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Let's consider a random quantum circuit $C$, applied to the $n$ qubit initial state $|0^{n}\rangle$, producing the state $|\psi\rangle$.

Consider a general efficiently implementable $m$-outcome POVM measurement $\{M_i : i = 0, 1, \ldots, m-1\}$. Let

\begin{equation} p_i = \text{Tr}\big(M_i |\psi\rangle \langle \psi|\big). \end{equation}

Is anything known, in general, about

\begin{equation} \mathbb{E}[p_i] ~~\text{and}~~\mathbb{E}[p_i^{2}] \end{equation} where the expectation is taken over the choices of the random circuit?

I am especially interested in the case when the POVM elements $M_i$ describe an entangled multi-qubit measurement (which is efficiently implementable).

Note that for the special case of when the POVM corresponds to an $2^{n}$-outcome standard basis measurement, we know that \begin{equation} \mathbb{E}[p_i] = \frac{1}{2^{n}}, ~~~ \mathbb{E}[p_i^{2}] = \frac{2}{2^{n}(2^{n} + 1)}. \end{equation}

Adam Zalcman
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BlackHat18
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2 Answers2

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In the absence of additional assumptions, $\mathbb{E}[p_i]$ can be any real number in $[0, 1]$. For example, let $a\in[0,1]$ and define the POVM as $M_0=aI$ and $M_1=(1-a)I$. Then

$$ \mathbb{E}[p_0] = \int \mathrm{tr}\left(aI|\psi\rangle\langle\psi|\right)d\psi = a \int \langle\psi|\psi\rangle d\psi = a $$

assuming the Haar measure is normalized. Similarly, $\mathbb{E}[p_1]=1-a$.

In the special case of the $2^n$-outcome measurement in the computational basis of all qubits, the fact that the respective POVM elements are orthogonal projectors is essential to obtaining $\mathbb{E}[p_i]=2^{-n}$.

For any candidate property of general POVMs, it pays to consider a number of simple special cases such as orthogonal projectors (as you did in your question), multiples of identity (as I did in my answer) and others such as SIC-POVMs. See also this question for additional inspiration.

Adam Zalcman
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  • One question out of curiosity, if we put one additional assumption --- that the POVMs are orthogonal projectors (not necessarily standard basis projectors but arbitrary ones) --- can we then say anything extra about the first and second moments? Perhaps, something like if it is an $m$ outcome POVM, then the first moment is ~ 1/m? – BlackHat18 Sep 13 '21 at 10:01
  • One observation: if the POVMs are orthogonal projectors and also unentangled across qubits, it follows directly from the translational invariance of the Haar measure that the second moment is the same. I am thinking about the entangled measurement case (let's say, for example, a Bell basis measurement on pairs of qubits): I have a hunch that the second moment should still be something like the unentangled case, but I could not show it. – BlackHat18 Sep 25 '21 at 05:50
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Using the formula mentioned in this question. We have $$\mathbb{E} [p_i]=\mathbb{E} \left[ \mathrm{Tr(}M_i|\psi \rangle \langle \psi |) \right] \\ =\mathrm{Tr(}M_i\int_{\mathbf{U}(d)}{d}\mu (U)U|0^n\rangle \langle 0^n|U^{\dagger}) \\ =\mathrm{Tr(}\int_{\mathbf{U}(d)}{d}\mu (U)U^{\dagger}M_iU|0^n\rangle \langle 0^n|) \\ =Tr\left( \frac{Tr\left( M_i \right)}{2^n}|0^n\rangle \langle 0^n| \right) \\ =\frac{Tr\left( M_i \right)}{2^n}$$

narip
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