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

I am referring to Nielsen and Chuang Quantum Computation and Quantum Information 10th Anniversary Edition Textbook, Chapter 8.3.

A 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:

$$\varepsilon(\rho\otimes|j\rangle\langle j'|)=\rho \space \delta_{j,j'}=tr_R(\rho\otimes|j\rangle\langle j'|)$$

Question: I do not understand how to arrive at $$\delta_{j,j'}$$, and what form will be the operator representation of $$E_i$$ 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. Help will be much appreciated.

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 $$E_i=I\otimes\langle j|,$$ 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 $$\mathcal{E}(\rho\otimes|j\rangle\langle j'|)=\sum_iE_i(\rho\otimes|j\rangle\langle j'|)E_i^\dagger.$$ Now, $$E_i\rho\otimes |j\rangle=\delta_{i,j}\rho$$ and $$\rho\otimes\langle j'|E_i^\dagger=\delta_{i,j'}\rho$$. Thus, $$\mathcal{E}(\rho\otimes|j\rangle\langle j'|)=\sum_i\rho\delta_{i,j}\delta_{i,j'}=\delta_{j,j'}\rho.$$

• Got it. Thanks!
– C.C.
May 7, 2020 at 15:16

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:
$$\rho = \sum_k p_k \left|q_k\right>\left so that
$$\varepsilon(\rho\otimes\left|j\right>\left\left|j\right>\left\left 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\left