How is the partial trace related to the operator sum representation?

Question

In Quantum Computation and Quantum Information by Nielsen and Chuang, the authors introduce operator sum representation in Section 8.2.3. They denote the evolution of a density matrix, when given an environment, is the following:

$$\varepsilon (\rho) = \mathrm{tr}_{\text{env}} [U(\rho\otimes \rho_{\text{env}})U^\dagger]$$

Where you are essentially taking the trace to discard the environment of the unitary evolution of the entire system. What I don't understand is how the operator sum representation is equivalent (Equations 8.9 and 8.11 in N&C)

$$\varepsilon (\rho) = \sum_k \langle \mathbf{e}_k|U[\rho \otimes |\mathbf{e}_0\rangle \langle \mathbf{e}_0|]U^\dagger|\mathbf{e}_k\rangle = \sum_k E_k\rho E_k^\dagger$$

In this equation, I take $|\mathbf{e}_k\rangle$ to represent the basis of the system and U to be a 4 x 4 unitary matrix governing the evolution. How is this equivalent to the first equation where you discard the trace? It seems like the second equation (equation 8.9 in N&C) above would yield a scalar quantity. What does this equation mean? I understand the first equation where you take the partial trace, but how does partial trace relate to the 2nd and 3rd equations? I'm a relative beginner in this field.

Answer 1

I think it helps here to write things explicitly.

Suppose $\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]$.

Pick a basis for the environment in which $|\mathbf e_0\rangle$ is the first element. Note that here $U$ is a unitary matrix in a bipartite system. The operator before taking the partial trace has matrix elements $$ [U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]_{ij,k\ell} = \sum_{\alpha,\gamma} U_{ij,\alpha 0} \rho_{\alpha\gamma} (U^\dagger)_{\gamma0,k\ell} = \sum_{\alpha,\gamma} U_{ij,\alpha0}\bar U_{k\ell,\gamma0} \rho_{\alpha\gamma}. $$ Now notice that the partial trace amounts here to make $j=\ell$ and sum over $j$ (because in this notation the indices $i,k$ refer to the system while $j,\ell$ to the environment), so that we get $$ [\mathcal E(\rho)]_{ik} = \sum_j [U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]_{ij,kj} = \sum_{\alpha\gamma j} U_{ij,\alpha0}\bar U_{kj,\gamma0}\rho_{\alpha\gamma}. $$ Notice how this is already essentially an operator sum representation: defining $(E_{j})_{i\alpha}\equiv U_{ij,\alpha0}$, we get $$[\mathcal E(\rho)]_{ik}=\sum_j (E_j)_{i\alpha} (\bar E_j)_{k\gamma}\rho_{\alpha\gamma} = \left[\sum_j E_j \rho E^\dagger_j\right]_{ik}.$$

Answer 2

$|e_k\rangle$ is the basis of the environment. Taking the sum of projections onto an orthonormal basis of one subsystem is the definition of the partial trace over that subsystem.

Answer 3

Let $\{|u_a\rangle\}_{a\in A}$ and $\{|v_e\rangle\}_{e\in E}$ be orthonormal bases for the space $A$ and the environment space $E$ resp. Now, if we express $\rho_A=\sum_A \alpha_a |u_a\rangle \langle u_a|$ and $\rho_E=|v_0\rangle \langle v_0|$ assuming that $\rho_E$ is some pure state on the environment. Then we can write

$$ \mathcal{E}(\rho_A)=tr_E\left(U\left(\rho_A\otimes \rho_E\right)U^*\right)\\ = \sum_{A}\alpha_a\, tr_E\left(\left(U|u_a\rangle |v_0\rangle\right)\left(U|u_a\rangle |v_0\rangle\right)^*\right)\\ = \sum_{A}\alpha_a\, \sum_E\langle v_e|U|v_0\rangle |u_a\rangle \langle u_a|\langle v_0|U^*|v_e\rangle\\ =\sum_E\langle v_e|U|v_0\rangle \left(\sum_A\alpha_a|u_a\rangle \langle u_a|\right)\langle v_0|U^*|v_e\rangle\\ =\sum_E\langle v_e|U|v_0\rangle \rho_A\left(\langle v_e|U|v_0\rangle\right)^*,$$

taking $E_e=\langle v_e|U|v_0\rangle$ be obtain the Kraus operators which act on the space $A$. I think its conceptually easier to see that the partial-trace unitary representation (or Stinespring representation) of a quantum channel is the purification of the Kraus representation, rather than the other way around.