# How is partial trace related to operator sum representation?

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.

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 $$[\mathcal E(\rho)]_{ik}=\sum_j (E_j)_{i\alpha} (\bar E_j)_{k\gamma}\rho_{\alpha\gamma} = \left[\sum_j E_j \rho E^\dagger_j\right]_{ik}.$$

$$|e_k\rangle$$ 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.

• I performed the calculation in MATLAB and was not able to multiply the result of $\langle \mathbf{e}_k|U$ with $[\rho \otimes |\mathbf{e}_0\rangle \langle \mathbf{e}_0|]$. This is because the result of $[\rho \otimes |\mathbf{e}_0\rangle \langle \mathbf{e}_0|]$ is an 8x8 matrix and the result of $\langle \mathbf{e}_k|U$ is a 1x4 matrix. Obviously these two cannot be multiplied (mathematically and in MATLAB), how would this be resolved? – C. Ardayfio Nov 25 '19 at 19:03
• Since you’ve not told me what the dimensions each of the systems is, it’s very hard for me to match up. But U should be a square matrix that is the same size as dim rho times dimension of environment. The the e_k should be a 1x dim of environment vector except that you have to remember that “do nothing on the system “ means tensoring with an identity matrix of the appropriate size. – DaftWullie Nov 25 '19 at 20:52
• In the example presented in N&C, the unitary matrix is a controlled NOT gate of dimension 4x4. From this, I'd assume |e_k> to be a 4x1 vector and <e_k| to be a 1x4 vector; however, when you multiply <e_k| with U, you get a 1x4 vector. This can't be multiplied by the tensor product of the system and the environment as this has a dimension of 8x8. Am I correct in saying that <e_k|*U is a 1x4 vector and this clearly can't be multiplied with the second part of the equation? – C. Ardayfio Nov 26 '19 at 3:57
• No, you've got things a bit jumbled. The system is a single qubit, so $\rho$ is $2\times 2$. The environment is also a qubit, so $|e_0\rangle\langle e_0|$ is also $2\times 2$. The unitary acts between the system and environment and so is $4\times 4$ Now, $\langle e_k|$ should be $1\times 2$ except that, implicitly, it's actually $I\otimes\langle e_k|$, which is $2\times 4$, and so all the mutliplications work. – DaftWullie Nov 26 '19 at 8:36

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.