Hermitian Operators and the Commutator

AI Thread Summary
If A is a Hermitian operator and [A, B] = 0, it does not necessarily imply that B is Hermitian. The discussion explores the relationship between eigenstates and eigenvalues of A and B, concluding that if B is an eigenstate of A, then B must be Hermitian. However, counterexamples are presented, such as U = e^{-iHt}, which commutes with H but may not be Hermitian. Additionally, examples like B = iA and B = iI illustrate that non-Hermitian operators can commute with Hermitian ones. The complexity of the problem highlights the nuances in operator relationships in quantum mechanics.
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Homework Statement


If A is a Hermitian operator, and [A,B]=0, must B necessarily be Hermitian as well?


Homework Equations





The Attempt at a Solution

 
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njcc7d said:

Homework Statement


If A is a Hermitian operator, and [A,B]=0, must B necessarily be Hermitian as well?


Homework Equations





The Attempt at a Solution


attempt at solution?
 
if Y is an eigenstate of both A and B with respective eigenvalues a and b and respective adjoints (A+) and (B+),

<Y|AB|Y> = <Y|BA|Y>
= <Y|Ab|Y> = <(B+)Y|A|Y>
= b<Y|A|Y> = (b*)<Y|A|Y>

Therefore, b=(b*), and so it follows that B=(B+), or B is Hermitian.
 
njcc7d said:
if Y is an eigenstate of both A and B with respective eigenvalues a and b and respective adjoints (A+) and (B+),

<Y|AB|Y> = <Y|BA|Y>
= <Y|Ab|Y> = <(B+)Y|A|Y>
= b<Y|A|Y> = (b*)<Y|A|Y>

Therefore, b=(b*), and so it follows that B=(B+), or B is Hermitian.

counter example:

consider a hermitian operator H. H commutes with any function of H.

For example, the function
<br /> U=e^{-iHt}\;.<br />

Does U commute with H?

Is U hermitian?
 
Much easier: how about B=iA?
 
fair enough. thank you for answering my question, though that makes the problem a little more complicated... i hate it when that happens.
 
Or the easiest of all: B=iI (with I the identity) :-)
 
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