Derivation of the density of states?

[patric44]

Homework Statement

i have a question concerning the derivation of the density of states

Relevant Equations

g(E) =sqrt(2)/pi^2*(m/hbar^2)^(3/2)*sqrt(E)

hi guys
i have a question about the derivation of the density of states , after solving the Schrodinger equation in the 3d potential box and using the boundary conditions ... etc we came to the conclusion that the quantum state occupy a volume of $\frac{\pi^{3}}{V_{T}}$ in k space
and to count the total number of quantum states its easier to count them in a shell with thickness dk then integrate , so we have :
$$N_{shell} =\frac{\frac{1}{8}4\pi*k^{2}dk}{\frac{\pi^{3}}{V_{T}}}$$
then integrating to get the total number of states give us :
$$N_{T} =\frac{V_{T}}{(2\pi^{2})}\frac{1}{3}k^{3}$$
and translating that expression in terms of the energy
$$N_{T} =\frac{V}{3}(\frac{\sqrt(2)}{\pi^{2}})(\frac{m}{\hbar^{2}})^{3/2}E^{3/2}$$
now dividing by V and E to get the number of quantum states per unit volume and energy give us :
$$g(E) =\frac{1}{3}(\frac{\sqrt(2)}{\pi^{}2})(\frac{m}{\hbar^{2}})^{3/2}E^{1/2}$$
but that expression doesn't look similer to the standerd one in textbooks
$$g(E) =(\frac{\sqrt(2)}{\pi^{}2})(\frac{m}{\hbar^{2}})^{3/2}E^{1/2}$$
what i am doing wrong here ?

 

[vanhees71]

Isn't $g(E)$ rather what you called $N_{\text{shell}}$ expressed in terms of $E=\hbar^2 k^2/(2m)$? Note that you also have to express also $\mathrm{d} k$ in terms of $\mathrm{d} E$. With this you get the textbook result, i.e.,
$$g(E)=\frac{\sqrt{m^3 E}}{\sqrt{2} \hbar^3 \pi^2}.$$

 

[patric44]

Isn't $g(E)$ rather what you called $N_{\text{shell}}$ expressed in terms of $E=\hbar^2 k^2/(2m)$? Note that you also have to express also $\mathrm{d} k$ in terms of $\mathrm{d} E$. With this you get the textbook result, i.e.,
$$g(E)=\frac{\sqrt{m^3 E}}{\sqrt{2} \hbar^3 \pi^2}.$$

i am a little bit confused here , isn't $N_{\text{shell}}$ is just the number of quantum states in the shell dk and the density of states is defined as the number of quantum sates per unit volume per unit energy , so i cannot just say that $\frac{N_{\text{shell}}}{V}$ is $g(E)$ since i need to get the total states by integration then divide by $V$ then divide by $E$ ? , replacing k and dk in $N_{\text{shell}}$ expression gives me :
$$ \frac{V}{2}*\frac{\sqrt{2}}{\pi^{2}}*(\frac{m}{\hbar^2})^{3/2}\sqrt{E}dE $$
which is not slimier to the expression i need :
$$g(E) = \frac{\sqrt{2}}{\pi^{2}}*(\frac{m}{\hbar^2})^{3/2}\sqrt{E}$$
am i misinterpreting something here ?

 

[vanhees71]

What's meant is by definition
$$g(E)=\frac{1}{V_T} \frac{\mathrm{d} N_{\text{shell}}}{\mathrm{d}E}=\frac{1}{V_T} \frac{\mathrm{d} N_{\text{shell}}}{\mathrm{d}k} \frac{\mathrm{d} k}{\mathrm{d} E}.$$
Your $N_{\text{shell}}$ should be rather $\mathrm{d} N_{\text{shell}}$.

Your additional factor 2 may be from a degeneracy of states, e.g., if you refer to the em. field in a cavity where for (almost all) $\vec{k}$ you have $g=2$ distinct polarization states.

 

[patric44]

What's meant is by definition
$$g(E)=\frac{1}{V_T} \frac{\mathrm{d} N_{\text{shell}}}{\mathrm{d}E}=\frac{1}{V_T} \frac{\mathrm{d} N_{\text{shell}}}{\mathrm{d}k} \frac{\mathrm{d} k}{\mathrm{d} E}.$$
Your $N_{\text{shell}}$ should be rather $\mathrm{d} N_{\text{shell}}$.

Your additional factor 2 may be from a degeneracy of states, e.g., if you refer to the em. field in a cavity where for (almost all) $\vec{k}$ you have $g=2$ distinct polarization states.

oh i think i got it now , so the density of states is just the number of energy states per unit volume confined to that specific dk shell , and the extra degeneracy will cancel the extra $\frac{1}{2}$ that i don't want , isn't that right ?
i am sorry but why is it defined that way ? its a bit confusing , isn't it just more intuitive to define the density of states as the total states in the 1/8 sphere per unit volume per unit energy ?!

 

[vela]

That would give you the average density whereas you want the density for a given $E$.

 

[mjohns92]

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