Fermionic Fock Space Homework: Solve Part E

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Homework Help Overview

The discussion revolves around proving properties related to Fermionic Fock space, specifically focusing on part e of a homework question. Participants are exploring the implications of the fermionic creation operator and its relation to the Fock states.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive the expression for the Fock state and are questioning the implications of the fermionic operators, particularly the behavior of the creation operator when applied multiple times.

Discussion Status

There is an ongoing exploration of the properties of the creation operator and its implications for the structure of the Fock space. Some participants have provided insights into the commutation relations and the nature of eigenstates, while others are questioning the completeness of the proposed expressions for the states.

Contextual Notes

Participants are discussing the constraints imposed by the fermionic nature of the operators, such as the condition that applying the creation operator twice yields zero. There is also mention of previous parts of the homework that may influence the current discussion.

samjohnny
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Homework Statement



Fermionic.png


Homework Equations



Given in question.

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

Then think about ##[H, c^\dagger] = \; ?##, etc.
 
strangerep said:
You know that ##c^2 = 0##. So what can you say about ##\left(c^\dagger \right)^2 = \; ?##

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

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?
 
samjohnny said:
If I remember correctly if two operators commute then they share the same eigenvalues.Is that correct?
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.
 
samjohnny said:
Then ##\left(c^\dagger \right)^2 = 0##.
Yes.
However I'm not picking up on the reasoning for determining the commutator. [...]
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## ...
 
strangerep said:
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## ...

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.
 
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).
 

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