Suppose we have the following Slater determinant: \begin{equation} | \Psi \rangle = \prod \limits_{i,i'} a^+_{i\alpha} a^+_{i'\beta} | \rangle \end{equation} where $a^+_{i\alpha}$ creates an electron in state $i$ with spin $\alpha$ and, in general, $i \neq i'$. I want to evaluate $\langle \Psi | S^2 | \Psi \rangle$ using second quantization. We can express the $S^2$ operator as \begin{equation} S^2 = S_- S_+ + S_z (S_z +1) \end{equation} with \begin{equation} S_- = \sum_p a^+_{p\beta} a_{p \alpha} \hspace{1.5cm} S_+ = \sum_p a^+_{p\alpha} a_{p \beta}. \end{equation} Since $|\Psi \rangle$ is an eigenfunction of $S_z$, evaluating $\langle \Psi | S_z | \Psi \rangle$ terms becomes trivial and the problem reduces to the evaluation of $\langle \Psi | S_- S_+ | \Psi \rangle$. In the standard restricted Hartree-Fock method, $i = i'$ and it is easy to show that $\langle \Psi | S_- S_+ | \Psi \rangle = 0$ using the canonical anticommutation relations. When $i \neq i'$ (unrestricted Hartree-Fock) we must have that $\langle \Psi | S_- S_+ | \Psi \rangle = N_\beta - \text{Tr}(PQ)$ (this I verified in books) where $N_\beta$ is the number of $\beta$ electrons and $P_{ij} = \langle \Psi | a^+_{j\alpha} a_{i\alpha} | \Psi \rangle$ and $Q_{ij} = \langle \Psi | a^+_{j\beta} a_{i\beta} | \Psi \rangle$.

My question is then, specifically, how do we obtain $\langle \Psi | S_- S_+ | \Psi \rangle = N_\beta - \text{Tr}(PQ)$ using the anticommutation relations of second quantization? If I try to use these anticommutation relations just in the way they are written in textbooks I do not get the right answer. Clearly, I am getting something wrong or some special consideration needs to be taken into account when $i \neq i'$. If someone can show me how to correctly use second quantization to get the right answer here, I would appreciate it a lot.


1 Answer 1


I don't have enough reputation to post this as a comment, which it should be.

Shouldn't the product in your definition of $\left|\Psi\right>$ go over $i,i'$? Is it restricted to $i>i'$? Is there any restriction that $\alpha\neq\beta$?

Could you provide details of the solution you have tried so far?


As you say in your comment, $$S_−S_+=a^\dagger_{p\beta}a_{p\alpha}a^\dagger_{q\alpha}a_{q\beta}.$$

Anticommute the $a^\dagger_{p\beta}$ through $a_{p\alpha}a^\dagger_{q\alpha}$. $$S_−S_+=a_{p\alpha}a^\dagger_{q\alpha}a^\dagger_{p\beta}a_{q\beta}.$$

Now anticommute $a^\dagger_{q\alpha}$ through $a_{p\alpha}$, but don't forget the possibility that $p=q$. $$S_−S_+=(\delta_{qp}-a^\dagger_{q\alpha}a_{p\alpha})a^\dagger_{p\beta}a_{q\beta},$$ and you've got, $$S_−S_+=a^\dagger_{p\beta}a_{p\beta}-a^\dagger_{q\alpha}a_{p\alpha}a^\dagger_{p\beta}a_{q\beta}.$$

And you're done.

  • $\begingroup$ Yes, the sum is over $i,i'$ (I edited to reflect that). We want the $\beta$ states to be different from the $\alpha$ states cause otherwise the $\Psi$ would be an eigenfunction of $S^2$. We can have $N_\alpha \neq N_\beta$ but if someone can show me how to work this out for the case $N_\alpha = N_\beta$ that is fine. What I tried is just evaluating the terms (summation over indices implied) $S_-S_+ = a^+_{p\beta} a_{p\alpha} a^+_{q\alpha} a_{q\beta} = P_{pq} Q_{pq}$ which yields only the Tr($PQ$) part, so I must be doing something wrong. $\endgroup$
    – Goku
    Commented Feb 15, 2013 at 3:16
  • $\begingroup$ sorry, I messed up the last part of my last comment and can't re-edit. If I use the anticommutator $\{a^+_{p\alpha}, a_{q\alpha} \} = P_{qp}$ I get $a^+_{p\beta} a_{p\alpha} a^+_{q\alpha} a_{q\beta} = 0$. $\endgroup$
    – Goku
    Commented Feb 15, 2013 at 3:39
  • $\begingroup$ +1. Useful answer. One question remains for me: In your third equation above, why can we write $\delta_{qp}$ rather than $P_{qp}$? I know there is a basis in which $P_{qp} = \delta_{qp}$ but then if use this basis to evaluate $a^+_{q\alpha} a_{p\alpha} a^+_{p\beta} a_{q\beta}$, wouldn't we just get $\delta_{qp}$? I believe the reason here is that alpha and beta states are different, so we cannot diagonalize them simultaneoulsy, I am right or is if because of another reason? $\endgroup$
    – Goku
    Commented Feb 15, 2013 at 4:46
  • $\begingroup$ The third line uses only the definition of the anticommutator, $\{a_{p\alpha},a^\dagger_{q\alpha}\} = \delta_{qp}$. This allows us to replace $a_{p\alpha}a^\dagger_{q\alpha}$ by $\delta_{pq}-a^\dagger_{q\alpha}a_{p\alpha}$. $\endgroup$
    – Flavin
    Commented Feb 15, 2013 at 12:50

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