Expectation value for non commuting operator

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  • #1
phyky
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if 2 hermitian operator A, B is commute, then AB=BA, the expectation value <.|AB|.>=<.|BA|.>. how about if A and B is non commute operator? so we can not calculate the exp value <.|AB|.> or <.|BA|.>?
 

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  • #2
Nugatory
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if 2 hermitian operator A, B is commute, then AB=BA, the expectation value <.|AB|.>=<.|BA|.>. how about if A and B is non commute operator? so we can not calculate the exp value <.|AB|.> or <.|BA|.>?

We can compute both, they just won't be equal.
 
  • #3
Ravi Mohan
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We can compute both, they just won't be equal.

In some cases (when the commutation is a projector to a particular eigen space) they might be equal.
 
  • #4
Nugatory
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In some cases (when the commutation is a projector to a particular eigen space) they might be equal.

Yes, of course. But in general they won't be.
 
  • #5
dextercioby
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We can compute both, they just won't be equal.

Generally we can't, because if the vector psi is in Dom(AB), it may not be in Dom(BA).
 
  • #6
phyky
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so we can only compute 1 of it? but how about <[A,B]>, we can compute it?
 
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  • #7
K^2
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Generally we can't, because if the vector psi is in Dom(AB), it may not be in Dom(BA).
Then the expectation is zero. Both operators map from Hilbert Space to Hilbert Space in this context. And you can always take an inner product between two vectors in Hilbert Space. They might be orthogonal, because they belong to different sub-spaces, but then the inner product is trivially zero and that's your expectation value.
 
  • #8
dextercioby
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I was simply arguing that <psi, AB psi> may be defined, while <psi,BA psi> not.
 
  • #9
phyky
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conclusion is if AB is non commute. we can only compute 1 of the expectation value, either <|AB|> or <|BA|>?
 
  • #10
Ravi Mohan
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conclusion is if AB is non commute. we can only compute 1 of the expectation value, either <|AB|> or <|BA|>?

In nearly all physical situations (without weird boundary conditions) you can compute both. One example, of weird boundary conditions, is particle in infinite square well where [itex]D(\hat{x}\hat{p}^\dagger) \neq D({\hat{p}^\dagger\hat{x}})[/itex] and you can't compute both (always).

If you are interested to know more about how to find domains of operators, you can check
http://arxiv.org/abs/quant-ph/9907069
and/or my blog post
http://ravimohan.me/2013/07/25/particle-in-a-box-infinite-square-well/ [Broken].
 
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  • #11
dextercioby
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Hi Ravi, just some points you left out on your blog article:

*the infinite sq. well is a non-physical example. There are no infinities in real life systems. Nonetheless, it serves as the perfect example to the fact that a theoretical resolution of a (non-)physical problem should always be done with the maximum knowledge of mathematics (in this case the theory of unbounded operators on Hilbert spaces).
*the reason that the momentum operator is symmetric and not self-adjoint relies with the physical boundary condition: psi (outside the box) = psi (left edge of the box) = psi (right edge of the box) = 0. If you relax the boundary condition to psi (left edge of the box) = psi (right edge of the box) =/= 0, then the momentum operator becomes self-adjoint. This would be the case for the finite square well, however (which is in fact a physically correct example).
*it's interesting to notice that 2 unphysical situations (the free particle on the real line and the free particle in an infinite square well) are usually presented in a non-mathematical manner to students to illustrate that the energy levels of the particle are either continous, or discrete. The real value of these 2 examples is actually to prove that functional analysis is the root of all quantum mechanical calculations.
 
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  • #12
phyky
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some inconsistency where since A and B is non commute, why we can compute <|AB|> or <|BA|>? so i think it is impossible from beginning because Heisenberg uncertainty forbid us to compute both A and B accurately.
 
  • #13
Jilang
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You can compute both. Consider xp-px . The answer is ih/2pi. What it's saying is that if you measure position and then momentum you won't get the same answer than if you measure momentum and then position.
 

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