Quantum Computing Asked by C.C. on April 13, 2021
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:
$$E_i bigg(sum_j lambda_j |q_jrangle|jrangle bigg)equiv lambda_i |q_irangle$$
whereby $|q_jrangle$ and $|jrangle$ 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:
$$varepsilon(rhootimes|jranglelangle j’|)=rho space delta_{j,j’}=tr_R(rhootimes|jranglelangle j’|)$$
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 $|jrangle$ 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 $$ E_i=Iotimeslangle j|, $$ which will certainly achieve the effect stated in your first equation. This lets us define the quantum operation $$ mathcal{E}(sigma)=sum_iE_isigma E_i^dagger $$ where $sigma$ is a density matrix on $QR$.
Now, let $rho$ be a density matrix on $Q$. We have $$ mathcal{E}(rhootimes|jranglelangle j'|)=sum_iE_i(rhootimes|jranglelangle j'|)E_i^dagger. $$ Now, $E_irhootimes |jrangle=delta_{i,j}rho$ and $rhootimeslangle j'|E_i^dagger=delta_{i,j'}rho$. Thus, $$ mathcal{E}(rhootimes|jranglelangle j'|)=sum_irhodelta_{i,j}delta_{i,j'}=delta_{j,j'}rho. $$
Correct answer by DaftWullie on April 13, 2021
Say $lambda_j=delta_{j,k}$ so that the first equation gives:
$$
E_ileft|q_kright>left|kright>=delta_{k,i}left|q_iright>
$$
Now, we can write $rho$ as:
$$
rho = sum_k p_k left|q_kright>left<q_kright|
$$
so that
$$
varepsilon(rhootimesleft|jright>left<j'right|)=sum_isum_k p_k E_i left|q_kright>left|jright>left<q_kright|left<j'right|E_idagger = sum_{i,k}p_kdelta_{j,i}delta_{j',i}left|q_kright>left<q_kright|
$$
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
$$
varepsilon(rhootimesleft|jright>left<j'right|)=delta_{j,j'} sum_k p_k left|q_kright>left<q_kright| = rho delta_{j,j'}
$$
Answered by SaumG on April 13, 2021
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