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.

  • $\begingroup$ 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 $\endgroup$
    – glS
    Jul 6, 2021 at 7:14

1 Answer 1


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, $\left\Vert | \psi \rangle \langle \psi | - | \phi \rangle \langle \phi | \right\Vert_{\mathrm{tr}}^{} = \sqrt{1 - \left| \left\langle \psi | \phi \right\rangle \right|^{2}}$.


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