In the Copenhagen interpretation of quantum mechanics (QM), if you make a measurement of a superposition state then it "collapses" into one of the constituent states. (I discussed superposition states briefly in my last post but one.) Physicists have spent decades discussing when the collapse occurs and even building models of the collapse process. This talk always brings to mind an image of a small boy with a screwdriver, standing next to a pile of dismantled chair parts, saying "It just collapsed!" In fact, there is no need for a collapse postulate if we properly understand the nature of a quantum state.
Ballentine defines a quantum state as follows:
Any repeatable process that yields well-defined probabilities for all observables may be termed a state preparation procedure. It may be a deliberate laboratory operation, ... or it may be a natural process not involving human intervention. If two or more procedures generate the the same set of probabilities, then these procedures are equivalent and are said to prepare the same state. (Quantum Mechanics, p. 33)
More simply, a state is "an ensemble of similarly prepared systems."
Think about a standard beam-splitting experiment. What emerges from the apparatus is a superposition of, say, left and right states: (|L> + |R>)/sqrt(2). Now let's say we measure the particle and find it's in the left beam. In the Copenhagen interpretation one would say the state has collapsed into pure |L>. But remember, a state is an ensemble of similarly prepared systems. One particle does not an ensemble make. To create a true |L> state, we would need to continue to measure the emerging particles and discard the result every time we measure an |R> particle. We can accomplish this easily - say by putting a brick in front of the right-hand beam. Now if someone says to me, "Hey, the wave function collapsed!" I'll respond, "Of course it 'collapsed' - you put a big honkin' brick into the apparatus!"
Now you are probably thinking, "It can't be that simple." But I think it really is that simple. Whenever you encounter wave function collapse in the standard interpretation, if you look at it closely, you will find the brick.
Let's look at a classic quantum experiment: the two-slit experiment. As is well known, if you allow a particle beam to pass through a barrier with two slits, you will see an interference pattern (resulting from the superposition of the two beams). However, if you place a detector near one of the slits so that you can tell which slit the particle passes through, then the interference pattern disappears. The standard explanation is to say that the detector collapses the wave function, so there is no longer a superpostion and no longer any interference pattern.
To see what's really going on, let's model the detector as an atom that is in its ground state, |0>. When the particle passes through the slit, the atom is put into its excited state, |1>.
Suppose we place this detector near the left-hand slit. Initially, the beam+atom system is in some state |i>. After the beam passes through the slit we have the (combined) final state
|f> = (|L>|1> + |R>|0>)/sqrt(2).
Now, suppose that the detector atom gets excited. Then we know that the beam particle is in the state |L>. But has the state "collapsed"? Remember that the initial state |i> represents an ensemble of similarly prepared systems. In that ensemble, some runs of the experiment will produce an excited detector atoms and some will not. So the apparently innocent words "suppose that the detector atom gets excited," actually represent our ignoring of all the parts of the ensemble in which the detector atom doesn't get excited.
Think of it this way: we run the experiment, and on the first trial the detector atom doesn't get excited. So we ignore that run. Second run, same result - so we ignore that one, too. Third run, the detector atom gets excited. "OK - that's the situation I was talking about. Now the wave function has collapsed!"
Nope - it only "collapsed" because you put a brick in the beam - by ignoring the results you weren't interested in.
That is to say, if we look only at the runs in which the atom gets excited, we will produce the |L> beam state. (Likewise if we keep only the runs in which the atom isn't excited - then we produce the |R> beam state.) Mathematically, this corresponds to projecting out one component of the wave function. That's the import of the words "suppose that the detector atom gets excited." They mean we have chosen to consider only a part of the total ensemble.
What if we keep both sets of results? There is a well-defined procedure for finding the state in cases like this. We start by forming the density matrix for the combined system:
Then we trace out the atom's state. This amounts to discarding terms that involve |0><1| or |1><0| in the expansion of the density matrix above. We end up with the reduced density matrix that describes the beam state:
(|L> <L| + |R> <R|)/2
This is what is called a mixed state. It cannot be written in terms of a wave function. It has a natural interpretation in the statistical interpretation: it is an equal mixture of the pure |L> state and the pure |R> state. But it is NOT a superposition state: there is no fixed phase between the two possibilities, so there is no interference.
If you didn't follow all of the above, here's the bottom line: normal QM leads to the conclusion that there won't be any interference when we place a detector near one of the slits - and we can come to this conclusion without any mention of wave function collapse, if we treat the detector in a properly quantum mechanical manner.
OK - this has already become a very long and technical post, so I'll save some final comments for next time.