0
$\begingroup$

When I define the partial trace as below, how can I prove it well-defined? I understand that I have to indicate $Tr_k(\rho)$ does not depend on how to take the ONB of $\mathbb{C}^2$ $$n\in \mathbb{Z}_{\ge 2}\\ k\in \{ 1,2,...,n\}\\ \{\mathbb{x_1},\mathbb{x_2}\}:ONB\ of \mathbb{C}^2\\ Tr_k:Mat_{2^n}(\mathbb{C})\rightarrow Mat_{2^{n-1}}(\mathbb{C)}\ be\ defined\ by\\ \rho\mapsto Tr_k(\rho):=\Sigma_{i=1}^2(\mathbb{I}\otimes \cdots\otimes\mathbb{I}\otimes\mathbb{x}_i^\dagger\otimes\mathbb{I}\otimes \cdots\otimes\mathbb{I})\rho (\mathbb{I}\otimes \cdots\otimes\mathbb{I}\otimes\mathbb{x}_i\otimes\mathbb{I}\otimes \cdots\otimes\mathbb{I})\ \ \Box$$

$\endgroup$
  • 1
    $\begingroup$ What does "ONB" mean? What is that square at the end of your last equation? $\endgroup$ – DanielSank Mar 25 at 8:21
  • $\begingroup$ $x_i$ should not be a basis of the vector space ($\mathbb{C}^2$), but of the operators onto it. $\endgroup$ – lcv Mar 25 at 9:21
  • $\begingroup$ Daniel->ONB is "orthonomal basis". $\endgroup$ – user225954 Mar 25 at 9:28
  • $\begingroup$ luv->What do you mean? $x_i$ is like $(1, 0)^t$ and $(0,1)^t$. $\endgroup$ – user225954 Mar 25 at 9:30
6
$\begingroup$

Several ways:

  1. The partial trace is the unique linear map such that $\mathrm{tr}_2(A\otimes B)=A\,\mathrm{tr}(B)$, and the trace is basis-independent.

  2. The partial trace is cyclic: $$\mathrm{tr}_2(X(I\otimes B))=\mathrm{tr}_2((I\otimes B)X)\ ,$$ as can be seen by inserting a resolution of the identity. The result then follows by considering $\mathrm{tr}_2((I\otimes U))X(I\otimes U^\dagger))$, with $U$ transforming between different bases of the 2nd system.

  3. Directly by inserting a resolution of the identity on the 2nd system: $$ \mathrm{tr}_2(X)=\sum_i\langle i|X|i\rangle = \sum \langle i|j\rangle\langle j|X|i\rangle = \sum_{ij}\langle j|X|i\rangle\langle i|j\rangle = \sum_j \langle j|X|j\rangle\ . $$ (Note that all bras/kets here are acting on the 2nd system only.)

$\endgroup$
  • $\begingroup$ When a matrix $A$ is decomposed into two ways such that $A=A_1 \otimes A_2=B_1\otimes B_2$, does a equation $A_2$tr$(A_1)=B_2$tr$(B_1)$ holds anytime? $\endgroup$ – user225954 Apr 8 at 7:47
  • 1
    $\begingroup$ Anytime. The decomposition $A=A_1\otimes A_2$ is unique, up to an overall complex factor. $\endgroup$ – Norbert Schuch Apr 8 at 8:22
  • $\begingroup$ When $A=A_1\otimes A_2$, $A=A_1\otimes A_2 + B_1\otimes B_2 - B_1\otimes B_2$ is also a decomposition of A, right? This is easy to show that the partial trace of the matrix is unique, but I wonder this holds at any time. $\endgroup$ – user225954 Apr 8 at 10:55
  • $\begingroup$ Any time. _______ $\endgroup$ – Norbert Schuch Apr 8 at 11:15

Your Answer

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