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

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.

• Consider to use \langle and \rangle instead of < and >; this looks much better IMO. Commented Jul 26, 2021 at 7:56
• Note: In general, the (non-interacting grand canonical) equilibrium expectation values of n-body operators can be evaluated with the help of Wick's theorem, c.f. e.g. Reichl: 'A Modern Course in Statistical Physics', 4th edition, Appendix D, section 2.3. Commented Jul 26, 2021 at 11:57

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.

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.

• Thanks, I am not very familiar with second quantization so I don't quite follow the first half of the derivation completely. Though I think I have some ideas now to try to reduce the operator expectation value to $tr\gamma o$ in 1st quantization. If I can get there, then as you have shown, the rest of the derivation follows from the fact that $\gamma_{ij} = \delta_{ij} \langle n_i \rangle$ Commented Jul 26, 2021 at 17:56
• @gigo318 Okay, great. A good introduction for the case of first quantization is given in Parr and Yang: 'Density-Functional Theory of Atoms and Molecules', 1989, section 2.3. The derivation of the fact that one can express expectation values with the help of the 1-RDM is similar as the derivation that for local one-body operators you can use the density to express the expectation value. Anyway, it is always good to learn some second quantization. Commented Jul 26, 2021 at 18:19
• @gigo318 Actually, there are better sources for the derivation. Take a look here or here, section 3.52. starting at p. 21. Hope this helps. Commented Jul 26, 2021 at 18:58
• Cool, I will definitely give these a read! I feel like many DFT sources I have read are either too cursory or jump straight into more advanced aspects of the theory. Commented Jul 26, 2021 at 19:06

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.