Value of tensor product of projectors If I have two projectors $\pi_1, \pi_2$ such that for some $|{\phi}\rangle$:
$\langle {\phi}| I \otimes \pi_1 |{\phi}\rangle \geq e$ and $\langle {\phi}| \pi_2 \otimes I | {\phi}\rangle \geq e$
What lower bound can I conclude about the following quantity?
$\langle {\phi} | \pi_2 \otimes \pi_1 |{\phi}\rangle$
 A: Summary:

*

*If the norm of $|\phi\rangle$ is not constrained, then the lower bound is zero.


*If $|\phi\rangle$ is required to be a unit vector, then the lower bound is either zero or $2e-1$, whichever is larger.
To derive these bounds, use the abbreviations
$$
\newcommand{\oP}{{\hat P}}
\newcommand{\oQ}{{\hat Q}}
\newcommand{\ra}{\rangle}
\newcommand{\la}{\langle}
 P \equiv I\otimes\pi_1
\hspace{2cm}
 Q \equiv \pi_2\otimes I.
$$
The operators $P$ and $Q$ are mutually commuting projection operators, and their product is the projection operator
$$
 PQ = \pi_2\otimes\pi_1.
$$
For any pair of mutually commuting projection operators $P$ and $Q$ and any vector $|\phi\ra$, we have the identity
$$
 \la\phi|\phi\ra  = A + B+C+D
\tag{1}
$$
with
\begin{align}
 A &= \la\phi|PQ|\phi\ra \\
 B &= \la\phi|P\oQ|\phi\ra \\
 C &= \la\phi|\oP Q|\phi\ra \\
 D &= \la\phi|\oP\oQ|\phi\ra
\end{align}
$$
 \oP\equiv 1-P
\hspace{2cm}
 \oQ\equiv 1-Q.
$$
Each of the quantities $A,B,C,D$ is between $0$ and $\la\phi|\phi\ra$.
Now suppose
$$
 \la\phi|P|\phi\ra = e
\hspace{2cm}
 \la\phi|Q|\phi\ra = e.
\tag{2}
$$
The goal is to derive a tight lower bound on $A$. Since $P$ and $Q$ are projection operators, we must have $0\leq e\leq\la\phi|\phi\ra$. Equations (2) are equivalent to
$$
 A+B = e
\hspace{2cm}
 A+C = e.
\tag{3}
$$
If $2e \leq \la\phi|\phi\ra$, then we can take
$$
 A = 0
\hspace{2cm}
 B=C=e
\hspace{2cm}
 D = \la\phi|\phi\ra-2e.
$$
This achieves $A=0$, which is the lowest we can go because $A$ cannot be negative. If $2e \geq \la\phi|\phi\ra$, then we can take
$$
 A = 2e-\la\phi|\phi\ra
\hspace{2cm}
 B=C=\la\phi|\phi\ra-e
\hspace{2cm}
 D = 0.
$$
This achieves $A = 2e-\la\phi|\phi\ra$, which is the lowest we can go because equations (1) and (3) imply
$$
 A = D + 2e-\la\phi|\phi\ra
$$
and $D$ cannot be negative. Altogether, this implies the results that were summarized above.
