- #1

rayveldkamp

- 60

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter rayveldkamp
- Start date

- #1

rayveldkamp

- 60

- 0

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

- #2

OlderDan

Science Advisor

Homework Helper

- 3,021

- 2

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.rayveldkamp said:

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

- #3

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

rayveldkamp said:

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

- #4

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

- 7,176

- 22

As far as I'm aware, thatHallsofIvy said:That's pretty much thedefinitionof Hermitian conjugate, isn't it?

I guess you

- #5

rayveldkamp

- 60

- 0

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

Share:

- Replies
- 1

- Views
- 368

- Replies
- 10

- Views
- 443

- Last Post

- Replies
- 4

- Views
- 343

- Replies
- 4

- Views
- 363

- Last Post

- Replies
- 3

- Views
- 334

- Last Post

- Replies
- 6

- Views
- 370

- Replies
- 2

- Views
- 149

- Last Post

- Replies
- 4

- Views
- 164

- Last Post

- Replies
- 15

- Views
- 314

- Replies
- 2

- Views
- 693