Density Operators

  • Thread starter Robert_G
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  • #26
Fredrik
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I might get a little closer,
we can just write

##\hat{\rho}=0.3|a_1\rangle \langle a_1|+0.7|a_2\rangle \langle a_2|##

and this ##\hat{rho}## is not an outer product of any vectors in the Hilbert space, because it is a classic statistic thing, right?
I think that's a valid way of looking at it, but if you want to be sure that the state you've written down isn't pure (i.e. equal to a projection operator for a 1-dimensional subspace), you should use the fact that if ##\rho=|\psi\rangle\langle\psi|##, then ##\rho^2=|\psi\rangle\langle\psi|\psi\rangle\langle\psi| =|\psi\rangle\langle\psi|=\rho##. Since your ##\rho## doesn't satisfy ##\rho^2=\rho##, it's not pure.

and How can we create a mix state with just one atom?
Flip a coin. If heads, use the preparation procedure that puts the atom in state ##|\alpha\rangle##. If tails, use the preparation procedure that puts the atom in state ##|\beta\rangle##. Now give the atom to your experimentalist friend and tell him that you flipped a coin to decide between the two procedures, but don't tell him the result of the flip. He should now use the state ##\frac 1 2|\alpha\rangle\langle\alpha|+\frac 1 2|\beta\rangle\langle\beta|##. (I explained these things in post #8 as well, but maybe you missed it).

Another option (more interesting, but also more difficult to understand) is to take an atom that's known to be entangled with another, as described by a 2-particle state. Then use the method of reduced states (reduced density matrices) to assign a 1-particle state to one of the atoms.
 
  • #27
atyy
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This is the one who start this discussion. After I read the above replies, I try to think it by myself. and let's see whether I got it right?

Density operators are used to describe system (A) that is composed by some subsystems (a1, a2, a3...ai....). The mechanics that the subsystems(...ai...) compose the whole system (A) obey the classic statistics. for example: If we just have two subsystems

[itex]|\Psi\rangle \langle \Psi|=0.3 |a_1\rangle \langle a_1| + 0.7 |a_2\rangle \langle a_2|[/itex] (*)

Well, the subsystems are the quantum systems.

So, for just one atom, The above state [itex]|\Psi\rangle \langle \Psi|[/itex] can not be realized, because, if [itex]|a_i\rangle[/itex] is the the eigenstate of same operator, then [itex]0.3^2+0.7^2\neq 1[/itex].

If we are considering a large number of atoms, this state can be realized. we can prepare those atoms in this particular way, which is given by the eq.(*). The 30% of the atoms are prepared in state [itex]|a_1\rangle[/itex], and 70% of the atoms are prepared in state[itex]|a_2\rangle[/itex]. For example, One can separate 30% of the atoms from the 70%, and excite them by laser, and make them all in same metastable state, and those 70% remain in the groud state, and mix them together.

Am i getting it correct?
That is essentially right (except for the LHS, as Fredrick points out above), but your terminology is very non-standard. Here, your subsystems are "non-interacting" subsystems - they are independent trials of the same "experiment". For this reason, people usually use the term "proper mixture", or "ignorance interpretable mixture". The term "subsystem" is not used, because each atom in the ensemble (independent trials of the same "experiment") is a whole system which is in a pure state that evolves according to Schroedinger's equation. The mixture is due to different atoms being distributed in different pure states according to some classical frequentist probability.

The "subsystem" terminology is usually reserved for a conceptually different density matrix, called the reduced density matrix. Here, the entire system is in a pure state, but we only measure a (interacting or non-interacting) subsystem of the entire system. The "state" of the subsystem is given by the reduced density matrix, which is also called an improper mixture. In contrast to a density matrix that represents a proper mixture, the reduced density matrix does not evolve according to Schroedinger's equation. Only the full system evolves according to Schroedinger's equation.
 
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  • #28
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Thank all of you. very happy to find this forums.
 

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