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I'm trying to derive equation $(1)$ on p.2 in Lloyd et al, 2013 which reads

$$ \text{Tr}_A\left[\exp(-i\theta S_{AB}) (\rho_A \otimes \sigma_B) \exp(i\theta S_{AB}) \right] = (\cos^2 \theta) \sigma_B + (\sin^2 \theta) \rho_B - i \sin \theta [\rho_B, \sigma_B] $$

where $S_{AB}$ is a swap operator for the two systems. My approach so far is to write the operator as $$ \exp(i\theta S_{AB}) = \begin{pmatrix} e^{i\theta} & 0 & 0 & 0 \\ 0 & \cos \theta & i \sin\theta & 0 \\ 0 & i \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & e^{i\theta} \end{pmatrix} \begin{matrix} \leftarrow\rho_A \otimes \rho_B \\ \leftarrow\rho_A \otimes \sigma_B\\ \leftarrow\sigma_A \otimes \rho_B \\ \leftarrow\sigma_A\otimes \sigma_B \end{matrix} $$

where the labels describe how each input is affected by this operation if the possible state configurations are represented as a column vector, e.g. $\rho\otimes \sigma \sim \hat{e}_1$ would become $\cos\theta (\rho\otimes \sigma) + \sin \theta (\sigma \otimes \rho) \sim \cos\theta \hat{e}_1 + \sin \theta \hat{e}_2$.

However, I haven't been able to carry out the calculation since I'm unsure how to apply the operator or partial trace on $(\rho\otimes \sigma)$ when I set up operators like this. Is there anything pathological about this representation of a SWAP, and if so what is the right approach to deriving the above equation?

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2 Answers 2

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For any $\theta\in\mathbb{R}$ and any operator $T$ such that $T^2=I$ we have

$$ \exp(i\theta T) = I\cos\theta + i T\sin\theta $$

(c.f. exercise 4.2 on p.175 in Nielsen & Chuang). Therefore,

$$ \exp(i\theta S_{AB}) = I \cos\theta + i S_{AB} \sin\theta $$

and we have

$$ \exp(-i\theta S_{AB}) (\rho_A \otimes \sigma_B) \exp(i\theta S_{AB}) = \\ \rho_A\otimes\sigma_B \cos^2\theta - i S_{AB} \rho_A\otimes\sigma_B \cos\theta\sin\theta + \\ + i \rho_A\otimes\sigma_B S_{AB} \cos\theta\sin\theta + \sigma_A\otimes\rho_B \sin^2\theta.\tag1 $$

By definition

$$ \mathrm{tr}_A(\rho_A\otimes\sigma_B) = \sigma_B \\ \mathrm{tr}_A(\sigma_A\otimes\rho_B) = \rho_B.\tag2 $$

The partial trace of $S_{AB}\rho_A\otimes\sigma_B$ is represented by the tensor network

tensor network for partial trace with swap

which shows that

$$ \mathrm{tr}_A(S_{AB}\rho_A\otimes\sigma_B) = \rho_B\sigma_B.\tag{3a} $$

Similarly,

$$ \mathrm{tr}_A(\rho_A\otimes\sigma_BS_{AB}) = \sigma_B\rho_B.\tag{3b} $$

Finally, tracing over $A$ in $(1)$ and substituting $(2)$, $(3a)$ and $(3b)$, we get

$$ \mathrm{tr}_A\left[\exp(-i\theta S_{AB}) (\rho_A \otimes \sigma_B) \exp(i\theta S_{AB})\right] = \\ \sigma_B\cos^2\theta + \rho_B\sin^2\theta - i\rho_B\sigma_B\cos\theta\sin\theta + i\sigma_B\rho_B\cos\theta\sin\theta = \\ \sigma_B\cos^2\theta + \rho_B\sin^2\theta - i[\rho_B, \sigma_B]\cos\theta\sin\theta.\tag4 $$


This result differs from the paper. I think the formula in the paper contains a mistake. Set $\theta=\frac{\pi}{2}$. In this case

$$ \exp\left(-\frac{i\pi S_{AB}}{2}\right) (\rho_A \otimes \sigma_B) \exp\left(\frac{i\pi S_{AB}}{2}\right) = S_{AB} (\rho_A \otimes \sigma_B) S_{AB} = \sigma_A \otimes \rho_B $$

and the partial trace over $A$ yields $\rho_B$. This special case agrees with $(4)$ and disagrees with the formula in the paper since the latter preserves the term with the commutator when $\theta=\frac{\pi}{2}$.

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Using the fact $ e^{i\theta S} = \text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S $, we calculate

\begin{align*} e^{-i\theta S} (\rho \otimes \sigma) e^{i\theta S} & = (\text{cos}(\theta) \cdot I - i \cdot \text{sin}(\theta) \cdot S) \big(\rho \otimes \sigma \big) (\text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S) \\ & = \text{cos}^2(\theta) (\rho \otimes \sigma) + \text{sin}^2(\theta) (\sigma \otimes \rho) - i \text{sin}(\theta) \text{cos}(\theta) \cdot \big( S(\rho \otimes \sigma) - (\rho \otimes \sigma)S \big) \end{align*}

Now we need only to calculate $ \text{Tr}_A S(\rho \otimes \sigma) $.

Let $ \rho = \sum_i p_i |x_i \rangle \langle x_i| $ and $ \sigma = \sum_i q_i |y_i \rangle \langle y_i| $.

\begin{align*} \text{Tr}_A S(\rho \otimes \sigma) &= \sum_{i,j} p_i q_j \text{Tr}_A [S \big(|x_i y_j \rangle \langle x_i y_j| \big)] \\ &= \sum_{i,j} p_i q_j \text{Tr}_A [\big(|y_j x_i \rangle \langle x_i y_j| \big)]\\ &= \sum_{i,j} p_i q_j \langle x_i | y_j \rangle \cdot|x_i \rangle \langle y_j| = \rho\sigma \end{align*} A similar calculation shows that $\text{Tr}_A (\rho \otimes \sigma)S = \sigma \rho$.

Finally: $$ \text{Tr}_A e^{-i\theta S} (\rho \otimes \sigma) e^{i\theta S} = \text{cos}^2(\theta) \sigma + \text{sin}^2(\theta) \rho - i \frac{\text{sin}(2\theta)}{2} \cdot [\rho, \sigma] $$

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