# Proof: 3 Reversible evolutions -- Hermitian Conjugate

1. Nov 4, 2015

### RJLiberator

1. The problem statement, all variables and given/known data
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)†

2. Relevant equations
† = hermitian conjugate

3. 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?

2. Nov 5, 2015

### Staff: Mentor

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

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.

3. Nov 5, 2015

### RJLiberator

Absolutely excellent point. I glossed over that.

Indeed, what you suggest seems to bring more clarity.