Derivation of the density of states?

In summary, the density of states is the number of energy states per unit volume confined to that specific dk shell. The extra degeneracy will cancel the extra 1/2 that you don't want.
  • #1
patric44
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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 ?
 
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  • #2
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.,
$$g(E)=\frac{\sqrt{m^3 E}}{\sqrt{2} \hbar^3 \pi^2}.$$
 
  • #3
vanhees71 said:
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 ?
 
  • #4
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.
 
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  • #5
vanhees71 said:
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 ?!
 
  • #6
That would give you the average density whereas you want the density for a given ##E##.
 
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Related to Derivation of the density of states?

1. How is the density of states defined?

The density of states is defined as the number of energy states per unit volume that are available to be occupied by particles in a given energy range.

2. What is the significance of the density of states?

The density of states is an important concept in understanding the behavior of particles in a material. It helps to determine the number of particles that can occupy a certain energy level, which in turn affects the material's physical and chemical properties.

3. How is the density of states related to energy levels?

The density of states is directly proportional to the number of energy levels available in a material. This means that as the energy levels increase, so does the density of states.

4. What is the formula for calculating the density of states?

The formula for calculating the density of states varies depending on the system being studied. In general, it can be expressed as dN/dE, where dN is the number of states in a given energy range and dE is the width of that energy range.

5. How does temperature affect the density of states?

Temperature has a significant impact on the density of states. As temperature increases, the number of available energy states also increases, resulting in a higher density of states. This can lead to changes in the material's properties, such as conductivity and thermal conductivity.

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