -655,52 +655,36 @@ Installing RDKit Module0.0/29.4 MB ? eta -:--:--
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
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+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.8/29.4 MB 17.6 MB/s eta 0:00:02
- ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.2/29.4 MB 32.2 MB/s eta 0:00:01
- ━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.8/29.4 MB 30.4 MB/s eta 0:00:01
+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 11.9/29.4 MB 72.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.4/29.4 MB 111.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━ 23.8/29.4 MB 110.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 26.5 MB/s eta 0:00:00
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 78.2 MB/s eta 0:00:00
?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions
+
Computing Fundamental Transition of CO under different levels of approximati
We will use the Morse potential as a model for the “exact” interatomic potential, and we will approximate this potential by different orders of a Taylor expansion: including up to quadratic (which is the harmonic oscillator approximation), cubic, and quartic terms. The harmonic and Morse potentials are exactly solvable, and the eigenfunctions and eigenvalues of the vibrational Hamiltonian with cubic and quartic potentials can be approximated using perturbation theory. Therefore, we will compare the fundamental transition computed exactly for harmonic and Morse potentials, and approximately at 2nd order of perturbation theory for cubic and quartic potentials to see the impact of various levels of potential truncation and approximation.
Within the Morse model, the vibrational Hamiltonian can be written as
-where
-The Morse parameters for \({\rm CO}\) are as follows: \(D_e = 11.225 \: {\rm eV}\), \(r_{eq} = 1.1283 \: {\rm Ang.}\), \(\beta = 2.5944 \: {\rm Ang.}^{-1}\), and \(\mu = 6.8606 \: {\rm amu}\).
The exact energy eigenvalues for Equation (1) can be written as
-where
-and
-The Morse potential can be approximated by a Taylor expansion as follows:
-where \(f^{(n)}(r_{eq})\) is the \(n^{th}\)-order derivative of the Morse potential evaluated at the equilibrium bondlength, e.g. \(f^{(1)}(r_{eq}) = \frac{d}{dr}V_{Morse}(r_{eq}).\)
We will define the Harmonic approximation to the potential as
-the cubic approximation to the potential as
-and the quartic approximation as
-Because we are using the Morse model as the “exact” interatomic potential in this notebook, we can compute these derivatives at \(r_{eq}\) analytically:
-However, in general we do not have an analytical form for the interatomic potential, so we must rely on numerical derivatives of the potential evaluated at the \(r_{eq}\). In the context of interatomic potentials computed by quantum chemistry methods (e.g. CCSD(T)), one must first identify the equilibrium geometry, and then compute derivatives by taking a number of single point calculations along all displacement coordinates to compute differences among. We will write the explicit expression for the second derivative using centered finite differences along the one displacement coordinate relevant for our \({\rm CO}\) molecule:
-where \(\Delta r\) represents a small displacement along the coordinate \(r\). Higher-order derivatives can also be computed, but will require larger numbers of displacements and therefore more energy evaluations by your quantum chemistry method. Expressions for higher-order derivatives along a single coordinate can be found here. Note that the number of displacement coordinates \(N\) grows linearly with the number of atoms, and that the number of displacements required to form the \(n^{{\rm th}}\)-order approximation to the potential grows as \(N^n\).
@@ -591,20 +591,20 @@Computing Fundamental Transition of CO under different levels of approximati
Perturbation Theory#
We can compute the exact vibrational transition energies for the Morse oscillator and the Harmonic oscillator using Equation (3), where the Harmonic oscillator transition energies come from Equation (3) with \(\chi_e = 0\). However, the transition energies when the potential is approximated as \(V_C(r)\) or \(V_Q(r)\) must be approximated. We will illustrate the use of Perturbation Theory approximate these transition energies.
Here we will consider the Hamiltonian
-where \(\hat{H}_0\) is exactly solved by the Harmonic oscillator energy eigenfunctions and eigenvalues (\(\psi^{(0)}_n(r)\), \(E^{(0)}_n\)), and \(V'(r)\) is the perturbation which will take the form of either \(V'(r) = \frac{1}{6}g(r-r_{eq})^3\) or \(V'(r) = \frac{1}{6}g(r-r_{eq})^3 + \frac{1}{24}h(r-r_{eq})^4\) in the cubic and quartic approximations, respectively.
We can calculate the energy of state \(n\) at 2nd order of perturbation theory as follows:
-Recall that for the zeroth-order functions have the form
-Numpy arrays: VectorsNumpy arrays are special types of variables that can make use of different mathematical operation in the numpy library. We will see that a lot of linear algebra operations can be performed with numpy arrays using very simple syntax. Numpy arrays can have an arbitrary number of dimensions, but we will use 2-dimensional numpy arrays with
a single column and multiple rows to denote a column vector. We can take the transpose of these numpy arrays to represent a row vector.
Here we will introduce the vector representation of special spin states that have precise value of z-spin, that is, they are the eigenstates of the \(\hat{S}_z\) operator:
-
-(35)#\[\begin{equation}
+
+(35)#\[\begin{equation}
|\chi_{\alpha}^{(z)} \rangle=
\begin{bmatrix}
1 \\
0 \\
\end{bmatrix}
\end{equation}\]
-
-(36)#\[\begin{equation}
+
+(36)#\[\begin{equation}
|\chi_{\beta}^{(z)}\rangle =
\begin{bmatrix}
0 \\
@@ -715,8 +715,8 @@ Numpy arrays: MatricesWe will use 2-dimensional numpy arrays with
a an equal number of rows and columns to denote square matrices.
Let’s use as an example matrix representation of the \(\hat{S}_z\) operator.
-
-(37)#\[\begin{equation}
+
+(37)#\[\begin{equation}
\mathbb{S}_z = \frac{\hbar}{2}
\begin{bmatrix}
1 & 0 \\
@@ -835,8 +835,8 @@ Eigenvalues and eigenvectors
Build matrix form of \(\mathbb{S}_x\) and \(\mathbb{S}_y\)#
The operator \(\hat{S}_x\) has the matrix form
-
-(38)#\[\begin{equation}
+
+(38)#\[\begin{equation}
\mathbb{S}_x = \frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
@@ -844,8 +844,8 @@ Build matrix form of \(\mathbb{S}
\end{bmatrix}
\end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-
-(39)#\[\begin{equation}
+
+(39)#\[\begin{equation}
\mathbb{S}_y = \frac{\hbar}{2}
\begin{bmatrix}
0 & -i \\
diff --git a/notebooks/chem/carbonyl_binary_classification.html b/notebooks/chem/carbonyl_binary_classification.html
index 6c98474..79c1c44 100644
--- a/notebooks/chem/carbonyl_binary_classification.html
+++ b/notebooks/chem/carbonyl_binary_classification.html
@@ -655,52 +655,36 @@ Installing RDKit Module0.0/29.4 MB ? eta -:--:--
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
- ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.5/29.4 MB 7.5 MB/s eta 0:00:04
+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
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- ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.2/29.4 MB 32.2 MB/s eta 0:00:01
- ━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.8/29.4 MB 30.4 MB/s eta 0:00:01
+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 11.9/29.4 MB 72.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.4/29.4 MB 111.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━ 23.8/29.4 MB 110.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 26.5 MB/s eta 0:00:00
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 78.2 MB/s eta 0:00:00
?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions
Numpy arrays: MatricesWe will use 2-dimensional numpy arrays with
a an equal number of rows and columns to denote square matrices.
Let’s use as an example matrix representation of the \(\hat{S}_z\) operator.
-
-(37)#\[\begin{equation}
+
+(37)#\[\begin{equation}
\mathbb{S}_z = \frac{\hbar}{2}
\begin{bmatrix}
1 & 0 \\
@@ -835,8 +835,8 @@ Eigenvalues and eigenvectors
Build matrix form of \(\mathbb{S}_x\) and \(\mathbb{S}_y\)#
The operator \(\hat{S}_x\) has the matrix form
-
-(38)#\[\begin{equation}
+
+(38)#\[\begin{equation}
\mathbb{S}_x = \frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
@@ -844,8 +844,8 @@ Build matrix form of \(\mathbb{S}
\end{bmatrix}
\end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-
-(39)#\[\begin{equation}
+
+(39)#\[\begin{equation}
\mathbb{S}_y = \frac{\hbar}{2}
\begin{bmatrix}
0 & -i \\
diff --git a/notebooks/chem/carbonyl_binary_classification.html b/notebooks/chem/carbonyl_binary_classification.html
index 6c98474..79c1c44 100644
--- a/notebooks/chem/carbonyl_binary_classification.html
+++ b/notebooks/chem/carbonyl_binary_classification.html
@@ -655,52 +655,36 @@ Installing RDKit Module0.0/29.4 MB ? eta -:--:--
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
- ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.5/29.4 MB 7.5 MB/s eta 0:00:04
+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.8/29.4 MB 17.6 MB/s eta 0:00:02
- ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.2/29.4 MB 32.2 MB/s eta 0:00:01
- ━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.8/29.4 MB 30.4 MB/s eta 0:00:01
+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 11.9/29.4 MB 72.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.4/29.4 MB 111.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━ 23.8/29.4 MB 110.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 26.5 MB/s eta 0:00:00
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 78.2 MB/s eta 0:00:00
?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions
Eigenvalues and eigenvectors
Build matrix form of \(\mathbb{S}_x\) and \(\mathbb{S}_y\)#
The operator \(\hat{S}_x\) has the matrix form
-
-(38)#\[\begin{equation}
+
+(38)#\[\begin{equation}
\mathbb{S}_x = \frac{\hbar}{2}
\begin{bmatrix}
0 & 1 \\
@@ -844,8 +844,8 @@ Build matrix form of \(\mathbb{S}
\end{bmatrix}
\end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-
-(39)#\[\begin{equation}
+
+(39)#\[\begin{equation}
\mathbb{S}_y = \frac{\hbar}{2}
\begin{bmatrix}
0 & -i \\
diff --git a/notebooks/chem/carbonyl_binary_classification.html b/notebooks/chem/carbonyl_binary_classification.html
index 6c98474..79c1c44 100644
--- a/notebooks/chem/carbonyl_binary_classification.html
+++ b/notebooks/chem/carbonyl_binary_classification.html
@@ -655,52 +655,36 @@ Installing RDKit Module0.0/29.4 MB ? eta -:--:--
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
- ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.5/29.4 MB 7.5 MB/s eta 0:00:04
+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.8/29.4 MB 17.6 MB/s eta 0:00:02
- ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.2/29.4 MB 32.2 MB/s eta 0:00:01
- ━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.8/29.4 MB 30.4 MB/s eta 0:00:01
+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 11.9/29.4 MB 72.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.4/29.4 MB 111.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━ 23.8/29.4 MB 110.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 26.5 MB/s eta 0:00:00
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 78.2 MB/s eta 0:00:00
?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions
Build matrix form of \(\mathbb{S} \end{bmatrix} \end{equation}\]
and the operator \(\hat{S}_y\) has the matrix form
-Installing RDKit Module0.0/29.4 MB ? eta -:--:--
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.1/29.4 MB 3.1 MB/s eta 0:00:10
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+ ╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 0.4/29.4 MB 11.5 MB/s eta 0:00:03
+ ━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.0/29.4 MB 14.3 MB/s eta 0:00:02
━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 1.8/29.4 MB 17.6 MB/s eta 0:00:02
- ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.2/29.4 MB 32.2 MB/s eta 0:00:01
- ━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.8/29.4 MB 30.4 MB/s eta 0:00:01
+ ━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 2.9/29.4 MB 20.8 MB/s eta 0:00:02
+ ━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 4.3/29.4 MB 24.7 MB/s eta 0:00:02
- ━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.3/29.4 MB 40.2 MB/s eta 0:00:01
+ ━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 6.1/29.4 MB 29.3 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 11.9/29.4 MB 72.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━╸━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 8.5/29.4 MB 35.0 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.4/29.4 MB 111.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━━ 20.9/29.4 MB 128.4 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━━━━━━━ 12.0/29.4 MB 57.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━━━━━━━━━━━ 16.3/29.4 MB 94.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━━━━━━ 23.8/29.4 MB 110.0 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━━━━━━━━━ 21.9/29.4 MB 139.9 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╺━━ 27.2/29.4 MB 102.8 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸━━ 27.6/29.4 MB 165.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 102.2 MB/s eta 0:00:01
-
-
- ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 26.5 MB/s eta 0:00:00
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━╸ 29.4/29.4 MB 163.1 MB/s eta 0:00:01
+ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ 29.4/29.4 MB 78.2 MB/s eta 0:00:00
?25h
@@ -714,8 +698,8 @@ Installing RDKit Module[notice] To update, run: pip install --upgrade pip
-2.36user 0.50system 0:03.74elapsed 76%CPU (0avgtext+0avgdata 79104maxresident)k
-0inputs+389744outputs (0major+52872minor)pagefaults 0swaps
+2.38user 0.42system 0:03.74elapsed 75%CPU (0avgtext+0avgdata 78980maxresident)k
+23560inputs+389744outputs (120major+53304minor)pagefaults 0swaps
diff --git a/notebooks/chem/gaussian_wavepacket.html b/notebooks/chem/gaussian_wavepacket.html
index dc3e71c..b6b648a 100644
--- a/notebooks/chem/gaussian_wavepacket.html
+++ b/notebooks/chem/gaussian_wavepacket.html
@@ -552,27 +552,27 @@ Learning Outcomes
Summary#
We will demonstrate the time-dependence of an electron initially described by a Gaussian wavepacket that is travelling between two infinitely tall potential walls located at \(x=0\) and \(x = L\). The initial state of this wavepacket will be given by
-
-(15)#\[\begin{equation}
+
+(15)#\[\begin{equation}
\Psi(x,t_0) = \frac{1}{\sigma \sqrt(2\pi)} {\rm exp}\left(\frac{-1}{2}\left(\frac{x-x_0}{\sigma}\right)^2\right) {\rm exp}(i k_0 x). \tag{1}
\end{equation}\]
Such a wavefunction describes a particle initially with a mean momentum and position of \(\hbar k_0\) and \(x_0\), respectively, and \(\sigma\) relates to the spread (or position uncertainty)
of the particle. In the code that follows, we will have \(x_0 = 200\), \(k_0 = 0.4\), and \(\sigma = 15\) atomic units. We will specify that the location of the left-most infinite potential is at \(L = 500\) atomic units. We will use the fact that we can we can expand this wavefunction in any complete basis, and we will use the basis of the energy eigenfunctions of the particle in a box with \(L = 500\) atomic units.
Hence, we will expand the wavefunction as follows:
-
-(16)#\[\begin{equation}
+
+(16)#\[\begin{equation}
\Psi(x,t) = \sum_{n=1}^N c_n \sqrt{\frac{2}{L}} {\rm sin}\left( \frac{n \pi x}{L} \right)
{\rm exp}\left(-\frac{i E_n t}{\hbar}\right), \tag{2}
\end{equation}\]
where we have used the known time-dependence of each energy eigenstate. The energy eigenvalues have the form
-
-(17)#\[\begin{equation}
+
+(17)#\[\begin{equation}
E_n = \frac{n^2 \pi^2 \hbar^2}{2 m L^2}. \tag{3}
\end{equation}\]
Since we are working in atomic units, \(\hbar\) and \(m\) will have the value of 1.
The coefficients for the above expansion can be determined simply by
-
-(18)#\[\begin{equation}
+
+(18)#\[\begin{equation}
c_n = \sqrt{\frac{2}{500}} \int_0^{500} {\rm sin}\left(\frac{n \pi x}{500}\right) \cdot \Psi(x,t_0) \: dx \tag{4}
\end{equation}\]
where we have set \(L = 500\). We will use the numpy function trapz(fx, x) to perform a trapezoid rule approximation to
@@ -729,7 +729,7 @@
Step 3: Set up plot and function parameters and form the Gaussian Wavepacket
return np.asarray(x, float)
-[<matplotlib.lines.Line2D at 0x7f7e74797880>]
+[<matplotlib.lines.Line2D at 0x7f8ac0f939a0>]
@@ -973,42 +973,42 @@ Step 4: Put it all together and animate!
- ![]()
+ ![]()
-
+ oninput="anim9edf39c4f83443e891aae338fb79e304.set_frame(parseInt(this.value));">
-
@@ -1018,9 +1018,9 @@ Step 4: Put it all together and animate!
-<matplotlib.legend.Legend at 0x7fc989752370>
+<matplotlib.legend.Legend at 0x7f8b4ff143d0>
-
-
+
+
@@ -771,8 +771,8 @@ Part 4, Histogram Position and VelocityText(0, 0.5, 'P(E)')
-
-
+
+
@@ -881,13 +881,13 @@ Part 5, run langevin verlet dynamics on the double well
-<matplotlib.legend.Legend at 0x7fc97eb770d0>
+<matplotlib.legend.Legend at 0x7f8b42f9de80>
-
-
-
+
+
+
diff --git a/notebooks/chem/linear_variational_method.html b/notebooks/chem/linear_variational_method.html
index 9cc068f..32399c2 100644
--- a/notebooks/chem/linear_variational_method.html
+++ b/notebooks/chem/linear_variational_method.html
@@ -599,15 +599,15 @@ Approach
Form of the trial wavefunction#
In particular, we will optimize the trial wavefunction given by
-
-(1)#\[\begin{equation}
+
+(1)#\[\begin{equation}
\Phi(x) = \sum_{n=1}^N c_n \psi_n(x)
\end{equation}\]
where the expansion coefficients \(c_n\) are real numbers
and \(\psi_n(x)\) are the energy eigenfunctions of the ordinary particle in a box that has no potential
between \(x=0\) and \(x=L\). In particular, these eigenfunctions have the form
-
-(2)#\[\begin{equation}
+
+(2)#\[\begin{equation}
\psi_n(x) = \sqrt{\frac{2}{10} } {\rm sin}\left(\frac{n \pi x}{10} \right).
\end{equation}\]
@@ -615,8 +615,8 @@ Form of the trial wavefunction#
We will seek to minimize the energy functional of the trial wavefunction through the expansion coefficients, where the
energy functional of the trial wavefunction can be written as
-
-(3)#\[\begin{equation}
+
+(3)#\[\begin{equation}
E_{trial} = \frac{\int_0^{10} \Phi^* (x) \: \hat{H} \: \Phi(x) dx }{\int_0^{10} \Phi^* (x) \: \Phi(x) dx },
\end{equation}\]
where we have recognized that the boundaries of the box are at \(x=0\) and \(x=10\), so our range of integration
@@ -626,15 +626,15 @@
Form of the energy functional of the trial wavefunctionAtomic Units#
We will express our Hamiltonian in atomic units where \(\hbar = 1\) and \(m = 1\), so we can write the Hamiltonian in
this unit system as:
-
-(4)#\[\begin{equation}
+
+(4)#\[\begin{equation}
\hat{H} = -\frac{1}{2} \frac{d^2}{dx^2} + \delta(x-5).
\end{equation}\]
We further note that the Hamiltonian can be written as a sum
of kinetic and potential operators, \(\hat{H} = \hat{T} + \hat{V}\)
where \(\hat{T} = -\frac{1}{2} \frac{d^2}{dx^2}\) and \(\hat{V} = \delta(x-5)\). For the ordinary particle in a box, the Hamiltonian only contains \(\hat{T}\), so we can note that the particle in a box energy eigenfunctions obey the following eigenvalue equation with the kinetic energy operator (in atomic units)
-
-(5)#\[\begin{equation}
+
+(5)#\[\begin{equation}
\hat{T} \psi_n(x) = \frac{\pi^2 n^2}{200} \psi_n(x) \equiv E_n \psi_n(x).
\end{equation}\]
@@ -643,24 +643,24 @@ Solving the Linear Variational Problem\(E_{trial}\) with respect to the expansion coefficients in \(\Phi(x)\). This leads
to the condition that the partial derivative of the trial energy with
respect to each expansion coefficient is zero:
-
-(6)#\[\begin{equation}
+
+(6)#\[\begin{equation}
\frac{\partial}{\partial c_m} E_{trial} = 0 \; \; \forall \; m.
\end{equation}\]
When this is true, the trial energy and the expansion
coefficients satisfy the following equations:
-
-(7)#\[\begin{equation}
+
+(7)#\[\begin{equation}
E_{trial} c_m = \sum_{n=1}^N H_{nm} c_n,
\end{equation}\]
where
-
-(8)#\[\begin{equation}
+
+(8)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx.
\end{equation}\]
This can be written as an eigenvalue equation
-
-(9)#\[\begin{equation}
+
+(9)#\[\begin{equation}
{\bf H} {\bf c} = E_{trial} {\bf c},
\end{equation}\]
where \({\bf H}\) is the matrix whose elements are given by \(H_{nm}\) and \({\bf c}\) is the vector of coefficients.
@@ -670,13 +670,13 @@ Questions Part 1:
The matrix element \(H_{nm}\) can be expressed as a sum of kinetic and potential matrix elements, \(T_{nm} + V_{nm}\) where
and
-
-(11)#\[\begin{equation}
+
+(11)#\[\begin{equation}
V_{nm} = \int_0^{10} \psi^*_n(x) \hat{V} \psi_m(x) dx.
\end{equation}\]
Write a general expression for the elements of \(T_{nm}\) and \(V_{nm}\)
@@ -928,7 +928,7 @@
Visualizing the variational ground-state
-<matplotlib.legend.Legend at 0x7f0b2fae8e50>
+<matplotlib.legend.Legend at 0x7f7dacfe6dc0>
@@ -951,18 +951,18 @@ Behavior of Total Energy, Kinetic Energy, and Potential Energy functionals w
trial wavefunction as a function of basis set size.
For a given trial wavefunction (as determined by the variationally determined ground-state eigenvector \({\bf c}\),
we can define the total energy as
-
-(12)#\[\begin{equation}
+
+(12)#\[\begin{equation}
E = {\bf c}^t {\bf H} {\bf c}
\end{equation}\]
the kinetic energy as
-
-(13)#\[\begin{equation}
+
+(13)#\[\begin{equation}
T = {\bf c}^t {\bf T} {\bf c},
\end{equation}\]
and the potential energy as
-
-<matplotlib.legend.Legend at 0x7f0b2fa09cd0>
+<matplotlib.legend.Legend at 0x7f7dacf60220>
diff --git a/notebooks/chem/pib_delta_potential_dynamics.html b/notebooks/chem/pib_delta_potential_dynamics.html
index 5e632c7..c2b1e53 100644
--- a/notebooks/chem/pib_delta_potential_dynamics.html
+++ b/notebooks/chem/pib_delta_potential_dynamics.html
@@ -601,8 +601,8 @@ Approach
\[ \psi_n(x) = \sqrt{ \frac{2}{10} } {\rm sin}\left( \frac{n \pi x}{10} \right), \tag{3} \]
and the elements of \( {\bf H}\) will be given by
-
-(19)#\[\begin{equation}
+
+(19)#\[\begin{equation}
H_{nm} = \int_0^{10} \psi^*_n(x) \hat{H} \psi_m(x) dx. \tag{4}
\end{equation}\]
Because the basis set comprised of all \(\psi_n(x)\) is complete, we can represent our wavefunction at any instant in time as a linear combination of these basis functions:
@@ -630,8 +630,8 @@ Numerically Solving the TDSE Part 2: Runge-Kutta Method
\[ {\bf c}(t_0 + \Delta t) = {\bf c}(t_0) + \frac{1}{6}\left(k_1 + 2k_2 + 2k_3 + k_4 \right)\Delta t \tag{9} \]
where
-
-(20)#\[\begin{align}
+
+(20)#\[\begin{align}
k_1 &= -\frac{i}{\hbar} {\bf H}(t_0) {\bf c}(t_0) \\
k_2 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_1 \frac{\Delta t}{2} \right) \\ \tag{10}
k_3 &= -\frac{i}{\hbar} {\bf H}\left(t_0 + \frac{\Delta t}{2}\right) \left( {\bf c}(t_0) + k_2 \frac{\Delta t}{2} \right) \\
@@ -1186,42 +1186,42 @@ Time-dependence with the delta potential
- ![]()
+ ![]()
-
+ oninput="anim34d5e4c23b7c435fb752a2abd681ec6b.set_frame(parseInt(this.value));">
-
@@ -1231,9 +1231,9 @@ Time-dependence with the delta potential
---2024-05-08 18:03:44-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
-Resolving drive.google.com (drive.google.com)... 172.217.4.78, 2607:f8b0:4009:81a::200e
-Connecting to drive.google.com (drive.google.com)|172.217.4.78|:443... connected.
+--2024-05-08 18:49:54-- https://drive.google.com/uc?export=download&id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG
+Resolving drive.google.com (drive.google.com)... 142.251.16.113, 142.251.16.138, 142.251.16.101, ...
+Connecting to drive.google.com (drive.google.com)|142.251.16.113|:443... connected.
+HTTP request sent, awaiting response...
-HTTP request sent, awaiting response... 303 See Other
+303 See Other
Location: https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download [following]
---2024-05-08 18:03:44-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
-Resolving drive.usercontent.google.com (drive.usercontent.google.com)...
+--2024-05-08 18:49:54-- https://drive.usercontent.google.com/download?id=1BWet8_t2gyNGtFNMP_1opqcYBHDinEfG&export=download
+Resolving drive.usercontent.google.com (drive.usercontent.google.com)... 172.253.122.132, 2607:f8b0:4004:c07::84
+Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|172.253.122.132|:443...
-142.250.191.97, 2607:f8b0:4009:81a::2001
-Connecting to drive.usercontent.google.com (drive.usercontent.google.com)|142.250.191.97|:443... connected.
-
-
-HTTP request sent, awaiting response...
+connected.
+HTTP request sent, awaiting response...
200 OK
@@ -602,7 +601,7 @@ Review: Plotting the dataThursday May 30, 2024Prof. Tom Holme, Iowa State University
10:15 am - 11:00 pm
Tour of ESCIP repository part 1 (escip.github.io)
“Tour of ESCIP repository part 1 (escip.github.io)”
Prof. Davit Potoyan
11 am - 11:45 pm
“Navigating ChatGPT: Empowering Education with the P.E.E.R.S. Approach”
Dr. Andrew Severin, Iowa State University
Thursday May 30, 2024Discussion
5:00-5:45 pm
Dr. Ardith Bravenec, University of Washington
“Inspiring Individual Intelligence: Constructivist Learning with Python Notebooks and
5:00-6:00 pm
AI”
Dr. Ardith Bravenec, University of Washington
5:00-6:00 pm
Dinner and lightning talks
5:00-5:15 pm
5:00-5:15 pm
“Implementing high-performance computing in undergraduate education”
Prof. Pavel Lukashev, University of Northern Iowa
5:20:5:35 pm
5:20:5:35 pm
“Exploring ChatGPT’s Role in STEM Education: Benefits and Challenges in Liberal Arts Classrooms”
Prof. Shanshan Rodriguez, Grinell College
5:40:5:55 pm
5:40:5:55 pm
Dr. Subhadip Biswas, Iowa State University
6:00 pm – 8:00pm
6:00 pm – 8:00pm
Social and Informal discussions