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.

  • $\begingroup$ 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 $||\cdot ||_2$. $\endgroup$
    – narip
    Commented Dec 24, 2022 at 6:57
  • $\begingroup$ I tried relating the euclidean norm to the trace norm with no success. $\endgroup$
    – Apo
    Commented Dec 25, 2022 at 4:04

1 Answer 1


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 $\frac{2-\epsilon ^2}{2}\le \mathrm{Re}\langle \psi |\phi \rangle \le \left| \langle \psi |\phi \rangle \right|$.


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