Consider an $n$-qubit Haar random unitary $U$.

I am trying to compute the expression

\begin{equation} \mathbb{E}\left[ \frac{\text{Tr}\left(|0^n\rangle \langle 0^n | ~U\rho U^*\right)}{\text{Tr}\left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^*\right)} \right], \end{equation}

where $\rho$ is an arbitrary $n$-qubit quantum state, the expectation is taken over the choice of $U$ and $\mathbb{I}$ is a $2 \times 2$ identity operator.

Note that each of the numerator and denominator can individually be computed in expectation, but I am not sure how to manipulate the ratio.


2 Answers 2


Here is a simpler way.

Since the Haar measure is left and right invariant, we are free to pull out a unitary to the left or right of $U$ and the expectation is left invariant.

In particular, let $U \mapsto (V \otimes \mathbb{I}_{2^{n-1}})U$, where $V$ is some 2-qubit unitary, and $\mathbb{I}_{2^{n-1}}$ the identity on the rest of the $n-1$ qubits. Your desired quantity can thus equivalently be written as

\begin{equation} \mathbb{E}_U\left[ \frac{\text{Tr}_{1,2,\cdots,n}\left(|0^n\rangle \langle 0^n | ~ (V \otimes \mathbb{I}_{2^{n-1}}) U\rho U^\dagger(V^\dagger \otimes \mathbb{I}_{2^{n-1}}) \right)}{\text{Tr}_{1,2,\cdots,n} \left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right]. \end{equation} Note the pleasing fact that $V$ appears in the numerator only, not in the denominator. I have also added subscripts in the traces, simply as book-keeping, in order to keep track of which spaces the trace is over: in both the numerator and denominator they are over qubits $1,2,\cdots,n$, which seems like redundant notation. But bear with me...

Since we get the same value for any choice of $V$, let's average over different instances of $V$, where $V$ is drawn uniformly over the single qubit unitary group. In other words, we Haar average $V$. Thus your desired value can be equivalently expressed \begin{equation} \mathbb{E}_U\mathbb{E}_V\left[ \frac{\text{Tr}_{1,2,\cdots,n}\left(|0^n\rangle \langle 0^n | ~ (V \otimes \mathbb{I}_{2^{n-1}}) U\rho U^\dagger(V^\dagger \otimes \mathbb{I}_{2^{n-1}}) \right)}{\text{Tr}_{1,2,\cdots,n} \left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right]. \end{equation} Now, the $V$-averaging we can do explicitly (using $\mathbb{E}_V[ V O V^\dagger] = \text{Tr}(O) \frac{\mathbb{I}}{2}$). This gives \begin{align} & \frac{1}{2}\mathbb{E}_U \left[ \frac{\text{Tr}_{1,2,\cdots,n}\left(|0^n\rangle \langle 0^n | \mathbb{I} \otimes \text{Tr}_1[ U\rho U^\dagger]) \right)}{\text{Tr}_{1,2,\cdots,n}\left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right] = \\ &\frac{1}{2}\mathbb{E}_U \left[ \frac{\text{Tr}_{2,\cdots,n}\left(|0^{n-1}\rangle \langle 0^{n-1} |\text{Tr}_1[ U\rho U^\dagger]) \right)}{\text{Tr}_{1,2,\cdots,n}\left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right] = \\ &\frac{1}{2}\mathbb{E}_U \left[ \frac{\text{Tr}_{2,\cdots,n}\left(|0^{n-1}\rangle \langle 0^{n-1} |\text{Tr}_1[ U\rho U^\dagger]) \right)}{\text{Tr}_{2,\cdots,n}\left(|0^{n-1}\rangle \langle 0^{n-1} |\text{Tr}_1[ U\rho U^\dagger]) \right)} \right] = \\ & \frac{1}{2}\mathbb{E}_U[1] = \nonumber \\ & \frac{1}{2}. \end{align}


Since the numerator and the denominator are (apparently) not independent, I'm not convinced that their expectation can be computed separately.

First of all, note that: $$\DeclareMathOperator{Tr}{Tr}\begin{align}\Tr\left(\mathbb{I} \otimes \left|0^{n-1}\middle\rangle\middle\langle 0^{n-1}\right|U\rho U^\dagger\right)&=\Tr\left(\left|0^n\middle\rangle\middle\langle0^n\right|U\rho U^\dagger + \left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)\\&=\Tr\left(\left|0^n\middle\rangle\middle\langle0^n\right|U\rho U^\dagger\right) + \Tr\left(\left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)\end{align}$$ Furthermore, note that: $$\Tr\left(\left|0^{n}\middle\rangle\middle\langle 0^{n}\right|U\rho U^\dagger\right)=a_{0,0}$$ and : $$\Tr\left(\left|1\middle\rangle\middle|0^{n-1}\middle\rangle\middle\langle1|\middle\langle0^{n-1}\right|U\rho U^\dagger\right)=a_{2^{n-1},2^{n-1}}$$ when denoting: $$U\rho U^\dagger=\begin{pmatrix}a_{0,0}&\cdots&a_{0, 2^{n}-1\\}\\\vdots & \ddots&\vdots\\a_{2^n-1,0}&\cdots&a_{2^n-1,2^n-1}\end{pmatrix}$$ Thus: $$\frac{\Tr\left(\left|0^{n}\middle\rangle\middle\langle 0^{n}\right|U\rho U^\dagger\right)}{\Tr\left(\mathbb{I} \otimes \left|0^{n-1}\middle\rangle\middle\langle 0^{n-1}\right|U\rho U^\dagger\right)}=\frac{a_{0,0}}{a_{0,0}+a_{2^{n-1},2^{n-1}}}$$ For now, suppose $\rho$ is pure. Thus, it can be written as $V|0\rangle$ without loss of generality. Since $UV$ is Haar-random for a Haar-random $U$, without loss of generality, we can assume that $\rho=|0\rangle\langle0|$. $a_{0,0}$ and $a_{2^{n-1},2^{n-1}}$ are then the probabilities of measuring the states $|0\rangle$ and $\left|2^{n-1}\right\rangle$ respectively. We know that for a Haar-random state, its coefficients are drawn independtly from a complex Gaussian and then normalized. That is, we can write $a_{0,0}$ as $\frac{A_{0,0}^2+B_{0,0}^2}{M^2}$ and $a_{2^{n-1},2^{n-1}}$ as $\frac{A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}{M^2}$, with $A_{i,i}$ and $B_{i,i}$ being independelty drawn from $\mathcal{N}(0;1)$. The expectation we want to compute is then: $$\mathbb{E}\left[\frac{\frac{A_{0,0}^2+B_{0,0}^2}{M^2}}{\frac{A_{0,0}^2+B_{0,0}^2}{M^2}+\frac{A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}{M^2}}\right]=\mathbb{E}\left[\frac{A_{0,0}^2+B_{0,0}^2}{A_{0,0}^2+B_{0,0}^2+A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}\right]$$ The interesting point here is that, contrarily to $a_{0,0}$ and $a_{2^{n-1},2^{n-1}}$, $A_{0,0}^2+B_{0,0}^2$ and $A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2$ are i.i.d. As such, we have: $$\mathbb{E}\left[\frac{A_{0,0}^2+B_{0,0}^2}{A_{0,0}^2+B_{0,0}^2+A_{2^{n-1},2^{n-1}}^2+B_{2^{n-1},2^{n-1}}^2}\right]=\frac12.$$

Indeed, for $X$ and $Y$ being i.i.d., the following holds: $$\mathbb{E}\left[\frac{X}{X+Y}\right]=\mathbb{E}\left[\frac{Y}{X+Y}\right]$$ but one also has that: $$\mathbb{E}\left[\frac{X}{X+Y}\right]+\mathbb{E}\left[\frac{Y}{X+Y}\right]=\mathbb{E}\left[\frac{X+Y}{X+Y}\right]=1$$ Thus, if $\rho$ is pure, this expectation equals $\frac12$. Suppose now that $\rho$ can be written as: $$\rho=\sum_kp_k\rho_k$$ with $\rho_k$ being pure for every $k$ and the $p_k$ summing to $1$. The quantity one wishes to take the expectation of is: $$\frac{\sum\limits_kp_ka_{0,0,k}}{\sum\limits_kp_ka_{0,0,k}+\sum\limits_kp_ka_{2^{n-1},2^{n-1},k}}$$ Using the same argument as before, these two random variables are i.i.d. Thus, the expectation value is still $\frac12$ in this case.

  • $\begingroup$ Thanks! One follow-up question: how general is the fact that $a_{0,0}$ and $a_{2^{n-1},2^{n-1}}$ are i.i.d? Is it true only for Haar random states or is it also true for $k$-designs? $\endgroup$
    – BlackHat18
    Jan 27, 2023 at 17:39
  • $\begingroup$ @BlackHat18 I think this is only true for Haar-random states. For instance, if I'm not mistaken, the Clifford group is a $2$-design, but $U|0\rangle$ for $U$ being sampled uniformly from the Clifford group doesn't yield a vector whose amplitudes are drawn from a complex Gaussian. $\endgroup$ Jan 27, 2023 at 21:13
  • $\begingroup$ how did you go from the expectation value of the ration to the ratio of the exp values? I mean the step $\mathbb{E}[\frac{1}{1+\frac{x}{y} }]=\frac{1}{1+\frac{\mathbb{E}[x]}{\mathbb{E}[y]}}$ $\endgroup$
    – glS
    Jan 27, 2023 at 23:17
  • $\begingroup$ @glS My reasoning was potentially flawed, but in fact you can simply use the fact that if $X$ and $Y$ are i.i.d., then $\mathbb{E}\left[\frac{X}{X+Y}\right]=\frac{\mathbb{E}[X]}{\mathbb{E}[X+Y]}=\frac12$. I've corrected it. $\endgroup$ Jan 28, 2023 at 0:07
  • $\begingroup$ @TristanNemoz I'm still not seeing it tbh. Assuming for simplicity a discrete prob distribution, and IID random variables, LHS is $\sum_{x,y} p_x p_y \frac{x}{x+y}$ while RHS is $\frac{\sum_x p_x x}{\sum_x p_x x+\sum_y p_y y}$. I don't see how these are equal in general. The IID assumption tells you $\mathbb{E}[f(X)g(Y)]=\mathbb{E}[f(X)]\mathbb{E}[g(Y)]$, but this is a bit more than that no? $\endgroup$
    – glS
    Jan 28, 2023 at 0:46

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