# Quantum Mechanis, matrix element, second quantization

Say that I have the following two-particle states:

$$|a\rangle = c_1^{(+)} c_2^{(+)} |0\rangle,$$ where $$c_1^{(+)}$$ and $$c_2^{(+)}$$ are two fermionic creation operators and

$$|b\rangle = c_3^{(+)} c_4^{(+)} |0\rangle,$$ where $$c_3^{(+)}$$ and $$c_4^{(+)}$$ are two fermionic creation operators.

I wish to find the matrix element: $$\langle b|V|a\rangle$$, where the operator $$V$$ is defined as:

$$V = \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k,$$ where the (+) corresponds to the first two being creation operators and the last two c’s are annihilation operators.

Furthermore, I know that $$\{c_i, c_j^{(+)}\} = c_i c_j^{(+)} + c_j^{(+)} c_i = \delta_{[ij]},$$ for fermionic creation and annihilation operators.

Now first, expressing the ket $$|b\rangle$$ as a bra gives me: $$\langle b| = \langle 0| c_3 c_4$$, with $$c_3$$ and $$c_4$$ now being annihilation operators (if I'm not mistaken).

So, I can write:

$$\langle b | V | a \rangle= \Bigg\langle 0 \Bigg| c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k c_1^{(+)} c_2^{(+)} \Bigg| 0 \Bigg\rangle .$$

1. An inner product is usually defined as something along the lines of: $$\langle b|V|a\rangle$$. Is it allowed to write: $$\langle b | V | a \rangle= \Bigg\langle 0 \Bigg| c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k c_1^{(+)} c_2^{(+)} \Bigg| 0 \Bigg\rangle .$$ and move $$c_3$$, $$c_4$$ etc. to the middle term?

2. Given that 1) is correct, how might I use the anti-commutation relation above to simplify the expression? I.e. how do I combine my $$1,2,3,4$$ operators with my $$i$$,$$j$$,$$k$$,$$l$$ operators?

• It is not clear how the $c_1^{(+)}$ and $c_2^{(+)}$ from $| a \rangle$ moved to the left of V. Can you clarify this in your question ? Jun 29 '20 at 13:52
• It's supposed to be to the right of V. :) Jun 29 '20 at 13:54
• 1) When conjugating $|b\rangle^\dagger$ you need to reverse the order of $c_3$ and $c_4$ so your later expressions are off by a minus sign. 2) You can use the latex command \dagger rather than the rather clunkier (+) notation you're using. Jun 29 '20 at 14:28

Use the fact that $$c_i | 0 \rangle=0,$$
$$\left( c_i | 0 \rangle \right)^{\dagger}= \langle0| c_i^{(+)}=0.$$
You can use the anti-commutation relations to move the annihilation operators to the right (or the creation operators to the left) one by one until you can annihilate the state. This would allow you to get a relation between the expectation value of 8 operators with the expectation value of 6 operators. If you repeat the trick enough times, you should get something that is proportional to one of the elements of $$V$$ (I expect it would be $$V_{i_1 i_2 i_3 i_4}$$, where $$i_1 i_2 i_3 i_4$$ form some permutation of $$1234$$.)
• I might be going off in the wrong direction here but using the anti-commutation relation I wrote in my question, would I be allowed to annihilate the following states in this manner: $$c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k c_1^{(+)} c_2^{(+)}$$ to: $$c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k c_1^{(+)} c_2^{(+)} + c_k c_1^{(+)}$$, where delta{k1}=0, so I am left with: $$c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_2^{(+)}$$ and then use the same on $c_l c_2^{(+)}$? Jul 15 '20 at 8:57
• @mayaper You need to be more explicit. $\delta_{k1}$ is not necessarily $0$. You are summing over all $k$, so for some of them it will be equal to $1$. For all of the other ones, you can commute through and annihilate, but you are left with a non-zero term after the commuting. Do it carefully to make sure you pick up all of the non-zero terms. Jul 15 '20 at 9:26
• Okay, so I end up with the following permutations: $$c_3 c_4 \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k c_1^{(+)} c_2^{(+)} + c_1^{(+)} c_4 + c_2^{(+)} c_4 + c_1^{(+)} c_3 + c_2^{(+)} c_3 + c_3 c_4 + c_2^{(+)} c_1^{(+)}$$. This then leaves me with: $$c_2^{(+)} c_1^{(+)} + \sum_{i,j,k,l} V_{[ijkl]} c_i^{(+)} c_j^{(+)} c_l c_k + c_3 c_4$$, as the last two terms are the only non-zero terms. If I then use your help with a creation operator acting on a ket ground-state equals zero and vice versa, I end up with simply the original sum. Have I made any assumptions that are incorrect? Jul 15 '20 at 10:35