Quantum Mutts

Let's say you have a thousand of your friends in a large field, and every one of them has exactly two coins - one penny and one dime - in his pocket. Or her pocket, as the case may be. No exceptions, they all have those and only those two coins. Now imagine nearly the same situation with a thousand friends, but this time five hundred of them have two pennies and five hundred of them have two dimes. You don't know which ones have which, but you do know for a fact that any given person has two of the same coin.

If you take a random sampling of people and ask them to randomly select exactly one of their coins to hand to you, will there be any statistical difference between the same-coin and different-coin groups? The answer is no - but as you know by now, quantum mechanics is weird.

Every quantum system can be described in terms of a state. The state completely describes the physics of the system. Some states are special; when we make a measurement of those states they always produce the same value. If we run electrons through a Stern-Gerlach device and then we measure the spin of those electrons we'll always get the same value. Or, if we happen to a laser crystal which has just flashed, we know (very loosely speaking) that there's been a sudden transition of all the atoms into the same energy state. These special states where we know for sure what the result of a measurement will be are called eigenstates of that particular measurement.

A person with two dimes is like a system in an eigenstate because you know for sure what the result of the "give me one coin" measurement will be - a dime. The group of a thousand people with matched coins is called a mixed state because the group as a whole is composed of people in different states. The group of people with mismatched coins is in a pure state because even though they're not in an eigenstate, every person is nonetheless in the same state. We'd say their state was a superposition of the dime and the penny eigenstate.

So in analogy to the people, let's say you do an experiment where you have electrons which are in a pure state where all of the electrons have a 50/50 chance of being in the up or down spin states and an experiment where half of the electrons have a 100% chance of being up and half have a 100% chance of being down. Is there going to be a difference between the situations?

There can be. If the two possible observable values of the pure state overlap in a quantum mechanical sense, then the expected number of pennies (or spin up states) you observe in a pure state might not be the the same as for a mixed state. So in quantum mechanics, if the system and observations satisfy certain conditions there can be statistical differences between 100 guys who have matching coins and a 50/50 mix of guys with identical coins. If you want to look at the math (mercifully forgoing the explanation), the expectation value of an observable X in a pure state Psi which is a linear combination of two eigenstates is:


That third term is not present in a mixed state, which is what causes the difference. Only if that third term happens to be zero will the two situations produce the same result.

This difference is formally described by the density matrix in quantum mechanics, and it's something that's vital in the quantum mechanical description of many-particle systems. They say you never really understand how weird quantum mechanics, you just get used to it. That's probably right.

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Now imagine nearly the same situation with a thousand friends, but this time five hundred of them have two pennies and five hundred of them have two pennies.

Don't you mean 'five hundred of them have two dimes'?
Why divide them into two groups if they both have two pennies?

Matt replies: Yes indeed, I've fixed it.

Since you're bringing up density matrices and Stern-Gerlach experiments, I'm going to provide a link to Schwinger's beautiful and easy to understand, but little known paper on the subject, The Algebra of Microscopic Measurement. In this paper, you can think of the "measurements" as "projection operators" or as "pure density matrix states", which are mathematically equivalent.

That paper has a second part, which is not quite so beautiful or simple, The Geometry of Quantum States. The second paper is interesting in that it provides a way of understanding the vacuum state that is more physical than mathematical; I think this is superior to the stuff they will teach you in your graduate QM classes. (Which are more oriented to getting you to calculate quickly and efficiently, rather than in letting you understand other interpretations of the foundations of quantum field theory.)

By Carl Brannen (not verified) on 17 Oct 2008 #permalink

That third term is not present in a mixed state, which is what causes the difference. Only if that third term happens to be zero will the two situations produce the same result.

The problem with classical models is that when you measure the state of a coin when it is a penny, it will always give the result penny; and when it is a dime, it will always give the result dime. Such models are a poor guide into the world of quantum physics. We should use models where the third term isn't zero. Such situations exist in the ordinary macroscopic world. For example, when we measure the location of an extended object, say an arrow, whose location is x1 (we may represent it by the ket |x1>), there is some probability to detect it at x2, simply because the object is not a point. Thus, if X is the location operator, the projection of X|x1> on |x2> is not equal to zero. Using such models helps a lot.

Schrodinger rules the waves...

The problem with your example is that its probability amplitudes are purely real. You need complex amplitudes to get the "weird" quantum effects that run counter to classical intuition.

By CCPhysicist (not verified) on 20 Oct 2008 #permalink