Conservation of the determinant of density matrix

Question

I was playing with Bloch's optical equations for a quantum system with two energetic levels in an external sinusoidal perturbation under Rotating Wave Approximation. Letting Feynman's vector $\vec R$ be $$\vec R=(2\Re\{\rho_{12}e^{i\omega t}\}, -2\Im\{\rho_{12}e^{i\omega t}\}, \rho_{11}-\rho_{22}),$$ one finds $$\dot{\vec R}=\vec R \times \vec B,$$ where $\vec B$ is a constant vector depending on $\omega$, on the eigenvalues of the unperturbed system, and on the strenght of the perturbation.

Clearly, the equation above expresses a rotation of $\vec R$, and, in particular, implies the conservation of its norm, so I evaluated it: $$\|\vec R\|^2=(\rho_{11}-\rho_{22})^2+4|\rho_{12}|^2=(\rho_{11}+\rho_{22})^2-4(\rho_{11} \rho_{22}-|\rho_{12}|^2).$$ The latter rearrangement clearly show that the conserved quantity is equal to $(\mathrm{tr} \rho)^2-4 \det(\rho)$. Now, of course one has $\mathrm{tr}\rho=1$, so I conclude that, in this case, $\det \rho$ is conserved.

Then I tried to figure out whether this is a general property of the density matrix. Given the usual Schrödinger Equation, it seems to me that it is verified at least in the finite-dimensional case. One in fact has $$i \hbar \dot \rho =[H, \rho],$$ whence, since the derivative of the determinant is given by $$\frac{\mathrm{d}}{\mathrm{d}t} A(t)=\det A(t) \mathrm{tr} (A(t)^{-1} \dot A(t)),$$ it follows that $$i \hbar \frac{\mathrm{d}}{\mathrm{d}t} \rho= \det \rho \mathrm{tr} (\rho^{-1} [H, \rho]) =\det\rho \mathrm{tr}(\rho^{-1}H\rho-H)=0,$$ the last equality following from ciclicity if trace.

Is it right what I've done here? Does the statement hold in more general conditions? Does anybody know where could I learn more about it or about related topics?

Answer 1

The determinant is the product of the eigenvalues. It is thus preserved under any unitary evolution (such as the one given by the Schrödinger equation). This holds independent of the dimension of the Hilbert space.

Answer 2

The density matrix for a $2\times 2$ system can be written as a linear combination of the identity and the Pauli matrices: $$ \rho= \frac12 \left(I+\boldsymbol{P}\cdot\boldsymbol{\sigma}\right) $$ where $\boldsymbol{P}=\langle \boldsymbol{\sigma}\rangle$ and $\boldsymbol{\sigma}=\left(\langle\sigma_x\rangle,\langle\sigma_y\rangle,\langle\sigma_z\rangle\right)$. The matrix $\boldsymbol{P}\cdot\boldsymbol{\sigma}$ has determinant $1$ and the (unitary) evolution obviously maps the identity to the identity and also does not change the determinant of $\boldsymbol{P}\cdot\boldsymbol{\sigma}$.