# Heisenberg Representation of Quantum Computers explain observable transformations

The Heisenberg Representation of Quantum Computers (Daniel Gottesman) http://arxiv.org/abs/quant-ph/9807006

Suppose we have a quantum computer in the state $|\psi\rangle$, and we apply the operator $U$. Then

$U N |\psi\rangle = U N U^\dagger U|\psi\rangle$

so after the operation, the operator $U N U^\dagger$ acts on states in just the way the operator $N$ did before the operation. Therefore, applying $U$ to the computer transforms an arbitrary operation $N$ by

$N \rightarrow U N U^\dagger$

For the first step, I understand that placing $U^\dagger U$ on the right hand side has no effect; it resolves to the identity matrix.

I follow what the text is saying about considering the the right hand side as applying $U N U^\dagger$ after $U$ has been applied to the state.

I don't quite follow the result for how the observable $N$ is changed when the operator $U$ is applied: $N \rightarrow U N U^\dagger$. I'd appreciate it if someone who "gets" it could put a little more explanatory text between the two steps.

## 1 Answer

The author shows that since $UN$ equivalent to $UNU^\dagger U$, that it is possible to change the order of the operations, if $N$ is replaced by $M=UNU^\dagger$. In that case, $UN$ = $MU$.

The remaining logic follows from this.

Note that $M$ is just $N$ in the $U$ basis; the choice of $U$ uniquely determines how $N$ changes to $M$ for this special commutation relation.

• That makes sense to me, but I'm missing what the ability to change the order of the operations implies. – David B May 6 '16 at 1:07
• @DavidB: I've included some analysis. Are there any U such that M=N? – Peter Diehr May 6 '16 at 2:26