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I've recently seen an example where a thermal average was carried using a plus sign instead of the usual minus sign inside the exponential.

$$\langle \mu \rangle = \frac{1}{Z} tr(e^{\beta H }\mu) \qquad\qquad Z = tr(e^{\beta H }) $$

I have simplified the equations they have used to keep it clear, the form of $\mu$ is not important but it is used to calculate the hole mobilities $\langle \mu \rangle $ of a crystal. $H$ is essentially a tight binding type Hamiltonian

$$H = \sum_{n}E_{n}\vert n \rangle\langle n \vert + \sum_{n,n'}V_{nn'}\vert n \rangle \langle n' \vert$$

where $\vert n \rangle$ are charge localised (hole) states, this is also written in second quantisation in some papers. I've heard some say that the sign change in the exponential is necessary so that you sum over the top of the band structure or something. However this sounds like a mistake because this idea of filling the top of band structure refers to hole occupations of the orbitals of the many-body wavefunction $\vert n \rangle$ right? So since we are taking the trace over eigenstates $\vert \psi_{n} \rangle$ of $H$ which are a linear-combination of $\vert n \rangle$ shouldn't we be using the usual $e^{-\beta H }$ factor?

Source: https://doi.org/10.1021/acs.jpcc.8b11916 Eq.~8 + Eq.~10

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  • $\begingroup$ Is it possible to know the $H(\mu)$ dependence? $\endgroup$
    – Javi
    Commented May 29, 2021 at 8:53
  • $\begingroup$ So the full equations is $L^{2}_{x} = \frac{1}{Z}\sum_{n,m} e^{\beta H}\langle \psi_{n} \vert j_{x} \vert \psi_{m} \rangle \langle \psi_{m} \vert j_{x} \vert \psi_{n} \rangle \frac{2}{(\hbar / \tau) - (E_{m} - E_{n})^{2}}$ and $\langle \mu \rangle = \frac{e}{kT}\frac{L^{2}_{x}}{2\tau}$ where $\psi_{m}$ are the eigenvectors of $H$ and $ j_{x} $ are current operators. $\endgroup$
    – Unskilled
    Commented May 29, 2021 at 9:07
  • $\begingroup$ doi.org/10.1021/acs.jpcc.8b11916 source for the above equation $\endgroup$
    – Unskilled
    Commented May 29, 2021 at 9:17
  • $\begingroup$ Also the system is put under periodic boundary conditions so the summations are large but finite. $\endgroup$
    – Unskilled
    Commented May 29, 2021 at 9:46

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If the number of states or the energy is not bounded, then $e^{\beta H}$ is not a well-defined probability distribution (cannot be normalized). Sometimes, the minus sign is just included in the definition of the Hamiltonian or in the $\beta$ parameter.

EDIT: Now that I remember this is the usual trick when dealing with holes. In the typical simplest model for particle absorption, in which particles have an energy $\epsilon_1=\epsilon$ if they are stuck to a lattice with $N$ nodes or energy $\epsilon_1=0$ otherwise. Then the number of absorbed particles is:

$N_{particles}=\frac{N}{1+e^{-\frac{\epsilon}{kt}}}$

You can compute from here the statistics for the holes using $N=N_{ particles}+N_{holes}$

$N_{holes}=\frac{N}{1+e^{\frac{\epsilon}{kt}}}$

So you could think of the holes as particles that have an energy spectrum that is the negative version of the original particles.

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  • $\begingroup$ ok so in this paper doi.org/10.1021/acs.jpcc.9b01902 they state that "Note that for hole carriers, the thermal weighting factor must be adapted by changing the sign in the exponent". $\endgroup$
    – Unskilled
    Commented May 29, 2021 at 9:15
  • $\begingroup$ Yes, reading the paper I remembered that you can do this kind of thing with the holes. Basically, if you were to compute the partition function for the particles, you would be using the proper $e^{-\beta H}$ weight. I edited my answer to give an example. $\endgroup$
    – Javi
    Commented May 31, 2021 at 7:45
  • $\begingroup$ Are you sure that is right? The $N_{particles}$ and $N_{holes}$ comes from the fermi-dirac statistics of electron and holes and is for the case of a noninteracting wavefunction of fermions right? However the wavefunctions $\vert \psi_{n} \rangle$ are the eigenstates of $H$ and would be formed from a basis set of noninteracting wavefunctions $\vert n \rangle$ . So you wouldn't use fermi-dirac statistics here because we run the thermal average over states $\vert \psi_{n} \rangle$ rather then electron/hole occupations. I'm using a slightly different notation to the paper. $\endgroup$
    – Unskilled
    Commented May 31, 2021 at 10:55
  • $\begingroup$ Honestly, I am not 100% sure. This is rather an intuition. However, I think that this result for the $N_{particles}$ and the $N_{holes}$ would stand if you use the thermal average over the states $|\psi_n>$, this is, the canonical ensemble ($<N_\epsilon>=\frac{Ne^{-\beta \epsilon}}{Z}$) $\endgroup$
    – Javi
    Commented Jun 1, 2021 at 9:11
  • $\begingroup$ Is that right? Isn't $\langle N_{\epsilon} \rangle = \frac{Ne^{-\beta \epsilon}}{Z}$ for a noninteracting case? $H$ in my example includes off diagonal terms so $\langle N_{\epsilon} \rangle = \frac{1}{Z} tr (e^{-\beta H} N_{\epsilon})$ which is not equal to the previous equation as far as I can tell. $\endgroup$
    – Unskilled
    Commented Jun 1, 2021 at 10:25

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