# What can I conclude about $\langle \phi|\pi_1\pi_2|\phi\rangle$ if $\langle \phi|\pi_i|\phi\rangle\ge e$?

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 $$\geq e$$?

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

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 $$2e^2-1$$.

• Thanks for explaining. I asked a related question at: physics.stackexchange.com/questions/594565/… Would appreciate it if you can share your insights. Nov 17, 2020 at 23:07
• @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. 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 $$2e^2-1$$. $$\langle\phi|\pi_1\pi_2|\phi\rangle\geq 2e^2-1$$

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

• Thanks. Can you please elaborate on how one can generalize from this example? Nov 17, 2020 at 17:45
• @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. Nov 17, 2020 at 19:47
• @KAJ226 Thanks. and are there other non-trivial counterexamples? Nov 17, 2020 at 20:57