Inner products and adjoint operators

In summary: An inner product is a function that takes two vectors and returns a scalar. There are two ways to write it in this case, but they mean the same thing. One is(A\phi, \psi) = \langle \phi | A^\dagger | \psi \rangleand the other is\langle A\phi | \psi\rangle = (A \phi, \psi)I don't understand what you mean by "the adjoint of the operator A".An inner product is a function that takes two vectors and returns a scalar. There are two ways to write it in this case, but they mean the same thing. One is(Aϕ,ψ)=⟨ϕ|A
  • #1
580
23
I'm trying to prove the following relation $$\langle\psi\lvert \hat{A}^{\dagger}\rvert\phi\rangle =\langle\phi\lvert \hat{A}\rvert\psi\rangle^{\ast}$$ where ##\lvert\phi\rangle## and ##\lvert\phi\rangle## are state vectors and ##\hat{A}^{\dagger}## is the adjoint of some operator ##\hat{A}## acting on those states.

As far as I understand, an Hermitian adjoint ##\hat{A}^{\dagger}## of an operator ##\hat{A}## is defined by $$\left(\hat{A}\lvert\psi\rangle ,\lvert\phi\rangle\right)=\left(\lvert\psi\rangle ,\hat{A}^{\dagger}\lvert\phi\rangle\right)$$ and the inner product of two kets defined such that $$\left(\lvert\psi\rangle ,\lvert\phi\rangle\right)\equiv\langle\psi\lvert\phi\rangle ,\qquad\langle\psi\lvert\phi\rangle =\langle\phi\lvert\psi\rangle^{\ast}$$ but using these two definitions I fail to arrive at the expression that I'm trying to prove, as it would appear from this that $$\langle\psi\lvert \hat{A}\rvert\phi\rangle =\langle\psi\lvert \hat{A}^{\dagger}\rvert\phi\rangle$$ If someone could explain my misunderstanding and the correct way to do it, I'd be grateful.
 
Physics news on Phys.org
  • #2
I think it's simple:
[itex]\langle \psi| A^\dagger \phi\rangle = \langle A \psi | \phi\rangle = \langle \phi| A \psi \rangle^*[/itex]

That doesn't imply that [itex]A = A^\dagger[/itex], as can be seen from the trivial example: [itex]A = i[/itex].
 
  • #3
stevendaryl said:
I think it's simple:
⟨ψ|A†ϕ⟩=⟨Aψ|ϕ⟩=⟨ϕ|Aψ⟩∗\langle \psi| A^\dagger \phi\rangle = \langle A \psi | \phi\rangle = \langle \phi| A \psi \rangle^*

Thanks, I thought I was just being stupid!

Just to check though, is ##\lvert A\psi\rangle =A\lvert\psi\rangle## and its dual given by ##\langle A\psi\rvert =\langle\psi\rvert A^{\dagger}## ? I'm still slightly confused by what went wrong with my definition for the inner product?!
 
  • #4
Frank Castle said:
Thanks, I thought I was just being stupid!

Just to check though, is ##\lvert A\psi\rangle =A\lvert\psi\rangle## and its dual given by ##\langle A\psi\rvert =\langle\psi\rvert A^{\dagger}## ? Should one really express the definiton a gave as $$\left(\lvert A\psi\rangle,\lvert\phi\rangle\right)=\left(\lvert \psi\rangle,\lvert A^{\dagger}\phi\rangle\right)$$ then (to avoid confusion)?

I think that [itex]A |\psi\rangle[/itex] means the same thing as [itex]|A \psi \rangle[/itex]. The only reason to write the [itex]A[/itex] sandwiched in between, as in [itex]\langle \phi|A|\psi \rangle[/itex] is because you can equally well think of [itex]A[/itex] as acting to the left, on [itex]\phi[/itex], or to the right, on [itex]\psi[/itex]. But that notation should (in my opinion) only be used when [itex]A[/itex] is self-adjoint.
 
  • #5
stevendaryl said:
The only reason to write the AA sandwiched in between, as in ⟨ϕ|A|ψ⟩\langle \phi|A|\psi \rangle is because you can equally well think of AA as acting to the left, on ϕ\phi, or to the right, on ψ\psi.

Would it not be in this case then that $$\langle\phi\rvert A\lvert\psi\rangle =\langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle$$ such that $$ \langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle =\langle A\psi\lvert\phi \rangle^{\ast}$$

I'm still a bit confused as to why my definition of the adjoint, i.e. $$\left(A^{\dagger}\lvert\psi\rangle ,\lvert\phi\rangle\right) =\left(\lvert\psi\rangle ,A\lvert\phi\rangle\right)$$ didn't work?!

Is it simply that I misinterpreted it slightly. Instead, would this be correct?!

$$(A\lvert\psi\rangle ,\lvert\phi\rangle)\equiv\left(\langle\psi\rvert A^{\dagger}\right)\lvert\phi\rangle =\langle\psi\rvert A^{\dagger}\lvert\phi\rangle$$ and $$(\lvert\psi\rangle ,A\lvert\phi\rangle)\equiv\langle\psi\rvert\left( A\lvert\phi\rangle\right) =\langle\psi\rvert A\lvert\phi\rangle$$ and furthermore, $$(\lvert\psi\rangle ,\lvert\phi\rangle)=(\lvert\phi\rangle ,\lvert\psi\rangle)^{\ast}.$$ We then have that $$(A\lvert\psi\rangle ,\lvert\phi\rangle)=\langle\psi\rvert A^{\dagger}\lvert\phi\rangle =(\lvert\psi\rangle ,A^{\dagger}\lvert\phi\rangle)=(A^{\dagger}\lvert\phi\rangle ,\lvert\psi\rangle)^{\ast}=(\lvert\phi\rangle ,A\lvert\psi\rangle)^{\ast}=\langle\phi\rvert A\lvert\psi\rangle^{\ast}$$ i.e. $$\langle\psi\rvert A^{\dagger}\lvert\phi\rangle =\langle\phi\rvert A\lvert\psi\rangle^{\ast}$$

stevendaryl said:
But that notation should (in my opinion) only be used when AA is self-adjoint.

I find it confusing that in so many quantum mechanics lecture notes/text books they pretty much exclusively use this notation.
 
Last edited:
  • #6
Frank Castle said:
Would it not be in this case then that $$\langle\phi\rvert A\lvert\psi\rangle =\langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle$$ such that $$ \langle A^{\dagger}\phi\lvert\psi\rangle =\langle\phi\lvert A\psi\rangle =\langle A\psi\lvert\phi \rangle^{\ast}$$

I'm still a bit confused as to why my definition of the adjoint, i.e. $$\left(A^{\dagger}\lvert\psi\rangle ,\lvert\phi\rangle\right) =\left(\lvert\psi\rangle ,A\lvert\phi\rangle\right)$$ didn't work?!

I don't see the difference. Relating your inner product with Dirac expressions,

[itex](\phi, \psi) = \langle \phi|\psi \rangle[/itex]

So

[itex](A^\dagger \phi, \psi) = \langle A^\dagger \phi | \psi \rangle[/itex]

The definition of [itex]\dagger[/itex] is that

[itex](A^\dagger \phi, \psi) = (\phi, A \psi) = \langle \phi | A \psi \rangle = \langle \phi | A | \psi \rangle[/itex]

or conversely

[itex](A \phi, \psi) = (\phi, A^\dagger \psi) = \langle \phi | A^\dagger \psi \rangle = \langle \phi | A^\dagger | \psi \rangle[/itex]

I don't see how you derived [itex]A = A^\dagger[/itex]

I find it confusing that in so many quantum mechanics lecture notes/text books they pretty much exclusively use this notation.

I guess it's unambiguous if you always assume that the operator acts to the right.
 
  • #7
stevendaryl said:
I don't see the difference. Relating your inner product with Dirac expressions,

(ϕ,ψ)=⟨ϕ|ψ⟩(\phi, \psi) = \langle \phi|\psi \rangle

So

(A†ϕ,ψ)=⟨A†ϕ|ψ⟩(A^\dagger \phi, \psi) = \langle A^\dagger \phi | \psi \rangle

The definition of †\dagger is that

(A†ϕ,ψ)=(ϕ,Aψ)=⟨ϕ|Aψ⟩=⟨ϕ|A|ψ⟩(A^\dagger \phi, \psi) = (\phi, A \psi) = \langle \phi | A \psi \rangle = \langle \phi | A | \psi \rangle

or conversely

(Aϕ,ψ)=(ϕ,A†ψ)=⟨ϕ|A†ψ⟩=⟨ϕ|A†|ψ⟩(A \phi, \psi) = (\phi, A^\dagger \psi) = \langle \phi | A^\dagger \psi \rangle = \langle \phi | A^\dagger | \psi \rangle

I don't see how you derived A=A†A = A^\dagger

Is the updated bit I added to the end of my previous post a correct derivation then?
 
  • #8
Frank Castle said:
Is the updated bit I added to the end of my previous post a correct derivation then?

It seems correct. Are you retracting your claim that it seems to show that [itex]A = A^\dagger[/itex]?
 
  • #9
stevendaryl said:
Are you retracting your claim that it seems to show that A=A†A = A^\dagger?

Yes. I just applied the definitions incorrectly in the first post and hence arrived at the incorrect result (I knew that it shouldn't be ##A=A^{\dagger}##, what I didn't understand straight away was the mistake that was making me obtain this erroneous result, but I think I've sorted it now).
 

1. What is an inner product in mathematics?

An inner product is a mathematical operation that takes two vectors and produces a scalar value. It is a generalization of the dot product between two vectors in Euclidean space. The result of an inner product is a measure of the similarity or angle between the two vectors.

2. How is an inner product used in linear algebra?

In linear algebra, inner products are used to define the notion of orthogonality between vectors. They also allow for the definition of norms, which are used to measure the length of a vector. Inner products are also used to define adjoint operators, which are important in functional analysis.

3. What is an adjoint operator?

An adjoint operator is a linear map that is the "dual" of another linear map. It is a generalization of the concept of a transpose for matrices. In functional analysis, adjoint operators are used to study the behavior of linear operators on infinite-dimensional vector spaces.

4. How are inner products and adjoint operators related?

The concept of an adjoint operator is closely related to the inner product. In fact, the adjoint operator is defined using the inner product. Given an inner product space, the adjoint operator of a linear map is the unique linear map that satisfies a certain property related to the inner product. This property is known as the adjointness property.

5. What are some real-world applications of inner products and adjoint operators?

Inner products and adjoint operators have many applications in various fields of science and engineering. Some examples include signal processing, quantum mechanics, and image processing. In these applications, inner products and adjoint operators are used to analyze and manipulate data, as well as to solve optimization problems.

Suggested for: Inner products and adjoint operators

Replies
7
Views
330
Replies
7
Views
961
Replies
2
Views
985
Replies
2
Views
539
Replies
9
Views
768
Replies
2
Views
144
Replies
0
Views
595
Replies
24
Views
424
Replies
3
Views
716
Back
Top