# Evolution of a state vector: Why is the action of $N$ equivalent to the action of $UNU^{†}$?

There is another question asked on this on stack exchange but I did not find any answers there that fully answered the question. In Gottesman's paper "The Heisenberg Representation of Quantum Computers", he says:

Suppose we have a quantum computer in the state $$| \psi \rangle$$, and we apply the operator $$U$$. Then $$UN |\psi \rangle = UNU^{†}U |\psi \rangle$$

The paper states that the operator $$UNU^{†}$$ acts on states in the same way that $$N$$ did before the operation.

I don't understand this. I understand that $$UN$$ and $$UNU^{†}U$$ act on the state in the same way. However, $$N$$ is acting on the state $$| \psi \rangle$$ whereas $$UNU^{†}$$ is acting on the state $$U |\psi \rangle$$.

Unless $$U$$ and $$N$$ are specifically commutative, I don't understand how the action of $$N$$ and $$UN^{†}U$$ are equivalent.

I think it probably helps to understand what Gottesman is trying to do with the operator $$N$$ (later in the paper). He wants to start with some state $$|\psi\rangle$$, but instead of directly describing the state $$|\psi\rangle$$, he wants to specify it in terms of some operators $$\{N_i\}$$ for which $$N_i|\psi\rangle=|\psi\rangle.$$ If you have enough $$N_i$$, then you no longer need to state $$|\psi\rangle$$, the set $$\{N_i\}$$ combined with the above relation is sufficient to implicitly define $$|\psi\rangle$$.

Then we want to ask what happens to the state $$|\psi\rangle$$ when it evolves under $$U$$. It becomes $$|\tilde\psi\rangle=U|\psi\rangle$$. How do we describe it in terms of some new operators $$\{\tilde N_i\}$$? $$\tilde N_i|\tilde\psi\rangle=|\tilde\psi\rangle$$ But how are the $$\tilde N_i$$ related to $$N_i$$ and $$U$$? We have \begin{align*} |\tilde\psi\rangle&=U|\psi\rangle \\ &=UN|\psi\rangle \\ &=(UNU^\dagger)U|\psi\rangle \\ &=(UNU^\dagger)|\tilde\psi\rangle. \end{align*} So, we see that $$\tilde N_i=UN_iU^\dagger$$. In the same way that we didn't need to write down $$|\psi\rangle$$, and instead relied on $$\{N_i\}$$, we never have to write down $$|\tilde\psi\rangle$$. We just update our description of the $$N_i$$ to $$\tilde N_i=UN_iU^\dagger$$.

Maybe it's easier to see when it's presented like this:

$$UN\left|\psi\right> = UNU^\dagger U\left|\psi\right>$$ means that $$UNU^\dagger$$ maps $$U\cdot \left|\psi\right>$$ to $$U \cdot N\left|\psi\right>$$. \begin{aligned} N \colon && \left|\psi\right> & \mapsto \phantom{U\cdot\ }N \left|\psi\right> \\ UNU^\dagger \colon&& U\cdot \left|\psi\right> & \mapsto U\cdot (N\left|\psi\right>) \end{aligned} So $$U$$ is the invertible transformation which transforms the action of $$N$$ to the action of $$UNU^\dagger$$.

Indeed mathematically $$N$$ and $$UNU\dagger$$ may be interpreted as 2 matrix representations of exactly the same physical operator $$n$$ (mind uppercase/lowercase), but in 2 different orthonormal basis.

Let us call $$(|e_n\rangle)_{n\in\mathbb{N}}$$ the orthonormal basis in which $$N$$ and $$|\psi\rangle$$ are expressed.

$$U^\dagger$$ is a unitary matrix: it maps the orthonormal basis $$(|e_n\rangle)_{n\in\mathbb{N}}$$ into another orthonormal basis $$(U^\dagger|e_n\rangle)_{n\in\mathbb{N}}=(|e'_n\rangle)$$; Recall now the change of basis formulas for both a vector and a linear operator: $$U|\psi\rangle$$ represents the vector $$|\psi\rangle$$ in the new basis $$(|e'_n\rangle)$$ $$UNU^\dagger$$ represents the operator $$N$$ in the new basis $$(|e'_n\rangle)$$ $$UNU^\dagger U|\psi\rangle=(UNU^\dagger )(U|\psi\rangle)=U(N|\psi\rangle)$$ represents the result of the operator $$N$$ applied to $$|\psi\rangle$$, expressed in the basis $$(|e'_n\rangle)$$

I think you're just confused by the wording.

$$U N | \psi \rangle$$ means apply N, then apply U.

$$UNU^\dagger U|𝜓 \rangle$$ means apply U, then apply $$UNU^\dagger$$

So $$U N | \psi \rangle = UNU^\dagger U|𝜓 \rangle$$ means, applying $$N$$ followed by $$U$$ is equal to applying $$U$$ followed by $$UNU^\dagger$$. Therefore Gottesman writes "$$UNU^\dagger$$ acts on states in the same way that $$N$$ did before the operation".

Indeed there is some unclarity in the wording "the operator $$UNU^\dagger$$ acts on states in the same way that $$N$$ did".

Better would be: "the operator $$UNU^\dagger$$ acts on states $$U|\psi\rangle$$ in the same way that $$N$$ acts on states $$|\psi\rangle$$".

We changed the operators and we changed the states. Only with both these changes will things be isomorphic to the original situation.