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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.

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

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$|e_k\rangle$ 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.

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  • $\begingroup$ 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? $\endgroup$ – C. Ardayfio Nov 25 '19 at 19:03
  • $\begingroup$ 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. $\endgroup$ – DaftWullie Nov 25 '19 at 20:52
  • $\begingroup$ 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? $\endgroup$ – C. Ardayfio Nov 26 '19 at 3:57
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    $\begingroup$ 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. $\endgroup$ – DaftWullie Nov 26 '19 at 8:36
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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.

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