2
$\begingroup$

In Dirac notation we can define the trace of an operator in Hilbert space $\rho$ as the follows,

$Tr(\rho)=\sum\limits_{|s\rangle \in B} \langle s| \rho |s\rangle$

where B is some orthonormal basis, and this quantity is basis independent.

If we swapped B with a non-orthogonal basis C, which, if any, of the properties of the trace will be preserved? In particular,

  • 1) Is it now basis dependent? (The intuitive answer is YES,)
  • 2) Under what conditions (on C and $\rho$) will this value exceed the value of the actual trace? I will settle for an answer assuming $\rho$ is a density operator (rank 1 projection).

Thanks!

$\endgroup$
3
$\begingroup$

If $X$ is the change of basis matrix, you have $$ \sum_{|t\rangle\in C}\langle t|\rho|t\rangle =\sum_{|s\rangle\in B}\langle Xs|\rho|Xs\rangle =\sum_{|s\rangle\in B}\langle s|X^*\rho X|s\rangle =\text{Tr}(X^*\rho X)=\text{Tr}(XX^*\rho). $$ In other words, by going through all bases you will obtain every possible positive linear functional. In particular, you can assign every positive value you want to $\rho$.

$\endgroup$
1
  • $\begingroup$ Thank you very much, this was the easy solution that I was failing to see. $\endgroup$ – zzz Apr 9 '14 at 4:57
1
$\begingroup$

It is basis dependent. Consider the operator identity operator on 2D Hilbert space. Let the basis by $s_1 = (1,0)$ and $s_2 = (0,2)$. Then $$ \sum_{n=1}^2 \langle s_n | I | s_n \rangle = 3, $$ whereas the trace computed in any orthonormal basis will be $2$.

Note - a mathematician will say that the trace of an operator IS basis independent. But their definition of "basis independent" will be subtly different from yours, and so you will be talking at cross purposes.

$\endgroup$
1
  • $\begingroup$ Thanks for the reply, my question was more concerned with non-orthogonal bases rather than non-normalized basis, where the basis dependence might be a little more involved depending on the transformation between the bases considered, as the other answer pointed out. $\endgroup$ – zzz Apr 9 '14 at 5:00

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.