# Eigenstates of Dirac Potential in Momentum Space

1. Aug 8, 2012

### Jolb

1. The problem statement, all variables and given/known data
Consider a particle moving in one dimension and bound to an attractive Dirac δ-function potential located at the origin. Work in units such that $m=\hbar=1$. The Hamiltonian is given, in real (x) space, by:
$$H=-\frac{1}{2}\frac{d^2}{dx^2}-\delta (x)$$

The (non normalized) wavefunction $\Psi (x)=<x|\Psi >=e^{-a|x|}$ is an eigenstate of this potential.

a) What is the corresponding wavefunction in wavevector space, $\tilde{\Psi}(k) = <k|\Psi>$? (Recall that $<x|k>=\frac{1}{\sqrt{2\pi}}e^{ikx}$)

b) Transform the Hamiltonian from the real space basis |x> to the wavevector space basis |k>. Observe that the kinetic energy is diagonal in k-space, but the δ-function potential scatters all wavevectors into all wavevectors with equal amplitude.

c) Now show, by direct calculation in wavevector space, that $\tilde{\Psi}(k)$ is an eigenstate of the Hamiltonian that you found in part (b). You may find the following integral of use: $$\int^\infty_{-\infty}\frac{dk}{1+k^2}=\pi$$

3. The attempt at a solution
a)
$$\tilde{\Psi}(k) = <k|\Psi>=\int^\infty_{-\infty}<k|x><x|\Psi>dx=\int^\infty_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-ikx}e^{-a|x|}dx$$

$$={Re}\left [2\int^\infty_{0}\frac{1}{\sqrt{2\pi}}e^{-ikx}e^{-ax}dx \right ]={Re}\left [\sqrt{\frac{2}{\pi}}\int^\infty_{0}e^{-(ik+a)x}dx \right ]={Re}\left [\sqrt{\frac{2}{\pi}}\left (\frac{1}{a+ik} \right ) \right ]=\sqrt{\frac{2}{\pi}}\frac{a}{a^2+k^2}$$

All this looks alright to me.

b) Since we're working with units where $\hbar=m=1$, then $\hat{p}=\hat{k}=-i\frac{d}{dx}$. So the derivative terms just substitute easily with k. All we need now is to Fourier transform the δ potential:
$$\tilde{\delta}(k)=\int^\infty_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-ikx}\delta(x)dx = \frac{1}{\sqrt{2\pi}}$$
$$\implies\hat{H}=\frac{1}{2}k^2-\frac{1}{\sqrt{2\pi}}$$

This looks okay: term on the left is "diagonal", whereas the term on the right scatters all wavenumbers to another wavenumber without a change in amplitude.

c) This is where I'm realizing something is really screwed up. Obviously we have just changed bases so the wavefunction is still an eigenstate of the potential. But I'm not getting that.
$$\hat{H}\tilde{\Psi}(k) = \left ( \frac{1}{2}k^2-\frac{1}{\sqrt{2\pi}} \right )\left (\sqrt{\frac{2}{\pi}}\frac{a}{a^2+k^2} \right )$$
But this looks clearly linearly independent from $\tilde{\Psi}(k)$.

Where have I gone wrong? I didn't use the integral they give you, and I bet I need to use it somewhere to get the right answer. But where?

Last edited: Aug 8, 2012
2. Aug 9, 2012

### Jolb

Maybe this should be moved to the Quantum forum, since it is a fundamental problem, just in momentum space rather than position space.

3. Aug 9, 2012

### Oxvillian

Hmm I think the mistake here is where you converted the delta potential to k-representation language. It's not just a Fourier transform - it should look more like
$$<k|V|\Psi> \; = \int dk' <k|V|k'><k'|\Psi> \; = -\frac{1}{2\pi}\int dk' <k'|\Psi>,$$
which is where their integral comes in. I think everything will then work out ok. Also you can probably fix the constant a right away by normalization (a=1?).

Last edited: Aug 9, 2012
4. Aug 10, 2012

### Jolb

Great! Thanks a lot. I need to remember to treat the potential like an operator, not a ket.

5. Aug 26, 2012

### WombRaider

Something seems a little off here, though: For the attractive dirac potential well, if E < 0, there is the bound state I believe you have here and a is real, but if E > 0, then the state isn't bound, and a is complex. Then, when you take the Re() part, it's not so simple.

Also, what exactly did you do to simplify the integral between when the limits were (-inf,inf) and (0,inf)?

Thanks!

6. Aug 28, 2012

### Oxvillian

Not sure why you would want to take the real part in this case

If E>0 we have another standard problem - try searching around for "scattering by a delta function potential".
Write eikx = cos(kx) + i sin(kx) and take another look at his/her integral. You should find that the imaginary part of the integrand is odd and that the real part of the integrand is even. So it does simplify as advertised.

7. Aug 31, 2012

### Jolb

I forgot to mention that this problem deals exclusively with bound states. The constant a is real, so the wavefunction given above part a is for a bound state.

Oxvillian is right about the integral trick: the i sin term is odd and the the e-a|x| is even, so whatever it contributes to the integral will be odd, so it cancels by symmetry. You only get a contribution from the cosine term, which is even, so notice the two. I guess you don't actually need to take the real part.

Last edited: Aug 31, 2012