# How does a projective measurement distinguish between two states in a $d$-dimensional Hilbert space?

Let $$\mathcal{H}$$ be a $$d$$ -dimensional Hilbert space, and let $$|\psi\rangle,|\phi\rangle \in \mathcal{H}$$ be two quantum states.

1. Show that if $$|\psi\rangle$$ and $$|\phi\rangle$$ are orthogonal, then there exists a projective measurement that distinguishes them. That is, there exists a two-outcome projective measurement $$\left\{P_{0}, P_{1}\right\}$$ such that $$p_{\psi}\left(P_{0}\right)=1 \quad \text { and } \quad p_{\phi}\left(P_{1}\right)=1$$
2. Show that if $$|\psi\rangle$$ and $$|\phi\rangle$$ are not orthogonal, then there is no projective measurement that distinguishes them.

Attemps: I form a general Projective operator an two General states, I tried it for a $$2*2$$ dimension and then I was wondering why do we say that $$|\psi\rangle$$ and $$|\phi\rangle$$ should be orthogonal?

• Hint for 1: what happens if you try to construct a measurement where one of the projectors is $|\psi \rangle \langle \psi |$? Oct 8, 2020 at 16:32

Let $$P_0=|\psi\rangle\!\langle\psi|$$ and $$P_1=I-P_0$$. This is a projective measurement which deterministically distinguishes the two orthogonal states.

More generally, consider a projective measurement with operators $$\newcommand{\ket}[1]{\lvert#1\rangle}\{P_i\}_{i=1}^d$$ and $$\newcommand{\braket}[2]{\langle #1\rvert #2\rangle}\newcommand{\ketbra}[1]{\lvert #1\rangle\!\langle #1\rvert} P_i\equiv\ketbra{\eta_i}$$ where $$\braket{\eta_i}{\eta_j}=\delta_{ij}$$, and a set of (not necessarily orthogonal) states $$\{\ket{\psi_i}\}_{i=1}^\ell$$ with $$\ell\le d$$. To distinguish the states deterministically we need $$\operatorname{Tr}(P_i \ketbra{\psi_j})=|\braket{\eta_i}{\psi_j}|^2=\delta_{ij}.\tag2$$ Define the matrices $$\Pi\equiv\sum_{i=1}^d |\eta_i\rangle\!\langle i|$$ and $$\Psi\equiv\sum_{i=1}^\ell|\psi_i\rangle\!\langle i|$$. Note that $$\Pi$$ is $$d\times d$$ and $$\Psi$$ is $$d\times\ell$$. Eq. (1) is thus equivalent to $$\Pi^\dagger \Psi=I_{d\times \ell}$$ (we can tune the definitions of the states $$\ket{\eta_i}$$ to have $$\braket{\eta_i}{\psi_i}\in\mathbb R$$ without any loss in generality). This is only possible if $$\Psi$$ is "maximally entangled", i.e. has rank $$\ell$$ and all its (nonzero) singular values equal $$1$$ (equivalently, iff $$\Psi^\dagger\Psi=I_{\ell}$$). This is true iff the states $$\ket{\psi_i}$$ are orthonormal.

The last statement follows from observing that $$\Psi^\dagger\Psi$$ has the same nonzero singular values/eigenvalues as $$\Psi\Psi^\dagger=\sum_{i=1}^\ell \ketbra{\psi_i}$$. The latter having an $$\ell$$-fold degenerate eigenvalue $$+1$$ means $$\Psi\Psi^\dagger = \sum_i\ketbra{\psi_i}=\sum_i\ketbra{\phi_i}$$ for some orthonormal set $$\{\ket{\phi_i}\}_{i=1}^\ell$$. This in turns implies that for some unitary $$U$$ we have $$\ket{\psi_i}=\sum_j U_{ij}\ket{\phi_j}$$, and thus $$\braket{\psi_i}{\psi_j}=\delta_{ij}$$.

This shows that non-orthogonal states cannot be distinguished deterministically via projective measurements (in fact, they cannot be distinguished deterministically by any measurement, but that's not what we are showing here).

• I take it $\Psi$ would need to be padded to make it equivalent in dimension to $\Pi$ before application of $\Pi$ to $Psi$? Oct 9, 2020 at 13:41
• @GaussStrife well $\Psi$ has dimension $d\times \ell$ and $\Pi$ is $d\times d$, so the dimensions match and the product $\Pi^\dagger\Psi$ is well defined. Explicitly, we simply have $(\Pi^\dagger\Psi)_{ij}=\langle \eta_i|\psi_j\rangle$
– glS
Oct 9, 2020 at 13:44