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.


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.

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    $\begingroup$ I have to disagree with this, strongly. The action of a linear transformation on a covector is defined to be compatible with its action on vectors; the only potential source of confusion is if you do things like mix up vectors and covectors or mess up the translation back to non-braket notation. And even then the notation helps you; when translating $\langle u | T | v \rangle$ only one of $uT$ and $Tv$ makes sense. (which one depends on your convention of whether linear transformations act on vectors from the right or left) $\endgroup$ – Hurkyl Nov 5 '15 at 18:27
  • $\begingroup$ @Hurkyl: For the case fo two different vectors, it gets even worse! The adjoint of an operator need not have the same domain of definition as the operator, so when writing $\langle \psi \vert A \vert \phi \rangle$ it can be that the expression $(\psi, A\phi)$ is not defined because $\phi$ isn't in $A$'s domain of definition, but $(A^\dagger \psi ,\phi)$ maybe a perfectly fine thing. In Dirac notation, both are the same. Dirac notation makes it in this way impossible to discuss the many subtle issues that arise when discussing the adjointness of operators on infinite-dimensional spaces. $\endgroup$ – ACuriousMind Nov 5 '15 at 18:38
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    $\begingroup$ I'm a little confused about this: the way I learned it, Dirac notation is defined such that $\langle\psi\rvert AB\lvert\psi\rangle$ can be interpreted as $(\psi, AB\psi)$ or as $(B^\dagger A^\dagger \psi, \psi)$, or even as $(A^\dagger\psi, B\psi)$, but not as $(BA\psi, \psi)$. $\endgroup$ – David Z Nov 5 '15 at 22:13
  • $\begingroup$ @DavidZ: That is the correct way, yes. However, many mistakes arise from forgetting that (because for self-adjoint operators, those are all the same). I'm not sure why, but I've seen quite often that people just "let the operators act to the left" without taking the adjoint, and always in Dirac notation. Although the notation may be defined correctly, it seems the way of writing it is suggestive of wrong manipulations to many. $\endgroup$ – ACuriousMind Nov 5 '15 at 22:19
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs: You are correct. I'm...bad at distinguishing left and right, apparently. $\endgroup$ – ACuriousMind Nov 5 '15 at 22:49

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.


$ \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.

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    $\begingroup$ The complex conjuguate of a complex number is not necessarily the same number you know it. $\endgroup$ – user97166 Nov 5 '15 at 16:54
  • $\begingroup$ I am not talking about complex conjugates, but matrix transpose $\endgroup$ – SuperCiocia Nov 5 '15 at 16:55
  • $\begingroup$ Ok then, I see! $\endgroup$ – user97166 Nov 5 '15 at 17:46

protected by Qmechanic Nov 5 '15 at 17:09

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