- #26

Fredrik

Staff Emeritus

Science Advisor

Gold Member

- 10,851

- 406

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.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?

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).and How can we create a mix state with just one atom?

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.