# A Using feedback in quantum measurements

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1. Jun 19, 2017

### Johny Boy

If we consider an efficient measurement performed on a system in a pure state. How would we use feedback (by applying to the system a unitary operator that depends upon the measurement result), to prepare the system in the same final state for every outcome of the measurement (this can be done for any efficient measurement).

So I think the idea is that say we start with a pure state $$\rho = | \psi \rangle \langle \psi|$$ and some desired final state after measurement $$\tilde{\rho}_n = \frac{A_n \rho A_n^{\dagger}}{p_n} = \frac{A_n | \psi \rangle \langle \psi | A_n ^{\dagger}}{p_n},$$ then we seek a unitary operator $$U_m$$ such that if $$m \neq n$$ then $$U_m A_m = A_n$$ thus resulting in $$\tilde{\rho}_m = \frac{U_mA_m | \psi \rangle \langle \psi | A_m^{\dagger}U^{\dagger}}{p_m} = \frac{A_n | \psi \rangle \langle \psi | A_n^{\dagger}}{p_n} = \tilde{\rho}_n.$$ I'm having difficulty thinking of how we could define this unitary operator $$U_m$$?

2. Jun 19, 2017

### Strilanc

Does it have to be efficient? Can the post-measurement states be non-orthogonal?

If not, and there are $N$ possible measurement results, then just use a permutation like $P = \sum_k^N | A_k \rangle \langle A_{k+1 \pmod{N}}|$. Then a valid corrective unitary operation is $U_m = P^{m} = \sum_k^N | A_k \rangle \langle A_{k+m \pmod{N}}|$.

3. Jun 19, 2017

### Johny Boy

Thanks for your response. There is no restriction on the post-measurement states, but it has to be efficient.
I'm not really following what you are doing, $A_n$ as I have it is a measurement operator, yet you have it as a ket and a bra? Could you please elaborate a bit on your idea.

4. Jun 19, 2017

### Strilanc

Well, I assumed that each measurement result corresponded to a particular vector in a basis (specifically the eigenbasis of the Hermitian operator defining the measurement). I used $A$ to refer to these vectors. And I know that there are valid easily-written unitary operations that simply permute the vectors of a basis, so I defined the operation that way.

5. Jun 19, 2017

### Johny Boy

Okay but how would this imply that $u_m A_m | \psi \rangle = A_n | \psi \rangle$? Are you famaliar with the fundamental theorem of quantum measurements? I think there might be a simpler way.

6. Jun 19, 2017

### Strilanc

The permutations I defined have $P^m$ send $a_m$ to $a_0$ for all $m$, where $a_m$ is the state of the system given that you measured $A_m$. To make it work for $a_{n \neq 0}$, just re-index the basis vectors.

7. Jun 19, 2017

### Johny Boy

I think I confused myself...
I think the statement implies something else, that given a state $\rho = | \psi \rangle \langle \psi|$ we could define a unitary operator $U_m$ (which depends on measurement result $m$ such that) such that for every $\tilde{\rho}_m = \frac{A_m \rho A^{\dagger}_{m}}{p_m}$ we have $U_m \rho U^{\dagger}_{m} = \tilde{\rho}_{m}$. An idea would be to consider the operator $V_m := U_{m}^{\dagger}A_m$ but I'm not sure we can
definitively state that this is a unitary operator?

8. Jun 19, 2017

### Johny Boy

I managed to resolve it I think. The idea is that since $\rho$ is pure, we can find a unitary operator such that $U_n | \psi \rangle = A_n | \psi \rangle$ the result follows easily from that observation.