Equation (1) in Seth Lloyd's paper on Quantum PCA says:

$\text{tr}_{p}\text{e}^{-iS\Delta t} \rho \otimes \sigma \text{e}^{iS\Delta t} = \cos^2(\Delta t)\sigma + \sin^2(\Delta t) \rho - i \sin(\Delta t)\cos(\Delta t) [\rho, \sigma]$

Where $S$ is the swap matrix, and $\Delta t$ is a small slice of time $t/n$, and $\sigma$, $\rho$ are density matricies (we wish to apply $\text{e}^{-i\rho t}$ to density matrix $\sigma$)

How would I go about proving this?


From Wikipedia, we have that

$\text{e}^{tA} = \text{e}^{st}[(\cosh(qt) - s \frac{\sinh(qt)}{q}) I + \frac{\sinh(qt)}{q} A]$ where $s = \frac{\text{tr}A}{2}$ and $q = \pm \sqrt{-\text{det}(A-sI)}$, by Cayley-Hamilton. Thus, we can expand:

$\text{tr}_{p} \text{e}^{-iS\Delta t} \rho \otimes \sigma \text{e}^{iS\Delta t} = \text{tr}_{p}(\text{e}^{-i\Delta t}(-I + S)\rho \otimes \sigma \text{e}^{i\Delta t} (-I + S)) = \text{tr}_{p}(\text{e}^{-i\Delta t}[\rho \otimes \sigma - S(\rho \otimes \sigma) - (\rho \otimes \sigma) S + \sigma \otimes \rho] \text{e}^{i\Delta t})$

However, I do not see how this simplifies to $\cos^2(\Delta t)\sigma + \sin^2(\Delta t) \rho - i \sin(\Delta t)\cos(\Delta t) [\rho, \sigma]$.

Am I making a mistake in my math, or is there a trick that I am not seeing to simplify the expression I obtained down to the one in the paper?

  1. Note that, for any pair of matrices $A,B$, you have $$e^A B e^{-A} = e^{{\rm ad}(A)} B \equiv \sum_{k=0}^\infty \frac{1}{k!}[\underbrace{A,[A,\cdots ,[A}_k,B]\cdots]] \equiv B + [A,B] + \frac12 [A,[A,B]] + \dots,$$ where ${\rm ad}(A)$ denotes the adjoint operator, ${\rm ad}(A):B\mapsto [A,B]$, and the complicated-looking object in the series is a repeated commutator with $k$ terms.

  2. Note that, if $S$ is the SWAP operator, then $$ \operatorname{Tr}_2\left(S(\rho\otimes\sigma)\right) = {\rm Tr}(\sigma\rho), \qquad \operatorname{Tr}_2\left((\rho\otimes\sigma)S\right) = {\rm Tr}(\rho\sigma). $$ Apart from directly showing this expliciting via the matrix elements of the components of the expression, you can see this identity quite nicely in diagrammatic notation.

I think this is pretty much all you need to get to the reported expression.


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