2
$\begingroup$

I know that the density of states $g(\epsilon)d\epsilon$ is the number of states in the energy range $[\epsilon, \epsilon + d\epsilon]$.

I considered a system of non-interacting free photons in 3 dimensions.

To calculate the density of states we just need:

1) The energy of our system. In this case we are dealing with the energy of photons, so:

$$\epsilon = \hbar \omega $$

2) The number of states with energy $\le \epsilon$ (let's call it $N(\epsilon))$. As 3 dimensions are considered, the number of states is enclosed in a sphere:

$$N(\epsilon) = \frac{4}{3} \pi R^3$$

Now I need to have the radius R in function of the energy so as to be able to compute the density of states:

$$g(\epsilon) = \frac{dN}{d\epsilon}$$

As I considered a free photon, its momentum is:

$$p = \hbar k$$

$$p = \frac{h}{L} (n_x + n_y + n_z)$$

Then, the energy can be rewritten as:

$$\epsilon (n_x, n_y, n_z) = \frac{p \omega}{k} = \frac{hc}{L}(n_x + n_y + n_z)$$

This expression should be correct (I checked dimensions), but it is not function of the radius.

But I suspect that $(n_x + n_y + n_z)$ has something to do with a geometric figure, which encloses the number of states...

Now I work out another example, completely independent of the one above.

COMPARISON WITH ANOTHER CASE TO SEE WHAT IS GOING ON

Actually I tested if $(n_x + n_y + n_z)$ had something to do with the radius of a sphere (I assumed 3D) using the free particle's case (without relativistic effects):

$$\epsilon = \frac{p^2}{2m}$$

For the energy, I got:

$$\epsilon (n_x, n_y, n_z) = \frac{h^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2)$$

At this point, I had the following thought:

$$(n_x^2 + n_y^2 + n_z^2) = R^2$$

And I gave it a shot... and ended up getting the correct density of states for the free particle (I also ignored spin as it is just about multiplying the final answer by a number).

So I would say that $(n_x + n_y + n_z)$ has something to do with the radius.

To sum up, I have the following troubles:

1') How can I get the radius from this equation $\epsilon (n_x, n_y, n_z) = \frac{p \omega}{k} = \frac{hc}{L}(n_x + n_y + n_z)$? Technically I cannot do $(n_x + n_y + n_z) = R$. Actually, I tried it but did not get it (at the end of my post I have written the density of states I should get).

2') In the $\epsilon = \frac{p^2}{2m}$ case I got:

$$\epsilon (n_x, n_y, n_z) = \frac{h^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2) = \frac{h^2 R^2}{2mL^2}$$

But this expression does not have dimensions of energy! I mean, if I regard $(n_x, n_y, n_z)$ as axis of my phase space I have no problems with dimensions but once I regard $(n_x^2 + n_y^2 + n_z^2)= R^2$ dimensions do not match in the equation for the above energy... but I get the right result! what is going on here?

3) Why the dimension of the density of states is time in photons and in the free particle case (assuming nonrelativistic quantum mechanics) is $\frac{T^2}{ML^2}$?

I know that the density of states of free photons (assuming Plank's energy; see Griffiths, page 244; EQ 5.112; second edition):

$$g(\omega) = \frac{V \omega^2}{\pi c^3}$$

Any help is appreciated.

$\endgroup$
  • 1
    $\begingroup$ First of all, momentum is a vector even in the photon case. That said $p = |\overrightarrow p| = \sqrt{n_x^2 + n_y^2 + n_z^2}/L$. $\endgroup$ – rnels12 Dec 13 '18 at 11:13
  • $\begingroup$ @rnels12 I think it should be $p = |\overrightarrow p| = h\sqrt{n_x^2 + n_y^2 + n_z^2}/L$. $\endgroup$ – JD_PM Dec 13 '18 at 17:59
1
$\begingroup$

The problem with your approach is that you ignore $\boldsymbol{k}$ being a vector. In fact

$$\boldsymbol{p}=\hbar \boldsymbol{k}=\frac{h}{L}\boldsymbol{n}$$

where $\boldsymbol{n}=\left(n_{x},n_{y},n_{z}\right)$. This gives you

$$R=\left|\boldsymbol{n}\right|=\frac{L}{h}\left|\boldsymbol{p}\right|=\frac{L}{hc}\varepsilon=\frac{L}{2\pi c}\omega$$

so (the factor of $2$ accounts for the two polarizations of the photon)

$$N\left(\omega\right)=2\cdot\frac{4}{3}\pi R^{3}=\frac{8\pi L^{3}}{3 \left(2\pi\right)^{3} c^3}\omega^{3}=\frac{V}{3 \pi^{2} c^3}\omega^{3}$$

$$g\left(\omega\right)=\frac{\partial N\left(\omega\right)}{\partial\omega}=\frac{V}{\pi^{2} c^3}\omega^{2}$$

$\endgroup$
  • $\begingroup$ thank you, your answer is helpful! Why $R=\left|\boldsymbol{n}\right|$ is the radius? $\endgroup$ – JD_PM Dec 13 '18 at 12:33
  • 1
    $\begingroup$ Because $\boldsymbol{n}=\left(n_{x},n_{y},n_{z}\right)$ satisfies $\left|\boldsymbol{n}\right|\leq \frac{L}{2\pi c}\omega\equiv R$, and $\boldsymbol{n}$ is like $\boldsymbol{r}=\left(x,y,z\right)$. $\endgroup$ – eranreches Dec 13 '18 at 12:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.