# What is the connection between an observable and a gate?

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?

• 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? Commented Jul 14 at 7:27
• 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? Commented Jul 14 at 9:06

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 $$\lambda=\lambda'$$.
• 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 $\rho\rightarrow\sum_\lambda P_\lambda\rho P_\lambda$? Commented Jul 14 at 12:23
• 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. Commented Jul 14 at 13:02