Number operator in the ground state

In summary, the given expression is a result of a bogulobov transformation involving fermionic creation and annihilation operators, where the change of integration measure is due to the shift to spherical coordinates in k-space. The operators in the integral cancel out, leaving a term that goes to zero and another that goes to 1 due to the ground state rules and the anticommutation relations.
  • #1
pleasehelpmeno
157
0

Homework Statement


Why does <0|[itex]\frac{1}{(2\pi)^3}[/itex]∫ [itex]\hat{a}^{\dagger}(t,r)[/itex] [itex]\hat{a}(t,r)[/itex] d[itex]^{3}[/itex] [itex]\textbf{k}[/itex] |0> = [itex]\frac{1}{\pi^2}∫[/itex]|β|^2 k^2 dk.

Where [itex]\hat{a} [/itex] and [itex]\hat{a}^{\dagger}[/itex] and its conjugate are bogulobov transformations given by:

[itex]\hat{a}[/itex](t,k) = [itex]\alpha[/itex](t)a(k) + β(t)[itex]b^{\dagger}[/itex](-k).

In the ground state a|0> =0 etc.


I am fairly certain it is some sort of table integral but i am not sure and want to prove it, any help or suggestions would be appreciated. I have taken the conjugate of the aforementioned a and multiplied it though but I don't understand why d[itex]^{3}[/itex] [itex]\textbf{k}[/itex] becomes k^2 dk. and how the pi factor changes, i.e. i get

[itex]\frac{1}{(2\pi)^3}[/itex]∫ [itex]b^{\dagger}[/itex](-k)b(-k)|β|^2 d[itex]^{3}[/itex] [itex]\textbf{k}[/itex]
 
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  • #2
pleasehelpmeno said:

Homework Statement


Why does <0|[itex]\frac{1}{(2\pi)^3}[/itex]∫ [itex]\hat{a}^{\dagger}(t,r)[/itex] [itex]\hat{a}(t,r)[/itex] d[itex]^{3}[/itex] [itex]\textbf{k}[/itex] |0> = [itex]\frac{1}{\pi^2}∫[/itex]|β|^2 k^2 dk.

Where [itex]\hat{a} [/itex] and [itex]\hat{a}^{\dagger}[/itex] and its conjugate are bogulobov transformations given by:

[itex]\hat{a}[/itex](t,k) = [itex]\alpha[/itex](t)a(k) + β(t)[itex]b^{\dagger}[/itex](-k).

In the ground state a|0> =0 etc.I am fairly certain it is some sort of table integral but i am not sure and want to prove it, any help or suggestions would be appreciated. I have taken the conjugate of the aforementioned a and multiplied it though but I don't understand why d[itex]^{3}[/itex] [itex]\textbf{k}[/itex] becomes k^2 dk. and how the pi factor changes, i.e. i get

[itex]\frac{1}{(2\pi)^3}[/itex]∫ [itex]b^{\dagger}[/itex](-k)b(-k)|β|^2 d[itex]^{3}[/itex] [itex]\textbf{k}[/itex]

The change of integration measure is just the standard shift to spherical coordinates in k-space:
[tex]d^3k=k^2 dk d\Omega[/tex]
Integrating over the angular integral [itex]d\Omega[/itex] is what changes the factors of Pi.
 
  • #3
excellent thankyou, why do the operators bb^\dagger disappear is it just that in the ground state they become equal to 1?
 
  • #4
Edit: I just caught your mistake. Check your work again. You should have [itex]<0|b b^{\dagger}|0>[/itex] not [itex]<0|b^{\dagger}b|0>[/itex] as you have in your first post.

Normal order all your operators and you should see how the operators drop out of the expression.
 
Last edited:
  • #5
sorry, [itex]\hat{a}[/itex](t,r) and its conjugate are bogulobov transofrmations of the fermionic creation and annihilation operators a(k) and b(-k).

The other bogulobv transformation although i didnt think it was needed is:

[itex]b^{\dagger}[/itex](t,k)=-β*(t)a(k)+ [itex]\alpha[/itex]* [itex]b^{\dagger}[/itex](k).

My method was to take the * of [itex]\hat{a}[/itex](t,r) and multiply it with [itex]\hat{a}[/itex](t,r) and then cancel out using the ground state rules and that <0| [itex]a^{\dagger}[/itex](t,r) =0. but it still leaves b [itex]b^{\dagger}[/itex](-r). In the integral, one of the 2's is canceled as there are also two degrees of freedom.
 
  • #6
G01 said:
Edit: I just caught your mistake. Check your work again. You should have [itex]<0|b b^{\dagger}|0>[/itex] not [itex]<0|b^{\dagger}b|0>[/itex] as you have in your first post.

Normal order all your operators and you should see how the operators drop out of the expression.

Thanks I didn't notice that, will that mean that it just goes to 1?
 
  • #7
pleasehelpmeno said:
Thanks I didn't notice that, will that mean that it just goes to 1?

Work it out. Use the anticommutation relations to move operators such that you end up with the annihilation operators to the right (this is called normal ordering). You should see that you are left with a term that goes to zero and another that goes to 1.
 
  • #8
yeah thought so thx
 

1. What is a number operator in the ground state?

The number operator in the ground state is a mathematical operator that measures the number of particles in the lowest energy level, or ground state, of a quantum system.

2. How is the number operator related to the energy of a system?

The number operator is related to the energy of a system through the energy eigenvalues, or allowed energy levels, of the system. The number operator determines the number of particles in each energy level, which in turn influences the overall energy of the system.

3. What is the significance of the ground state in quantum mechanics?

The ground state is the lowest energy state that a quantum system can exist in. It is significant because it serves as a reference point for measuring the energy of other states and provides important insights into the behavior of quantum systems.

4. How does the number operator behave under the process of particle creation and annihilation?

Under particle creation and annihilation, the number operator changes the number of particles in each energy level. When a particle is created, the number operator increases the number of particles in the corresponding energy level, and when a particle is annihilated, the number operator decreases the number of particles in that energy level.

5. Can the number operator be used to predict the behavior of a quantum system?

Yes, the number operator is a key tool in predicting the behavior of a quantum system. By measuring the number of particles in each energy level, the number operator can help determine the probabilities of different outcomes and provide valuable information about the system's overall energy and dynamics.

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