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.


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