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Problem:

Calculate the density of states of a particle with mass $m$ in a 3D-harmonic oscillator with frequency $\omega$. $$ \rho(E) = \frac{m}{2\pi^2\hbar^3} \int d^3r \sqrt{2m(E-V(\vec{r}))}\Theta(E-V(\vec{r})) $$ with the heavy-side function $\Theta(x)$


Solution (so far):

To answer the question I have to calculate the integral. I'm unsure if the steps I did so far are correct.

1.

I tried to calculate the integral using spherical coordinates: $$ \int_0^{2\pi} \int_0^\pi \int_0^{r_0} \sqrt{2m(E-V(\vec{r}))} \, r^2 \sin(\theta) \, dr \, d\theta \, d\phi $$ The integral is $0$ for $r>r_0$ because of the heavy-side function.

2.

As long as $E$ is greater than $V(r)$ the heavy-side function is equal to $0$. $$\Rightarrow E= V(\vec{r_0}) \rightarrow E=\frac{1}{2} m \omega^2 r_0^2 \rightarrow r_0 = \sqrt{\frac{2E}{m\omega^2}} $$ This is the upper limit of the integration over $r$.


Edit 1:

I assumed that my solution is correct so far and tried to solve the integral $$\int_0^{r_0} \sqrt{2m(E-\frac{1}{2} m \omega^2 r^2))} \, r^2 \, dr$$ but I don't know how to do it. We got following hint on our exercise sheet: $$ \int_0^1 x^2\sqrt{1-x^2}dx = \int_0^1 x^2 \sqrt{\frac{1}{x}-1} dx = \frac{\pi}{16} $$

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  • $\begingroup$ Hint: 6-d phase space integral proportional to $\hbar^6$ $\endgroup$ – IntuitivePhysics Dec 1 '16 at 23:16

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