Properties of Hermitian operators in complex vector spaces

[MSUmath]

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 $\left[A,B\right]$ = $\lambda$B 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.

 

[MSUmath]

I should point out that the end of the above problem is to show that B$\left|a\right\rangle$ = const. $\left|a +\lambda\right\rangle$

 

[vela]

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 $\left[A,B\right]$ = $\lambda$B show that B$\left|a\right\rangle$ = const. $\left|a\right\rangle$

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.

 

[MSUmath]

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

 

[paranormal]

The property described in the homework statement is a well-known and fundamental property of Hermitian operators in complex vector spaces. It is known as the eigenvalue-eigenvector property and is a result of the spectral decomposition of Hermitian operators.

To prove this property, we can start by using the definition of the commutator and expanding it out:

\left[A,B\right] = AB - BA = \sum \left|a\right\rangle a \left\langle a\right|B - B\sum \left|a\right\rangle a \left\langle a\right|

Next, we can use the fact that A is a Hermitian operator, meaning that A = A^\dagger, where A^\dagger is the Hermitian conjugate of A. This allows us to rewrite A as:

A = A^\dagger = \sum \left|a\right\rangle a^* \left\langle a\right|

where a^* denotes the complex conjugate of a. Substituting this into the commutator, we get:

\left[A,B\right] = \sum \left|a\right\rangle a^* \left\langle a\right|B - B\sum \left|a\right\rangle a \left\langle a\right|

= \sum \left|a\right\rangle \left(a^*B - Ba\right) \left\langle a\right|

Next, we can use the fact that \left[A,B\right] = \lambdaB to rewrite the commutator as:

\left|a\right\rangle \left(a^*B - Ba\right) \left\langle a\right| = \lambdaB\left|a\right\rangle \left\langle a\right|

Now, we can use the eigenvalue definition to write \lambdaB as \lambda\left|a\right\rangle \left\langle a\right|. This allows us to rewrite the commutator as:

\left|a\right\rangle \left(a^*B - Ba\right) \left\langle a\right| = \lambda\left|a\right\rangle \left\langle a\right|

Finally, we can rearrange this equation to get:

\left(a^*B - Ba\right) \left|a