Sunday 21 August 2016

quantum mechanics - Why does the density matrix $rho$ obey a wrong-signed Heisenberg equation of motion?


The density matrix is defined as $$ \rho_\psi ~:=~ \frac{\lvert\psi(t)\rangle \langle \psi(t)\vert}{ \langle \psi(t) |\psi(t)\rangle }$$ in the Schrödinger picture. $\rho_\psi$ is obviously a time dependent projector, and the equation of motion on these projectors become:
$$ i\hbar\frac{d}{dt} \rho_\psi ~=~ [H,\rho_\psi] \tag{S} $$ but my book also reports that the Heisenberg equation of motion on the operators/observables is: $$ \mathrm{i}\hbar\frac{d}{dt} A ~=~ [A,H] . \tag{H} $$
Why are the signs in eqs. (S) and (H) opposite?


Isn't $A$ an operator like $ \rho_\psi$, although time independent? They belong to the same operator space, so I don't think I can apply duality, but I know that $A$ operate on the states to give us the expectation value through the relation $$ \mathrm{Tr}( \rho_\psi A)$$ so it should be in the dual space of the observables.



Answer




Actually, you can use duality:


the normal states of quantum mechanics are objects of the (unique) predual of the von Neumann algebra of quantum observables.


Using a concrete example: if the algebra of observables are the bounded operators on a Hilbert space, the predual are the trace class operators. Of them, the normal states are the ones positive, self-adjoint and of trace norm one.


It is then clear that by mutual duality the evolution on observables/states induces the evolution of states/observables; and that takes into account of the "minus sign" in the generator that is different between the two.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...