Dirac notation - specific acting orientation for operators

Question

I have this doubt:

Imagine two operators $A$ and $B$ and the state $\psi$.

I know that the following statement is true:

$$\langle\psi| A|\psi\rangle^*=\langle\psi| A^\dagger|\psi\rangle$$

But is it correct to write: $$ \langle\psi|AB|\psi\rangle^*=\langle\psi|B^\dagger A^\dagger|\psi\rangle=\langle\psi|B^\dagger A^\dagger\psi\rangle\hspace{15pt}?$$

This doubt came to me, because I was doing some execises and applied this identity. I found out that what I reached was wrong. I tried to find the error and only this came to my head.

Answer 1

Dirac notation is ill-suited for non-self-adjoint operators. Here's why:

Let $(-,-)$ be the inner product on our Hilbert space. The expectation value of $AB$ is then $$ \langle AB \rangle_\psi = (\psi,AB\psi)$$ by definition, and Dirac notation writes $\langle \psi \vert AB \vert \psi \rangle$. for this. But, in this notation, it is no longer clear to which side the operator $AB$ acts - one could as well interpret this expression as meaning $(BA\psi,\psi)$, which is not the same if $A,B$ are not self-adjoint. So, by $$ \langle \psi \vert A \vert \psi \rangle^\ast = \langle \psi \vert A^\dagger \vert \psi \rangle$$ you really mean $$ (\psi,A\psi)^\ast = (A\psi,\psi) = (\psi,A^\dagger\psi)$$ where the last equality is by definition of the adjoint.

So, examining the expression with $AB$, we find $$ (\psi,AB\psi)^\ast = (AB\psi,\psi) = (\psi,(AB)^\dagger\psi) = (\psi,B^\dagger A^\dagger\psi)$$ and thus $$ \langle \psi \vert AB \vert \psi \rangle^\ast = \langle\psi\vert B^\dagger A^\dagger \vert \psi \rangle $$ if all operators are interpreted as acting on the states to their right. However, since this is not usually understood - for self-adjoint operators it doesn't matter, and many texts freely switch the direction of the action of the operators whenever convenient - you should refrain from using Dirac notation for operators which are not self-adjoint.

Answer 2

Yes it is wrong because multiplication of matrices, you know it, gives matrices and I don't think it makes sense to put a matrice inside a ket or even a bra vector. Actually even with a constant (complex number) if you have $k |v\rangle$ it does not make any sense to put it inside the ket vector like $|kv \rangle$. However if you have a constant $k$ in the following position $$\langle u| k | v\rangle$$ you can write $k \langle u | v \rangle$ now in your case you have knowing $A$ and $B$ are hermitian $$A^\dagger|\psi \rangle= a|\psi \rangle$$ and the same for $B^\dagger$ and where the little $a$ is an eigenvalue, which is a complex number. In this case, you will be able to take them in front of the inner product $\langle \psi | \psi \rangle$ which is equal to one for $|\psi \rangle$ normalized.

Hope it helps.

Answer 3

$ \langle\psi|AB|\psi\rangle $ is a complex number (as opposed to a matrix), so taking its transpose gives you back the same thing, i.e. $$ \langle\psi|AB|\psi\rangle^{\dagger} = \langle\psi|AB|\psi\rangle^*, $$ and therefore $$ = \langle\psi|B^\dagger A^\dagger|\psi\rangle . $$

EDIT I just realised that you then equated this to $ \langle\psi|B^\dagger A^\dagger\psi\rangle$, sorry I did not catch that - see ACuriousMind's answer above.