5
$\begingroup$

So I have this problem where a particle is on a ring of perimeter L, and the coordinate of the particle is denoted by $s$, $0<s<L$. There is a nonzero potential which varies with $s, V(s)$, and is always smaller than the energy of the particle. For $E>V$, the WKB wave function is $$\psi(x) = \frac{1}{\sqrt{p(x)}}(C_+e^{i\phi(x)} + C_-e^{-i\phi(x)}),$$ $$ \phi(x) := \frac{1}{\hbar}\int_{x_0}^{x}\sqrt{2m(E - V(x'))}dx',$$ with $$p(x) := \sqrt{2m(E - V(x))}.$$

What should I do if I want to get the quantization condition for the energies $E_n$? I've tried doing $\psi(s) = \psi(s + nL),$ $n$ being an integer, like with regular "particle on a ring" problems, but since we don't have any other boundary conditions, and $V(s + nL) = V(s)$ (because all we did was go around the loop $n$ times), I just get $0 = 0$ and can't seem to get anywhere with this. How could I get an expression for the quantized energies?

$\endgroup$
0
7
$\begingroup$

You don't use $\psi(s) = \psi(s + n L)$ for $n$ integer, because the ring always has length $L$, its length has nothing to do with $n$. Instead you impose the condition $\psi(s) = \psi(s + L)$, which implies that $$\phi(L) - \phi(0) = 2 \pi n$$ which $n$ is the energy level. Then you get $$n h = \int_0^L \sqrt{2 m (E_n - V(x))} \, dx$$ which is a typical WKB quantization integral, from which you compute the $E_n$ in the usual way.

$\endgroup$
1
  • 1
    $\begingroup$ @LudwigvanDirac Yes; I used $2 \pi \hbar = h$. $\endgroup$ – knzhou Apr 1 '20 at 2:31
1
$\begingroup$

Let us modify OP's notation to acknowledge the dependence of the lower integration bound $$\phi(x_2,x_1) ~:=~ \frac{1}{\hbar}\int_{x_1}^{x_2}\!dx\sqrt{2m(E - V(x))}.\tag{1}$$ Then $$\phi(x_3,x_1)~=~\phi(x_3,x_2)+\phi(x_2,x_1).\tag{2}$$ From the periodicity of the potential $V$ we have $$\phi(L+x,x)~=~\phi(L,0).\tag{3}$$ From the single-valueness of the wavefunction, we get $$ \psi(x)=\psi(x+L),\tag{4}$$ or equivalently, $$\begin{align} \sum_{\pm} C_{\pm}e^{\pm i\phi(x,x_0)} ~\stackrel{(4)}{=}~& \sum_{\pm} C_{\pm}e^{\pm i\phi(x+L,x_0)} ~\stackrel{(2)}{=}~ \sum_{\pm} C_{\pm}e^{\pm i\phi(x+L,x)}e^{\pm i\phi(x,x_0)}\cr ~\stackrel{(3)}{=}~& \sum_{\pm} C_{\pm}e^{\pm i\phi(L,0)}e^{\pm i\phi(x,x_0)}.\end{align}\tag{5}$$ Eq. (5) are infinitely many equations for 2 unknowns $e^{\pm i\phi(L,0)}$. By picking at least 2 values of $x$, it becomes clear that the only solution to (5) is $$ e^{\pm i\phi(L,0)}~=~1, \tag{6}$$ or equivalently, $$ \phi(L,0)~\in~2\pi\mathbb{Z},\tag{7}$$ which leads to the well-known WKB quantization rule.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.