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Archive for October 12th, 2011

phy456 Problem set 4, problem 2 notes

Posted by peeterjoot on October 12, 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.)]

Problem 2.

I was deceived by an incorrect result in Mathematica, which led me to believe that the second order energy perturbation was zero (whereas part (c) of the problem asked if it was greater or lesser than zero). I started starting writing this up to show my reasoning, but our Professor quickly provided an example after class showing how this zero must be wrong, and I didn’t have to show him any of this.


Recall first the one dimensional particle in a box. Within the box we have to solve

\begin{aligned}\frac{P^2}{2m} \psi = E\psi\end{aligned} \hspace{\stretch{1}}(1.1)

and find

\begin{aligned}\psi \sim e^{\frac{i}{\hbar} \sqrt{2 m E} x} \end{aligned} \hspace{\stretch{1}}(1.2)


\begin{aligned}k = \frac{\sqrt{2 m E}}{\hbar}\end{aligned} \hspace{\stretch{1}}(1.3)

our general state, involving terms of each sign, takes the form

\begin{aligned}\psi = A e^{ i k x } +B e^{ -i k x }\end{aligned} \hspace{\stretch{1}}(1.4)

Inserting boundary conditions gives us

\begin{aligned}\begin{bmatrix}\psi(-L/2) \\ \psi(L/2)\end{bmatrix}\begin{bmatrix}e^{ -i k \frac{L}{2} } +e^{ i k \frac{L}{2} } \\ e^{ i k \frac{L}{2} } +e^{ -i k \frac{L}{2} }\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.5)

The determinant is zero

\begin{aligned}e^{-i k L} - e^{i k L} = 0,\end{aligned} \hspace{\stretch{1}}(1.6)

which provides our constraint on k

\begin{aligned}e^{2 i k L} = 1.\end{aligned} \hspace{\stretch{1}}(1.7)

We require 2 k L = 2 \pi n for any integer n, or

\begin{aligned}k = \frac{\pi n}{L}.\end{aligned} \hspace{\stretch{1}}(1.8)

This quantizes the energy, and inverting 1.3 gives us

\begin{aligned}E = \frac{1}{{2m}} \left( \frac{\hbar \pi n }{L} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.9)

To complete the task of matching boundary value conditions we cheat and recall that the particular linear combinations that we need to match the boundary constraint of zero at \pm L/2 were sums and differences yielding cosines and sines respectively. Since

\begin{aligned}{\left.{{\sin\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \pm \sin\left(\frac{\pi n}{2}\right)\end{aligned} \hspace{\stretch{1}}(1.10)

So sines are the wave functions for n = 2, 4, ... since \sin(n \pi) = 0 for integer n. Similarly

\begin{aligned}{\left.{{\cos\left( \frac{\pi n x }{L} \right) }}\right\vert}_{{x = \pm L/2}} = \cos\left(\frac{\pi n}{2}\right).\end{aligned} \hspace{\stretch{1}}(1.11)

Cosine becomes zero at \pi/2, 3\pi/2, \cdots, so our wave function is the cosine for n = 1, 3, 5, \cdots.

Normalizing gives us

\begin{aligned}\psi_n(x) = \sqrt{\frac{2}{L}}\left\{\begin{array}{l l}\cos\left(\frac{\pi n x}{L}\right) & \quad n = 1, 3, 5, \cdots \\ \sin\left(\frac{\pi n x}{L}\right) & \quad n = 2, 4, 6, \cdots \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.12)

Two non-interacting particles. Three lowest energy levels and degeneracies

Forming the Hamiltonian for two particles in the box without interaction, we have within the box

\begin{aligned}H = \frac{P_1^2}{2m} +\frac{P_2^2}{2m} \end{aligned} \hspace{\stretch{1}}(1.13)

we can apply separation of variables, and it becomes clear that our wave functions have the form

\begin{aligned}\psi_{nm}(x_1, x_2) = \psi_n(x_1) \psi_m(x_2)\end{aligned} \hspace{\stretch{1}}(1.14)

Plugging in

\begin{aligned}H \psi = E \psi,\end{aligned} \hspace{\stretch{1}}(1.15)

supplies the energy levels for the two particle wavefunction, giving

\begin{aligned}\begin{aligned}H \psi_{nm} &= \frac{\hbar^2}{2m}\left(\left(\frac{\pi n}{L}\right)^2+\left(\frac{\pi m}{L}\right)^2\right)\psi_{nm} \\ &= \frac{1}{2m} \left(\frac{\hbar \pi}{L}\right)^2 ( n^2 + m^2 ) \psi_{nm}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.16)

Letting n, m each range over [1,3] for example we find

\begin{aligned}\begin{array}{l l l}n & m & n^2 + m^2 \\ 1 & 1 & 2 \\ 1 & 2 & 5 \\ 1 & 3 & 10 \\ 2 & 1 & 5 \\ 2 & 2 & 8 \\ 2 & 3 & 13 \\ 3 & 1 & 10 \\ 3 & 2 & 13 \\ 3 & 3 & 18\end{array}\end{aligned} \hspace{\stretch{1}}(1.17)

It’s clear that our lowest energy levels are

\begin{aligned}\frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2  \\ \frac{5}{2m} \left(\frac{\hbar \pi}{L}\right)^2  \\ \frac{4}{m} \left(\frac{\hbar \pi}{L}\right)^2 \end{aligned}

with degeneracies 1, 2, 1 respectively.

Ground state energy with interaction perturbation to first order.

With c_0 positive and an interaction potential of the form

\begin{aligned}U(X_1, X_2) = - c_0 \delta(X_1 - X_2)\end{aligned} \hspace{\stretch{1}}(1.18)

The second order perturbation of the ground state energy is

\begin{aligned}E = E_{11}^{(0)} + H_{11;11}' + \sum_{nm \ne 11} \frac{{\left\lvert{H_{11;11}' }\right\rvert}^2}{E_{11} - E_{nm}}\end{aligned} \hspace{\stretch{1}}(1.19)


\begin{aligned}E_{11}^{(0)} = \frac{1}{m} \left(\frac{\hbar \pi}{L}\right)^2,\end{aligned} \hspace{\stretch{1}}(1.20)


\begin{aligned}H_{nm;ab}' = -c_0 {\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}\end{aligned} \hspace{\stretch{1}}(1.21)

to proceed, we need to expand the matrix element

\begin{aligned}{\langle {\psi_{nm}} \rvert} \delta(X_1 - X_2) {\lvert {\psi_{ab}} \rangle}&=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle {\langle {x_1 x_2} \rvert} \delta(X_1 - X_2) {\lvert {y_1 y_2 } \rangle} \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 dy_1 dy_2\left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \delta^2(\mathbf{x} - \mathbf{y}) \left\langle{{y_1 y_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int dx_1 dx_2 \left\langle{{\psi_{nm}}} \vert {{x_1 x_2}}\right\rangle \delta(x_1 - x_2) \left\langle{{x_1 x_2}} \vert {{\psi_{ab}}}\right\rangle \\ &=\int_{-L/2}^{L/2} dx\psi_{nm}(x, x)\psi_{ab}(x, x)\end{aligned}

So, for our first order calculation we need

\begin{aligned}H_{11; 11}' &= - c_0\int_{-L/2}^{L/2} dx\psi_{11}(x, x)\psi_{11}(x, x) \\ &=\frac{4}{L^2}\int_{-L/2}^{L/2} dx\cos^4( \pi x /L ) \\ &=- \frac{3 c_0}{2 L}\end{aligned}

For the second order perturbation of the energy, it is clear that this will reduce the first order approximation for each matrix element that is non-zero.

Attempting that calculation with \href{\

This worksheet can be seen to be giving misleading results, by evaluating

\begin{aligned}\int_{-\frac{L}{2}}^{\frac{L}{2}} \left(\frac{2}{L}\right)^2 \cos ^2\left(\frac{\pi  x}{L}\right) \cos ^2\left(\frac{3 \pi  x}{L}\right) \, dx = \frac{1}{L}\end{aligned} \hspace{\stretch{1}}(1.22)

Yet, the FullSimplify gives

\begin{aligned}\text{FullSimplify}\left[\int_{-\frac{L}{2}}^{\frac{L}{2}} \text{Cos}\left[\frac{\pi  x}{L}\right]^2 \left(\frac{2}{L}\right)^2 \text{Cos}\left[\frac{(2 n+1) \pi  x}{L}\right] \text{Cos}\left[\frac{(2 m+1) \pi  x}{L}\right] \, dx,\{m,n\}\in \text{Integers}\right] = 0\end{aligned} \hspace{\stretch{1}}(1.23)

I’m hoping that asking about this on stackoverflow will clarify how to use Mathematica correctly for this calculation.

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