Properties of Hermitian operators in complex vector spaces

AI Thread Summary
The discussion centers on proving a property of Hermitian operators in complex vector spaces, specifically regarding the operator B and its relationship with a Hermitian operator A. The initial attempt to show that B|a⟩ = const.|a⟩ is deemed incorrect, as it conflicts with the goal of demonstrating B|a⟩ = const.|a + λ⟩. The inconsistency arises from misunderstanding the implications of the commutation relation [A, B] = λB. Participants clarify that the correct approach involves using the commutation relation to derive the second statement. Ultimately, the focus is on resolving the confusion surrounding the application of operator B to the state |a⟩.
MSUmath
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Homework Statement



Given a Hermitian operator A = \sum \left|a\right\rangle a \left\langle a\right| and B any operator (in general, not Hermitian) such that \left[A,B\right] = \lambdaB show that B\left|a\right\rangle = const. \left|a\right\rangle

Homework Equations



Basically those listed above plus possibly the Hermitian condition and eigenvalue definition which I will not list since they are well known.

The Attempt at a Solution



I have tried expanding it out in terms of the commutator, but this seems like the wrong approach. I am not sure that there is a way to calculate it directly. I do not think the proof is very involved but I am approaching it the wrong way and can't seem to get anywhere. This is listed as a fundamental property of a Hermitian operator in a complex vector space in my text, though it does not prove it and other references seem unconcerned. I am also not very comfortable using this kind of spectral decomposition, which is a bit of a problem.
 
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I should point out that the end of the above problem is to show that B\left|a\right\rangle = const. \left|a +\lambda\right\rangle
 
MSUmath said:

Homework Statement



Given a Hermitian operator A = \sum \left|a\right\rangle a \left\langle a\right| and B any operator (in general, not Hermitian) such that \left[A,B\right] = \lambdaB show that B\left|a\right\rangle = const. \left|a\right\rangle
MSUmath said:
I should point out that the end of the above problem is to show that B\left|a\right\rangle = const. \left|a +\lambda\right\rangle
These two statements are inconsistent. In both cases, you're applying B to |a>, but you're getting different answers.

You can prove the second statement using the commutation relation.
 
They are inconsistent because the statement in the first post is incorrect. The correct relation to prove is the second.
 
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