Quantum Mechanics, commutators and Hermitian Operators

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Homework Statement



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

Homework Equations



Let
A=â
B=[tex]\hat{}b[/tex]

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.
 

Answers and Replies

  • #2
vela
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Homework Statement



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

Homework Equations



Let
A=â
B=[tex]\hat{}b[/tex]

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.
 
  • #3
dextercioby
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problem said:
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.
 
  • #4
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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.
 
  • #5
dextercioby
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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.
 
  • #6
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you should be concentrating on the lhs rather than the rhs
 
  • #7
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you should be concentrating on the lhs rather than the rhs

[a,b]+=(ab-ba)+=ba-ab=[b,a]?
 
  • #8
vela
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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].
 
  • #9
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so using [b,a] = - [a,b] what can you deduce about lambda
 
  • #10
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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?
 
  • #11
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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?
 
  • #12
vela
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Either is fine.
 

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