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Suppose we express the phase space volume occupied by all mechanical states $(\mathbf{q},\mathbf{p})$ with energy equal to or less than mechanical energy $E$ as:

\begin{equation}\Phi(E,\lambda)=\int\frac{d^{n}qd^{n}p}{h^n}\Theta(E-H(\mathbf{q},\mathbf{p}));\lambda)\end{equation}

Where $n$ is the degs of freedom, $H$ is the Hamiltonian for the system,$(\mathbf{q},\mathbf{p})$ the coordinates and generalized momentum, $\lambda$ a parameter of $H$, and $h$ has dimensions of action, leaving $\Phi$ dimensionless. Note that the density of states is then simply $\dfrac{\partial\Phi}{\partial E}$.

Suppose the simplest case, a 1-d particle in a box of length $L$. The Hamiltonian is simply $H=\frac{\mathbf{p}^2}{2m}$, and the energy levels:

\begin{equation}E_n=\frac{n²\hbar^2\pi^2}{2mL^2}\end{equation}

So, if $p\leq\frac{n\pi\hbar}{L}$: \begin{equation}\Phi(E)=\int\frac{dqdp}{h}\end{equation}

And 0 otherwise. So,as far as calculating D.O.S goes, seems like I got nowhere. Of course you could just say: \begin{equation}\omega(E)=\int dqdp=h\Phi(E)\end{equation}

But that's pretty circular. How can I get actual results from this definition?

Adrien-Marie Deschamps
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  • The question is not formulated entirely clearly. It is not difficult to verify that the number of states $\Phi(E)$ calculated quasi-classically agrees with the number calculated from the quantum spectrum if the energy $E$ is large enough, $E\gg\hbar^2/mL^2$. – Gec Mar 12 '24 at 04:41
  • What do you mean with the theta function acting on an operator in $\Theta(E-H(\mathbf{q},\mathbf{p}))$? – Jos Bergervoet Mar 12 '24 at 08:52
  • There is some liberal switching between classical and quantum here... also, $\lambda$ in the first question seems in a wrong place. Perhaps these answers could help: https://physics.stackexchange.com/a/640519/247642, https://physics.stackexchange.com/a/653648/247642 – Roger V. Mar 12 '24 at 09:14
  • @JosBergervoet the geometrical intuition behind $\Phi$ is that it's the hypervolume in phase-space, with the region determined by the energies given in the hamiltonian (we suppose a hamiltonian system). Then the theta function is simply putting a lower bound on mechanical energy, asserting that we integrate over the correct region – Adrien-Marie Deschamps Mar 12 '24 at 13:48

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As pointed out in the comments, there's some semiclassical taste that seems to be more motivated by confusion, rather than being an actual approach. Using quantized energy levels won't make much sense for what I think you're trying to do.

Of course, there's nothing wrong with your equation: \begin{equation}\Phi(E)=\frac{1}{h}\int dqdp \end{equation}

Since the movement is 1-d within a box of lenght $L$, and using classical momentum $p=\sqrt{2mE}$: \begin{equation}\Phi(E)=\frac{1}{h}\int_{0}^{L}dx\int_{-\sqrt{2mE}}^{\sqrt{2mE}}dp=\frac{2}{h}L\sqrt{2mE}\end{equation}

Now you can just take the partial derivative wrt energy and you have your answer. Note, as you said, $\Phi$ is associated with hypervolumes delimited by the hypersurfaces of energy. Here the system is 1-d, so the phase space is easy is a 2-d diagram, and $\Phi$ is just the area:

enter image description here

Where $p(E)=\sqrt{2mE}$. Note that the area in white is equal to $\Phi(E)$, minus the factor $h$, which is probably used to reduce dimensions.

Johann Wagner
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