# How can we upper bound the norm of a partial trace?

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 $$\| \mathrm{Tr}_{B} (|\phi_{1}\rangle\langle\phi_{2}| )\|_{1}$$?

• The $1$-norm decreases under partial trace and so there is an upper bound of $1$ if the states are normalized. Nov 23 '20 at 21:13
• @Rammus, is it possible to get a stricter upper bound than 1? What I am looking for is to prove $|| tr_{B} (|\phi_{1}\rangle \langle \phi_{2}|) || < 1$ Nov 23 '20 at 21:18

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, $$\|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 = \||0 \rangle \langle 1|\|_1 = 1.$$