# Weird state density calculation

1. Apr 28, 2005

### Palindrom

Hey everyone,
I'm having some trouble understanding this, any help would be appreciated:

Calculate the density of states for a free particle with momentum $$$\hbar k$$$ for the angles between $$$\left[ {\theta _0 ,\theta _0 + d\theta } \right]$$$ relative to an electric field $$$\vec \varepsilon$$$, in the ultra-relative limit $$$E \cong pc$$$.

In my solutions, the first thing they do is to say: Well, as usual, first find $$$N\left( E \right)$$$ and then take the derivative with respect to $$$E$$$, but that's OK. The problem is how they calculate $$$N\left( E \right)$$$:

$$$N\left( E \right) = \frac{V}{{h^3 }}\int\limits_{\theta \in \left[ {\theta _0 ,\theta _0 + d\theta } \right]} {d^3 p} = \frac{V}{{h^3 }}2\pi \sin \left( {\theta _0 } \right)d\theta \int\limits_0^{p_{\max } } {p^2 d^2 p}$$$

I can live with the first move. But I don't understand where this sine comes from, or $$$2\pi$$$, and that other integral... help?

2. Apr 28, 2005

### seratend

dp^3= 2.pi. sin(theta) p^2dp (spherical coordinates with pre integration over phi angle)
Therafter, you have the integration domain of the integral: [theta, theta+dtheta] that forces the final result (to be more correct, the dtheta should appear).

Seratend.
P.S. I think your rhs integral should be p^2dp and not p^2dp^2 (otherwise I have not well understood your pb)

3. Apr 28, 2005

### Palindrom

Yes but you see, this is what I don't understand. The integral is in momentum space, not coordinates. So you can use "spherical coordinates", but why are the angles here the same angles as in space?
You see what's giving me trouble?
Thanks!

4. Apr 28, 2005

### Palindrom

OK, let me reformulate my question: is the angle between the velocity (and therefore the momentum) and the field, or between the position and the field?
Come to think of it, the particles position can't be defined, right? So it should be between momentum and field, and so the calculation would be clear.
Did I understand that correctly?

5. Apr 29, 2005

### seratend

Ok, let's try to stay simple.
You know that momentun is linked to the position through the commutator relation :
[q_i,p_j]=i.hbar delta_{ij}. [1]

Now, if you change the direction of the position coordinates, let's call it q_theta for the new direction, you always may define the conjugate momentum for this direction: [q_theta,p_theta]= i.hbar.

In the relation of my previous post, we have just have defined the momentum along the direction theta, p_theta. It is associated therefore the momentum associated to the position of the theta direction through the commutator relation.

If you are note sure of my answer, just write the p_theta momentum on the (px,py,pz) coordinates, and then use the relation [1] to find the conjugate observable q_theta that gives the result [q_theta,p_theta]= i.hbar. You will find that it is the theta direction position axis.

Seratend.

Last edited: Apr 29, 2005
6. Apr 29, 2005

### Palindrom

OK great, thanks a lot.