How to Calculate the Sum of Fermionic Operator Products?

[LagrangeEuler]

Homework Statement

How to calculate?
$\sum _{i,j} \langle 0|\prod_n \hat{C}_n \hat{C}^+_i\hat{C}_j \prod_n \hat{C}^+_n|0 \rangle$

Homework Equations

$\hat{C}^+, \hat{C}$ are fermionic operators.
$\{\hat{C}_i,\hat{C}^+_j\}=\delta_{i,j}$

The Attempt at a Solution

I have a question. What is $|0 \rangle$? Is that maybe $|0 0\rangle$?

 

[PhysicsGente]

Are the \hat{C}'s ladder operators?

 

[Mute]

Presumably you have some number of Fermions, $N$. Then, $\left| 0 \right\rangle$ represents the vacuum state which contains no Fermions - you can take it as a shorthand for

$$\underbrace{\left| 0 0 \dots 0 \right\rangle}_{N~\rm{terms}}.$$

So, the Fermion creation operator for the kth fermion when acting on this state will create a Fermion in the kth slot of the vacuum state (changing the 0 to a 1).

That is,

$$\hat{c}_k^\dagger \left| 0 0 \dots 0_k \dots 0 \right\rangle = \left| 0 0 \dots 1_k \dots 0 \right\rangle,$$
where I used a subscript $k$ to denote that that is the kth entry in the state.

 

[LagrangeEuler]

Tnx.

$c^+_k|00…0k…0\rangle=|00…1k…0\rangle$
and is maybe
$c^+_k|00…1k…0\rangle=0?$
Then
$\prod_n c^+_n |000...0\rangle=|111...1\rangle$

 

[LagrangeEuler]

So result is
$\sum_{i,j}\langle 1111...|\hat{C}^+_i\hat{C}_j|111...\rangle=\sum_{i,j}\delta_{i,j}?$

 

[Mute]

So result is
$\sum_{i,j}\langle 1111...|\hat{C}^+_i\hat{C}_j|111...\rangle=\sum_{i,j}\delta_{i,j}?$

That's what it looks like to me. Note that since you're summing over i and j you should get a number out at the end.

 

[LagrangeEuler]

And is it necessarily to $\langle 1...1|1...1 \rangle$ be $1$?

 

[sgd37]

this is a problem in anti commutation. Anti commute the annihilation operators past all the other operators so that they may annihilate the vacuum, in other words you have to normal order the operators

 

[Mute]

And is it necessarily to $\langle 1...1|1...1 \rangle$ be $1$?

While not strictly necessary, typically the wavefunctions are defined to be normalized, no?