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In the class notes and the text [1] the Von Neumann entropy is defined as

\begin{equation}\label{eqn:densityMatrixEntropy:20}

S = -\textrm{Tr} \rho \ln \rho.

\end{equation}

In one of our problems I had trouble evaluating this, having calculated a density operator matrix representation

\begin{equation}\label{eqn:densityMatrixEntropy:40}

\rho = E \wedge E^{-1},

\end{equation}

where

\begin{equation}\label{eqn:densityMatrixEntropy:60}

E = \inv{\sqrt{2}}

\begin{bmatrix}

1 & 1 \\

1 & -1

\end{bmatrix},

\end{equation}

and

\begin{equation}\label{eqn:densityMatrixEntropy:100}

\wedge =

\begin{bmatrix}

1 & 0 \\

0 & 0

\end{bmatrix}.

\end{equation}

The usual method of evaluating a function of a matrix is to assume the function has a power series representation, and that a similarity transformation of the form \( A = E \wedge E^{-1} \) is possible, so that

\begin{equation}\label{eqn:densityMatrixEntropy:80}

f(A) = E f(\wedge) E^{-1},

\end{equation}

however, when attempting to do this with the matrix of \ref{eqn:densityMatrixEntropy:40} leads to an undesirable result

\begin{equation}\label{eqn:densityMatrixEntropy:120}

\ln \rho =

\inv{2}

\begin{bmatrix}

1 & 1 \\

1 & -1

\end{bmatrix}

\begin{bmatrix}

\ln 1 & 0 \\

0 & \ln 0

\end{bmatrix}

\begin{bmatrix}

1 & 1 \\

1 & -1

\end{bmatrix}.

\end{equation}

The \( \ln 0 \) makes the evaluation of this matrix logarithm rather unpleasant. To give meaning to the entropy expression, we have to do two things, the first is treating the trace operation as a higher precedence than the logarithms that it contains. That is

\begin{equation}\label{eqn:densityMatrixEntropy:140}

\begin{aligned}

-\textrm{Tr} ( \rho \ln \rho )

&=

-\textrm{Tr} ( E \wedge E^{-1} E \ln \wedge E^{-1} ) \\

&=

-\textrm{Tr} ( E \wedge \ln \wedge E^{-1} ) \\

&=

-\textrm{Tr} ( E^{-1} E \wedge \ln \wedge ) \\

&=

-\textrm{Tr} ( \wedge \ln \wedge ) \\

&=

– \sum_k \wedge_{kk} \ln \wedge_{kk}.

\end{aligned}

\end{equation}

Now the matrix of the logarithm need not be evaluated, but we still need to give meaning to \( \wedge_{kk} \ln \wedge_{kk} \) for zero diagonal entries. This can be done by considering a limiting scenerio

\begin{equation}\label{eqn:densityMatrixEntropy:160}

\begin{aligned}

-\lim_{a \rightarrow 0} a \ln a

&=

-\lim_{x \rightarrow \infty} e^{-x} \ln e^{-x} \\

&=

\lim_{x \rightarrow \infty} x e^{-x} \\

&=

0.

\end{aligned}

\end{equation}

The entropy can now be expressed in the unambiguous form, summing over all the non-zero eigenvalues of the density operator

\begin{equation}\label{eqn:densityMatrixEntropy:180}

\boxed{

S = – \sum_{ \wedge_{kk} \ne 0} \wedge_{kk} \ln \wedge_{kk}.

}

\end{equation}

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.

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