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I have encountered a problem in quantum field theory which is the vacuum expectation value of a bunch of creation and annihilation operators. For instance, I have the following vacuum expectation value: \begin{equation} \langle 0| a_{k1} a_{k2} a^{\dagger}_{k3} a^{\dagger}_{k4} |0 \rangle = \langle a^{\dagger}_{k2} a^{\dagger}_{k1} 0 | a^{\dagger}_{k3}a^{\dagger}_{k4}0 \rangle \end{equation} Then, we can use the notation of multiple-states $ a^{\dagger}_{k1}... a^{\dagger}_{kn}|0\rangle = |k1, ...,kn\rangle$ to simplify the above equation: \begin{equation} \langle 0| a_{k1} a_{k2} a^{\dagger}_{k3} a^{\dagger}_{k4} |0 \rangle = \langle k2, k1| k3,k4 \rangle \end{equation} Therefore, my question is that whether this expectation value $\langle k2, k1| k3,k4\rangle $ obeys the orthogonality of states such that: \begin{equation} \langle k2, k1| k3,k4 \rangle = \delta^{(3)}(k2-k3) \delta^{(3)}(k1-k4) \end{equation}

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Assuming we are working with bosons, then: $$ \langle k_1,k_2|k_3,k_4\rangle=\delta^{(3)}(k_1-k_3)\delta^{(3)}(k_2-k_4)+\delta^{(3)}(k_1-k_4)\delta^{(3)}(k_2-k_3) $$ To see this, you can use the commutation relations to move all the creation operators to the left and annihilation operators to the right. Those terms with annihilation operators acting on the vacuum on the right vanish (similarly for creation operators acting on the vacuum on the left) and you are left with combinations of delta functions: \begin{align} \langle 0| a_{k_1}a_{k_2}a^\dagger_{k_3}a^\dagger_{k_4}|0\rangle &=\langle 0| a_{k_1} a^\dagger_{k_3}a_{k_2}a^\dagger_{k_4}|0\rangle + \delta^{(3)}(k_2-k_3)\langle 0| a_{k_1}a^\dagger_{k_4}|0\rangle \\ &=\delta^{(3)}(k_2-k_4)\langle 0| a_{k_1}a^\dagger_{k_3}|0\rangle + \delta^{(3)}(k_2-k_3)\delta^{(3)}(k_1-k_4) \\ &= \delta^{(3)}(k_1-k_3)\delta^{(3)}(k_2-k_4)+\delta^{(3)}(k_1-k_4)\delta^{(3)}(k_2-k_3) \end{align}

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Yes, your reasoning us correct. Two comments though:

  • it is often more practical to act with all the operators on the right, rather than separate them in two groups.
  • you need to account for the exchange effects - your answer is partial ($k2=k3, k1=k4$ and $k1=k3, k2=k4$ with appropriate signs).
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