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\exnumber{6511}
\heading{Finding energy eigenstates in a semi-classical approximation} 
\auname{Roni Ben-Maimon and Omri Davidson}


{\bf The problem:}

\Dn

A particle of mass ${M}$ is free to move in 1 Dimension. 
Using the WKB approximation find the energy eigenstates ${E_n}$ of the particle in the following cases:

\Dn

(1) The potential is $V(x)=c|x|^\alpha$ where $\alpha>0$. use the following notation: $B(\alpha)=\int_{0}^{1} \sqrt{1-x^\alpha} dx$.

(2) A potential well in the interval $[-a,a]$ with a potential floor of $V(x)=V_0sign(x)$. It is enough to write the answer for energy levels above ${V_0}$.

(3) Bonus question: in section (2) calculate the eigenenergies using perturbation theory up to second order.



\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
In the WKB approximation bounded states satisfy the condition:
\[
\int_{x_1}^{x_2} p(x) dx = \pi(n+\frac{1}{2})
\]

The turning points are:
\[
x_1=-\big(\frac{E}{c}\big)^\frac{1}{\alpha}
\]
\[
x_2=\big(\frac{E}{c}\big)^\frac{1}{\alpha}
\]
And the momentum:
\[p(x)=\sqrt{2M(E-V(x))}=\sqrt{2M(E-c|x|^\alpha)}\]
Calculating the integral:
\[
\int_{-(\frac{E}{c})^\frac{1}{\alpha}}^{(\frac{E}{c})^\frac{1}{\alpha}}\sqrt{2M(E-c|x|^\alpha)} dx =\sqrt{2ME}\bigg[\int_{-(\frac{E}{c})^\frac{1}{\alpha}}^{0}\sqrt{(1-\frac{c}{E}(-x)^\alpha)} dx + \int_{0}^{(\frac{E}{c})^\frac{1}{\alpha}}\sqrt{1-\frac{c}{E}x^\alpha} dx\bigg] 
\]
\[
=2B(\alpha)\big(\frac{E}{c}\big)^\frac{1}{\alpha}\sqrt{2ME}
\]

So we get:
\[
2B(\alpha)\big(\frac{E}{c}\big)^\frac{1}{\alpha}\sqrt{2ME} = \pi(n+\frac{1}{2})
\]
Solving for ${E_n}$:
\[
E_n= \bigg[\frac{\pi(n+\frac{1}{2})}{\sqrt{8M}B(\alpha)}c^\frac{1}{\alpha}\bigg]^\frac{2\alpha}{\alpha+2}
\]


%%%%%%%%%%%%%%%
\Dn

(2)
In the case of a potential well the walls are "hard" therefore the equation for the bound energies is:

\[
\int_{x_1}^{x_2} p(x) dx ={\pi}n
\]

\[
\int_{-a}^{a} \sqrt{2M(E-V(x))} dx
= \int_{-a}^{0} \sqrt{2M(E+V_0)} dx + \int_{0}^{a} \sqrt{2M(E-V_0)} dx
\]
\[
a\sqrt{2M(E+V_0)} + a\sqrt{(2M(E-V_0)} 
= {\pi}n
\]


Solving the equation for ${E}$ gives:

\[
E_n 
= \epsilon_n + \frac{V_0^2}{4\epsilon_n}
\]

where ${\epsilon_n = \frac{\pi^2n^2}{8Ma^2}}$

%%%%%%%%%%%%%%
\Dn

(3)
We look at a symmetrical potential well with a perturbation of ${V(x)=V_0sign(x)}$.

The eigenstates of the unperturbated Hamiltonian:

 odd n: ${\Psi_n= \frac{1}{\sqrt{a}}\cos{k_nx}}$ 

 even n: ${\Psi_n= \frac{1}{\sqrt{a}}\sin{k_nx}}$ 
 
where:
${k_n=\frac{{\pi}n}{2a}}$ and the energy is ${\epsilon_n=\frac{{\pi}^2n^2}{8Ma^2}}$

 Finding the matrix elements of the perturbation:

On the diagonal:
\[
V_{nn}= \langle n|v|n\rangle 
=\frac{1}{a}\int_{-a}^{a}\cos^2{(k_nx)}V(x)dx
= \frac{1}{a}\int_{-a}^{a}\sin^2{(k_nx)}V(x)dx
=0\] because ${\cos^2{(k_nx)}V(x)}$ and ${\sin^2{(k_nx)}V(x)}$ are odd functions. 

Therfore:
\[E_n^{[1]}=V_{nn}=0\]

Off diagonal elements:
\[
m=odd, n=odd:
\frac{1}{a}\int_{-a}^{a}\cos{(k_mx)}\cos{(k_nx)}V(x)dx=0
\]
\[
m=even, n=even:
\frac{1}{a}\int_{-a}^{a}\sin{(k_mx)}\sin{(k_nx)}V(x)dx=0
\]
Because the integrands in both integrals are odd functions.
\[
m=odd, n=even: 
\frac{1}{a}\int_{-a}^{a}\cos{(k_mx)}\sin{(k_nx)}V(x)dx=
\]
\[
=\frac{V_0}{2a}\bigg[-\int_{-a}^{0}\sin{((k_n+k_m)x)}+\sin{((k_n-k_m)x)}dx + \int_{0}^{a}\sin{((k_n+k_m)x)}+\sin{((k_n-k_m)x)}dx\bigg]
\]
\[
=\frac{-V_0}{a}\bigg[\frac{\cos{((k_n+k_m)x)}}{k_n+k_m}+\frac{\cos{((k_n-k_m)x)}}{k_n-k_m}\bigg]_0^a
\]
We notice that for ${x=a}$ the expression equals 0 because ${(k_n{\pm}k_m)a=\frac{2l+1}{2}\pi}$ for an integer ${l}$.

So we are left with:
\[
V_{mn}=
\frac{2V_0}{\pi}\Big(\frac{1}{n+m}+\frac{1}{n-m}\Big)
=\frac{4V_0}{\pi}\Big(\frac{n}{n^2-m^2}\Big)
\]
For ${m=even, n=odd}$, we get the same result but with a replacement of ${m}$ with ${n}$:
\[
V_{mn}=
\frac{2V_0}{\pi}\Big(\frac{1}{n+m}+\frac{1}{m-n}\Big)
=\frac{4V_0}{\pi}\Big(\frac{m}{m^2-n^2}\Big)
\]
The second order correction of the energy:
\[
E_n^{[2]}=\sum_{m (\neq n)} \frac{|V_{mn}|^2}{\epsilon_n -\epsilon_m}
\]
For the symmetrical states (odd ${n}$) we get:
\[
E_n^{[2]} = \sum_{m(\neq n)} \Big (\frac{4V_0m}{\pi(m^2 -n^2)}\Big)^2\frac{8Ma^2}{\pi^2(n^2-m^2)}
=\frac{V_0^2}{\epsilon_n}\frac{16}{\pi^2}\sum_{l=1}^{\infty} \frac{n^2(2l)^2}{(n^2-(2l)^2)^3}
=\frac{V_0^2}{\epsilon_n}C_s(n)
\]
For the antisymmetrical states (even ${n}$) we get:
\[
E_n^{[2]} = \sum_{m(\neq n)} \Big (\frac{4V_0n}{\pi(n^2 -m^2)}\Big)^2\frac{8Ma^2}{\pi^2(n^2-m^2)}
=\frac{V_0^2}{\epsilon_n}\frac{16}{\pi^2}\sum_{l=0}^{\infty} \frac{n^4}{(n^2-(2l+1)^2)^3}
=\frac{V_0^2}{\epsilon_n}C_{as}(n)
\]

It is clear that the most significant contribution to the sum is when ${m=n\pm1}$. Therfore, we calculate the contribution of these elements for ${C_{s}(n)}$ and ${C_{as}(n)}$ in the limit ${n\rightarrow \infty}$:

For ${C_{s}(n)}$:
\[
\lim_{n\to\infty} \frac{16}{\pi^2}\Big(\frac{n^2(n+1)^2}{(n^2-(n+1)^2)^3}+\frac{n^2(n-1)^2}{(n^2-(n-1)^2)^3}\Big)
\cong -0.203
\]
For ${C_{as}(n)}$:
\[
\lim_{n\to\infty} \frac{16}{\pi^2}\Big(\frac{n^4}{(n^2-(n+1)^2)^3}+\frac{n^4}{(n^2-(n-1)^2)^3}\Big)
\cong 0.608
\]

In order to improve the accuracy, we approximate the sums using Matlab for different energy levels. The results are shown in the following graph:

\begin{figure}[htbp]
	\centering
		\includegraphics[scale=0.36]{graph.png}
	\label{fig:graph}
\end{figure}

We get ${C_{s}(n)=-\frac{1}{4}}$ and ${C_{as}(n)=\frac{3}{4}}$ for all values of ${n}$.

The energy up to second order perturbation is:

for the symmetrical states: 
\[
E_n 
= \epsilon_n - \frac{V_0^2}{4\epsilon_n}
\]
for the antisymmetrical states: 
\[
E_n 
= \epsilon_n + \frac{3V_0^2}{4\epsilon_n}
\]

Comparing the WKB approximation to perturbation theory, we see that it is the algebraic average of the symmetric and antisymmetric states.

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\end{document} 
