Couple Of Quick Linear Operators Questions

Sorry about the lack of formating, _{abc} means subscript.

I don't understand the question below from a textbook, (I'll copy out a bit from before it so it makes sense)any explaination would be appreciated.:

If (under a linear operator A) the basis vectors suffer a change:
A|i> =|i'>
then any vector, V, in the space undergos a change that is readily calculable:
<i|V'>=<i|A|V>
=v_{j}<i|A|j>
=A_{ij}v_{j} where a sum over j is implied and v_j are the original components of V in the original coordinates.
Which can be cast in matrix form.

The book then says: Convince youself that the same matrix A_{ij}/latex] acting to the left on the row vector corresponding to any <v'| gives the row vector corresponding to <v''|=<v'|A.
This is the bit I don't understand.

Second question is, given A and B are Hermitian operators what can you say about [A,B] (the commutator)? If it's possible to give me a hint rather than a whole answer that would be great but if that's not possible an answer would be fine as well.

Thanks (and sorry this is really basic).

1. If A|v'> = |v''> then taking the Hermitian conjugate gives <v''| = <v'|A^(dagger). Since A is Hermitian, A^(dagger) = A, so <v''| = <v'|A.

2. If you know that A = A^dagger and B = B^dagger, then that's ([A,B])^dagger?

Thank you.

So [AB] is always real for hermitian A, B and imaginary for anti-hermitian A,B.

Last post should obviously have said pure imaginary for hermitian A,B...

Sorry I should have been more specific, in the first part of the question I don't think they are assuming A is hermitian.

On the very off chance you happen to have the book on a shelf next to your computer its p22 in Shankar's book on QM (though of course I don't expect you do).
Thank you

This post has been edited by plmokn on Jul 1 2007, 09:23 PM

I'm really not having a good day: post above should have said:
I don't think A is necessarily hermitian

<j|A|i> can be split up as (<j|) (A|i>) or (<j|A) (|i>). This is because the inner product is preserved by such a decomposition. However, if A is not hermitian then A|j> and <j|A are not (co)vector equivalents. Generally you're working over a complex vector space so to get the conjugate of things, you take the complex conjuage, so you'd have A|i> <--> <i|A^(dagger). Therefore <i|A^(dagger) = <i|A if and only if either A is Hermitian over a complex vector space or A is symmetric over a Real vector space (which reduces it to the equivalent of hermitian)

No. If A and B are Hermitian then you have the following :

[A,B] = AB-BA
([A,B])^(dagger) = (AB-BA)^(dagger) = (AB)^(dagger) - (BA)^dagger = B^dag A^dag - A^dag B^dag = BA-AB = [B,A] = -[A,B].

Therefore, if A,B are Hermitian, then [A,B] is antihermitian.

Thank you.