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I am wondering if a random unitary matrix taken from a Haar measure (as in, it is uniformly sampled at random) can yield a uniformly sampled random state vector.

In section 3 of this paper it says "It is worthwhile mentioning that, although not advantageous, it is possible to use the rows or columns of such a random unitary matrix as random state vector" and also says in the previous section that " Another manner of obtaining samples with similar properties is by using the rows or columns of random unitary matrices, which we shall discuss in the next section."

I am a bit confused by the wording of this paper. Is it explicitly saying that taking a column or row from a random unitary matrix sampled uniformly will in fact give a random state vector with respect to the Haar measure?

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Yes.

A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as likely as $|\psi\rangle$ for any unitary $U$.

On the other hand, a Haar random unitary $V$ is defined the same way: "$UV$ is just as likely as $V$, for any $U$."

Thus, a column of $V$, let's call is $|\chi\rangle$, has again the property that $|\chi\rangle$ and $U|\chi\rangle$ for any $U$ are "equally likely", that is, $|\chi\rangle$ is distributed Haar random.

(As stated in the beginning, the colloquial "equally likely" should be understood as an invariance property of the measure.)

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  • $\begingroup$ can you elaborate more on: "Thus, a column of $V$, let's call is $|\chi\rangle$, has again the property that $|\chi\rangle$ and $U|\chi\rangle$ for any $U$ are "equally likely", that is, $|\chi\rangle$ is distributed Haar random." Why is $|\chi\rangle$ and $U|\chi\rangle$ for any $U$ "equally likely"? $\endgroup$
    – Quantum Guy 123
    Jan 19, 2021 at 16:16
  • $\begingroup$ given a column vector, does that map to a unique unitary? not sure how to show that. $\endgroup$
    – Quantum Guy 123
    Jan 19, 2021 at 16:17
  • $\begingroup$ @QuantumGuy123 No, obviously not, just by parameter counting. $\endgroup$
    – Norbert Schuch
    Jan 19, 2021 at 17:36
  • $\begingroup$ @QuantumGuy123 With regard to the first question: If the probability of picking a unitary U or V*U are the same, then also the probability of picking their respective first columns, which are chi (for U, that's how I define chi) and U*chi, are the same. $\endgroup$
    – Norbert Schuch
    Jan 19, 2021 at 17:41
  • $\begingroup$ ok thanks. also can you provide a reference/source for the two claims you stated: 1) A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as likely as $|\psi\rangle$ for any unitary $U$. and 2) On the other hand, a Haar random unitary $V$ is defined the same way: "$UV$ is just as likely as $V$, for any $U$." $\endgroup$
    – Quantum Guy 123
    Jan 19, 2021 at 20:32
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Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state.

That means that there is some state, call it $|\psi\rangle$, that is relatively more likely to be found when sampling states with such procedure. But that would mean that the unitaries whose first column is $|\psi\rangle$ are relatively more likely to be found than others. This is clearly against the assumption that the unitaries are uniformly sampled. We conclude that taking the first (or any other) column of a uniformly sampled unitary matrix gives a uniformly sampled quantum state.

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  • $\begingroup$ "This pretty much just boils down to observing that there is a bijection between (pure) quantum states and the k-th columns of unitary matrices." that is not true. two different unitary matrices can have the same k-th column. $\endgroup$
    – Quantum Guy 123
    Jan 22, 2021 at 16:25
  • $\begingroup$ @QuantumGuy123 true, I removed that statement $\endgroup$
    – glS ♦
    Jan 23, 2021 at 0:13

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