Suppose I have
$$\rho=\sum p_k \vert \phi_k \rangle \langle \phi_k\vert = \sum \lambda_k \vert \lambda_k \rangle \langle \lambda_k\vert$$
where I call the second decomposition "orthonormal", meaning $\langle \lambda_l\vert \lambda_m\rangle =\delta_{lm}$, and the numbers before the projectors are the probabilities.
Why is the orthonormal decomposition more special than a generic one? By "special" I am referring to, for example, the fact that the Von Neumann entropy $S$ of the density operator and the Shannon entropy $H$ of the state distribution coincide only in the orthonormal decomposition $$H(\lbrace p_i \rbrace)=\sum p_i \log \frac{1}{p_i} \geq H(\lbrace \lambda_i \rbrace) = S(\rho)=-\operatorname{tr}\rho \log \rho.$$
Can it be said that othornomal states are more "independent" than those which are not? Can this be made mathematically precise?
Best Answer
Well, this seems to be a linear algebra question.
Firstly, mathematically, when you want to calculate the trace, i.e. \begin{align} tr(\cdot) = \sum_{\lambda}\langle\lambda|\cdot|\lambda\rangle \end{align} This calculation of course requires the $|\lambda\rangle$-basis is an orthogonal one -- think about in the original matrix language in math, any matrix, when assigned with a linear space, implicitly uses a orthogonal basis.
As you have already mentioned, the definition of entanglement entropy uses trace: \begin{align} S &= -tr\big[\rho\log{\rho}\big] \\ &= -\sum_{\lambda}\langle\lambda|\rho\log{\rho}|\lambda\rangle \end{align} Firstly, for sure you could use any basis $|\phi\rangle$, orthogonal or un-orthogonal, to calculate $S$, as long as you can simply calculate each: \begin{align} \langle\lambda|\phi\rangle \end{align} And there are more than one orthogonal-basis, for sure, related by orthogonal transformations. But if you are not using the "eigen-basis" in which $\rho$ has a diagonal form, then you'll have to calculate the above quantity anyway -- that means difficulty of calculation has not been reduced at all. But, say, if you could use the diagonal basis, then the trace calculation would be simply a summation of diagonal elements of $\rho\log{\rho}$.