MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


Suppose I have a density matrix $\rho$ which is proportional to a projector $P$ formed by tensoring together $N$ small projectors $P^{(i)}$ of rank 2:

$P^{(i)} = |a\rangle_i\langle a| + |b\rangle_i\langle b|$

$P=P^{(1)}\otimes \cdots \otimes P^{(N)} = \bigotimes_{i=1}^N P^{(i)}$

$\rho = P/2^n \quad , \quad \mathrm{Tr} \rho = 1 $.

(I'm a physicists, and I don't really know how to express $P^{(i)}$ explicitly without using bra-ket notation.) The small projectors $P^{(i)}$ act on their own (potentially very large) Hilbert space $H^{(i)}$ spanned by, say, $\{|a\rangle_i, |b\rangle_i, |c\rangle_i,...\}$, and $P$ acts on $H = \bigotimes_{i=1}^N H^{(i)}$.

Then since the $2^N$ non-zero eigenvalues of $\rho$ are all the same, the entropy is obviously

$H[\rho] = -\sum_{s} \lambda_s \ln \lambda_s = N \ln 2 $.

Now, suppose I split the density matrix in half, and operate on each half with either a local set of unitaries (all identical except that the act on different $H^{(i)}$) or their inverses. That is, suppose

$U = \bigotimes_{i=1}^N U^{(i)}$

$U^{(i)} |a\rangle_i = |a+\rangle_i \quad , \quad {U^{(i)}}^{-1} |a\rangle_i = |a-\rangle_i \quad , \quad U^{(i)} |b\rangle_i = |b+\rangle_i \quad , \quad {U^{(i)}}^{-1} |b\rangle_i = |b-\rangle_i $

with a new density matrix $\eta$:

$\eta = (U P U^{-1} + U^{-1} P U)/2^{N+1}$.

Assume we know the inner products

$\langle a-|b+\rangle \quad, \quad \langle a+|b-\rangle \quad, \quad \langle a+|a-\rangle \quad, \quad \langle b+|b-\rangle \quad ,$

(which are the same for all $i$ because the $U^{(i)}$ are all identical).


Is it possible to calculate the entropy $H[\eta]$?

If useful, it's fine to assume that $U^{(i)}$ is infitesimal so that

$\langle a-|b+\rangle \sim 0 \quad, \quad \langle a+|b-\rangle \sim 0 \quad, \quad \langle a+|a-\rangle \sim 1 \quad, \quad \langle b+|b-\rangle \sim 1 \quad .$

Simpler case

It might be helpful to see the solution to the (much) simpler case where the $P^{(i)}$ are rank 1. In this case, we can write

$P^{(i)} = |a\rangle_i\langle a|$

$P = |a\rangle_1\langle a| \otimes \cdots \otimes |a\rangle_N\langle a| = |A\rangle\langle A|$

$\rho = P \quad , \quad \mathrm{Tr} \rho = 1$

$\eta = (U P U^{-1} + U^{-1} P U)/2 $

$\quad = (|A+\rangle\langle A+| + |A-\rangle\langle A-|)/2$

and get that the two non-zero eigenvalues are

$\lambda_\pm = [1 \pm |\langle A+|A-\rangle_i|]/2$


$\langle A+|A-\rangle = \prod_{i=1}^N \langle a+|a-\rangle_i = \langle a+|a-\rangle^N$.

Elaboration of Answer

Steve Flammia graciously provided the answer below, which hinges on the relationship between the eigenvalues of the two operators $P+Q$ and $PQP$, where $P$ and $Q$ are orthogonal projectors. (The eigenvalues of $PQP$ are easy to compute because the matrix is diagonal in the same basis as $P$ and the matrix elements are expressed in terms of the known inner products.) As an exercise for myself, and to help out anyone else reading this question, the relationship is proved here.

Consider two orthogonal projectors $P$ and $Q$ with respective images $M$ and $N$. We can assume the intersections

$ \quad M \cap N \quad , \quad M \cap N^\perp \quad , \quad M^\perp \cap N \quad , \quad M^\perp \cap N^\perp$

are all {0}, i.e. $M$ and $N$ are in "generic position". (If not, then the Hilbert space can be decomposed into $H=H' \oplus H_0$, where $M$ and $N$ leave both $H'$ and $H_0$ invariant, $M$ and $N$ are simultaneously diagonal on $H_0$, and the above intersections are trivial on $H'$. We then restrict our attention $H'$.) As shown by Halmos [1], for any pair of orthogonal projectors $P$ and $Q$ of equal rank we can decompose the Hilbert space into two subspaces of equal dimension, $H=K \oplus K$, in which

$P = \pmatrix{1& 0 \\\ 0& 0}$

$Q = \pmatrix{C^2 & CS \\\ CS & S^2}$

where $C$ and $S$ are positive matrices on $K$ which commute and for which $C^2 + S^2 =1$. If we let $c_i$ be the eigenvalues of $C$, then the non-zero eigenvalues of

$PQP = \pmatrix{C^2& 0 \\\ 0& 0}$

are $c_i^2$ and the eigenvalues of

$P+Q=\pmatrix{1 + C^2& CS \\\ CS & S^2}$

can be shown to be $1 \pm c_i$.


[1] Halmos, P.R. "Two Subspaces". Transactions of the American Mathematical Society, Vol. 144 (Oct., 1969), pp. 381-389.

share|cite|improve this question
May I suggest a few changes to your post? You should correct the typo in the title, and perhaps either make it more precise (eliminating words like "split" and "tweak" which are not very precise), or just shorten it. – Steve Flammia Jun 17 '10 at 17:16
Yup, I couldn't think of precise statement for the title at the time, but I think it's better now. Many thanks. – Jess Riedel Jun 21 '10 at 19:40
up vote 2 down vote accepted

First observation, you can unitarily rotate the first term without changing the spectrum so you can just consider a single unitary with out loss of generality. Call the rotated term $Q$.

Next, it helps to know about the canonical form for two projectors. Given two projectors P and Q in general position, you can always compute the eigenvalues of P+Q as follows. They are simply $1 \pm x_j$ where the $x_j$ are the positive square roots of the eigenvalues of the operator $PQP$ (or $QPQ$... it doesn't matter).

For your problem, because of the tensor product structure, this is actually quite easy to do. The eigenvalues $x_j$ will be all possible products of the local eigenvalues for $P^{(i)} Q^{(i)} P^{(i)}$, with a square root. The entropy is then a sum over these configurations, suitably normalized.

share|cite|improve this answer
Note that this formula agrees with the (trivial) case where $P$ is rank 1. – Steve Flammia Jun 17 '10 at 17:21
Wow, this is exactly what I needed. Thanks for teaching me about the generic position for projectors. – Jess Riedel Jun 21 '10 at 19:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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