I am having some trouble answering the following question in my "Advanced Quantum Mechanics" course:

Using the integral equation: $$\psi(x) = Ae^{ikx} + Be^{-ikx} - \int_{-\infty}^{\infty}G^{\pm}(x,x^{\prime})U(x^{\prime})\psi(x^{\prime})\:\mathrm{d}x^{\prime}$$ With the potential $U(x) = q\delta(x)$, where $q > 0$. Show that $\psi(0) = (1 + \frac{iqm}{k\hbar^{2}})$.

We can find that the Green's function for $E > 0$ is:

$$G^{\pm}(x,x^{\prime}) = \pm\frac{im}{k\hbar^{2}}\exp(\pm ik|x-x^{\prime}|)$$

Noting that we are scattering from the left (not stated in the problem but assumed from earlier problems), we can write:

$$\psi(x) = e^{ikx} - \frac{imq}{k\hbar^{2}}\int_{-\infty}^{\infty}\exp(ik|x-x^{\prime}|)\delta(x^{\prime})\psi(x^{\prime})\:\mathrm{d}x^{\prime}$$

Evaluating the integral:

$$\psi(x) = e^{ikx} - \frac{imq}{k\hbar^{2}}\exp(ik|x|)\psi(0)$$

This allows us to solve our scattering problem:

$$\psi(x) = \begin{cases}\left(1-\frac{imq}{k\hbar^{2}}\psi(0)\right)e^{ikx} & x > 0 \\ e^{ikx} - \frac{imq}{k\hbar^{2}}e^{-ikx}\psi(0) & x < 0\end{cases}$$

So we can find the transmission and reflection amplitudes:

$$T = \left|1-\frac{imq}{k\hbar^{2}}\psi(0)\right|^{2},\qquad R = \left|\frac{imq}{k\hbar^{2}}\psi(0)\right|^{2}$$

However, I am not sure how to show that $\psi(0)$ is the expression given?

  • $\begingroup$ hint: you already know $\psi(x)$. What do you get if you set $x=0$? $\endgroup$ – AccidentalFourierTransform May 7 '16 at 15:46
  • $\begingroup$ Ahh yes, that makes sense $\endgroup$ – Thomas Russell May 7 '16 at 15:49

Your Answer

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

Browse other questions tagged or ask your own question.