The Uncertainty of Particle Positions and its Effect on Probability Densities

[mrspeedybob]

Suppose you have 2 neutrons. There is a probability cloud associated with each neutron's position. Neutrons are indistinguishable from each other so really all you can do is add the 2 probability clouds together and state that it represents 2 particles, you can't say which neutron is which.

Now consider 6 points in a straight line. A B and C are very close together, there is some distance, and then D E and F are very close. The center of 1 neutron's cloud is at B, the other is at E. When the clouds are added together the points of maximum probability shift from B and E to C and D. In essence the 2 neutrons are drawn to each other as a result of the uncertainty of their positions.

Is this a derivation of 1 of the 4 fundamental forces, if so which 1? If not does this effect have a name?

 

[Chopin]

The probability densities for separate particles don't add together in that way. Each particle has its own density throughout space, so the presence of other particles doesn't change the measurements for anyone of them.

This misconception comes from incorrectly trying to extend the concept of the Schrodinger wavefunction to multiple particles. A single particle has a probability density indexed by three variables--the three spatial dimensions. However, a double-particle system's wavefunction is indexed by six variables--three dimensions for each of the two particles. This wavefunction cannot be plotted on a graph the same way a single particle's function can, so it's quite a bit more difficult to visualize. For more information on how to discuss the wavefunctions of multi-particle states, look up the term 'Fock Space'.

 

[The_Duck]

As Chopin says, you can't just add single-particle wave functions to get multiparticle wave functions. However there is an effect like what you are thinking of, the http://en.wikipedia.org/wiki/Exchange_interaction" or exchange "force" which comes from the symmetrization or antisymmetrization required for wave functions describe multiple identical bosons or fermions.

 

[Chopin]

Oh right, I forgot about that. However, in mrspeedybob's example, the particles are fermions, so the exchange force will actually be repulsive (the Pauli Exclusion Principle), correct?

 

[The_Duck]

I believe it works like this: If the spin state of two fermions is symmetric (triplet state), then the spatial wave function must be antisymmetric and the exchange effect means you will tend to find them farther apart. But if the spin state is antisymmetric (singlet state) then the spatial wave function is symmetric and you will tend to find them closer together.

 

[Chopin]

Huh, I didn't realize it was that complicated. I thought if the spin states were opposite, the two particles are guaranteed to be distinguishable, so the exchange interaction didn't take place. Guess I need to read more about that.

In any case, as I understand it, the exchange interaction isn't really a "force" in the sense of the real forces. It doesn't compel the particles to move together or apart as time passes, like say the electromagnetic force would. It just says that, at all times, there's a very high/very low chance of the particles being close together.

 

[K^2]

No, you still don't know which particle is which. In the simplest scenario, with no regard to interactions, if you have two fermions with wave functions f₁(x) and f₂(x), the total wave function of the two of them would be something like F(x₁,x₂) = f₁(x₁)f₂(x₂) - f₁(x₂)f₂(x₁). This will result in Pauli repulsion between two particles. Note that the functions f₁ and f₂ include both the probability density and spin information, as well as any other quantum numbers.

 

[The_Duck]

This isn't something I've taken the time to understand well in the past. Playing with it a bit I see that it is possible to construct an antisymmetric position+spin state that looks like

$$|\psi \rangle = \frac{1}{\sqrt{2}}(\psi_a(x_1) \psi_b(x_2) |\uparrow\downarrow\rangle - \psi_b(x_1)\psi_a(x_2)|\downarrow\uparrow\rangle)$$

which you might say describes a spin-up particle with spatial wave function psi_a and a spin-down particle with wave function psi_b. If you calculate the expectation value of (x1-x2)^2 in this state I believe you get the same answer as if you were dealing with two distinguishable particles described by a multiparticle wave function

$$\psi_a(x_1) \psi_b(x_2)$$

So there seems to be no exchange force. Compare

$$\frac{1}{2}(\psi_a(x_1) \psi_b(x_2) - \psi_b(x_1) \psi_a(x_2))(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)$$

where the expectation of (x1-x2)^2 is larger and

$$\frac{1}{2}(\psi_a(x_1) \psi_b(x_2) + \psi_b(x_1) \psi_a(x_2))(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$$

where it is smaller. In these two states the spins are in "opposite directions" but there is an exchange force. The original state I wrote down is the sum of these last two (up to normalization).

 

[mrspeedybob]

The probability densities for separate particles don't add together in that way. Each particle has its own density throughout space, so the presence of other particles doesn't change the measurements for anyone of them.

This misconception comes from incorrectly trying to extend the concept of the Schrodinger wavefunction to multiple particles. A single particle has a probability density indexed by three variables--the three spatial dimensions. However, a double-particle system's wavefunction is indexed by six variables--three dimensions for each of the two particles. This wavefunction cannot be plotted on a graph the same way a single particle's function can, so it's quite a bit more difficult to visualize. For more information on how to discuss the wavefunctions of multi-particle states, look up the term 'Fock Space'.

I am supposing no interaction between the probability fields. If we suppose for a second that the 2 particles are distinguishable we can call them neutron-B and neutron-E. If the probability of finding neutron-B at point C is X and the probability of finding neutron-E and point C is Y then the probability of finding a neutron at C is 1-(1-X)(1-Y). The same logic applies to point D. The probabilities are less at points A and F.