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Say $M$ is a matrix acting on $C^r \otimes C^s$. $X$ is the system of dimension $r$, and $Y$ is the system of dimension $s$.

With $|\psi\rangle$ sampled from Haar, how can we show that $$ \int (I_r \otimes \langle\psi|) A (I_r \otimes |\psi\rangle) \, \mathrm{d}\psi = s^{-1} \, Tr_Y(A) $$

where $I_r$ denotes the $r \times r$ identity matrix, and $Tr_Y(\cdot)$ denotes tracing out the $Y$ system.

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1 Answer 1

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I would choose to think of $|\psi\rangle$ as $U|0\rangle$ where $U$ is any unitary. But, I can also think of it as $U'|1\rangle$, or $U''|2\rangle,\ldots$. Hence, I can write this as \begin{align*} \int(I_r\otimes\langle\psi|)A(I_r\otimes|\psi\rangle)d\psi&=\frac{1}{s}\sum_{i=1}^s\int(I_r\otimes\langle i|U^\dagger)A(I_r\otimes U|i\rangle)dU \\ &=\frac{1}{s}\sum_{i=1}^s\int\text{Tr}_Y\left(A(I_r\otimes U|i\rangle)(I_r\otimes\langle i|U^\dagger)\right)dU \\ &=\frac{1}{s}\sum_{i=1}^s\int\text{Tr}_Y\left(A(I_r\otimes U|i\rangle\langle i|U^\dagger)\right)dU \\ &=\frac{1}{s}\int\text{Tr}_Y\left(A(I_r\otimes UIU^\dagger)\right)dU \\ &=\frac{1}{s}\int\text{Tr}_Y\left(A\right)dU \\ &=\frac{1}{s}\text{Tr}_Y\left(A\right) \end{align*}

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