If I have two projectors $\pi_1, \pi_2$ such that for some $|{\phi}\rangle$:

$\langle {\phi}| \pi_1 |{\phi}\rangle \geq e$ and $\langle {\phi}| \pi_2 | {\phi}\rangle \geq e$

What can I conclude about the following quantity?

$\langle {\phi} | \pi_1 \pi_2 |{\phi}\rangle$

Is it also $\geq e$?

  • $\begingroup$ What if $\pi_1 \pi_2=0$? $\endgroup$
    – peterh
    Nov 17, 2020 at 21:46

3 Answers 3


Based on those relations there's nothing more that you can conclude. Consider the two extremes.

At one extreme $\pi_1$ and $\pi_2$ project onto the same subspace, in which case: $$\langle {\phi} | \pi_1 \pi_2 |{\phi}\rangle = \langle {\phi} | \pi_1 |{\phi}\rangle = \langle {\phi} | \pi_2 |{\phi}\rangle \ge e, \;\; \pi_1=\pi_2=\pi_1 \pi_2.$$

At the other extreme $\pi_1$ and $\pi_2$ project onto perpendicular subspaces, in which case $$\langle {\phi} | \pi_1 \pi_2 |{\phi}\rangle = 0, \;\; \pi_1 \pi_2 = 0.$$

Based on the stated relationships, the projectors could exist anywhere between and including these two extremes. If $e>0$, you could at least say that $\vert \phi \rangle$ is not perpendicular to either the $\pi_1$ or $\pi_2$ subspaces, but that alone still doesn't tell you any more about the value of $\langle {\phi} | \pi_1 \pi_2 |{\phi}\rangle$.

EDIT: DaftWullie's answer makes an important point that I missed. If $e>\frac{1}{\sqrt{2}}$ (with $\vert \phi \rangle$ normalized), $\pi_1$ and $\pi_2$ cannot be orthogonal projectors, in which case $e$ imposes the lower bound $2e^2-1$.

  • $\begingroup$ Thanks for explaining. I asked a related question at: physics.stackexchange.com/questions/594565/… Would appreciate it if you can share your insights. $\endgroup$ Nov 17, 2020 at 23:07
  • $\begingroup$ @islamfaisal My pleasure. Looks like Chiral Anomaly provided a solid answer while I was thinking it through. That's similar to the example I had in mind. $\endgroup$ Nov 18, 2020 at 2:09

Let's assume $$ \pi_i|\phi\rangle=e_i|\phi\rangle+\sqrt{1-e_i^2}|\phi_i^\perp\rangle, $$ where $\langle\phi|\phi_i^\perp\rangle=0$ and, for simplicity, let's assume the $e_i$ are real. We can immediately expand $$ \langle\phi|\pi_1\pi_2|\phi\rangle=\left(f_1\langle\phi|+\sqrt{1-f_1^2}\langle\phi_1^\perp|\right)\left(f_2|\phi\rangle+\sqrt{1-f_2^2}|\phi_2^\perp\rangle\right) $$ This simplifies to $$ f_1f_2+\sqrt{1-f_1^2}\sqrt{1-f_2^2}\langle\phi_1^\perp|\phi_2^\perp\rangle, $$ which is bounded between $$ f_1f_2\pm\sqrt{1-f_1^2}\sqrt{1-f_2^2}. $$ Thus, we can say $$ \min_{f_1,f_2\geq e}f_1f_2-\sqrt{1-f_1^2}\sqrt{1-f_2^2}\leq\langle\phi|\pi_1\pi_2|\phi\rangle\leq\max_{f_1,f_2\geq e}f_1f_2+\sqrt{1-f_1^2}\sqrt{1-f_2^2}. $$ Clearly the right-had side of this bound is just 1. But it's the left side we're really interested in. To perform the minimisation, let $f_i=\cos\theta_i$. Then, we want $$ \min_{\theta_1,\theta_2}\cos(\theta_1+\theta_2). $$ If we assume that $\theta_1+\theta_2<\pi$, the minimum is achieved by setting both $\theta_1$ and $\theta_2$ as large as possible, corresponding with $f_i=e$. Thus, provided $e>0$, the minimum value is $2e^2-1$. $$ \langle\phi|\pi_1\pi_2|\phi\rangle\geq 2e^2-1 $$


$$ \pi_1 = \begin{pmatrix} 1&0\\ 0&0\\ \end{pmatrix}\\ \pi_2 = \begin{pmatrix} 0&0\\ 0&1\\ \end{pmatrix}\\ \rho = \frac{1}{\sqrt{2}} \begin{pmatrix} 1\\ 1\\ \end{pmatrix}\\ e = \frac{1}{2} $$

  • $\begingroup$ Thanks. Can you please elaborate on how one can generalize from this example? $\endgroup$ Nov 17, 2020 at 17:45
  • $\begingroup$ @islamfaisal If you calculate $\langle \phi | \pi_1 \pi_2 | \phi \rangle $ in this case, what would you get? This is an example to show the last statement in your question is not true. $\endgroup$
    – KAJ226
    Nov 17, 2020 at 19:47
  • $\begingroup$ @KAJ226 Thanks. and are there other non-trivial counterexamples? $\endgroup$ Nov 17, 2020 at 20:57

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