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When we calculate the density of states for an electron, in the standard way as done in statistical mechanics textbooks ( by integrating once over space, then integrating over $\theta$ and $\phi$ of the k-space,etc), we finally multiply a '2' as the spin degeneracy.

My question is, since the electron can be in an infinite number of states on the Bloch sphere, why are we just taking two states? I know that only two of the infinite number of states are orthogonal, but even if the huge number of states are non-orthogonal, why should not they contribute to an increase in the number of states?

I did not want to use the term degeneracy in the title, because degeneracy is defined as the dimension of the eigenspace of the degenerate eigenvalue, so that is of course two. What I want to know is, why are we taking only the orthogonal states?

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  • $\begingroup$ Because there are only two states that the spin can take , with or against the direction of motion? The infinity you are talking about is the infinity of possible orientations/coordiante-systems.not degrees of freedom. $\endgroup$ – anna v Apr 29 at 13:54
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    $\begingroup$ Why should I take only the degrees of freedom? The state of the electron can be a|up>+b|down>, where a and b can take any possible values and |up> and |down> are the orthogonal kets. SInce these states are all, at the least, 'distinct' (not necessarily orthogonal), why shouldn't I count one state for each of these states? $\endgroup$ – Abhirup Mukherjee Apr 29 at 14:04
  • $\begingroup$ @b can be +1/2 or -1/2 only as far as spin goes. $\endgroup$ – anna v Apr 29 at 15:12
  • $\begingroup$ The spin state can be a superposition of those states. $\endgroup$ – Abhirup Mukherjee Apr 29 at 15:32
  • $\begingroup$ No , quantum numbers are just that, numbers. $\endgroup$ – anna v Apr 29 at 16:43
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Good question! Really what you care about when calculating the density of states is the number of orthogonal states, whether you are talking about spin, spatial states, or whatever. In other words, you want to know density of Hilbert space dimension. To give a non-spin example if you are looking at the density of states in an $L\times L\times L$ box of Fermi gas, you get the momentum states by enforcing periodic boundary conditions... but by your argument, as soon as you have two momentum eigenfunctions, you should have infinite density of states, because each single particle could be in any superposition of those two states. As you know, this is not actually the case: the same applies for spins. Count the number of orthogonal states, not the total number of vectors in the Hilbert space (usually infinite!)

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    $\begingroup$ Yeah I know that. But suddenly it feels like I have been assuming it all the time. By 'it', I mean the fact that we take only orthogonal states while calculating the density of states. Could you shed some light as to why we take only orthogonal states? I have a gut feeling its something really basic and simple. $\endgroup$ – Abhirup Mukherjee Apr 29 at 19:30
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    $\begingroup$ To clarify as to why I feel that we should take all states (including non-orthogonal), my reasoning is that since non-orthogonal states are also states in which the electron can reside, they should increase the probability of the electron having that energy. Consequently, those non-orthogonal states should increase the count of the density of states. $\endgroup$ – Abhirup Mukherjee Apr 29 at 19:35
  • $\begingroup$ I think the problem is hiding in saying things like "the probability of the electron having that energy," which is not an operationally well-defined idea. What we can talk about it is "the probability of finding the electron with that energy." This is what the density of states really refers to, but the probability of finding an electron with a given energy is a property of a particular measurement, and we know that any measurement of the spin (of an electron) has two possible outcomes, not infinite. This is a subtle point -- thanks for making me think about it :) $\endgroup$ – Will Apr 29 at 19:47
  • $\begingroup$ Does that answer your question? Basically, you are correct that the density of states is ultimately about probabilities, but these depend on numbers of measurement outcomes satisfying some criterion (e.g., returning some given energy), not on the actual number of state vectors. $\endgroup$ – Will Apr 29 at 19:51
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    $\begingroup$ What you mean is that the density of states g(k) will be more if there are more number of states available in which we can 'detect' the electron such that it has wave-vector k. And since detectable states have to be orthogonal, the non-orthogonal states dont matter. The non-orthogonal states dont influence the dimension of the Hilbert space. Does that sound okay? $\endgroup$ – Abhirup Mukherjee Apr 29 at 20:11
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I like Will's answer, but I thought I would also add one showing where the factor of $2$ (and, in general, the dimension of the degenerate Hilbert space) actually comes from in the formalism.

In the canonical formulation, if we decompose the Hilbert space in question as $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$, where you can think of $\mathcal{H}_1$ as describing the space component and $\mathcal{H}_2$ as describing spin, then the partition function for the system at temperature $\beta=1/kT$ is

$$Z\equiv\sum_{E}e^{-\beta E}=\text{Tr}\left(e^{-\beta H}\right),$$

where $H$ is the Hamiltonian acting on $\mathcal{H}$, and the $\text{Tr}$ trace is acting on the whole Hilbert space as well.

Now, if all spins are degenerate at each energy level, then the Hamiltonian naturally decomposes as $H=H_{s}\otimes\textbf{1}$, where $H_{s}$ can be thought of as the spatial Hamiltonian. Then the partition function will be

$$Z=\text{Tr}\left(e^{-\beta H_s\otimes\textbf{1}}\right)=\text{Tr}\left(e^{-\beta H_s}\otimes\textbf{1}\right)=\text{tr}_1\left(e^{-\beta H_s}\right)\text{tr}_2(\textbf{1})=\text{dim}(\mathcal{H}_2)Z_{s},$$

where $Z_{s}$ is defined to be the partition function for $H_s$. This is schematically where the dimension of the degenerate Hilbert space comes from in these calculations. In particular, since the density of states for a continuous spectrum is defined so that

$$Z=\int\mathrm{d}E\,g(E)\,e^{-\beta E},$$

it is clear that having a degenerate subspace $\mathcal{H}_2$ corresponds to multiplying the density of states by $\dim(\mathcal{H}_2)$, or, in the case of spin, multiplying by $2$ (or $2s+1$ for a massive spin-$s$ particle).

I hope this helps!

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  • $\begingroup$ Thanks for giving a more formal perspective. $\endgroup$ – Abhirup Mukherjee Apr 30 at 9:19
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    $\begingroup$ This is a unequivocally a better answer than mine... $\endgroup$ – Will Apr 30 at 14:25

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