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

  • Thread starter rayveldkamp
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In summary, The conversation discusses the relationship between a vector |g> and its corresponding operator A, and the relationship between the dual vectors <g| and <f|B. The problem is with the notation, but the solution involves using matrix representation and the definition of Hermitian conjugate. The conversation also mentions the derivation of this result and the use of Dirac notation in linear analysis.
  • #1
rayveldkamp
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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
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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
 

1. What is the meaning of "Proof of |g> = A|f> implies

This statement is a mathematical expression that refers to the proof that the inner product of |g> and |f> is equal to the inner product of A|f> and B|g>. In other words, it shows that the inner product of two vectors is preserved under linear transformations.

2. How is this proof relevant in science?

This proof is relevant in many areas of science, including physics, engineering, and computer science. It is commonly used to prove theorems and solve problems in linear algebra and quantum mechanics.

3. Can you provide an example of how this proof is used?

Sure! For example, in quantum mechanics, this proof is used to show that the expectation value of an observable operator is independent of the choice of basis. This is a fundamental concept in quantum mechanics and is crucial in understanding the behavior of particles at the microscopic level.

4. What are the key components of this proof?

The key components of this proof are the properties of inner products, specifically the bilinearity and conjugate symmetry properties. These properties allow us to manipulate the inner product of vectors and show that it is preserved under linear transformations.

5. Is this proof considered a basic or advanced concept in mathematics?

This proof is considered a basic concept in mathematics, specifically in the field of linear algebra. It is often taught in introductory courses and is essential for understanding more advanced concepts in linear algebra and quantum mechanics.

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