# Perfectly distinguishable states are mutually orthogonal

$$\newcommand{\bra}{\left<#1\right|}\newcommand{\ket}{\left|#1\right>}\newcommand{\bk}{\left<#1\middle|#2\right>}$$

Let $$H$$ be a Hilbert space. We call states $$\ket{\psi_1}, \dots, \ket{\psi_n} \in H$$ perfectly distinguishable if there exists a measurement system $$\big\{M_i\big\}_{i=1}^{m}$$, with $$m\geq n$$, such that $$|| M_j\ket{\psi_i}||^2 = 1$$ if $$i = j$$ and $$|| M_j\ket{\psi_i}||^2 = 0$$ otherwise . Show that the states $$\ket{\psi_1}, \dots, \ket{\psi_n}$$ are perfectly distinguishable exactly when they are mutually orthogonal.

It is not too hard to see that you can construct a measurement system from mutually orthogonal states that satisfy the desired property. (As Nielsen & Chuang outline in Quantum Computation and Quantum Information).

However, I am not really sure how one might go about showing the other direction, i.e. why perfectly distinguishable states are necessarily mutually orthogonal. I am assuming I would have to apply $$M_j\ket{\psi_i}$$ to $$\ket{\psi_j}$$ and use properties of the measurement system, along with taking advantage of the Dirac notation, but I have not been able to get myself anywhere useful by doing this.

Edit: The definition of "(finite outcome) measurement" I am using is: a set of operators $$\big\{M_i\big\}_{i=1}^{n}$$ such that $$\sum_{i=1}^{n}M_i^*M_i = I$$.

• What is a measurement system? A set of projection operators? – doetoe Feb 10 at 18:53
• @doetoe I just added the definition I am using to the question. I don’t think it’s quite the same as a set of projection operators. – user222378 Feb 10 at 19:32

Here's a proof for 2 states and the argument can easily be extended for more states. You are given that

$$\langle \psi_1\vert M_1\vert \psi_1\rangle = 1\\ \langle \psi_2\vert M_1\vert \psi_2\rangle = 0 \\ \langle \psi_1\vert M_2\vert \psi_1\rangle = 0 \\ \langle \psi_2\vert M_2\vert \psi_2\rangle = 1$$

Next insert the identity into $$\langle \psi_1\vert M_2\vert \psi_1\rangle = 0$$ to obtain $$\langle \psi_1\vert\psi_1\rangle\langle\psi_1\vert M_2\vert \psi_1\rangle + \langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$

The first term is zero so we have $$\langle \psi_1\vert\psi_2\rangle\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$.

Assume $$\langle \psi_1\vert\psi_2\rangle \neq 0$$ which means that $$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0$$. Since $$\langle \psi_1\vert\psi_2\rangle \neq 0$$, we may write $$\vert\psi_1\rangle = \sqrt{\lambda}\vert\psi_2\rangle + \sqrt{(1-\lambda)}\vert\psi_2^\perp\rangle$$ for some nonzero $$\lambda$$. Then we have

$$\langle\psi_2\vert M_2\vert \psi_1\rangle = 0 \implies \sqrt{\lambda}\langle\psi_2\vert M_2\vert\psi_2\rangle + \sqrt{(1-\lambda)}\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = 0$$

But $$\langle\psi_2\vert M_2\vert\psi_2\rangle = 1$$, so we must have $$\langle\psi_2\vert M_2 = \langle\psi_2\vert$$ and hence, $$\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = 0$$. Therefore, $$\sqrt{\lambda}\langle\psi_2\vert M_2\vert\psi_2\rangle + \sqrt{(1-\lambda)}\langle\psi_2\vert M_2\vert\psi_2^\perp\rangle = \sqrt{\lambda} \neq 0$$

We have arrived at a contradiction so we conclude that our assumption was wrong and $$\langle\psi_1\vert\psi_2\rangle = 0$$

Note that $$\| M_j|\psi_i\rangle\|^2 = 0$$ is equivalent to $$M_j|\psi_i\rangle = 0$$. Consider any $$\psi_i,\psi_j$$. Since by definition

$$\sum_{k=1}^{n}M_k^*M_k = I$$

we have

$$\langle\psi_i|\psi_j\rangle = \sum_{k=1}^{n}\langle\psi_i M_k^*M_k\psi_j\rangle$$

(and note that $$\langle\psi_i M_k^*|$$ really is $$|M_k\psi_i\rangle^\dagger$$). The terms of this sum can only be nonzero if both $$|M_k\psi_i\rangle$$ and $$|M_k\psi_j\rangle$$ are nonzero, which is the case when $$i = k$$ and $$j = k$$, giving

$$\langle\psi_i|\psi_j\rangle = \sum_{k=1}^{n}\langle\psi_i M_k^*M_k\psi_j\rangle = \delta_{ik}\delta_{kj}\| M_i|\psi_j\rangle\|^2 = \delta_{ij}.$$