# Why is it not easy to distinguish $U|\psi\rangle$ and $U'|\psi\rangle$ if $\|U-U'\|<\epsilon$?

So I am currently working on an assignment, which is about the induced Euclidian norm

$$||A||:= \max_{v\in\mathbb{C}^d\text{ s.t. }||v||_2=1} ||Av||_2$$

for some $$A\in\mathbb{C}^{d\times d}$$.

For given unitaries $$U,U'\in\mathbb{C}^{d\times d}$$ s.t. $$||U-U'||<\epsilon$$ for some small $$\epsilon$$ and an arbitrary state $$|\psi\rangle\in\mathbb{C}^d$$, my task is to explain, why it is not that easy to distinguish via any measurement the output of $$U|\psi\rangle$$ and $$U'|\psi\rangle$$.

I don't have any idea how to explain this statement because I don't get the intuition behind it. Maybe someone can help me/at least give me a form of intuition, why this statement holds.

• intuitively, well, $\|U-U'\|<\epsilon$ means that $U$ and $U'$ are "almost the same", and thus the states $U|\psi\rangle$ and $U'|\psi\rangle$ are also "almost the same", and thus harder to tell apart
– glS
Jul 6, 2021 at 7:14

Since this an assignment problem,'' here are a few hints and you should be able to fill in the gaps. Let $$\mathcal{H} \cong \mathbb{C}^{d}$$ and the Euclidean or $$2$$-norm of a vector $$| v \rangle \in \mathcal{H}$$ is $$\left\Vert | v \rangle \right\Vert_{2}^{} := \sqrt{\left\langle v|v \right\rangle}$$. There are several ways to define norms on the operator space, one of which is to use a vector norm to induce an operator norm, which is the case here. Define $$\left\Vert A \right\Vert_{2}^{} := \max_{| \psi \rangle \in \mathcal{H}} \left\Vert A | \psi \rangle \right\Vert_{2}^{} = \max_{| \psi \rangle \in \mathcal{H}} \sqrt{\left\langle \psi|A^{\dagger}A |\psi \right\rangle}$$.
For $$U,V$$ unitaries, let us compute what $$\left\Vert U-V \right\Vert_{2}^{} < \epsilon$$ is equal to using the $$\left\Vert \cdot \right\Vert_{2}^{}$$ norm. A quick calculation shows that this is equivalent to $$\mathrm{Re} \left\langle \phi | V^{\dagger} U | \phi \right\rangle \geq 1 - \epsilon/2$$, where $$| \phi \rangle$$ is the state for which the maximum is achieved (in the definition of the operator norm).
Now, recall that if two pure states $$| \alpha \rangle, | \beta \rangle$$ are orthogonal, that is, $$\left\langle \alpha | \beta \right\rangle = 0$$ then they are perfectly distinguishable. This is because one can find a projective measurement, $$\Pi_{\alpha} \equiv | \alpha \rangle \langle \alpha | , \Pi_{\beta} \equiv | \beta \rangle \langle \beta |$$ to distinguish them. On the other hand, if their inner product is close to one, then they are difficult to distinguish via a projective measurement.
To compare the outputs of channels $$U,V$$, we can ask, what is the maximum distinguishability of their outputs, namely, the maximal inner product between the states $$U | \psi \rangle$$ and $$V | \psi \rangle$$, that is, $$\max_{\psi \in \mathcal{H}} \left\langle \psi | V^{\dagger} U |\psi \right\rangle$$, which is related to the $$\left\Vert U-V \right\Vert_{2}^{}$$ above.
As a final comment, note that, given two states $$\rho,\sigma$$, the trace norm, $$\left\Vert \rho - \sigma \right\Vert_{\mathrm{tr}}^{} := \frac{1}{2} \left\Vert \rho - \sigma \right\Vert_{1}^{}$$ has an operational interpretation as the maximum success probability in distinguishing two quantum states $$\rho,\sigma$$ via a POVM measurement (and not just projective measurements). Moreover, for pure states, $$\left\Vert | \psi \rangle \langle \psi | - | \phi \rangle \langle \phi | \right\Vert_{\mathrm{tr}}^{} = \sqrt{1 - \left| \left\langle \psi | \phi \right\rangle \right|^{2}}$$.