0
$\begingroup$

The definition of the von Neumann entropy of a mixed state says that it can be calculated as the Shannon entropy of coefficients of the decomposition of the state into a sum of projectors.

My question: Are we assuming this decomposition is a family of orthogonal states or can it be anything?

$\endgroup$

1 Answer 1

1
$\begingroup$

Yes, the projectors must be orthogonal and rank one. The coefficients are actually the eigenvalues of the density matrix, so orthogonality follows from the spectral theorem. More precisely, if $$ \rho=\sum_{i=1}^n\lambda_iP_i=\sum_{i=1}^n\lambda_i|\psi_i\rangle\langle\psi_i|\tag1 $$ is an eigendecomposition of the mixed state $\rho$, then its von Neumann entropy is the Shannon entropy $H$ of the eigenvalues $\lambda_i$ $$ S(\rho)=H(\lambda_1,\dots,\lambda_n)=-\sum_{i=1}^n\lambda_i\log\lambda_i.\tag2 $$ The spectral theorem guarantees that $\rho$ can be written as in $(1)$ and that $|\psi_i\rangle$ are an orthonormal basis.


Incidentally, the Shannon entropy of the coefficients of an expression of $\rho$ as a convex combination of arbitrary projectors is not a valid definition of anything. For example, the state $|0\rangle\langle 0|$ can be written as $$ |0\rangle\langle 0|=\sum_{i=1}^mp_iP\tag3 $$ where $P=|0\rangle\langle 0|$ is a projector and $p_1,\dots,p_m$ is any valid probability distribution. Clearly, if we allow the set of (non-orthogonal) projectors $\{P,\dots,P\}$ then $S(|0\rangle\langle 0|)=H(p_0,\dots,p_n)$ would be equal to every non-negative real number.

Thus, if the projectors in $(1)$ were not orthogonal then the definition $(2)$ would make no sense.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.