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?
Best Answer
The density of states is defined so that $\nu(E) dE$ is the number of states with energy in the small interval $(E, E + dE)$. In other words, the total number of states up to some energy $E_0$ (not just at $E_0$) is \begin{equation} N(E_0) = \int_0^{E_0} \nu(E) dE \quad\quad (1). \end{equation} However, another way of writing this number is \begin{align} N(E_0) &= \sum_{|\vec{p}| < \sqrt{2m E_0}} 1 \\ &= \sum_{\vec{p}} \theta \left ( E_0 - \frac{|\vec{p}|^2}{2m} \right ) \\ &\approx \frac{V}{h^3} \int \theta \left ( E_0 - \frac{|\vec{p}|^2}{2m} \right ) d^3 p. \quad\quad (2) \end{align} Something which should be clear from (1) is that $\nu(E) = \frac{dN}{dE}$. Applying this to (2), we can use the fact that Heaviside functions differentiate to a Dirac delta to get \begin{equation} \nu(E) = \frac{V}{h^3} \int \delta \left ( E_0 - \frac{|\vec{p}|^2}{2m} \right ) d^3 p \end{equation} which looks like what you wrote. So there is no contradiction here. Letting infinite momenta appear in the domain is fine because (for finite $E_0$) the delta will kill their contribution.
Now to answer your first question, an integral of 1 over the momenta that satisfy $|\vec{p}| < \sqrt{2mE}$ is nothing but the volume of a sphere having radius $\sqrt{2mE}$. Therefore we get \begin{align} \nu(E) = \frac{V}{(2\pi \hbar)^3} \frac{d}{dE} \left [ \frac{4}{3} \pi \sqrt{2mE}^3 \right ] = V \frac{\sqrt{m^3 E}}{\sqrt{2} \pi^2 \hbar^3}. \end{align}