stevendaryl said:
Which postulate are you calling the "measurement postulate"? There are basically four assumptions in the standard, collapse interpretation of quantum mechanics:
- Observables correspond to Hermitian operators.
- When you measure an observable, you get an eigenvalue of the corresponding operator.
- The probability of getting eigenvalue o is the absolute square of the wave function projected onto the eigenstate corresponding to o.
- Afterwards, the system is in the eigenstate corresponding to the eigenvalue obtained.
The use of density matrices certainly doesn't require assumption 4. It's not clear to me that it actually requires assumption 2, either.
Let's see what we need to construct the density representation.
By definition, an ensemble is a list of states with the probability of finding them. Only one of the states is realized, but we do not know which one. So our fundamental ensemble is the set \{(|\psi_n\rangle,p_n):n\in\mathbb{N}\} where the |\psi_n\rangle are a family of (in general nonorthogonal) states and p_n are the associated probabilities. Now suppose that our Hilbert space is only 2 dimensional, that is we have a basis \{|0\rangle,|1\rangle\}.
A density operator in this Hilbert space is a hermitian 2x2 matrix. Clearly, this matrix encodes much less information than the (possibly very long) list of states in the ensemble. Since there are infinitely many possible realized (mutually non-orthogonal) states in this Hilbert space, there is no way of reducing the information down to a density matrix without throwing some of the information away.
The idea for reduction is that not all of that information is observable when we perform a measurement according to the measurement postulate. Let's pick an orthonormal measurement basis \{|a\rangle,|b\rangle\} and perform a measurement. Depending on which of the ensemble states is realized, we get an outcome |a\rangle with the measurement probability \langle\psi_n|a\rangle\langle a|\psi_n\rangle and the corresponding expression for |b\rangle. Since all single ensemble states are mapped to the same two measurement outcomes, we can collect the results and multiply both probabilities, the measurement outcome probability and the ensemble state probability. That means we have |a\rangle with probability \sum_n p_n \langle\psi_n|a\rangle\langle a|\psi_n\rangle and |b\rangle with probability \sum_n p_n \langle\psi_n|b\rangle\langle b|\psi_n\rangle. Both probabilities sum to 1 if the states are normalized and the ensemble proabilities sum to 1. That means \sum_n p_n \langle\psi_n|a\rangle\langle a|\psi_n\rangle+\sum_n p_n \langle\psi_n|b\rangle\langle b|\psi_n\rangle=1. We can rewrite this as \mathrm{tr}_{a,b}\sum_n|\psi_n\rangle p_n \langle\psi_n| = 1, which works in any basis, not just the indicated measurement basis. Let's call the matrix we trace over \rho. We can also read off the probability expressions, that for a projector P onto any subspace the associated probability of finding a measurement outcome in this subspace is \mathrm{tr}(\rho P). It's straight forward to find the expectation value of any operator A from here to be \langle A \rangle = \mathrm{tr}(\rho A). Other similar results follow.
So we have seen, that \rho encodes all possible measurement outcomes in any possible measurement basis, with potentially a lot less required information than encoded in the original ensemble description. But we had to use assumptions from the measurement postulate for this. Explicitly in your labeling I used 1) when I assumed that the measurement basis was orthogonal and again for the expectation value, where I also used 2). The remaining 3) and 4) were used in the measurement process of the single ensemble states.
Now you may know of a way to derive the same result without making these references. I would be curious to learn how. But unless such a way exists, the use of a density matrix requires the measurement postulate.
Cheers,
Jazz