Is a projective measurements over a superposition of eigenstates possible?

Question

All observables admit a spectral decomposition in terms of projectors $P_m$ into the eigenspace corresponding to the eigenvalue $m$. So given for example a collection of kets $|0\rangle, |1\rangle,..., |n\rangle$ with eigenvalues $a_0,a_1,...,a_n$ I can always perform a measurement on a state $|\psi\rangle=\alpha_0|0\rangle+...+\alpha_n|n\rangle$ by a projection such as $$P_0|\psi\rangle=| 0\rangle\langle 0|\psi\rangle=\alpha_0|0\rangle$$ (to be eventually renormalized) with outcome $a_0$. But what if I wanted to see if the system is in a superposition of certain eigenstates? For example, I may wish to apply the projector $$P=\frac{1}{2}(|0\rangle+|1\rangle)(\langle 0|+\langle 1|)=\frac{P_0+|0\rangle\langle1|+|1\rangle\langle 0|+P_1}{2}.$$ Clearly $P$ doesn't belong to the set $\{P_m\}$ as $|0\rangle$ and $|1\rangle$ are associated to different eigenvalues.

My question is: is such a measurement possible, and what would be the outcome of the measurement?

Answer 1

Yes, such a measurement is possible and the outcomes can be mapped to the outcomes of a computational basis measurement.

This is accomplished by the use a unitary to transform from the basis we wish to measure in to the basis we know how to measure in. In the example from the question, we wish to measure in the $X$ basis

$$ |+\rangle = \frac{|0\rangle + |1\rangle}{\sqrt{2}} \\ |-\rangle = \frac{|0\rangle - |1\rangle}{\sqrt{2}} $$

corresponding to the projector $P=|+\rangle\langle+|$ and its orthogonal complement $|-\rangle\langle-|$. We can perform this measurement by applying the Hadamard gate

$$ H = |0\rangle\langle +| + |1\rangle\langle -| $$

followed by the computational basis measurement. If we care about obtaining the appropriate post-measurement state, then after the computational basis measurement we apply $H^\dagger$ to the collapsed state.

In general, any measurement specified by a complete set of orthogonal projectors $Q_k = |\psi_k\rangle\langle\psi_k|$ can be implemented using computational basis measurement $P_k = |k\rangle\langle k|$ and the unitary $U = \sum_k |k\rangle\langle\psi_k|$. This follows from the identity

$$ Q_k = U^\dagger P_k U. $$

Answer 2

Given an arbitrary state $|\psi\rangle$, be it a superposition of computational basis states or not, you can always (in principle) measure in a basis that contains $|\psi\rangle$. Mathematically, a trivial example is using the pair of projectors $(P,I-P)$ with $P=|\psi\rangle\!\langle\psi|$.