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Bras and Kets

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
 

Answers and Replies

OlderDan
Science Advisor
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rayveldkamp said:
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
Science Advisor
Homework Helper
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rayveldkamp said:
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
Staff Emeritus
Science Advisor
Gold Member
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HallsofIvy said:
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.
 
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
 

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