# Imaginary Momentum

1. Nov 12, 2005

### emob2p

Hi,
When solving the delta potential Schrod. eq in momentum space, one finds that the poles of the wave function correspond to the bound states. This is the same result when solving the hydrogen atom in momentum space. However, the poles are when the momentum is pure imaginary. My question is what's the physical interpretation of imaginary momentum? Does this mean bound electrons in hydrogen have imaginary momentum? Thanks.

2. Nov 12, 2005

### inha

The momentum of an (bound) electron in a hydrogen atom is an observable. It's most definetly real.

3. Nov 12, 2005

### emob2p

This is know...that's why I'm confused.

4. Nov 12, 2005

Staff Emeritus
The wave function is not anything you measure; it is called amplitude and is a complex function. The poles are where it goes to infinity. When you observe the momentum, a Hermitian operator corresponding to your observation acts on the amplitude to produce real eigenvalues (this is what a Hermitian operator is defined to do). One of these eigenvalues is the momentum you observe.

5. Nov 12, 2005

### emob2p

But the wave function is a function of momentum. Hence it goes to infinity (a pole) for a certain value of imaginary momentum. What is the physical interpretation of this imaginary quantity?

6. Nov 12, 2005

### Physics Monkey

For the delta function potential, the position space wavefunction is the exponential of an absolute value $$\psi(x) \sim e^{- \beta |x|}$$, and the momentum space wavefunction is a Lorentzian $$\psi(p) \sim \frac{1}{p^2 + \hbar^2 \beta^2}$$. The momentum space wave function only has physical meaning when the momentum is real, but the imaginary poles do have mathematical content. They tell you that the position space wavefunction has an exponential decay. This information is obtained from the theory of residues in complex analysis.

This is an example of the use of complex analysis in the study of quantum mechanics and hints at a deeper connection. For example, the operator function $$G(z) = \frac{1}{H - z}$$ is called the resolvent of the Hamiltonian. This operator function has many interesting properties, it has a pole whenever z = B, the energy of a bound state. In addition, the "residue" at such a pole is nothing but the operator $$|\psi \rangle \langle \psi |$$, the projector corresponding to the bound state! It also has a branch cut along the values of E corresponding to the continuous spectrum, the scattering states. Though it may seem like this object is highly formal (and it is!), useful information can be obtained from it.

That may be more than you wanted to know, but I can probably provide some links if you're interested.

Last edited: Nov 12, 2005
7. Nov 13, 2005

### emob2p

Thanks Monkey. It's really cool how such a result falls out of the mathematics. I'd be interested in checking out those links to see the general argument.
A few questions:

1) Im not really sure what it means to have an operator in the denominator as you give for G(x).

2) I'm curious why you used the absolute value sign on $$\psi(x) \sim e^{- \beta |x|}$$. Now we have $$\psi(p) \sim \frac{1}{p^2 + \hbar^2 \beta^2}$$ singular at $$\p = \pm \i \hbar \beta$$. And if I'm using the techniques of residues correctly, shouldn't this correspond to the wave function $$\psi(x) \sim e^{- \beta x} + e^{\beta x}$$ since we're supposed to sum Res f(z) at each singularity. Now written like this, we don't have a bound state because this thing blows up regardless if $$x \rightarrow \pm \infty$$. What gives?

Last edited: Nov 13, 2005
8. Nov 13, 2005

### Physics Monkey

Regarding you first question, you can think of $$G(z)$$ as simply the inverse of $$H - z$$ for each z. In other words, $$G(z) = (H-z)^{-1},$$ that is the meaning of the operator in the denominator. Of course, these operators are infinite dimensional objects in general and thus rather formal and perhaps unfamiliar. Of course, the resolvent is well defined for finite dimensional systems as well. It is an interesting exercise to work out the resolvent for a two level spin system since it allows you to see the poles come out naturally as the eigenvalues and get some intuition for what G(z) looks like.

Regarding question two, the momentum space wavefunction does have two imaginary poles at $$\pm i \hbar \beta$$, but only one of these two poles contributes to the integral depending on whether you close your contour in the upper or lower half plane. Which plane you close the contour in depends on the sign of x (and how you have defined the fourier transform), and this is how the absolute value appears. For positive x you should get only $$e^{-\beta x}$$ and for negative x you should get only $$e^{\beta x}$$ which can be put together as $$e^{-\beta |x| }$$ for all x. This is the answer to your question about the bound state blowing up.