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I wanted to find the density of states of a 1D ideal, noninteracting Fermi gas. My workings are below:

$$D(\epsilon) = \frac{1}{2\pi}\int_{0}^{\infty}\delta(\epsilon-\epsilon_k)dk \times2$$ $$\epsilon_k=\frac{\hbar^2k^2}{2m} \rightarrow dk=\frac{\sqrt{2m}}{\hbar}\cdot\frac{1}{2\sqrt{\epsilon_k}}d\epsilon_k$$

Which means that: $$D(\epsilon)=\frac{1}{\pi}\frac{\sqrt{2m}}{2\hbar}\int_0^{\infty}\frac{\delta(\epsilon-\epsilon_k)}{\sqrt{\epsilon_k}}d\epsilon_k=\frac{1}{2\pi\hbar}\sqrt{\frac{2m}{\epsilon}}$$ The irritating thing is, on sources I've checked, the answer "should" be: $$D(\epsilon)=\frac{1}{\pi\hbar}\sqrt{\frac{2m}{\epsilon}}$$ I'm not too sure as to where the additional factor of 2 may be coming from, as to me my logic (including the spin degeneracy of 2, etc) seems to make sense.

As an additional point to this, I had to do a similar procedure for 2D, and I got the correct answer that I anticipated I should have.

Am I missing something that would yield this factor of 2?

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It looks to me like you're only integrating over positive values of $k$. If you want to take account of electrons going the other way, you'll need an additional factor of $2$ (which is taken care of by the polar coordinate angular integral in $>1$ dimensions).

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    $\begingroup$ I didn't even think of that, but that makes a whole lot of a sense. Thank you! $\endgroup$ Oct 6, 2020 at 14:31

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