I am currently reading the paper quantum principal component analysis from Seth Lloyd's article Quantum Principal Component Analysis There is the following equation stated.

Suppose that on is presented with n copies of $\rho$. A simple trick allows one to apply the unitary transformation $e^{-i\rho t}$ to any density matrix $\sigma$ up to $n$th order in $t$. Note that \begin{align}\label{eq1}\tag{1} \text{tr}_pe^{-iS\Delta t}\rho\otimes\sigma e^{iS\Delta t} &= \left(\cos^2\Delta t\right)\sigma + \left(\sin^2 \Delta t\right)\rho - i\sin\Delta t\left[\rho, \sigma\right] \\ &= \sigma - i\Delta t\left[\rho, \sigma\right] + \mathcal O\left(\Delta t^2\right) \end{align} Here $\text{tr}_p$ is the partial trace over the first variable and $S$ is the swap operator. $S$ is a sparse matrix and so $e^{-iS\Delta t}$ can be performed efficiently [6-9]. Repeated application of \eqref{eq1}

I know from the qiskit website, that we can express $\mathrm{e}^{i\gamma B}$ as $\cos(\gamma)I + i\sin(\gamma)B $ with $\gamma$ being some real number and $B$ is an involutory matrix. Can someone explain me why there is only a single $\sigma$ and no $\rho$ in $(\cos^2\triangle t)\sigma$ and a single $\rho$ and no $\sigma$ in $(\sin^2\triangle t)\rho$? Is it because of the partial trace?

Any intuition or approach is welcome. Many thanks in advance.


1 Answer 1


As you say, start by expanding $e^{-iS\Delta t}=\cos(\Delta t)I-i\sin(\Delta t)S$, so you'd be calculating $$ (\cos(\Delta t)I-i\sin(\Delta t)S)\rho\otimes\sigma(\cos(\Delta t)I+i\sin(\Delta t)S). $$ If you multiply out all the terms, then the $\cos^2(\Delta t)$ comes from the two $I$ terms, leaving you with $\rho\otimes\sigma$. If you trace out the first term, you're left with $\sigma$.

Similarly, $S(\rho\otimes\sigma)S=\sigma\otimes\rho$, so if you trace out the first system, you're left with $\rho$.

It's actually the cross-terms that are the trickier ones. Take, for example, $S(\rho\otimes\sigma)$. If I try to trace out the first system, I have \begin{align} \sum_{i,j,k}|j\rangle\langle k|\langle i,j|S\rho\otimes\sigma|i,k\rangle&=\sum_{i,j,k}|j\rangle\langle k|\langle j,i|\rho\otimes\sigma|i,k\rangle \\ &=\sum_{i,j,k}|j\rangle\langle k|\langle j|\rho|i\rangle\langle i|\sigma|k\rangle \\ &=\sum_{j,k}|j\rangle\langle k|\langle j|\rho\sigma|k\rangle \\ &=\rho\sigma. \end{align}

  • $\begingroup$ Thank you. This was very helpful and insightful. $\endgroup$
    – Nap0leon
    Jan 17, 2020 at 10:09

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