Inner products and adjoint operators

[Frank Castle]

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.

 

[stevendaryl]

I think it's simple:
$\langle \psi| A^\dagger \phi\rangle = \langle A \psi | \phi\rangle = \langle \phi| A \psi \rangle^*$

That doesn't imply that $A = A^\dagger$, as can be seen from the trivial example: $A = i$.

 

[Frank Castle]

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?!

 

[stevendaryl]

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 $A |\psi\rangle$ means the same thing as $|A \psi \rangle$. The only reason to write the $A$ sandwiched in between, as in $\langle \phi|A|\psi \rangle$ is because you can equally well think of $A$ as acting to the left, on $\phi$, or to the right, on $\psi$. But that notation should (in my opinion) only be used when $A$ is self-adjoint.

 

[Frank Castle]

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. $$\langle\psi\rvert A^{\dagger}\lvert\phi\rangle =\langle\phi\rvert A\lvert\psi\rangle^{\ast}$$

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.

 

[stevendaryl]

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,

$(\phi, \psi) = \langle \phi|\psi \rangle$

So

$(A^\dagger \phi, \psi) = \langle A^\dagger \phi | \psi \rangle$

The definition of $\dagger$ is that

$(A^\dagger \phi, \psi) = (\phi, A \psi) = \langle \phi | A \psi \rangle = \langle \phi | A | \psi \rangle$

or conversely

$(A \phi, \psi) = (\phi, A^\dagger \psi) = \langle \phi | A^\dagger \psi \rangle = \langle \phi | A^\dagger | \psi \rangle$

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

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.

 

[Frank Castle]

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?

 

[stevendaryl]

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 $A = A^\dagger$?

 

[Frank Castle]

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).