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Let $\rho = \sum_{x \in \{0,1\}^n} P_x |x \rangle\langle x|$ be a separable mixed state over bit strings $x$ of size $n$. Suppose also that $U = U_1 \otimes \cdots \otimes U_n$ is a product of local unitaries. I am interested in calculating the probability that the first qubit of the state $U\rho U^\dagger$ is one. Call this probability $P(\text{first qubit = $|1 \rangle$})$. I approached this in two different ways, but the answers don't seem to agree, so I am confused which is right and why the other is wrong.

Approach One: Writing $x = x_1 x_2 \dots x_n$ with each $x_i \in \{0,1\}$, we have $$ P(\text{first qubit = $|1 \rangle$}) = \sum_{x \in \{0,1\}^n} P(\text{first qubit = $|1 \rangle$} \mid x)P_x. $$ Here, the conditional probability is $P(\text{first qubit = $|1 \rangle$} \mid x) = ||\Pi_1U|x \rangle||^2$, where $\Pi_1 = |1 \rangle\langle 1| \otimes I_{n-1}$ is the projector onto the first qubit being $|1 \rangle$. Evaluating this probability, I obtain that $$ P(\text{first qubit = $|1 \rangle$}) = \sum_{x \in \{0,1\}^n} |\langle 1 |U_1 |x_1 \rangle|^2 P_x. $$ Approach Two: The probability is $$ P(\text{first qubit = $|1 \rangle$}) = \text{tr}(\Pi_1 U \rho U^\dagger). $$ Expanding this I obtain $$ P(\text{first qubit = $|1 \rangle$}) = \sum_{x \in \{0,1\}^n} \sum_{z \in \{0,1\}^{n-1}}|\langle 1 | U_1 |x_1\rangle|^2 |\langle z |U_2 \otimes \cdots \otimes U_n |x_2x_3\dots x_n \rangle|^2 P_x. $$ It is not obvious to me that these two probabilities are the same. If they are not, what is my mistake?

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2 Answers 2

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They're the same because $$\sum_z |\langle z|\tilde U|\tilde x \rangle|^2=1$$ for any unitary $\tilde U$ and input $|\tilde x\rangle$. This follows directly from the normalization of probabilities: $|\langle z|\tilde U|\tilde x \rangle|^2$ is the probability of finding the outcome $z$ after evolving $|\tilde x\rangle$ through $\tilde U$.

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Another reason why they are the same, equivalent to @glS's answer: resolution of identity $$\sum_{z \in \{0,1\}^{n-1}}|z\rangle\langle z|=I_{n-1}.$$ We can expand the absolute square in the second approach to show it agrees with the first: \begin{aligned} P(\text{first qubit = $|1 \rangle$}) &= \sum_{x \in \{0,1\}^n} \sum_{z \in \{0,1\}^{n-1}}|\langle 1 | U_1 |x_1\rangle|^2 |\langle z |U_2 \otimes \cdots \otimes U_n |x_2x_3\dots x_n \rangle|^2 P_x\\ &= \sum_{x \in \{0,1\}^n} \sum_{z \in \{0,1\}^{n-1}}|\langle 1 | U_1 |x_1\rangle|^2 \langle x_2x_3\dots x_n |U_n^\dagger \otimes \cdots \otimes U_2^\dagger |z \rangle\langle z |U_2 \otimes \cdots \otimes U_n |x_2x_3\dots x_n \rangle P_x\\ &= \sum_{x \in \{0,1\}^n} |\langle 1 | U_1 |x_1\rangle|^2 \langle x_2x_3\dots x_n |(U_n \otimes \cdots \otimes U_2)^\dagger I_{n-1} (|U_2 \otimes \cdots \otimes U_n) |x_2x_3\dots x_n \rangle P_x\\ &= \sum_{x \in \{0,1\}^n} |\langle 1 | U_1 |x_1\rangle|^2 \langle x_2x_3\dots x_n |x_2x_3\dots x_n \rangle P_x\\ &= \sum_{x \in \{0,1\}^n} |\langle 1 | U_1 |x_1\rangle|^2 P_x. \end{aligned}

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