In quantum circuits, why does $UNU^\dagger$ act on states in the same way $N$ acts before the operation?

I understand that the Schrodinger picture changes the quantum states, while the Heisenberg picture changes the operators. In this paper The Heisenberg Representation of Quantum Computers, in equations 1,2, what does it mean that "so after the operation, the operator $$UNU^\dagger$$ acts on states in just the way the operator $$N$$ did before the operation"?

• StackExchange sites are best suited for posts asking individual, laser-focused questions. I took the liberty of removing the second part of the post, as it was asking a seemingly separate question. Feel free to ask that one in a separate post
– glS
May 28 at 8:06
• Thank you for the edits! May 29 at 16:08

Suppose you have some state $$\newcommand{\ket}[1]{\lvert#1\rangle}\ket\psi$$ (though note that you could make the identical argument using an arbitrary vector rather than a state), and you consider the action of some operator $$N$$ on it. Here, $$N$$ could be another gate, an operator representing some measurement, or just an arbitrary abstract operation that you are interested in for whatever reason. You thus get the state (or vector) $$N\ket\psi$$.

If you now ask how this state evolves through some gate $$U$$, you get $$UN\ket\psi$$. There is, however, a different (but equivalent) way of thinking about this: because $$U$$ is unitary, you can write $$UNU^\dagger U\ket\psi$$. This means that rather than thinking about $$\ket\psi$$ evolving through the two operations, you can interpret this as applying the operation $$UNU^\dagger$$ to the evolved vector $$U\ket\psi$$.

In some cases, this might make for an easier analysis of how states evolve through a circuit. Case in point, you might find that if $$N\in\mathcal P$$ for some group $$\mathcal P$$, then the map $$N\mapsto UNU^\dagger$$ is a group homomorphism, provided $$U$$ has specific properties (namely, in general, provided that $$U$$ belongs to the normaliser of $$\mathcal P$$ in the relevant group of unitaries).

If you then consider such "effective evolution" for sufficiently many $$N$$, you end up effectively offloading the task of calculating the evolution of $$\ket\psi$$ into that of calculating the evolution of a set of $$N$$, which might be easier if you can show that you need a relatively small number of operations $$N$$ to do this, and that such set of operations is invariant under the action of some subgroup of unitaries.

• Thanks! According to Nielsen&Chuang, section 10.5.2, it seems that $N$ is assumed to be a stabilizer of $|\psi>$ for this Heisenberg interpretation to work? May 29 at 16:07
• I'm not making any assumption on $N$ here. I'm not sure what you are referring to exactly without knowing what the text says
– glS
May 29 at 16:43
• I guess I'm still confused about exactly what it means for operators $UNU^\dagger$ and $N$ to act on states in the same way. Before applying $U$, the operator $N$ takes $\newcommand{\ket}[1]{\lvert#1\rangle}\ket\psi$ to $N\ket{\psi}$. After applying $U$, the operator $UNU^\dagger$ takes the state $U\ket{\psi}$ to $UN\ket{\psi}$. How are these two behaviors considered the 'same'? May 29 at 19:28
• @FSeed those two do not act in the same on the state, that's not what is being said. It's $UN$ and $UNU^\dagger U$ which act in the same way
– glS
May 29 at 20:22