I am stuck calculating a simple expectation value for an operator, which is expressed in second quantization. I know the result, but I fail to proof it.

Lets say I have one-particle wave function $|\phi_n\rangle$ given by $|\phi_n\rangle=\sum_{j=1}^K |\alpha_j\rangle A_{j,n}$, where $K$ is the number of orbitals/sites in the system and the $A_{j,n}$ are the probability amplitudes of the oribitals $|\alpha_j\rangle$. The index $n$ labels the particles in the system, of which we have $N$.

The orbitals are orthonormal, i.e. $$\sum_j A_{j,m}^* A_{j,n} = \delta_{m,n}.$$

Let's ignore any spin degrees of freedom. The many particle wave function is now given by $$|\Psi\rangle = \left(\prod_{n=1}^N \sum_{j=1}^K \hat{c}^\dagger_j A_{j,n}\right) |\text{vac}\rangle,$$ where the $\hat{c}^\dagger_j$ is the usual creation operator on site $j$.

What I now want to calculate, is the expectation value $$\langle \Psi|H_\text{hop}|\Psi\rangle$$ with $$H_\text{hop}=-t \sum_{j=1}^K \hat{c}^\dagger_{j+1}\hat{c}_j + h.c.$$ the usual hopping Hamiltonian. I have the strong feeling (and one example calculation supported this), that the result is just $$\langle \Psi|H_\text{hop}|\Psi\rangle = -t \sum_{n=1}^N \sum_{j=1}^K ( A_{j,n}^* A_{j+1,n} + A_{j+1,n}^* A_{j,n} )$$

I think this result is trivially related to the Slater-Condon rules, but I fail to see the connection. Additionally, I fail to explicitly calculate the expectation value, which contains the sum and products of the creation/annihilation operators.

What is a good way to prove my result?

  • $\begingroup$ Hint: use the CCR-s associated with the ${\hat c}_j$-s. Doesn't your sample calculation already suggest something like this? $\endgroup$
    – udrv
    May 31, 2016 at 23:39
  • $\begingroup$ Well, I bet I have to use the commutation relations at some point. But I really don't see, how I can arrive at the general result. As the scalar product is linear, it is surely enough to just consider one term from $H_\text{hop}$, so maybe one can set w.l.o.g. $j=1$. But then I have still lots of $\hat{c}^\dagger$ from the $|\Psi\rangle$ (Slater determinants). How do I proceed? $\endgroup$
    – Merlin1896
    Jun 1, 2016 at 7:47

1 Answer 1


For future convenience denote $$ {\hat \phi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_j} $$ $$ {\hat \chi}^\dagger_n = \sum_j{A_{j,n} {\hat c}^\dagger_{j+1}} $$ The average you want to calculate reads then $$ \langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\langle 0 | \prod_n{{\hat \phi}_n} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} |0\rangle $$ Starting with the regular CCR-s, $$ \left[ {\hat c}_j, {\hat c}^\dagger_k \right]_\mp = \delta_{jk},\;\;\;\;\; \left[ {\hat c}_j, {\hat c}_k\right]_\mp = \left[ {\hat c}^\dagger_j, {\hat c}^\dagger_k\right]_\mp = 0 $$ obtain $$ \left[ \sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j}, {\hat \phi}^\dagger_n\right] = {\hat \chi}^\dagger_n $$ and $$ {\hat \phi}^\dagger_n {\hat \chi}^\dagger_m = \pm {\hat \chi}^\dagger_m {\hat \phi}^\dagger_n $$ $$ {\hat \phi}_n {\hat \chi}^\dagger_m = \pm {\hat \chi}^\dagger_m {\hat \phi}_n + \sum_j{A^*_{j+1, n}A_{j,m}} $$ Now, in the average you need to calculate, use the above to successively move $\left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)$ past the orbital operators to its right: $$ \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right)\prod_n{{\hat \phi}^\dagger_n} = {\hat \phi}^\dagger_1 \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>1}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} = $$ $$ = \prod_{m=1}^{m=2} {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} + {\hat \phi}^\dagger_1 {\hat \chi}^\dagger_2 \prod_{n > 2} {{\hat \phi}^\dagger_n} = $$ $$ = \prod_{m=1}^{m=2} {\hat \phi}^\dagger_m \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) \prod_{n>2}{{\hat \phi}^\dagger_n} + {\hat \chi}^\dagger_1 \prod_{n\neq 1} {{\hat \phi}^\dagger_n} \pm {\hat \chi}^\dagger_2 \prod_{n \neq 2} {{\hat \phi}^\dagger_n} = \dots = $$ $$ = \prod_m {{\hat \phi}^\dagger_m} \left(\sum_j{{\hat c}^\dagger_{j+1}{\hat c}_j} \right) + \sum_m{ (\pm 1)^{m-1} {\hat \chi}^\dagger_m \prod_{n \neq m} {{\hat \phi}^\dagger_n}} $$ The first term will annihilate the rhs vacuum, so only the sum remains. Now bring in the orbital operators on the left and flip each ${\hat \chi}^\dagger_m$ past them: $$ \left(\prod_n{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m = \pm \left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m {\hat \phi}_1 + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} = $$ $$ \left(\prod_{n>1}{{\hat \phi}_n}\right) {\hat \chi}^\dagger_m \prod_{l=1}^{l=2} {{\hat \phi}_l} + \left(\prod_{n\neq 1}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, 1}A_{j,m}} \pm \left(\prod_{n\neq 2}{{\hat \phi}_n}\right)\sum_j{A^*_{j+1, 2}A_{j,m}} = \dots = $$ $$ = {\hat \chi}^\dagger_m \left(\prod_n{{\hat \phi}_n}\right) + \sum_l{(\pm 1)^{l-1}\left(\prod_{n\neq l}{{\hat \phi}_n}\right) \sum_j{A^*_{j+1, l}A_{j,m}} } $$ The first term now annihilates the lhs vacuum, and after substituting everything the desired average becomes $$ \langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1} A^*_{j+1, l}A_{j,m} \langle 0 |\left(\prod_{n\neq l}{{\hat \phi}_n}\right)\left( \prod_{n' \neq m} {{\hat \phi}^\dagger_{n'}}\right) |0\rangle } = $$ $$ \langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, l,m}{ (\pm 1)^{m-1} (\pm 1)^{l-1} A^*_{j+1, l}A_{j,m}\delta_{l,m} } $$ and finally $$ \langle \Psi | {\hat H}_0| \Psi \rangle = - t\;\sum_{j, m}{ A^*_{j+1, m}A_{j,m} } $$ There may be some bugs I missed, but this is the general idea.


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