Operator expectation value for system of non-interacting particles (Fermions)

Question

I am reading the book "Electronic Structure" by Richard Martin which poses the following problem:

Show that the expectation value of an operator $\hat O$ in a system of identical, non-interacting Fermions (i.e. electrons with the independent particle approximation) has the following form:

$$ \left<\hat O\right> = \sum_{i,\sigma} f_i^\sigma \left<\psi_i^\sigma|\hat O|\psi_i^\sigma\right>$$

Where, $f_i^\sigma = \frac{1}{e^{\beta(\epsilon_i^\sigma - \mu)} + 1}$ and $\psi_i^\sigma$ is the eigenvector of the ith single particle state with spin $\sigma$.

I have begun the derivation thusly (j is the jth eigenstate of the multibody system):

$$ \left<\hat O\right> = Tr\left(\hat \rho \hat O \right) = \sum_j \left<\Psi_j|\hat \rho \hat O |\Psi_j \right>$$

Since $\hat \rho$ is Hermitian:

$$ \sum_j \left<\Psi_j|\hat \rho \hat O |\Psi_j \right> = \sum_j \left<\hat \rho\Psi_j| \hat O |\Psi_j \right> $$

In the Grand Canoncial Ensemble: $\hat \rho = \frac{1}{Z}e^{-\beta(\hat H - \mu\hat N)} $

Further, considering that the jth eigenstate of the multibody system will be composed of N single particle wave functions, one can write:

$$ \sum_j \left<\hat \rho\Psi_j| \hat O |\Psi_j \right> = \sum_{\{n_i,\sigma_i\}} \frac{1}{Z}e^{-\beta(\sum_i n_i\epsilon_i^{\sigma_i}-\mu \sum_i n_i)}\left<\Psi_j| \hat O |\Psi_j \right> $$

Based on my understanding of the other quantum mechanics reference I have been using, if $\hat O$ was simply the operator which returns the number of particles in the ith single particle state with spin $\sigma$, the sum reduces to the $f_i^\sigma$. So my guess as how to obtain the general result is to suppose that $\hat O$ is the combination of N single particle operators $\hat O = \sum_i \hat O_i$.

Hence,

$$\left<\Psi_j| \hat O |\Psi_j \right> = \left<\Psi_j| \sum_i \hat O_i |\Psi_j \right> = \sum_i \left<\psi_i^\sigma | \hat O_i | \psi_i^\sigma \right>$$

After this point, I am stuck and I do not see how this could reduce down to the given result. Further it would seem to me that the author should have placed a restriction on $\hat O$ to be only a single particle operator on the rhs of the target result.

Answer 1

I think this is a situation where it is much easier to derive the desired result using second quantization. To start, let us consider a system of identical particles and let $O = \sum\limits_{k} o_k $ denote a generic one-body operator on the respective Fock space. In the language of second quantization, we can express this operator as $$ O = \sum\limits_{ij} \langle i|o|j\rangle \,a_i^\dagger a_j \quad . $$ The expectation value of an (one-body) operator in the state $\rho$ is defined by $\langle O \rangle_\rho \equiv \mathrm{Tr} \, \rho \, O$, where the trace is performed on the Fock space. By defining the elements of the one-body reduced density matrix in the state $\rho$ as $$\gamma_{ij} \equiv \mathrm{Tr}\, \rho\, a_j^\dagger a_i \quad ,$$ we see that we can write the expectation value of $O$ as $$\langle O \rangle _\rho = \mathrm{tr}\, \gamma \,o \quad , $$ where now the trace is performed on the single-particle Hilbert space.

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In the following, $i,j$ denote elements of the basis in which the single-particle Hamiltonian is diagonal.

For a system of non-interacting fermions and in equilibrium in the grand canonical ensemble, it follows that

$$ \gamma_{ij} = \delta_{ij} \,\langle n_i \rangle_\rho \quad . $$ This can be derived by e.g. applying Wick's theorem. Here, $$\langle n_i \rangle_\rho = \frac{1}{e^{(\epsilon_i - \mu)/{k_{\mathrm{B}}T}}+1} $$ is the well-known expression for the average occupation number of the single-particle state $i$.

Finally, this shows that indeed $$ \langle O \rangle_\rho = \sum\limits_i \frac{1}{e^{(\epsilon_i - \mu)/{k_{\mathrm{B}}T}}+1}\, \langle i|o|i\rangle \quad \quad ,$$ for a system of non-interacting fermions in grand canonical equilibrium.

As a last point, note that an analogous result holds for the case of non-interacting identical bosons.

Answer 2

So I believe it can also be derived using 1st quantization as follows:

$$\langle \hat O \rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \langle \Psi_j|\hat \rho \hat O |\Psi_j\rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \langle \hat \rho \Psi_j|\hat O |\Psi_j\rangle =\sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j \langle \Psi_j| \hat O |\Psi_j\rangle$$

Where $\tilde \rho_j$ is the eigenvalue of $\hat \rho$ associated with state j.

Presuming that $\hat O$ can be written as the sum of one body operators:

$$\hat O = \sum_k \hat o_k$$

Where k is the index of particle k. One can now write:

$$\langle \hat O \rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j \langle \Psi_j| \sum_k \hat o_k |\Psi_j\rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j \sum_i n_i\langle \psi_i| \hat o |\psi_i\rangle$$

Where I have used the fact that the expectation value of the single body operator $\hat o_k$ acting on the the multibody wave function $\Psi_j$ is: $$ \langle \Psi_j | \hat o_k | \Psi_j \rangle = \sum_i n_i \frac{(N-1)!}{N!} \langle\psi_i|\hat o |\psi_i \rangle$$

Where i sums over all the single-body eigenstates present in $\Psi_j$, the $N!$ in the denominator comes from the fact that $\Psi_j$ is represented by a Slater determinant of single-body eigenstates. The $(N-1)!$ in the numerator represents the fact that each particle index is repeated $N!/N$ times. Finally, $n_i$ represents the number of times that eigenstate i is repeated.

Thus when this sum is performed N times, the factorials are eliminated.

Returning to $\langle \hat O \rangle$, one has:

$$\langle \hat O \rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j \sum_i n_i\langle \psi_i| \hat o |\psi_i\rangle = \sum_{\left\{ n \right\}_j \in N^{\infty}} \sum_i \tilde \rho_j n_i\langle \psi_i| \hat o |\psi_i\rangle = \sum_{i} \sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j n_i\langle \psi_i| \hat o |\psi_i\rangle $$

Which can be achieved by re-indexing the sum appropriately (I think?).

Continuing:

$$\langle \hat O \rangle = \sum_i \left( \sum_{\left\{ n \right\}_j \in N^{\infty}} \tilde \rho_j n_i \right) \langle \psi_i| \hat o |\psi_i\rangle = \sum_i \langle n_i \rangle \langle \psi_i| \hat o |\psi_i\rangle$$

Using the fact that: $$ \langle n_i \rangle = \frac{1}{e^{\beta(\epsilon_i-\mu)} \pm 1} = f_i$$

One has:

$$\langle \hat O \rangle = \sum_i f_i \langle \psi_i| \hat o |\psi_i\rangle $$

As originally desired.