I Applying Heisenberg picture to density operator

Haorong Wu
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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.
 
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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.
 
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atyy said:
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.
 
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If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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