# Fermionic Fock Space

1. Jan 16, 2016

### samjohnny

1. The problem statement, all variables and given/known data

2. Relevant equations

Given in question.

3. The attempt at a solution

Hello

I'm having some difficulty with part e of this question. Not sure how to go about proving that. Would a possibility would be deriving the following equation:

$|n⟩= \frac{1}{\sqrt{n!}}(c^†)^n |0⟩$ ?

2. Jan 16, 2016

### strangerep

You know that $c^2 = 0$. So what can you say about $\left(c^\dagger \right)^2 = \; ?$

Then think about $[H, c^\dagger] = \; ?$, etc.

3. Jan 17, 2016

### samjohnny

Thanks for the reply. Then $\left(c^\dagger \right)^2 = 0$. However I'm not picking up on the reasoning for determining the commutator. If I remember correctly if two operators commute then they share the same eigenvalues, is that correct?

4. Jan 17, 2016

### vela

Staff Emeritus
No, it's a bit more subtle. For example, for a free particle, the momentum $\hat p$ and Hamiltonian $\hat H = \frac{\hat p^2}{2m}$ commute, but the eigenvalues of $\hat p$ have units of momentum and the eigenvalues of $\hat H$ are energies. Clearly, they can't be the same values. Moreover, you can have eigenstates of the Hamiltonian which are not momentum eigenstates.

5. Jan 17, 2016

### strangerep

Yes.
You gave an equation for $|n\rangle$. So what is $|2\rangle, |3\rangle, \dots$ ? From this what can you conclude about the structure of the Fock space?

Then evaluate $H |0\rangle$ and $H |1\rangle$ ...

6. Jan 18, 2016

### samjohnny

Right, so $|n\rangle = 0$ when $n=2,3,4...$, in which case the only non vanishing ones are when $n=0,1$ as required. And then evaluating $H |0\rangle$ and $H |1\rangle$ is straight forward using the relations given in part d and before.

7. Jan 18, 2016

### vela

Staff Emeritus
Is your expression for $\lvert n \rangle$ justified? Does applying $c^\dagger$ repeatedly generate all possible states?

I think it would be better if you considered the fact that you can write N in terms of H. Then use what you showed in part (c).