Proof: 3 Reversible evolutions -- Hermitian Conjugate

[RJLiberator]

Homework Statement

Consider a qubit whihc undergoes a sequence of three reversible evolutions of 3 unitary matrices A, B, and C (in that order). Suppose that no matter what the initial state |v> of the qubit is before the three evolutions, it always comes back to the sam state |v> after the three evolutions.
Show that we must have C=(BA)†

Homework Equations

† = hermitian conjugate

The Attempt at a Solution

The diagram of the reversible evolution allows us to see that the process |v> --> A --> B --> C = |v> results in the equation:

C(BA)|v> = |v>
From here we see that: C(BA) = I (where I is the identity matrix)
We multiple both sides by (BA)†
C(BA)(BA)†=I(BA)†
By definition of unitary we see
C=(BA)†This was quite easy, we see it only took 3-4 steps.

Have I successfully completed this proof? Recall, I needed to show that this works for all possible |v>.
|v> seems irrelevant, however, since it could be 'cancelled' out in the second step.

Success? Or failure?

 

[DrClaude]

C(BA)(BA)†=I(BA)†
By definition of unitary we see
C=(BA)†

I would give more details here, as you can only take that A is unitary and B is unitary, not suppose it for BA.

Have I successfully completed this proof? Recall, I needed to show that this works for all possible |v>.
|v> seems irrelevant, however, since it could be 'cancelled' out in the second step.

Again, to give more details: $C(BA)|v\rangle = |v\rangle \, \forall \, |v\rangle \Rightarrow C(BA) = I$. You can only go from the first to the second line if you say that this has to hold for all states.

 

[RJLiberator]

I would give more details here, as you can only take that A is unitary and B is unitary, not suppose it for BA.

Absolutely excellent point. I glossed over that.

Again, to give more details: C(BA)|v⟩=|v⟩∀|v⟩⇒C(BA)=I. You can only go from the first to the second line if you say that this has to hold for all states.

Indeed, what you suggest seems to bring more clarity.