# Computing a ratio involving Haar random unitaries

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

I am trying to compute the expression

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

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.

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

$$$$\mathbb{E}_U\left[ \frac{\text{Tr}_{1,2,\cdots,n}\left(|0^n\rangle \langle 0^n | ~ (V \otimes \mathbb{I}_{2^{n-1}}) U\rho U^\dagger(V^\dagger \otimes \mathbb{I}_{2^{n-1}}) \right)}{\text{Tr}_{1,2,\cdots,n} \left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right].$$$$ Note the pleasing fact that $$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 $$$$\mathbb{E}_U\mathbb{E}_V\left[ \frac{\text{Tr}_{1,2,\cdots,n}\left(|0^n\rangle \langle 0^n | ~ (V \otimes \mathbb{I}_{2^{n-1}}) U\rho U^\dagger(V^\dagger \otimes \mathbb{I}_{2^{n-1}}) \right)}{\text{Tr}_{1,2,\cdots,n} \left(\mathbb{I} \otimes |0^{n-1}\rangle \langle 0^{n-1} | ~U\rho U^\dagger \right)} \right].$$$$ Now, the $$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.

• 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? Jan 27 at 17:39
• @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. Jan 27 at 21:13
• how did you go from the expectation value of the ration to the ratio of the exp values? I mean the step $\mathbb{E}[\frac{1}{1+\frac{x}{y} }]=\frac{1}{1+\frac{\mathbb{E}[x]}{\mathbb{E}[y]}}$
– glS
Jan 27 at 23:17
• @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. Jan 28 at 0:07
• @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?
– glS
Jan 28 at 0:46