Physics, asked by ananddevtiwari606, 1 year ago

What is Merzbacher's justification in this step? (Second-Quantization)?

Answers

Answered by Anonymous
0
[Long Version]

In chapter 20 of Merzbacher's "Quantum Mechanics", the algebra of the annihilation and creation operators is derived. Specifically,

a†ia†j±a†ja†i=0ai†aj†±aj†ai†=0

where, of course, the plus sign corresponds to operators on fermions and the opposite for bosons. However, there is one particular step in the derivation that I can't understand. I've quoted it below, shortening some parts for brevity.

... Hence, for any state many-particle state |ψ⟩|ψ⟩,

a†ia†j|ψ⟩=λa†ja†i|ψ⟩ai†aj†|ψ⟩=λaj†ai†|ψ⟩

To show that λλ doesn't depend on |ψ⟩|ψ⟩ or the indices i,ji,j, we switch to another basis through a unitary transformation.

(a†ia†j−λa†ja†i)|ψ⟩=∑k,lckiclj(b†kb†l−λb†lb†k)|ψ⟩=0(1)(1)(ai†aj†−λaj†ai†)|ψ⟩=∑k,lckiclj(bk†bl†−λbl†bk†)|ψ⟩=0

where cijcij are the matrix elements of the unitary transformation, constrained by the unitarity-condition,

∑kc∗ikckj=δij∑kcik∗ckj=δij

These matrix elements are just complex numbers. If the theory is to have the same form in any representation (unitary symmetry), equation (1) above holds for every |ψ⟩|ψ⟩ only if for each pair k,lk,l,

b†ib†j−λb†jb†i=0bi†bj†−λbj†bi†=0

The bold text is what I don't understand. I obviously know that if a state is equal to zero, you can expand that state in terms of a linearly independent basis and equate each component to zero (by the definition of linear independence). I also know that any creation operator merely "bumps up" the state it acts on, apart from a constant. But a priori we don't know whether this normalization depends on the specific representation we use. That's what we're trying to deduce! This is how I think the derivation above should have actually gone.


Answered by shreeya589
0
hope this answer may help you
Attachments:
Similar questions