Proof of |g> = A|f> implies <g| = <f|B

[rayveldkamp]

Hi,
I am trying to show that |g> = A|f> implies
<g| = <f|B

where A is an operator and B is its Hermitian conjugate.
I think my problem is with notation, but i have not been able to show this as yet.
Thanks

Ray

 

[OlderDan]

Hi,
I am trying to show that |g> = A|f> implies
<g| = <f|B

where A is an operator and B is its Hermitian conjugate.
I think my problem is with notation, but i have not been able to show this as yet.
Thanks

Ray

In a matrix representation, you can write the original equation as a sum of products using matrix multiplication rules. Take the complex conjugate, and replace the conjugates of the elements of A with elements of B. Then from the relationship between elements of <g| and |g>, <f| and |f> you have all you need.

 

[HallsofIvy]

Hi,
I am trying to show that |g> = A|f> implies
<g| = <f|B

where A is an operator and B is its Hermitian conjugate.
I think my problem is with notation, but i have not been able to show this as yet.
Thanks

Ray

That's pretty much the definition of Hermitian conjugate, isn't it?

 

[Gokul43201]

That's pretty much the definition of Hermitian conjugate, isn't it?

As far as I'm aware, that is how the hermitian conjugate (or adjoint) is defined - though the dual correspondence for A|a> !

I guess you could take the matrix operation of finding the adjoint as definition, and "derive" this result as Older Dan suggests.

 

[rayveldkamp]

Hi,
We have only just been introduced to Dirac notation, and have not had a lot of experience in linear analysis and dual vector spaces etc... I have figured out how to do it, i just contract |g> with an arbitrary bra <h|, then do the same with <g| and an arbitrary bra |h>, show the two are equal and hence the expression for <g| must be correct.
Thanks for the help guys, much appreciated