# Random quantum circuits and general efficient POVM measurement

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

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

Is anything known, in general, about

$$$$\mathbb{E}[p_i] ~~\text{and}~~\mathbb{E}[p_i^{2}]$$$$ 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 $$$$\mathbb{E}[p_i] = \frac{1}{2^{n}}, ~~~ \mathbb{E}[p_i^{2}] = \frac{2}{2^{n}(2^{n} + 1)}.$$$$

## 1 Answer

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.

• 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? Sep 13 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. Sep 25 at 5:50