Applying Heisenberg picture to density operator

[Haorong Wu]

TL;DR

How density operator evolves in the Heisenberg picture?

Suppose that a particle evolves from point A to point B. The state of the particle can be written as $$\rho=\sum \left | m\right >\rho_{mn}\left< n\right | .$$ Because the basis is evolving as the particle travels, I am considering applying the Heisenberg picture to the density operator.

Let the initial state of the particle be $$\rho_0=\sum \left | m,0\right >\rho^0_{mn}\left< n,0\right |,$$ and the selecting operator at $z$ be $$\left |n,z \right > \left <m,z \right | $$. Then the element of $\rho$ at $z$ could be extracted as $$\rho(z)_{mn}=tr(\rho_0 \left |n,z \right > \left <m,z \right |)= \sum_{m'n'}\left <m,z \right | \left | m',0\right >\rho^0_{m'n'}\left< n',0\right | \left |n,z \right > . $$

I am not sure whether this is correct or not. The density operator after all is not a state vector.

 

[atyy]

The Hilbert space does not evolve in either the Schroedinger or Heisenberg pictures. There is no concept of evolving basis. The basis is something you are always free to choose, and is purely a matter of convenience.

The quantum state (ie, the state vector or density operator) is constant in the Heisenberg picture.

 

[Haorong Wu]

The Hilbert space does not evolve in either the Schroedinger or Heisenberg pictures. There is no concept of evolving basis. The basis is something you are always free to choose, and is purely a matter of convenience.

The quantum state (ie, the state vector or density operator) is constant in the Heisenberg picture.

Thanks! I will reconsider it.