# Bounding inner product of states with distance

Suppose we are given two states quantum states $$|{\psi}\rangle$$ and $$|{\phi}\rangle$$ over $$n$$ qubits. We know that the distance between the states is bounded by $$\epsilon$$:

$$|| |{\psi}\rangle- |{\phi}\rangle||_2 \leq \epsilon$$

Where $$||. ||_2$$ denotes the euclidean metric over the space of $$n$$ qubits. Intuitively, I would expect that some kind of lower bound can be given for $$|\langle \phi | \psi \rangle|$$ in terms of $$\epsilon$$ but I haven't bee able to find one.

I tried using a known inequality for the fidelity but unfortunately depends on the trace distance. The inequality I'm referring to is

$$1-\sqrt{F(\phi,\psi)}\leq \frac{1}{2}|| |\phi \rangle\langle\phi | - |\psi \rangle\langle\psi | ||_{tr}$$.

We know that $$F(\phi,\psi)=|\langle \phi | \psi \rangle|^2$$ and thus we can bound the inner product of both states if the trace distance is bounded. Unfortunately I don't see how can this help if the euclidean distance is bounded instead.

• Can you expand $|| |{\psi}\rangle- |{\phi}\rangle||_2$ to see what it is? By saying expand I mean, write the expression out which does not contain $||\cdot ||_2$. Commented Dec 24, 2022 at 6:57
• I tried relating the euclidean norm to the trace norm with no success.
– Apo
Commented Dec 25, 2022 at 4:04

If I understand your problem correctly, you can just expand $$|| |{\psi}\rangle- |{\phi}\rangle||_2$$ to get the result, i.e. $$|||\psi \rangle -|\phi \rangle ||_2=\sqrt{\left( \langle \psi |-\langle \phi | \right) \left( |\psi \rangle -|\phi \rangle \right)}$$, if $$|\psi \rangle$$ and $$|\phi \rangle$$ are normalized, we have $$\sqrt{\left( \langle \psi |-\langle \phi | \right) \left( |\psi \rangle -|\phi \rangle \right)}=\sqrt{2-\langle \psi |\phi \rangle -\langle \phi |\psi \rangle}=\sqrt{2-2\mathrm{Re}\langle \psi |\phi \rangle}\le \epsilon \tag1$$ hence from eq(1) we have $$2-2\mathrm{Re}\langle \psi |\phi \rangle \le \epsilon ^2$$, and finally we have $$\frac{2-\epsilon ^2}{2}\le \mathrm{Re}\langle \psi |\phi \rangle \le \left| \langle \psi |\phi \rangle \right|$$.