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I am reading some introductory quantum mechanics and I don't understand the connection between an observable and a gate. I thought a gate just applies a rotation to a state while a measurement gives you some information about the state itself so the fact that these are related is unclear.

For example, what is the connection between the Pauli-$Z$ gate, which is given by the matrix $diag[1, -1]$ in the standard basis and measurement in the $Z$ basis? Does every gate, which is represented by some unitary $U$ have a corresponding observable, which represents measurement in its eigenbasis?

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  • $\begingroup$ While a measurement does give you some information about the state it also – much like a gate – changes the state in the process. So after both a gate and a measurement you will end up with a new state of the system where both can be described via the quantum channel formalism. Is this what you are asking about? $\endgroup$ Commented Jul 14 at 7:27
  • $\begingroup$ Yes, I understand the gate in the channel formalism e.g. action of an $X$ gate is given by $\rho\rightarrow X\rho X^\dagger$. What is the $X$-basis measurement in the same formalism? $\endgroup$
    – Katie
    Commented Jul 14 at 9:06

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Let us start from some unitary $U$ with corresponding spectral decomposition $U=\sum_{\lambda\in\sigma(U)}\lambda P_\lambda$. Here, $\sigma(U)$ denotes the spectrum (i.e. set of eigenvalues) of $U$ and $P_\lambda$ are the spectral projections, that is, $P_\lambda^\dagger=P_\lambda$ and $P_\lambda P_{\lambda'}=\delta_{\lambda,\lambda'}P_\lambda$ for all $\lambda,\lambda'\in\sigma(U)$, as well as $\sum_{\lambda\in\sigma(U)}P_\lambda={\bf1}$. To put this into a more familiar quantum info context the spectral projections are of the form $P_\lambda=|g_\lambda\rangle\langle g_\lambda|$ or, if $\lambda$ has multiplicity larger than one, $P_\lambda=\sum_k|g_\lambda^{(k)}\rangle\langle g_\lambda^{(k)}|$.

In any case the corresponding channel induced by the gate $U$ can be written as $$U(\cdot)U^\dagger=\sum_{\lambda,\lambda'\in\sigma(U)}\lambda(\lambda')^*P_\lambda(\cdot)P_{\lambda'}\,.\tag1$$

On the other hand $U$ (more precisely: the Hamiltonian $H$ which generates $U$ via $U=e^{iH}$) induces a projective measurement through its spectral decomposition. Thus if some outcome ($\log\lambda$) is measured on some system in the initial state $\rho$, then the corresponding post-measurement state is $$ \frac{P_\lambda\rho P_\lambda}{{\rm tr}(\rho P_\lambda)}\,. $$ Thus these formalisms are connected by the fact both of them are determined by the spectral decomposition—and, in particular, the spectral projections $P_\lambda$—of the unitary $U$.

In addition, there is a scenario where a measurement can be expressed as a quantum channel (just like the gate transformation $\rho\mapsto U\rho U^\dagger$ is a quantum channel). To every measurement one can associate a "expected density operator" which can be interpreted as the state of the system after measurement if the experimenter does not have access to the measurement outcome. For a projection-valued measurement this channel takes the form $ \rho\mapsto\sum_{\lambda\in\sigma(U)}P_\lambda\rho P_\lambda\,. $ In some sense these are then the "diagonal terms" of the channel $U(\cdot)U^\dagger$, i.e. only those terms in (1) for which $\lambda=\lambda'$.

For more on measurement, (expected) post-measurement, and the like, cf., e.g., this other answer of mine or Chapter 3.3 in "Principles of Quantum Communication Theory: A Modern Approach" by Khatri & Wilde.

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  • $\begingroup$ Can I ask how this relates to the Born rule where we pick a Hermitian operator $A$ and are interested in the expectation value $Tr(A\rho)$? Is that also not a measurement and if so, how is this related to the measurement channel $\rho\rightarrow\sum_\lambda P_\lambda\rho P_\lambda$? $\endgroup$
    – Katie
    Commented Jul 14 at 12:23
  • $\begingroup$ On the one hand there is the probability ${\rm tr}(P_\lambda\rho)$ of measuring the outcome $\lambda$ (Born rule) and on the other hand there is the corresponding post-measurement state $\frac{P_\lambda\rho P_\lambda}{{\rm tr}(P_\lambda\rho)}$. Now if we multiply state with associated probability and we sum over all outcomes $\lambda$ we end up with an expectation "value"/state given by $\rho\mapsto\sum_\lambda P_\lambda\rho P_\lambda$. And this is where the measurement channel comes from. $\endgroup$ Commented Jul 14 at 13:02

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