# I Inner products and adjoint operators

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1. Apr 25, 2016

### 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.

2. Apr 25, 2016

### stevendaryl

Staff Emeritus
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: $A = i$.

3. Apr 25, 2016

### Frank Castle

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. Apr 25, 2016

### stevendaryl

Staff Emeritus
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.

5. Apr 25, 2016

### Frank Castle

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

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

Last edited: Apr 25, 2016
6. Apr 25, 2016

### stevendaryl

Staff Emeritus
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 guess it's unambiguous if you always assume that the operator acts to the right.

7. Apr 25, 2016

### Frank Castle

Is the updated bit I added to the end of my previous post a correct derivation then?

8. Apr 25, 2016

### stevendaryl

Staff Emeritus
It seems correct. Are you retracting your claim that it seems to show that $A = A^\dagger$?

9. Apr 25, 2016

### Frank Castle

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).