1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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, $$ \mathcal{E}(\rho\otimes|j\rangle\langle j'|)=\sum_i\rho\delta_{i,j}\delta_{i,j'}=\delta_{j,j'}\rho. $$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Got it. Thanks! $\endgroup$ – C.C. May 7 at 15:16
-1
$\begingroup$

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
$$ \varepsilon(\rho\otimes\left|j\right>\left<j'\right|)=\delta_{j,j'} \sum_k p_k \left|q_k\right>\left<q_k\right| = \rho \delta_{j,j'} $$

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.