Quantum Computing Asked by Nap0leon on September 5, 2021
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 $e^{-irho t}$ to any density matrix $sigma$ up to $n$th order in $t$. Note that
begin{align}label{eq1}tag{1}
text{tr}_pe^{-iSDelta t}rhootimessigma e^{iSDelta t} &= left(cos^2Delta tright)sigma + left(sin^2 Delta tright)rho – isinDelta tleft[rho, sigmaright]
&= sigma – iDelta tleft[rho, sigmaright] + mathcal Oleft(Delta t^2right)
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 $e^{-iSDelta t}$ can be performed efficiently [6-9]. Repeated application of eqref{eq1}
I know from the qiskit website, that we can express $mathrm{e}^{igamma B}$ as $cos(gamma)I + isin(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 $(cos^2triangle t)sigma$ and a single $rho$ and no $sigma$ in $(sin^2triangle t)rho$? Is it because of the partial trace?
Any intuition or approach is welcome. Many thanks in advance.
As you say, start by expanding $e^{-iSDelta t}=cos(Delta t)I-isin(Delta t)S$, so you'd be calculating $$ (cos(Delta t)I-isin(Delta t)S)rhootimessigma(cos(Delta t)I+isin(Delta t)S). $$ If you multiply out all the terms, then the $cos^2(Delta t)$ comes from the two $I$ terms, leaving you with $rhootimessigma$. If you trace out the first term, you're left with $sigma$.
Similarly, $S(rhootimessigma)S=sigmaotimesrho$, 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, $S(rhootimessigma)$. If I try to trace out the first system, I have begin{align} sum_{i,j,k}|jranglelangle k|langle i,j|Srhootimessigma|i,krangle&=sum_{i,j,k}|jranglelangle k|langle j,i|rhootimessigma|i,krangle &=sum_{i,j,k}|jranglelangle k|langle j|rho|iranglelangle i|sigma|krangle &=sum_{j,k}|jranglelangle k|langle j|rhosigma|krangle &=rhosigma. end{align}
Correct answer by DaftWullie on September 5, 2021
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