Quantum Mechanics, commutators and Hermitian Operators

[umagongdi]

Homework Statement

Suppose that the commutator between two Hermitian operators â and \hat{}b is [â,\hat{}b]=λ, where λ is a complex number. Show that the real part of λ must vanish.

Homework Equations

Let
A=â
B=\hat{}b

The Attempt at a Solution

AΨ=aΨ BΨ=bΨ
ABΨ=AbΨ=bAΨ=baΨ=abΨ=aBΨ=BaΨ=BAΨ

AB=BA
[A,B]=0

This can't be the answer since we need the complex part.

or
[A,B]Ψ=ABΨ-BAΨ=Re(λ)Ψ+Im(λ)Ψ, i have no idea how to get rid of Re(λ). Any ideas? Thanks.

 

[vela]

Homework Statement

Suppose that the commutator between two Hermitian operators â and \hat{}b is [â,\hat{}b]=λ, where λ is a complex number. Show that the real part of λ must vanish.

Homework Equations

Let
A=â
B=\hat{}b

The Attempt at a Solution

AΨ=aΨ BΨ=bΨ
ABΨ=AbΨ=bAΨ=baΨ=abΨ=aBΨ=BaΨ=BAΨ

AB=BA
[A,B]=0

This can't be the answer since we need the complex part.

You're assuming ψ is an eigenstate of both A and B. That can't be true unless A and B commute, which isn't necessarily true.

[A,B]Ψ=ABΨ-BAΨ=Re(λ)Ψ+Im(λ)Ψ, i have no idea how to get rid of Re(λ). Any ideas? Thanks.

You need to use the fact that A and B are Hermitian. A common tactic is to take the adjoint of both sides of an equation. Try that.

 

[dextercioby]

Suppose that the commutator between two Hermitian operators a and b is [a,b]=λ, where λ is a complex number.

I hate when textbooks are not rigurous. The λ should be multiplied by the unit operator on the Hilbert space which 'carries' the 2 operators a and b.

 

[umagongdi]

You need to use the fact that A and B are Hermitian. A common tactic is to take the adjoint of both sides of an equation. Try that.[/QUOTE]

We know that A is Hermitian if A=(A*)^T. So maybe,

[A,B]=Re(λ)+Im(λ)

[A*,B*]^T=(Re(λ)+Im(λ))*^T
=(Re(λ)-Im(λ))^T
=Re(λ)^T-Im(λ)^T

Is the transpose of a real number=0? Thanks for taking the time.

 

[dextercioby]

It makes no sense to speak about transposing a real number, or a complex number. Transposition is an operation one applies to matrices. And even if a number is a matrix 1x1, the "transpose" of it can't be 0 unless you're "transposing" the number 0.

 

[sgd37]

you should be concentrating on the lhs rather than the rhs

 

[umagongdi]

you should be concentrating on the lhs rather than the rhs

[a,b]⁺=(ab-ba)⁺=ba-ab=[b,a]?

 

[vela]

As bigubau noted above, the RHS is really λ multiplied by the unit operator. In matrix form, it would be λ times the identity matrix I, so its adjoint is λ^*I or, more sloppily, just λ^*.

And yes, [A,B]^† = [B,A].

 

[sgd37]

so using [b,a] = - [a,b] what can you deduce about lambda

 

[umagongdi]

so using [b,a] = - [a,b] what can you deduce about lambda

Oh i get it now :)

First you need to adjoint to obtain an equation, which you can compare to the original.

LHS
[A,B]⁺=(AB-BA)⁺=BA-AB=[B,A]=-[A,B]

RHS
λ⁺= Re(λ)-iIm(λ)

[A,B]= -Re(λ)+iIm(λ)

comparing this to the original

[A,B]= Re(λ)+iIm(λ)

Therefore

Re(λ)=-Re(λ)

and hence λ must be zero. Is this right?

 

[umagongdi]

or you can just add them together

[A,B]+[A,B] = Re(λ)+iIm(λ)-Re(λ)+iIm(λ)

2[A,B] = 2iIm(λ)

[A,B] = iIm(λ)

Is it better to write this?

 

[vela]

Either is fine.