# Peeter Joot's (OLD) Blog.

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## PHY456H1F: Quantum Mechanics II. Lecture 2 (Taught by Prof J.E. Sipe). Approximate methods.

Posted by peeterjoot on September 15, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Peeter’s lecture notes from class. May not be entirely coherent.

# Approximate methods for finding energy eigenvalues and eigenkets.

In many situations one has a Hamiltonian $H$

\begin{aligned}H {\lvert {\Psi_{n \alpha}} \rangle} = E_n {\lvert {\Psi_{n \alpha}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

Here $\alpha$ is a “degeneracy index” (example: as in Hydrogen atom).

Why?

\begin{itemize}
\item Simplifies dynamics

take

\begin{aligned}{\lvert {\Psi(0)} \rangle}= \sum_{n\alpha}{\lvert {\Psi_{n \alpha}} \rangle}\left\langle{{\Psi_{n \alpha}}} \vert {{\Psi(0)}}\right\rangle&= \sum_{n\alpha} c_{n \alpha} {\lvert {\Psi_{n \alpha}} \rangle}\end{aligned}

Then

\begin{aligned}{\lvert {\Psi(t)} \rangle}&=e^{-i H t/\hbar}{\lvert {\Psi(0)} \rangle} \\ &=\sum_{n\alpha} c_{n \alpha}e^{-i H t/\hbar}{\lvert {\Psi_{n \alpha}} \rangle} \\ &=\sum_{n\alpha} c_{n \alpha}e^{-i E_n t/\hbar}{\lvert {\Psi_{n \alpha}} \rangle}\end{aligned}

\item “Applied field”‘ can often be thought of a driving the system from one eigenstate to another.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig1}
\caption{qmTwoL2fig1}
\end{figure}

\item Stat mech.

In thermal equilibrium

\begin{aligned}\left\langle{{\mathcal{O}}}\right\rangle =\frac{\sum_{n \alpha} {\langle {\Psi_{n\alpha}} \rvert} \mathcal{O} {\lvert {\Psi_{n \alpha}} \rangle} e^{-\beta E_n}}{Z}\end{aligned} \hspace{\stretch{1}}(1.2)

where

\begin{aligned}\beta = \frac{1}{{k_B T}},\end{aligned} \hspace{\stretch{1}}(1.3)

and

\begin{aligned}Z = \sum_{n \alpha} e^{-\beta E_n}\end{aligned} \hspace{\stretch{1}}(1.4)

\end{itemize}

# Variational principle

Consider any ket

\begin{aligned}{\lvert {\Psi} \rangle} = \sum_{n \alpha} c_{n \alpha} {\lvert {\Psi_{n \alpha}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.5)

(perhaps not even normalized), and where

\begin{aligned}c_{n \alpha} = \left\langle{{\Psi_{n \alpha}}} \vert {{\Psi}}\right\rangle\end{aligned} \hspace{\stretch{1}}(2.6)

but we don’t know these.

\begin{aligned}\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle = \sum_{n \alpha} {\left\lvert{c_{n \alpha}}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.7)

\begin{aligned}\frac{{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle}&=\frac{\sum_{n \alpha} {\left\lvert{c_{n \alpha}}\right\rvert}^2 E_n}{\sum_{m \beta} {\left\lvert{c_{m \beta}}\right\rvert}^2} \\ &\ge\frac{\sum_{n \alpha} {\left\lvert{c_{n \alpha}}\right\rvert}^2 E_0}{\sum_{m \beta} {\left\lvert{c_{m \beta}}\right\rvert}^2} \\ &=E_0\end{aligned}

So for any ket we can form the upper bound for the ground state energy

\begin{aligned}\frac{{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle}\ge E_0\end{aligned} \hspace{\stretch{1}}(2.8)

There’s a whole set of strategies based on estimating the ground state energy. This is called the Variational principle for ground state. See section 24.2 in the text [1].

We define the functional

\begin{aligned}E[\Psi] =\frac{{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle}\ge E_0\end{aligned} \hspace{\stretch{1}}(2.9)

If ${\lvert {\Psi} \rangle} = c {\lvert {\Psi_0} \rangle}$ where ${\lvert {\Psi_0} \rangle}$ is the normalized ground state, then

\begin{aligned}E[ c \Psi_0 ] = E_0\end{aligned} \hspace{\stretch{1}}(2.10)

## Examples.

### Hydrogen atom

\begin{aligned}{\langle {\mathbf{r}} \rvert} H {\lvert {\mathbf{r}'} \rangle} = \mathcal{H} \delta^3(\mathbf{r} - \mathbf{r}')\end{aligned} \hspace{\stretch{1}}(2.11)

where

\begin{aligned}\mathcal{H} = -\frac{\hbar^2}{2 \mu} \boldsymbol{\nabla}^2 - \frac{e^2}{r}\end{aligned} \hspace{\stretch{1}}(2.12)

Here $\mu$ is the reduced mass.

We know the exact solution:

\begin{aligned}H {\lvert {\Psi_0} \rangle}\end{aligned} \hspace{\stretch{1}}(2.13)

\begin{aligned}E_0 = -R_y\end{aligned} \hspace{\stretch{1}}(2.14)

\begin{aligned}R_y = \frac{\mu e^4}{2 \hbar^2} \approx 13.6 \text{eV}\end{aligned} \hspace{\stretch{1}}(2.15)

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\Psi_0}}\right\rangle = \Phi_{100}(\mathbf{r}) = \left( \frac{1}{{\pi a_0^3}}\right)^{1/2} e^{-r/a_0}\end{aligned} \hspace{\stretch{1}}(2.16)

\begin{aligned}a_0 = \frac{\hbar^2}{\mu e^2} \approx 0.53 \text{\AA}\end{aligned} \hspace{\stretch{1}}(2.17)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig2}
\caption{qmTwoL2fig2}
\end{figure}

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig3}
\caption{qmTwoL2fig3}
\end{figure}

estimate

\begin{aligned}{\langle {\Psi} \rvert} H {\lvert {\Psi} \rangle} &= \int d^3 \mathbf{r} \Psi^{*}(\mathbf{r}) \left( -\frac{\hbar^2}{2 \mu} \boldsymbol{\nabla}^2 - \frac{e^2}{r} \right) \Psi(\mathbf{r}) \\ \left\langle{{\Psi}} \vert {{\Psi}}\right\rangle &= \int d^3 \mathbf{r} {\left\lvert{\Psi(\mathbf{r})}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.18)

Or guess shape

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig4}
\caption{qmTwoL2fig4}
\end{figure}

Using the trial wave function $e^{-\alpha r^2}$

\begin{aligned}E[\Psi] \rightarrow E(\alpha)\end{aligned} \hspace{\stretch{1}}(2.20)

\begin{aligned}E(\alpha) =\frac{\int d^3 \mathbf{r} e^{-\alpha r^2} \left( -\frac{\hbar^2}{2 \mu} \boldsymbol{\nabla}^2 - \frac{e^2}{r} \right) e^{-\alpha r^2}}{\int d^3\mathbf{r} e^{-2 \alpha r^2}}\end{aligned} \hspace{\stretch{1}}(2.21)

find

\begin{aligned}E(\alpha) = A \alpha - B \alpha^{1/2}\end{aligned} \hspace{\stretch{1}}(2.22)

\begin{aligned}A &= \frac{3 \hbar^2}{2\mu} \\ B &= 2 e^2 \left( \frac{2}{\pi} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.23)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig5}
\caption{qmTwoL2fig5}
\end{figure}

Minimum at

\begin{aligned}\alpha_0 =\left( \frac{\mu e^2}{ \hbar^2 } \right) \frac{8 }{9 \pi}\end{aligned} \hspace{\stretch{1}}(2.25)

So

\begin{aligned}E(\alpha_0) =- \frac{\mu e^4 }{2 \hbar^2} \frac{8 }{3 \pi} = -0.85 R_y\end{aligned} \hspace{\stretch{1}}(2.26)

### Helium atom

Assume an infinite nuclear mass with nucleus charge $2 e$

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL2fig6}
\caption{qmTwoL2fig6}
\end{figure}

ground state wavefunction

\begin{aligned}\Psi_0(\mathbf{r}_1, \mathbf{r}_2)\end{aligned} \hspace{\stretch{1}}(2.27)

The problem that we want to solve is

\begin{aligned}\left(-\frac{\hbar^2}{2 m} \boldsymbol{\nabla}_1^2-\frac{\hbar^2}{2 m} \boldsymbol{\nabla}_2^2- \frac{2 e}{r}+\frac{e^2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}\right)\Psi_0(\mathbf{r}_1, \mathbf{r}_2) = E_0 \Psi_0(\mathbf{r}_1, \mathbf{r}_2)\end{aligned} \hspace{\stretch{1}}(2.28)

Nobody can solve this problem. It is one of the simplest real problems in QM that cannot be solved exactly.

Suppose that we neglected the electron, electron repulsion. Then

\begin{aligned}\Psi_0(\mathbf{r}_1, \mathbf{r}_2)=\bar{\Phi}_{110}(\mathbf{r}_1)\bar{\Phi}_{110}(\mathbf{r}_2)\end{aligned} \hspace{\stretch{1}}(2.29)

where

\begin{aligned}\left( -\frac{\hbar^2}{2 m} \boldsymbol{\nabla}^2- \frac{2 e}{r} \right)\bar{\Phi}_{100}(\mathbf{r}) = \epsilon \bar{\Phi}_{100}(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.30)

with

\begin{aligned}\epsilon = - 4 R_y\end{aligned} \hspace{\stretch{1}}(2.31)

\begin{aligned}R_y = \frac{m e^4}{2 \hbar^2}\end{aligned} \hspace{\stretch{1}}(2.32)

This is the solution to

\begin{aligned}\left(-\frac{\hbar^2}{2 m} \boldsymbol{\nabla}_1^2-\frac{\hbar^2}{2 m} \boldsymbol{\nabla}_2^2- \frac{2 e}{r}\right)\Psi_0^{(0)}(\mathbf{r}_1, \mathbf{r}_2) = E_0 \Psi_0(\mathbf{r}_1, \mathbf{r}_2)=E_0^{(0)} \Psi_0^{(0)}(\mathbf{r}_1, \mathbf{r}_2)\end{aligned} \hspace{\stretch{1}}(2.33)

\begin{aligned}E_0^{(0)} = - 8 R_y.\end{aligned} \hspace{\stretch{1}}(2.34)

Now we want to put back in the electron electron repulsion, and make an estimate.

Trial wavefunction

\begin{aligned}\Psi(\mathbf{r}_1, \mathbf{r}_2, Z) =\left(\left(\frac{Z^3}{ \pi a_0^3 }\right)^{1/2} e^{-Z r_1/a_0}\right)\left(\left(\frac{Z^3}{ \pi a_0^3 }\right)^{1/2} e^{-Z r_2/a_0}\right)\end{aligned} \hspace{\stretch{1}}(2.35)

expect that the best estimate is for $Z \in [1,2]$.

This can be calculated numerically, and we find

\begin{aligned}E(Z) = 2 R_Y \left( Z^2 - 4 Z + \frac{5}{8} Z \right)\end{aligned} \hspace{\stretch{1}}(2.36)

The $Z^2$ comes from the kinetic energy. The $-4 Z$ is the electron nuclear attraction, and the final term is from the electron-electron repulsion.

The actual minimum is

\begin{aligned}Z = 2 - \frac{5}{16}\end{aligned} \hspace{\stretch{1}}(2.37)

\begin{aligned}E(2 - 5/16) = -77.5 \text{eV}\end{aligned} \hspace{\stretch{1}}(2.38)

Whereas the measured value is $-78.6 \text{eV}$.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.