1
$\begingroup$

For $d<\infty$ consider a $d$-dimensional Hilbert space with ONB's $\{e_j\}_{j=1}^d,\{e_j^\prime\}_{j=1}^d$ and arbitrary matrices $A,B$. Show that $\text{tr}(A^*B)=\langle\eta|(A\otimes B)\eta\rangle$ with $\eta=\sum_{j=1}^d e_j\otimes e_j\in\mathcal{H}\otimes\mathcal{H}$ the maximial entangled state.

The idea of the proof is quite easy. Choose a unitary matrix $U$ so that $Ue_j=e_j^\prime$. Then $$ \text{tr}(A^*B)=\text{tr}(UA^*U^*B)=\sum_{i,j}\langle e_j'|UA^*U^*|e_i'\rangle\langle e_i'|B|e_j'\rangle=\sum_{i,j}\langle e_j|A^*|e_i\rangle\langle e_i'|B|e_j'\rangle $$ and on the other side $$ \langle\eta|(A\otimes B)|\eta\rangle=\sum_{ij}\langle e_i\otimes e_i'|(A\otimes B)|e_j\otimes e_j'\rangle=\sum_{ij}\langle e_i|A|e_j\rangle\langle e_i'|B|e_j'\rangle. $$ The paper of Mr. Jinchuan Hou (https://arxiv.org/pdf/1107.0354.pdf, page 4) claims $$ \sum_{ij}\langle e_i|A|e_j\rangle\langle e_i'|B|e_j'\rangle=\sum_{i,j}\langle e_j|A^*|e_i\rangle\langle e_i'|B|e_j'\rangle. $$ I do not grasp this last equality. Thank you for your help.

$\endgroup$
3
  • $\begingroup$ It holds if $A$ is self-adjoint, do you assume that? $\endgroup$ – Kiryl Pesotski Mar 20 '17 at 17:13
  • $\begingroup$ No, in this cases it is pretty obvious. Namely $A=U^*V$ with $U,V$ unitary is the case I am interested in. $\endgroup$ – julian Mar 20 '17 at 17:19
  • 1
    $\begingroup$ My question contains an error. It seems that the problem is an exercise from the textbook "Quantum Computation and Quantum Information" written by Michael A. Nielsen and Isaac L. Chuang. In their errata list I found the correction $\operatorname{tr}(A^tB)=\langle\eta|(A\otimes B)|\eta\rangle$ and thus it's trivial. Astonishing that Mr. Jinchuan Hou was able to prove it. Thank you for your replies and sorry for wasting your time. $\endgroup$ – julian Mar 20 '17 at 18:07
0
$\begingroup$

Take $$A=\left( \array{1 & 1 \\ 0 & 1}\right)$$ and $$B=\left( \array{a & b \\ c & d}\right)$$

with respect to the standard basis and calculate the sums in the equality in question. You get $a+d+c$ in one case and $a+d+b$ in the other case. This means some additional property is needed for this equality.

(this does not qualify as answer, but I hate writing matrices in comments)

$\endgroup$
2
  • $\begingroup$ Thanks for the counter example. In Mr. Jinchuan's lemma are no further requirements listed. $\endgroup$ – julian Mar 20 '17 at 17:57
  • $\begingroup$ @julian **** happens $\endgroup$ – Thomas Mar 20 '17 at 20:49

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.