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

Question

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.

Answer 1

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 $\tilde N_i=UN_iU^\dagger$.

Answer 2

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

The identity $UN\left|\psi\right> = UNU^\dagger U\left|\psi\right>$ 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 $UNU^\dagger$.

Answer 3

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 $(|e'_n\rangle)$

Answer 4

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".

Answer 5

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.