# Weird state density calculation

• Palindrom
In summary, the density of states for a free particle with momentum is calculated by finding N(E), taking the derivative with respect to E, but then getting stuck because they need to find 2pi sin(theta) which they can't seem to find. The problem is resolved by finding the integration domain of the integral and then using the relation [1] to find the conjugate momentum.

#### 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?

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)

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!

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?

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:
OK great, thanks a lot.

## 1. What is weird state density calculation?

Weird state density calculation is a method used to determine the number of energy states that are available to a system at a given temperature. It takes into account the quantum mechanical properties of particles and their interactions.

## 2. How is weird state density calculation different from regular state density calculation?

Weird state density calculation takes into account the effects of quantum mechanics, such as the Pauli exclusion principle, on the number of energy states available to a system. Regular state density calculation does not consider these effects and is only applicable to classical systems.

## 3. What are the applications of weird state density calculation?

Weird state density calculation is used in various fields of physics, such as statistical mechanics, condensed matter physics, and nuclear physics, to study the behavior of particles at the quantum level. It is also used in chemistry to understand the properties of molecules and their interactions.

## 4. How is weird state density calculation used in thermodynamics?

In thermodynamics, weird state density calculation is used to determine the partition function, which is a key quantity that relates the thermodynamic properties of a system to its microscopic properties. It is also used to calculate the entropy and free energy of a system.

## 5. Are there any limitations to weird state density calculation?

Yes, there are limitations to weird state density calculation. It is most accurate for systems with a large number of particles and at low temperatures. It also does not take into account the effects of relativity and can only be used for particles that obey the laws of quantum mechanics.