Prove that the partial trace is a quantum operation, finding its Kraus representation

Question

I am referring to Nielsen and Chuang's "Quantum Computation and Quantum Information" (10th Anniversary Edition), Chapter 8.3. The linear operator $E_i:H_{QR}\longrightarrow H_Q $ is defined by: $$E_i \bigg(\sum_j \lambda_j |q_j\rangle|j\rangle \bigg)\equiv \lambda_i |q_i\rangle$$ whereby $|q_j\rangle$ and $|j\rangle$ are arbitrary states of system Q and the basis of system R, respectively. Define $\varepsilon$ to be the quantum operation with the operation elements $\{E_i\}$: $$\varepsilon(\rho)\equiv \sum_i E_i \rho E_i^{\dagger}$$ The text went on to say: $$\varepsilon(\rho\otimes|j\rangle\langle j'|)=\rho \space \delta_{j,j'}={\rm tr}_R(\rho\otimes|j\rangle\langle j'|)$$ Question: I do not understand how to arrive at $\delta_{j,j'}$, and what form the operator representation of $E_i$ will take? From what I've observed, system Q and R are not entangled in the last equation and $E_i$ seems to disregard whatever $|j\rangle$ basis of system R.

Answer 1

I think the presentation in N&C is a little confusing because $\rho$ is used in two contexts. I'll substitute one of those for a $\sigma$.

You can define $$ E_i=I\otimes\langle i|, $$ which will certainly achieve the effect stated in your first equation. This lets us define the quantum operation $$ \mathcal{E}(\sigma)=\sum_iE_i\sigma E_i^\dagger $$ where $\sigma$ is a density matrix on $QR$.

Now, let $\rho$ be a density matrix on $Q$. We have \begin{align} \mathcal{E}(\rho\otimes|j\rangle\langle j'|)&=\sum_iE_i(\rho\otimes|j\rangle\langle j'|)E_i^\dagger\\ &=\sum_i(I\otimes\langle i|)(\rho\otimes|j\rangle\langle j'|)(I\otimes|i\rangle)\\ &=\sum_i\rho\otimes\langle i|j\rangle\langle j'|i\rangle\\ &=\sum_i\rho\otimes\delta_{i,j}\delta_{j',i}\\ &=\rho\otimes\delta_{j,j'}\\ &=\rho\delta_{j,j'} \end{align}

Answer 2

Say $\lambda_j=\delta_{j,k}$ so that the first equation gives:
$$ E_i\left|q_k\right>\left|k\right>=\delta_{k,i}\left|q_i\right> $$ Now, we can write $\rho$ as:
$$ \rho = \sum_k p_k \left|q_k\right>\left<q_k\right| $$ so that
$$ \varepsilon(\rho\otimes\left|j\right>\left<j'\right|)=\sum_i\sum_k p_k E_i \left|q_k\right>\left|j\right>\left<q_k\right|\left<j'\right|E_i\dagger = \sum_{i,k}p_k\delta_{j,i}\delta_{j',i}\left|q_k\right>\left<q_k\right| $$ This term is non-zero only when both kronecker deltas are 1 which happens only when $i=j$ and $i=j'$, which is only possible when $j=j'$. This gives us the required
$$ \varepsilon(\rho\otimes\left|j\right>\left<j'\right|)=\delta_{j,j'} \sum_k p_k \left|q_k\right>\left<q_k\right| = \rho \delta_{j,j'} $$