Perfectly distinguishable states are mutually orthogonal

Question

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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 $\sum_{i=1}^{n}M_i^*M_i = I$.

Answer 1

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 $\langle\psi_1\vert\psi_2\rangle = 0$

Answer 2

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

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