# Trace in non-orthogonal basis

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!

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 $\rho$.
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 $2$.