# Two particles in superposition

1. Nov 4, 2007

### freerangequark

If you had an isolated system consisting of two particles both of which were in superposition, would they interact with each other and collapse each other's wave function?

Thanks,
FRQ

2. Nov 5, 2007

### f95toli

Whether or not they would interact depends on what kind of partcle it is. There needs to be some sort of "coupling mechanism" , e.g. some electromagnetic dipole-dipole interaction. If there is a "classical" interaction there will also be a "QM" interaction.

The word "collapse" is somewhat missleading. Wha happens if you have e.g. two coupled spin 1/2 systems is essentially that a "hybrid" system is created with new energy levels, wavefunctions etc. However, it is still very much a QM system.

3. Nov 5, 2007

### freerangequark

Not being anything more than a reader of popular physics books, I'm trying to understand what you're saying.

Are you talking about an entangled pair?

In my example, I was wondering what would happen if the isolated system contained two of the same type of particle... lets say a pair of non-entangled photons in superposition. (when I say non-entangled, I mean not entangled with each other.)

Thanks,
FRQ

4. Nov 6, 2007

### OOO

One could also consider the two electrons in a helium atom (assuming that the core is approximately fixed due to its large mass). Without interaction the electrons would remain in a product state (in an antisymmetrized form, to be precise) for all time if they were initially. A product state is a state where the measurement of position of electron 1 yields independent results of the measurement of position of electron 2:

$$\psi(x_1,x_2) = \psi_1(x_1)\cdot\psi_2(x_2)$$

(for simplicity I've neglected fermionic antisymmetrization)

But if the electrons do interact, an initial product state will get mixed with other product states during the time evolution. Therefore the measurement of electron 2 will almost always depend on measurement of electron 1:

$$\psi(x_1,x_2) \not= \psi_1(x_1)\cdot\psi_2(x_2)$$

Of course even a system without interaction need not always be in a product state. For example, it may have started as a product state in the past (e.g. after you have simultaneously measured the position of both electrons), then you "turned" the interaction on, which has mixed product states, then you turn the interaction off. The system will then remain in its current mixed state, although there is no interaction any more. Thus the superposition of pure product states can be considered as a kind of memory of past interactions.

Last edited: Nov 6, 2007
5. Nov 6, 2007

### blechman

The problem is that something "external" to the system might be able to collapse the wavefunction of said system, but something "internal" can not. So for example, if you had one photon, another photon can come along and potentially "collapse" the wavefunction. HOWEVER: the system of two photons has a wavefunction that has not collapsed. Another photon can come along and collapse the wavefunction of the 2-photon system, but the wavefunction of the 3-photon system remains uncollapsed. This chain can go on forever!

This is a problem that lives in the heart of the physics subfield known as "Quantum Cosmology", where people have tried to construct the "wavefunction of the universe" (Wheeler-DeWitt Equation). What is not clear is that even if such a thing exists (and I'm not saying that it does or doesn't) - can it ever be "collapsed"?

Another problem that has been popularized by Roger Penrose is: how can you ever isolate a system, since gravity couples to everything? Therefore, isn't the (**incredibly** tiny!) gravitational pull from the Andromeda galaxy "collapsing the wavefunction" of every isolated system we construct?

As you can see, FRQ, this is a very deep problem!

6. Nov 6, 2007

### cesiumfrog

What happens if two particles (such that the state of each one individually could be expressed as a superposition of basis states, but which aren't initially entangled) interact? Don't they just become entangled (such that the state of the pair is not yet "collapsed" but could be expressed as a superposition....)?

Wouldn't that only be true if the interaction were strong enough that you could determine something about the "isolated system" by measuring changes over in Andromeda? (Lately, Penrose has been known to publicise some questionable ideas.)

7. Nov 7, 2007

### blechman

Perhaps. I'm not saying I necessarily agree with it. I only throw it out there as a possible added confusion. Penrose has said some outlandish things in his time, but he's a very smart guy! I think it's still an open debate as to how serious to take such issues. I am not a QC expert, and every time I try to get into it, I come out with a splitting headache!