Vacuum expectation values of combinations of ##a^\dagger## and ##a##

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SUMMARY

The discussion focuses on calculating vacuum expectation values of products of creation and annihilation operators for bosons, specifically the expression ##\langle 0| a_{k_1} a^\dagger_{k_2} a_{k_3} a^\dagger_{k_4} |0 \rangle##. The participant correctly identifies that commuting operators ##k_3## and ##k_4## leads to two terms, where the second term vanishes due to the property that ##a |0 \rangle = 0##. This confirms that the vacuum expectation value simplifies to the first term involving the delta function, ##\delta(k_3-k_4)##.

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LayMuon
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I am slightly confused on how do we calculate vacuum expectation values of product of creation and annihilation operators for bosons, e.g. ##\langle 0| a_{k_1} a^\dagger_{k_2} a_{k_3} a^\dagger_{k_4} |0 \rangle##

If i commute ##k_3## and ##k_4##:

$$\langle 0| a_{k_1} a^\dagger_{k_2} a_{k_3} a^\dagger_{k_4} |0 \rangle = \langle 0| a_{k_1} a^\dagger_{k_2} |0 \rangle \delta(k_3-k_4) + \langle 0| a_{k_1} a^\dagger_{k_2} a^\dagger_{k_4} a_{k_3} |0 \rangle $$
Wouldn't the second term give zero automatically because ## a |0 \rangle = 0## ?
 
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LayMuon said:
Wouldn't the second term give zero automatically because ## a |0 \rangle = 0## ?
Yes.
 
thanks.
 

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