I've come across this problem in Nielsen & Chuang's Quantum Information book (problem 2.64)

Suppose Bob is given a quantum state chosen from a set $|ψ_1 \rangle, . . . , |ψ_m\rangle$ of linearly independent states. Construct a POVM $\{E_1, E_2, . . ., E_{m+1}\}$ such that if outcome $E_i$ occurs, $1 ≤ i ≤ m$, then Bob knows with certainty that he was given the state $|ψ_i\rangle$. (The POVM must be such that $\langle ψ_i|E_i|ψ_i \rangle > 0$ for each $i$.)

This is my proposed solution:

Denote by $|\phi_i\rangle$ the (unique? I guess it doesn't matter) vector orthogonal to the subspace spanned by $\{ | \psi_j \rangle \}_{j \neq i}$ and define

$$E_i = \sum_{i\neq j} | \phi_j \rangle \langle\phi_j |$$

Then $\langle \psi_j | E_i | \psi_j\rangle = 0$ by construction, and $\langle \psi_i | E_i | \psi_i\rangle > 0$. The last operator is defined to satisfy completeness:

$$E_{m+1} = \mathbb{I} - \sum_{j=1}^m E_j.$$

So, when get gets outcome $i$, he knows it can't have been any of the other $\psi_j$'s, so it must have been $\psi_i$ for sure. If he gets outcome $m+1$, he doesn't know anything. Is this correct?

What happens if now we introduce another vector to the set: $\psi \rangle = a |ψ_1 \rangle + b |ψ_2 \rangle$, i.e. drop the linear independence condition (just on a simple example here). How would that affect the $E_j$'s, is it still possible to construct a POVM like that?


2 Answers 2


A crucial hypothesis is missed in your construction.

Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false.

This point provides an answer to your last question as well.

If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be re-proposed as the constraint I pointed out cannot be satisfied. Indeed, even if the correspondingly added normalized vector $\phi$ is orthogonal to all $\psi_i$, it is (evidently) impossible that $\phi \not \perp \psi = \sum_{i=1}^n c_i \psi_i$.

  • $\begingroup$ So how do I guarantee that $\phi_i \not \perp \psi_i$ in my construction? I thought this was always the case - I was visualizing the situation as a plane in 3D space (subspace spanned by $i \neq j$) and a perpendicular vector to it ($\phi_i$). This vector can never be itself perpendicular to any other vector that's not contained inside the plane. I thought this generalized to the case of a $m-1$ dimensional subspace of an $m$ dim. space? $\endgroup$ May 1, 2016 at 13:17
  • $\begingroup$ Suppose that the space has $m+q$ dimensions with $q>0$ (as you generally assume when supposing that $E_{m+1}\neq 0$). In this case there necessarily exist a vector $\phi$ which is simultaneously orthogonal to the space spanned by all the $m$ vectors $\psi_i$, thus it is orthogonal to all every $\psi_i$ with $i=1,\ldots, m$. It happens even if they are linearly independent. Therefore it is not always the case and you should pick out the $\phi_i$ imposing also the hypothesis I pointed out. It is always possible. However it is not possible, if the $\psi_i$ are not linearly independent. $\endgroup$ May 1, 2016 at 13:31
  • $\begingroup$ But in this construction, the space has dimension $m$ and all of these subspaces are $m-1$ dimensional, so I don't see how this applies here. $\endgroup$ May 1, 2016 at 13:33
  • $\begingroup$ Are you saying that you also assume that the space has dimension $m$ from scratch? $\endgroup$ May 1, 2016 at 13:35
  • $\begingroup$ It seems that I was indeed assuming that, and for no reason. The question doesn't mention the dimensionality of space at all (though presumably it should be finite-dimensional)... now I understand your point $\endgroup$ May 1, 2016 at 13:38

The implicit context of this exercise is some Hilbert space $H$ of dimension at least $m$. Let $W$ be the ($m$-dimensional) subspace of $H$ spanned by the set $\{|\psi_1\rangle, ..., |\psi_m\rangle\}$ and let, for each $i=1,...,m$, $V_i$ be the ($m-1$ dimensional) subspace of $W$ spanned by the set $\{|\psi_{j}\rangle \mid j \neq i\}$.

From elementary Hilbert space theory we know that each of the vectors $|\psi_i\rangle$ can be split up into a sum $|\psi_i\rangle = |v_i\rangle + |o_i\rangle$, where $|v_i\rangle \in V_i$ and $|o_i\rangle \in V_i^\perp \cap W$ ($|v_i\rangle$ is the orthogonal projection of $|\psi_i\rangle$ onto $V_i$ and $|o_i\rangle = |\psi_i\rangle - |v_i\rangle)$. Note that $|o_i\rangle \neq 0$ for each $i$, as $|\psi_i \rangle$ by the assumption of linear independence does not belong to $V_i$.

Now let $E_i=\frac{|o_i\rangle\langle o_i|}{m+1}$, for $i=1,...,m$, and let $E_{m+1} = I - \sum_{j=1}^m E_j$. Then it's an easy check that set of operators $\{E_1, ...,E_{m+1}\}$ satisfies the requirements for a POVM (the factors $\frac{1}{m+1}$ are to make sure that $E_{m+1}$ stays positive). Furthermore, for $1\leq i, k \leq m$ with $i\neq k$, we have $\langle\psi_k|E_i|\psi_k\rangle = 0$, and

$$\langle\psi_i|E_i|\psi_i\rangle = \frac{1}{m+1}\big( \langle v_i|o_i\rangle + \langle o_i|v_i\rangle + 2\langle o_i|o_i\rangle \big) = \frac{2}{m+1}\langle o_i|o_i\rangle>0,$$ as wanted.

  • $\begingroup$ There is something that I think is necessary still. How does Bob know which $E_i$ took? Also if $\langle o_i|o_i\rangle$ are equal for two values of $i$ (the extreme case would be that the set $\lbrace |\Psi_i\rangle\rbrace$ is orthonormal), how can Bob determine which $|\Psi_i\rangle$ he has measured? $\endgroup$ Apr 19, 2021 at 13:30
  • $\begingroup$ I think one needs instead $E_i=\frac{|o_i\rangle \langle o_i|}{i+1}$. $\endgroup$ Apr 19, 2021 at 13:32
  • $\begingroup$ To make sure $E_{m+1}$ stays positive, $E_i$ needs to be defined as $E_i=\frac{|o_i\rangle\langle o_i|}{m\langle o_i|o_i\rangle}$. $\endgroup$
    – QNA
    Nov 8, 2021 at 21:39
  • $\begingroup$ Let me show why this is so. Assume $w\in W$ is an arbitrary vector from $W$. Then $\langle w|E_i|w\rangle=\frac{\langle w|o_i\rangle\langle o_i|w\rangle}{m\langle o_i|o_i\rangle}=\frac{|\langle o_i|w\rangle|^2}{m\langle o_i|o_i\rangle}$. This is obviously $>0$, so $E_i$ for $i \le m$ is positive, as required. Now, $\langle w|E_{m+1}|w\rangle=\langle w|w\rangle-\sum_{i=1}^m\frac{|\langle o_i|w\rangle|^2}{m\langle o_i|o_i\rangle}$. $\endgroup$
    – QNA
    Nov 8, 2021 at 21:40
  • $\begingroup$ By Cauchy-Schwarz inequality $|\langle o_i|w\rangle|^2\le\langle o_i|o_i\rangle\langle w|w\rangle$, so $\langle w|E_{m+1}|w\rangle\ge\langle w|w\rangle-\sum_{i=1}^m\frac{\langle o_i|o_i\rangle\langle w|w\rangle}{m\langle o_i|o_i\rangle}=\langle w|w\rangle-\langle w|w\rangle=0$, so $E_{m+1}$ is positive, as required. $\endgroup$
    – QNA
    Nov 8, 2021 at 21:40

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