The purity of a quantum state is defined as $$ \gamma = \mathrm{Tr}(\rho^2) $$ In the Schrödinger picture, it's easy to use this formula to see how the purity of the system changes as the system evolves. In the Heisenberg picture, the density matrix $\rho$ is constant. How does one calculate the purity?

  • $\begingroup$ I believe it is the same as above, $\text{Tr}(\rho^2)$ (or calculate the von Neumann entropy $-\rho\ln\rho$) except in the case of the heisenberg picture it is constant. One can of course change this by the insertion of decoherence effects/Lindblad operators $\endgroup$
    – MKF
    Commented Jun 12, 2018 at 16:07

1 Answer 1


For a closed system, whose time-evolution is unitary and given by the Schrodinger equation, any quantity $S$ quantifying a state's "mixed-ness" (or entanglement entropy, depending on your perspective) that takes the form $S = \text{Tr}\left[ f(\rho) \right]$ for some analytic function $f$ has a value that is identical in the Schrodinger and Heisenberg pictures. You can see this by the cyclic property of the trace and the equation $\rho_\text{Schro}(t) = U(t) \rho_\text{Heis} U^\dagger(t)$. $S$ could be the purity, the von Neumann entropy, or the Renyi entropy. Moreover, $S$ is constant in either picture because the Schmidt weights (the eigenvalue spectrum of $\rho$) do not change over time in either picture.

For an open system, which is directly coupled to its environment, $\rho$ is not constant in either picture. It doesn't evolve unitarily but instead via Kraus maps, which are basically what you get when you take Schrodinger time evolution and trace out part of the Hilbert space.

  • $\begingroup$ Thank you! I was indeed thinking about an open system for one and a closed system for the other. In an open system, is there a "short-cut" of calculating Kraus operators instead of tracing over the whole density matrix? Or is it just something formal? $\endgroup$ Commented Jun 14, 2018 at 12:30
  • $\begingroup$ @YantingTeng Oh sure, you don't need to actually manually trace out the environment degrees of freedom in order to use Kraus maps. That's just the physical motivation justifying their mathematical properties. In practice, they're just mathematical (super-)operators that act directly on the system's density matrix, with no reference to the environment. It's just like how you could form a reduced density matrix by starting with the "universe's" pure state vector and tracing out the environment Hilbert space, but you don't have to - you could also just start with the density matrix directly. $\endgroup$
    – tparker
    Commented Jun 14, 2018 at 14:25

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.