1
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Hint for 1: what happens if you try to construct a measurement where one of the projectors is $|\psi \rangle \langle \psi |$? $\endgroup$ – Rammus Oct 8 '20 at 16:32
1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ I take it $\Psi$ would need to be padded to make it equivalent in dimension to $\Pi$ before application of $\Pi$ to $Psi$? $\endgroup$ – GaussStrife Oct 9 '20 at 13:41
  • 1
    $\begingroup$ @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 $(\Pi^\dagger\Psi)_{ij}=\langle \eta_i|\psi_j\rangle$ $\endgroup$ – glS Oct 9 '20 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.