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2003 II 5

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This diagram is the same as in White, for the bound-state problem.

By the Bohr-Sommerfeld condition, if we want the solution that goes to zero at both $z=\pm\infty$ , we must have:

$\int_{a}^{b}Q^{1/2}dx=\left(n+1/2\right)\pi$

Then:

$\int_{-E^{1/2}}^{E^{1/2}}\left(E-x^{2}\right)^{1/2}dx=\frac{E\pi}{2}$

So that $E = 2 n + 1$ .

The solutions in region 1 are (using $b = E 1 / 2$ ):

$\left(b,z\right)=\frac{1}{\left(E-z^{2}\right)^{1/4}}\exp\left[-i\int_{b}^{z}\left(E-x^{2}\right)^{1/2}dx\right]$

It is clear that this is unbounded, and is the dominant solution for $z > b$ . Then to get the solution that is zero at $z = a$ :

$y\left(z=a\right)=0=\frac{A}{\left(E-a^{2}\right)^{1/4}}\exp\left[-i\int_{b}^{a}\left(E-x^{2}\right)^{1/2}dx\right]+\frac{B}{\left(E-a^{2}\right)^{1/4}}\exp\left[i\int_{b}^{a}\left(E-x^{2}\right)^{1/2}dx\right]$

This relation can be simplified:

$\frac{A}{B}=-\exp\left[2i\int_{b}^{a}\left(E-x^{2}\right)^{1/2}dx\right]$

Then:

$y_{1}=A\left(b,z\right)_{d}+B\left(z,b\right)_{s}$

Crossing to region 2, we hop off the stokes line and so we pick up half of the stokes constant:

$y_{2}=A\left(b,z\right)_{d}+\left[B+\frac{i}{2}A\right]\left(z,b\right)_{s}$

Then we cross an anti-stokes:

$y_{3}=A\left(b,z\right)_{s}+\left[B+\frac{i}{2}A\right]\left(z,b\right)_{d}$

Now we must cross a stokes:

$y_{4}=\left[\frac{1}{2}A+iB\right]\left(b,z\right)_{s}+\left[B+\frac{i}{2}A\right]\left(z,b\right)_{d}$

This will then be the solution on the axis. Now if we were to connect with the zero at $- b$ :

$y_{5}=\left[\frac{1}{2}A+iB\right]\left[b,-b\right]\left(-b,z\right)_{s}+\left[B+\frac{i}{2}A\right]\left(z,-b\right)_{d}\left[-b,b\right]$

Using $\left[-b,b\right]=\exp\left(-i\int_{-b}^{b}Q^{1/2}dz\right)=e^{-iW}$ :

$y_{5}=\left[\frac{1}{2}A+iB\right]e^{iW}\left(-b,z\right)_{s}+\left[B+\frac{i}{2}A\right]e^{-iW}\left(z,-b\right)_{d}$

Now requiring that $y\left(z\right)=y\left(-z\right)$ :

$\left[\frac{1}{2}A+iB\right]\left(b,z\right)_{s}+\left[B+\frac{i}{2}A\right]\left(z,b\right)_{d}=\left[\frac{1}{2}A+iB\right]e^{iW}\left(-b,-z\right)_{s}+\left[B+\frac{i}{2}A\right]e^{-iW}\left(-z,-b\right)_{d}$

Since $Q 1 / 2$ is even, $\left(-b,-z\right)=\left(z,b\right)$ :

$\left[\frac{1}{2}A+iB\right]\left(b,z\right)+\left[B+\frac{i}{2}A\right]\left(z,b\right)=\left[\frac{1}{2}A+iB\right]e^{iW}\left(z,b\right)+\left[B+\frac{i}{2}A\right]e^{-iW}\left(b,z\right)$

This gives an equation for $W$ :

$B+\frac{i}{2}A=\left[\frac{1}{2}A+iB\right]e^{iW}$

Using the condition for $A / B$ :

$e^{iW}=\frac{2-ie^{2iF}}{2i-e^{2iF}}$

Where:

$\begin{array}{rcl} W = \int_{-E^{1/2}}^{E^{1/2}}\left(E-x^{2}\right)^{1/2}dz=\frac{E\pi}{2}\\ F = \int_{E^{1/2}}^{a}\left(E-x^{2}\right)^{1/2}dx\end{array}$

which will then give a condition on $E$ . If $a\rightarrow\infty$ , $F\rightarrow\infty$ , and: