In our lecture today, the professor introduced the concept of density of states. We found the expression of it, for the 3D case, but no steps were shown and also we did not specify the system at hand and what type of ensemble we are considering. With that said, the following was written:
$$\Sigma_{\vec p} \rightarrow V\int \frac{d^3p}{(2\pi\hbar)^3}\rightarrow V\int \nu(E)dE$$
where $\nu(E)=\int \frac{\delta(\epsilon-\epsilon_p)}{(2\pi\hbar)^3}d^3p$ is the density of states.
Then considering that $\epsilon_{\vec p}=\frac{\vec p^2}{2m}$ we have ultimately:
$$\nu (E)= \frac{m^{\frac 3 2} \epsilon^{\frac 1 2}}{\sqrt{2} \pi^2 \hbar^3}$$.
EDIT: I assume that we are considering the case of 1 particle in a box.
I am interested in two things:
- How did we reach the above formula. I tried to derive it but I couldn't.
- In this formula $\nu(E)=\int \frac{\delta(\epsilon-\epsilon_p)}{(2\pi\hbar)^3}d^3p$ , while the boundaries for the momentum are from $-\infty$ to $\infty$, for the energy we have a finite specific value, in this case $\epsilon_{\vec p}=\frac{\vec p^2}{2m}$. How is this possible? When we know from this: $\epsilon_{\vec p}=\frac{\vec p^2}{2m}$ that the momentum cannot have infinite values, but values which satisfy this equation?