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

In summary, the vacuum expectation value is the average value of an operator in a quantum state. In the case of combinations of <code>a^\dagger</code> and <code>a</code>, the vacuum expectation value is zero because both <code>a^\dagger</code> and <code>a</code> lower the energy of a state by one unit in the vacuum state, which is defined as the state with the lowest possible energy. The formula for calculating the vacuum expectation value of a combination of <code>a^\dagger</code> and <code>a</code> is <code> <math>\langle 0| a^\dagger a |0 \rangle = 0</math> </code>, which
  • #1
LayMuon
149
1
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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  • #2
LayMuon said:
Wouldn't the second term give zero automatically because ## a |0 \rangle = 0## ?
Yes.
 
  • #3
thanks.
 

What are the vacuum expectation values of combinations of a^\dagger and a?

The vacuum expectation value is the average value of an operator in a quantum state. In the case of combinations of a^\dagger and a, the vacuum expectation value is zero. This is because the vacuum state is defined as the state with the lowest possible energy, and a and a^\dagger both lower the energy of a state by one unit. Thus, in the vacuum state, both a and a^\dagger will have an expectation value of zero.

How do you calculate the vacuum expectation value of a combination of a^\dagger and a?

The vacuum expectation value of a combination of a^\dagger and a can be calculated using the formula \langle 0| a^\dagger a |0 \rangle = 0 . This formula represents the average value of a^\dagger a in the vacuum state |0\rangle. Since the vacuum state has zero particles, the operator a^\dagger a will have an expectation value of zero.

What is the physical significance of the vacuum expectation value of combinations of a^\dagger and a?

The vacuum expectation value of combinations of a^\dagger and a is important because it represents the lowest possible energy state in a quantum system. This value can be used to calculate the ground state energy and other properties of the system. In addition, it can also be used to determine the probabilities of different energy states being occupied by particles in the system.

How do the vacuum expectation values of a^\dagger and a change in different quantum states?

The vacuum expectation values of a^\dagger and a will vary depending on the quantum state. In general, the expectation value of a^\dagger will be higher in states with more particles, while the expectation value of a will be higher in states with fewer particles. This is because a^\dagger creates particles, while a destroys them.

What is the relationship between the vacuum expectation values of a^\dagger and a?

The vacuum expectation values of a^\dagger and a are related by the commutation relation [a, a^\dagger] = 1. This means that the two operators do not commute, and the order in which they are applied matters. This relationship also implies that the expectation values of a^\dagger and a cannot both be zero at the same time.

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