# Some help on second-quantization formalism?

I'm reading Schwabl's Advanced Quantum Mechanics and I'm having issues with understanding this line:

$$\langle\psi|a_i^\dagger a_j^\dagger a_ma_k|\psi\rangle=\langle\psi|(\delta_{im}\delta_{jk}a_m^\dagger a_k^\dagger+\delta_{ik}\delta_{jm}a_k^\dagger a_m^\dagger)a_ma_k|\psi\rangle$$

I don't get the reason in the right-hand side we have a sum, I understand the purpose of the delta's in forming number operators but no idea about the sum. I would thank you, if you give me some light about this.

• Do I get the question right that it is about the identity $v_i = \sum_j \delta_{ij} v_j$? Commented Jul 13, 2021 at 18:16
• Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! Commented Jul 13, 2021 at 18:17
• @Photon it's more than that, because the left hand side depends on more indices than the right hand side Commented Jul 13, 2021 at 20:10
• I suspect the state $|\psi\rangle$ has a fixed number of particles of each type (ie, in each mode) - is this correct? Then the expression will vanish unless the annihilation of a particle is paired with the creation of the same particle, hence the delta functions Commented Jul 13, 2021 at 20:12
• @QuantumMechanic I'm quite confused, the r.h.s seems ill-defined as $m$ and $k$ appear three times there... Not really compatible with the Einstein summation convention. I'm not even sure which indices are to summed over on the r.h.s. Commented Jul 13, 2021 at 20:27

The state mentioned in the book is specified to be of the form $$|\psi\rangle=a_1^\dagger a_2^\dagger\cdots a_N^\dagger|0\rangle;$$ i.e., this is a state of $$N$$ electrons in $$N$$ different modes labeled from $$1$$ to $$N$$ (there could be other modes not occupied, and by convention we may have simply chosen to label the occupied modes as $$1$$ to $$N$$).

The overall equality follows from boolean logic coupled with the realizations: \begin{aligned} &a_i|\psi\rangle=0\quad\forall\, i>N\\ &a_i^\dagger a_i|\psi\rangle=|\psi\rangle\quad\forall\,i\leq N\\ &\langle\psi|a_i^\dagger a_i=\langle\psi|\quad\forall\,i\leq N\\ &\langle \psi|a_i|\psi\rangle=\langle \psi|a_i^\dagger|\psi\rangle=0\quad \forall\, i\\ &\langle\psi|a_i^\dagger a_j|\psi\rangle=0\quad\forall\, i\neq j\\ &a_ia_j|\psi\rangle=\langle\psi|a_j^\dagger a_i^\dagger=0\quad\forall\, i=j. \end{aligned} We also will make use of the fermionic relations $$a_ia_j+a_ja_i=0\qquad a_i^\dagger a_j^\dagger +a_j^\dagger a_i^\dagger =0\qquad a_i a_j^\dagger +a_j^\dagger a_i =\delta_{ij}$$ From this, we find that the left-hand side vanishes unless either [$$k=j$$ and $$m=i$$] or [$$k=i$$ and $$m=j$$]. This is actually an exclusive or, because of the fermionic nature of the particles, and the right-hand side immediately follows from boolean logic.

To make things explicitly clear, we will evaluate both the right- and left-hand sides of the desired equation for all possible combinations of subscripts.

1. $$i>N$$ or $$j>N$$ or $$k>N$$ or $$m>N$$: both sides vanish because of annihilation operators acting on $$|\psi\rangle$$ or creation operators acting on $$\langle \psi|$$.

For the rest we can thus assume all indices to be $$\leq N$$.

1. $$k=m$$. Then both sides vanish because $$a_m^2|\psi\rangle=0$$.
2. $$i=j$$. Then the left-hand side vanishes because $$\langle\psi|a_i^{\dagger\,2}=0$$. The delta functions on the right hand side vanish unless $$k=m=i$$ (because $$i=j$$). But if $$k=m$$ then we again have $$a_m^2|\psi\rangle=0$$ on the right-hand side so both sides vanish.

For the rest we can now assume that $$i\neq j$$ and $$k\neq m$$.

1. $$i=m$$. We know that $$j\neq i$$ so we can use the final commutation relation to rewrite the left-hand side as $$\langle \psi| a_i^\dagger(-a_i a_j^\dagger)a_k|\psi\rangle=-\langle\psi|a_j^\dagger a_k|\psi\rangle=-\delta_{jk}$$, where we have used the number operator acting on the left. This vanishes when $$j\neq k$$ and equals $$-1$$ when $$j=k$$ by normalization of the state. As for the right-hand side, we know that $$i=m$$ and assumed $$j\neq i$$ so $$j\neq m$$. The delta functions on the right-hand side thus become \begin{aligned} \langle\psi|\left(1\delta_{jk}a_m^\dagger a_k^\dagger+\delta_{ik}0a_k^\dagger a_m^\dagger\right)a_ma_k|\psi\rangle=&\delta_{jk}\langle\psi|a_m^\dagger a_k^\dagger a_ma_k|\psi\rangle\\ =&-\delta_{jk}\langle\psi|a_m^\dagger a_m a_k^\dagger a_k|\psi\rangle\\ =&-\delta_{jk} \end{aligned}

For the rest we can now assume $$i\neq m$$.

1. If $$i=k$$ and $$j=m$$, we get that the second-term on the right-hand side equals the entire left-hand side. We also know that $$i\neq m\Rightarrow i\neq j$$, so the other term on the right-hand side vanishes, and we have equality.
2. If $$i=k$$ and $$j\neq m$$, the right-hand side again vanishes. The left-hand side also vanishes, as you can verify by equating \begin{aligned} \langle\psi|a_i^\dagger a_j^\dagger a_m a_k|\psi\rangle&=(-1)^2\langle\psi| a_j^\dagger a_m a_i^\dagger a_k|\psi\rangle\\ &=\langle\psi| a_j^\dagger a_m |\psi\rangle=0. \end{aligned}

I hope that's sufficiently explicit! The sum is required to make sure all of the cases are simultaneously satisfied. Without the sum, you would not have a general expression. There is never an actual case where both terms in the sum are nonzero. There are lots of quicker ways to do this, but showing the explicit calculations will hopefully reassure you of the equality.

• Here you show that the different cases match with the expression of the right but it is still not clear for me how to derive the right side. I mean, here you verified the validity of the right side but it is far of being a derivation. However, I find the part of "an exclusive or" illuminating. Since we have a probability amplitude (the expectation value), should we take the "or" as a sum, like in standard probability problems? Commented Jul 17, 2021 at 21:31
• In principle, showing that two quantities are always equal to each other is equivalent to a derivation, so I'll take your question as "how could I have come up with this myself?" The answer is that yes, we take the "or" as a sum, like in standard probability problems. The only caveat here is that the $i=j=k=m$ scenario must yield a zero value - but that's no problem, because both $a_m^2|\psi\rangle=0$ ensures that both sides vanish. So the overall sum is an "or" (not an exlusive or) and we simply add the two expectation values Commented Jul 17, 2021 at 21:54
• (the exclusive or is what is required for the right-hand side to be nonzero, but we are okay with a regular or overall because the left-hand side vanishes when the right-hand side vanishes) Commented Jul 17, 2021 at 21:55