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I was watching a presentation and I run into this example.

What does $\phi$ with the (x)t on the exponent mean?, similarly for the other symbol

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This presentation seems to be about pseudorandom quantum states.

A set of quantum states $\left\{\phi_k\right\}_k$ indexed by some key values is said to be pseudorandom if no adversary can distinguish polynomially many copies of a state sampled by choosing a random key $k$ from as many copies of a Haar-random state $|\psi\rangle$.

In cryptography, such properties are often modeled via some games, which is the case here. The challenger will choose a bit $b\in\{0,1\}$ at random.

  • If $b=0$, then they will choose a random key $k$ and give the adversary $t$ copies of $\phi_k$. This is what $\phi_k^{\otimes t}$ represents, as mentioned by @DaftWullie in his answer.
  • If $b=1$, then they will sample a Haar-random state $|\psi\rangle$ and give the adversary $t$ copies of the same Haar-random state, which is represented by $|\psi\rangle^{\otimes t}$.

(By the way, if you don't know what an Haar-measure is, picking a Haar-random state means here to pick a quantum state uniformly at random among the set of all states.)

The adversary, upon being given $t$ copies of a state, must tell whether $b=0$ or $b=1$. If they cannot win with probability significantly larger than $\frac12$, which is just the probability of guessing, then the set $\left\{\phi_k\right\}_k$ is pseudorandom.

This is exactly what's depicted here. Note that what's potentially misleading is that both $\varphi_k$ and $|\psi\rangle$ are quantum states, even though the first one is not written using ket notation. It is usually written $\left|\phi_k\right\rangle$, which means that $t$ copies of it are usually denoted $\left|\phi_k\right\rangle^{\otimes t}$.

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This is the tensor product symbol. Thus, something like $\phi^{\otimes t}$ means that you have $t$ copies of the state $\phi$.

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  • $\begingroup$ Could you comment also on what $\phi^{\otimes t}_k$ might mean? $\endgroup$ Commented Jul 24, 2023 at 14:00
  • $\begingroup$ Without more context, not really. I assume it just means that $\phi$ depends on the parameter $k$ that was the input to the StateGen function, and you have $t$ copies of it. $\endgroup$
    – DaftWullie
    Commented Jul 24, 2023 at 14:27

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