Suppose we have the normalised states $|\phi_{1}\rangle,|\phi_{2}\rangle \in A \otimes B$ where $A$ and $B$ are $d$-dimensional complex vector spaces.

Suppose $|\langle\phi_{2}|\phi_{1}\rangle| < 1$.

Can we say what is the upper bound of $\| \mathrm{Tr}_{B} (|\phi_{1}\rangle\langle\phi_{2}| )\|_{1}$?

  • $\begingroup$ The $1$-norm decreases under partial trace and so there is an upper bound of $1$ if the states are normalized. $\endgroup$
    – Rammus
    Nov 23, 2020 at 21:13
  • $\begingroup$ @Rammus, is it possible to get a stricter upper bound than 1? What I am looking for is to prove $|| tr_{B} (|\phi_{1}\rangle \langle \phi_{2}|) || < 1$ $\endgroup$
    – user07
    Nov 23, 2020 at 21:18

1 Answer 1


The $1$-norm decreases under partial trace and so we have an upper bound of $1$ when the states are normalized, $$ \|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 \leq \||\psi_1\rangle \langle \psi_2|\|_1 = 1. $$

This bound cannot be improved upon without extra information about the states. Here is a counterexample. Take $|\psi_1 \rangle = |00\rangle$ and $|\psi_2 \rangle = |10\rangle$. Then we have $\langle \psi_1 |\psi_2\rangle = 0$. Moreover we have, $$ \|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 = \||0 \rangle \langle 1|\|_1 = 1. $$


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