Question: Using the fact that if $|\phi\rangle = (AB) |\psi\rangle $ then $\langle \phi| = (AB)^{\dagger}\langle\psi|$ prove that $$(AB)^{\dagger} = B^{\dagger}A^{\dagger}.$$


I assume that the kets are normalized. Then, $$\langle\phi|\phi\rangle = \langle\psi| (AB)^{\dagger} AB |\psi\rangle = 1 $$ necessarily implies that $(AB)^{\dagger}AB$ is equivalent to the identity matrix. Then $$(AB)^\dagger AB = I$$ $$(AB)^\dagger AB \, B^\dagger A^\dagger = B^\dagger A^\dagger$$ $$(AB)^\dagger = B^\dagger A^\dagger$$ as desired.

Not very happy with how I deduced that $(AB)^\dagger AB$ must be equal to the identity matrix (not even sure if its correct!). Just wanted to get some opinions on the line of reasoning - is it correct, is it clear, etc…

  • $\begingroup$ It is definitely not true that $(AB)^\dagger AB$ is the identity matrix. Not sure how you got there from what you wrote. But just take $A=B=\hat{X}$, that's an immediate counterexample. $\endgroup$ Mar 8, 2018 at 4:43
  • $\begingroup$ My reasoning: $<\psi| (AB)^{\dagger} AB |\psi> = 1$ so $<\psi| (AB)^{\dagger} AB |\psi> = <\psi|\psi>$. Thus, whatever is in between the bra and ket cannot alter them - hence, identity matrix. Can you help me find the mistake? $\endgroup$
    – talrefae
    Mar 8, 2018 at 4:45
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    $\begingroup$ Who says $\langle \psi|(AB)^\dagger AB|\psi\rangle=1$? Just because $|\psi\rangle$ is normalized doesn't mean $AB|\psi\rangle$ will be. In fact, usually $AB|\psi\rangle$ will NOT be normalized! $\endgroup$ Mar 8, 2018 at 4:48
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    $\begingroup$ To get further insight into @JahanClaes's comments, look up the notion of an operator norm. You don't get to choose what $\langle \phi| \phi\rangle$ is if you normalize $|\psi\rangle$; the value of the former lies between the minimum and maximum singular values of $A\,B$ if we're talking about the finite dimensional case. Otherwise, in the infinite dimensional case, if $A\,B$ is unbounded, $\langle \phi| \phi\rangle$ has a minimum value given by the minimum magnitude of an eigenvalue of $A\,B$, and this may or may not be less than unity .... $\endgroup$ Mar 8, 2018 at 7:44
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    $\begingroup$ ... and, even if it is unity for some $|\psi\rangle$, it won't in general be for all $|\psi\rangle$. $\endgroup$ Mar 8, 2018 at 7:45

2 Answers 2


Let's consider the operators $A$ and $B$. We want to show that $(AB)^\dagger=B^\dagger A^\dagger$. Hopefully you'll agree that if we can show $\langle\psi|(AB)^\dagger|\chi\rangle=\langle\psi|B^\dagger A^\dagger|\chi\rangle$ for every $|\chi\rangle$ and $|\psi\rangle$, we've proven the two operators are equal.

So, let's set $|\phi\rangle\equiv AB|\psi\rangle$, and consider $\langle\psi|(AB)^\dagger|\chi\rangle=\langle\phi|\chi\rangle$. This is equal to $\langle\chi|\phi\rangle^*=(\langle\chi|AB|\psi\rangle)^*$

Now, let's set $|\eta\rangle\equiv B|\psi\rangle$, $|\xi\rangle\equiv A^\dagger|\chi\rangle$. Then we can write $\langle\psi|B^\dagger A^\dagger|\chi\rangle=\langle \eta|\xi\rangle=\langle\xi|\eta\rangle^* =(\langle\chi|A B|\psi\rangle)^*$.

We thus see that $\langle\psi|(AB)^\dagger|\chi\rangle=\langle\psi|B^\dagger A^\dagger|\chi\rangle$ for every $|\chi\rangle$ and $|\psi\rangle$.

Note based on confusion in your question/comments: It's not true that if $|\psi\rangle$ is a normalized vector, then $\hat{A}|\psi\rangle$ is necessarily a normalized vector. In QM, you'll always want to START with a normalized vector, but you don't usually end up with one after applying operators to it. So, if you consider $|\psi\rangle$ to be a normalized vector in your proof above, you can't assume $AB|\psi\rangle$ is normalized, and if you assume $AB|\psi\rangle$ is normalized it follows that you can't demand $|\psi\rangle$ is normalized.

This is all equivalent to saying that you can have a pair of operators $AB$ such that $(AB)^\dagger AB$ has an expectation value besides 1. For example, taking $A=B=\hat{X}$, we have that $(AB)^\dagger AB=\hat{X}^4$. If it were true that $\langle\psi|\hat{X}^4|\psi\rangle=1$ for every normalized $|\psi\rangle$, we'd have discovered a truly surprising fact about the universe: every particle, in every possible state, in every possible coordinate system, all have the same average value of $x^4$. This is unlikely to be true!

  • $\begingroup$ If $|\psi\rangle$ is normalized and $ A|\psi\rangle$ is also normalized, then A is an isometric operator. $\endgroup$
    – DanielC
    Mar 8, 2018 at 8:29

You're confusing two different concepts:

  • A the Hermitian conjugate of an operator $A$ is the (provably unique) operator $A^\dagger$ such that for all states $\phi,\psi\in\mathcal H$, $$\langle \phi, A\psi\rangle = \langle A^\dagger\phi,\psi\rangle.$$
  • An operator $U$ is unitary iff $U ^\dagger U=\mathbb I $.

You're trying to use the fact that $AB$ is unitary (which is not guaranteed, and which is false in general) to prove something much more basic.

Working with Dirac notation at this stage generally does more harm than good, so I'll use mathematicians' notation for the inner product, $\langle\cdot, \cdot\rangle$. To prove that $(AB)^\dagger=B^\dagger A^\dagger$, you need to prove that for all $\phi,\psi\in\mathcal H$, the conjugate property holds as $$\langle \phi, AB\psi\rangle = \langle B^\dagger A^\dagger\phi,\psi\rangle,$$ i.e. that the operator $B^\dagger A^\dagger$ obeys the conjugate's property that $\langle \phi, AB\psi\rangle = \langle (AB)^\dagger\phi,\psi\rangle$. To prove that, first you use the conjugate property of $A$ acting on $\phi$ and $\chi=B\psi$, $$ \langle \phi, AB\psi\rangle = \langle \phi, A\chi\rangle = \langle A^\dagger\phi,\chi\rangle = \langle A^\dagger\phi,B\psi\rangle , $$ and then you do the same with $B$ acting on $\bar\chi= A^\dagger\phi$ and $\psi$, $$ \langle \phi, AB\psi\rangle = \langle A^\dagger\phi,B\psi\rangle = \langle \bar\chi,B\psi\rangle = \langle B^\dagger\bar\chi,\psi\rangle = \langle B^\dagger A^\dagger\phi,\psi\rangle , $$ and you're done.

(The proof you've given, on the other hand, is much closer to a proof of the fact that if $U$ and $V$ are unitaries, then $UV$ is also unitary.)


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