Ytukyg

This topic is from chapter 11 of Kitaev, Shen and Vyalyi's Classical and Quantum Computation. If $V$ is an isometric (i.e. preserves the inner product) embedding $V: mathcal{N} rightarrow mathcal{N} otimes mathcal{F}$, and we define the superoperator $V cdot V^dagger: rho mapsto Vrho V^dagger$, what is the trace over $mathcal{F}$ of $Vrho V^dagger$? I did a computation which seems to show that $Tr_{mathcal{F}}(Vrho V^dagger) = Tr_mathcal{F}(rho)$, but I don't think this result is right. Where did I go wrong in the following argument?

Let ${phi_i}$ be an orthonormal basis for $mathcal{F}$. Then

begin{equation*}
begin{split}
Tr_{mathcal{F}}(Vrho V^dagger) & = Tr_{mathcal{F}}(V^dagger Vrho) \
& = sum_ilanglephi_i,V^dagger Vrhophi_irangle \
& = sum_ilangle Vphi_i,Vrhophi_irangle \
& = sum_ilanglephi_i,rhophi_irangle \
& = Tr_mathcal{F}(rho)
end{split}
end{equation*}
where the first step is from the invariance of trace under cyclic permutations and the removal of the $V$s from the inner product is because $V$ is an isometry.

There are a number of things wrong with your computation.

1. The expression "$operatorname{Tr}_{mathcal{F}}(rho)$" doesn't make any sense, as $rho$ is an operator on $mathcal{N}$ and not on $mathcal{N}otimesmathcal{F}$. The expressions "$rhophi_{i}$" likewise are meaningless, as $phi_iinmathcal{F}$.




2. The partial trace is not invariant under cyclic permutations, so your first step is incorrect.




3. The operator $V^dagger Vrho$ isn't square (it is an operator from $mathcal{N}rightarrowmathcal{Notimesmathcal{F}}$) so it doesn't make sense to take its partial trace.




4. While the trace of an operator $Xin operatorname{L} (mathcal{F})$ is computed as $operatorname{Tr}(X)=sum_{i}langle phi_i, Xphi_irangle$, the partial trace is not computed this way. Indeed, if $Xin operatorname{L} (mathcal{Notimes F})$, the expression "$langle phi_i, Xphi_irangle$" doesn't even make sense.

In general, the expression "$operatorname{Tr}(Vrho V^dagger)$" may not be simplified further.