In Nielsen and Chuang exercise 2.64, the following problem is given:

Suppose Bob is given a quantum state chosen from a set $\{ \lvert \psi_1 \rangle, \ldots , \lvert \psi_m \rangle \}$ of linearly independent states [in some Hilbert space]. Construct a POVM $\{E_1, E_2, \ldots, E_{m+1} \}$ such that if outcome $E_i$ occurs, $1 \le i \le m$, then Bob knows with certainty that he was given the state $\lvert \psi_i \rangle$. (The POVM must be such that $\langle \psi_i | E_i | \psi_i\rangle > 0$ for each i.)

Here a POVM is a positive operator-valued measurement, or a set $\{E_i\}$ of positive (semidefinite) operators satisfying the completeness condition $$ \sum_i E_i = I. $$ Note that in this problem $\lvert \psi_i \rangle$ are not assumed to be orthogonal.

My approach to this problem is to construct each $E_i$ for $i \le m$ as a scaled projector: $$ E_i = \alpha_i P_i, $$ where $P_i$ is the projector onto the orthogonal complement of $\mathrm{Span}(\lvert \psi_1 \rangle, \ldots, \lvert \hat{\psi_i} \rangle, \ldots, \lvert \psi_m \rangle)$ in $\mathrm{Span}(\lvert \psi_1 \rangle, \ldots, \lvert \psi_m \rangle)$ satisfying $\langle \psi_i | P_i | \psi_i \rangle > 0$, and $\alpha_i$ is some positive real number.

Then we have the desired properties that each $E_i$ is positive, and $\langle \psi_i | E_j | \psi_i \rangle = \delta_{ij} \beta_i$ for some real positive $\beta_i$. Indeed, $E_i$ has eigenvalues $\alpha_i$ and $0$; $E_j | \psi_i \rangle = 0$ if $i \ne j$ as $P_j$ is a projector onto a space orthogonal to $\lvert \psi_i \rangle$; and $$\langle \psi_i | E_i | \psi_i \rangle = \alpha_i \cos \theta_i \langle \psi_i | \psi_i \rangle = \beta_i > 0, $$ where $\theta_i$ is the angle between $\lvert \psi_i \rangle$ and $P_i \lvert \psi_i \rangle$.

Then for any choice of $\{\alpha_i\}$ we have constructed positive $E_i$ for $i \le m$, but this set of positive operators is not necessarily complete. To complete the set, let $E_{m+1} = I - \sum_{i \le m} E_i$. It seems that as long as we can ensure $E_{m+1}$ is positive, then we have constructed the desired POVM. However, I am not sure how to assign $\{\alpha_i\}$ in order to do so.

Naively, I can't see why we don't have a great deal of freedom over $\alpha_i$. As a sum of Hermitian operators, $E_{m+1}$ is automatically Hermitian. And $\langle \psi_i | E_{m+1} | \psi_i \rangle = \langle \psi_i | \psi_i \rangle - \beta_i \ge 0$ so long as $\alpha_i \cos \theta_i \le 1$. However, this so-called POVM would be able to distinguish the states $\lvert \psi_i \rangle$ perfectly if we were allowed to choose $\alpha_i = 1/\cos \theta_i$, in violation of the no-cloning theorem (as shown in Nielsen–Chuang, exercise 1.2, with stronger bounds implicit in Box 2.3).

I feel I must be mistaken about the criterion for $E_{m+1}$ to be positive—it does not seem sufficient to ensure that $\langle \psi_i | E_{m+1} | \psi_i \rangle \ge 0$ for all $i$, even though they are linearly independent.

Is this a fruitful approach for this problem? If so, what conditions on $\{\alpha_i\}$ would ensure $E_{m+1}$ is positive? What have I missed?


1 Answer 1


Your approach seems perfectly fine and probably what the authors intended.

Note that there is no need for us to find the "best" (i.e. largest) $\alpha_i$ such that $I - \sum_i E_i$ is positive definite. Note that $\lambda$ is an eigenvalue of $M$ if and only if $1 - \lambda$ is an eigenvalue of $I - M$. Thus, for a positive semidefinite matrix $M$, $I - M$ will be positive definite if and only if all eigenvalues of $M$ are at most equal to $1$. Moreover, if $M$ is positive semidefinite, then the largest eigenvalue of $M$ is equal to $\|M\|$, the "spectral norm" of $M$. Now, note that $$ \left\|\sum_i P_i \right\| \leq \sum_{i} \|P_i\| = \sum_i 1 = m. $$ Thus, if we take $\alpha_i = \frac 1m$ for all $i$, we find that $$ \left\|\sum_i E_i \right\| = \left\|\frac 1m \sum_i P_i \right\| = \frac 1m \left\|\sum_i P_i \right\| \leq 1, $$ So that $I - \sum_i E_i$ is positive semidefinite, as desired.

For another perspective, note that for a Hermitian matrix $M$, $I - M$ is positive definite iff for all $|\phi\rangle$, we have $$ \langle \phi|I - M|\phi \rangle \geq 0 \implies\\ \langle \phi|I|\phi \rangle - \langle \phi|M|\phi \rangle \geq 0 \implies\\ \langle \phi |M|\phi \rangle \leq \langle \phi |\phi \rangle. $$ On the other hand, note that $\langle \phi |P_i |\phi\rangle \leq \langle \phi |\phi \rangle$, so that $$ \left\langle \phi \left| \sum_i P_i \right| \phi \right\rangle \leq \sum_i \langle \phi |P_i|\phi \rangle \leq m \cdot \langle \phi|\phi \rangle. $$ By a similar argument to the previous approach, we can conclude that taking $\alpha_i = \frac 1m$ ensures that $I - \sum_i E_i$ is positive semidefinite.

If you were interested in finding the maximal values of $\alpha_i$ (say, in the sense of maximizing $\sum_i \alpha_i$) such that $E_{m+1}$ is still positive semidefinite, you could do so via a dual semidefinite program. In particular, if $\alpha$ and $b$ are the column-vectors $\alpha = (\alpha_1,\alpha_2,\dots,\alpha_m)$ and $b = (1,1,\dots,1)$, then we aim to solve the optimization problem $$ \max_{\alpha \in \Bbb R^n} b^\top \alpha \quad \text{subject to } \quad \sum_{i=1}^m \alpha_i P_i \preceq I. $$ Suffice it to say, such problems are highly non-trivial and typically done with computer assistance.

This of course is much simpler in the case where the $|\psi_i\rangle$ are mutually orthogonal, which would imply that the $E_i$ are mutually orthogonal projections, which would imply that the constraint is equivalent to $\alpha_i \leq 1$ for all $i$.

  • $\begingroup$ @jackson Of course, I should read more carefully. Thanks! $\endgroup$ Commented Jan 25, 2023 at 17:45
  • $\begingroup$ @jackson See my edited answer. $\endgroup$ Commented Jan 25, 2023 at 17:56
  • $\begingroup$ I wish I could upvote again for the edits! Just to add something of substance to avoid the comment getting deleted, I noticed there is also a principle bounding $\beta_i$ from above in Box 2.3 of Nielsen and Chuang, though in practice I'm not sure if this (not obviously sharp) upper bound is any easier to calculate than the DPSD program $\endgroup$
    – jackson
    Commented Jan 25, 2023 at 18:33
  • 1
    $\begingroup$ @jackson Happy to see my work appreciated! I wouldn't be surprised if your observation led to some non-trivial upper bound in this case; keep in mind that we do have a little bit of extra "structure" here relative to a generic problem of this kind in the sense that the $E_i$ all have rank 1. $\endgroup$ Commented Jan 25, 2023 at 18:40
  • $\begingroup$ @jackson We also get an interesting family of upper bounds from the "dual" nature of this optimization problem, in particular using "weak duality". For any positive semidefinite matrix $X$ such that $\operatorname{Trace}(XE_i) = b_i$ for $i = 1,\dots,m$, $\operatorname{Trace}(X)$ is an upper bound to the optimal $b^\top \alpha$. We could use the low-rank structure here to make that constraint a bit "nicer" here, if we wanted. $\endgroup$ Commented Jan 25, 2023 at 18:44

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