Everyone - Ben M, Reality Check, anyone else here who understands physics - Pedrone said the following:
Again, as a layman, this couldn't be any more Greek to me if it walked out of the local Hellenistic restaurant and handed me a gyro. So to this well-ignorant layman: Why is the notion of spin-interaction (spin-spin interaction, perhaps, is the best way to put it?) incorrect?
In quantum mechanics, the Hamiltonian is what describes the energy of a system. It's also what tells us how our system evolves with time. When you talk about "interactions", what you frequently mean are terms in your Hamiltonian which depend on the state of the two things which are interacting. So if we have a term in our Hamiltonian which depends on the spin states of two particles, we can call that a spin-spin interaction.
The problem here is that "exchange" interactions don't actually show up in the Hamiltonian. More on that below.
I understand that the Pauli exclusion principle prevents particles from having the same state, but does it mean that they cannot influence each other?
Not at all.
Let's actually back up a bit and discuss where Pauli exclusion comes from: exchange symmetry.
Let's say I have two distinguishable particles. We'll label the position of particle 1 as x1, and the position of particle 2 as x2. Let's say particle 1 is in state a, with wave function A(x1), and particle 2 is in state b with wave function B(x2). I can then write their
combined wave function F(x1,x2) as a simple product F(x1,x2) = A(x1)B(x2). More complex wave functions (ie, sums of products) are possible, but this will do for our example.
Now, what if I have two
indistinguishable particles? Well, I should still be able to write a wave function of the form F(x1,x2). But the particles are identical, so it shouldn't matter if I switch places. So F(x1,x2) = F(x2,x1). But actually, that's not quite right. It shouldn't matter to any
observation if I switch places, but since observables depend on F^2, not F, then I should really have F(x1,x2) = +/- F(x2,x1). It turns out that for bosons, it's always a +, and for fermions it's always a -. This is known as exchange symmetry: when I exchange the particles, I get either the same wave function or its negative.
Now, I still want one particle to be in state a and 1 particle to be in state b. So I'd like something along the lines of F(x1,x2) = A(x1)B(x2). But that doesn't satisfy the above exchange symmetry. So instead I need something like this:
F(x1,x2) = A(x1)B(x2) +/- B(x1)A(x2)
where again, the sign difference is for bosons vs. fermions. Now this difference may not at first glance look like it matters much, but it actually matters a great deal. For example, if we're dealing with bosons, then A = B is an acceptable solution. But if we're dealing with fermions, then obviously A = B produces F = 0, which is not a valid solution. That's where Fermi exclusion comes from.
So how does spin enter into this? Well, if A and B include spin, then I can make the
spin parts different but the
spatial part the same. Conversely, if the spatial parts are different then I can have the spin parts equal.
So now let's bring this back to our previous discussion. Let's say I have a Hamiltonian for two identical charged particles which repel each other. The "interaction" is purely Coulomb repulsion, meaning that the Hamiltonian itself has no terms involving either spin. The Hamiltonian only deals with the spatial wave function. But spin is still very relevant in such a case, because spin constrains the possible spatial wave states we can have. So in a sense, the energy depends on the spin state even though the spin is not part of the Hamiltonian. So in one sense, there's no spin-spin interaction. But one could still
call this a spin-spin interaction. At this point it's a semantic issue. The original Hamiltonian contains no spin-dependent terms but the energies of our states are still different for different spins. But generally, this wouldn't be called a spin interaction, it would be called an exchange interaction. And that is different from something like a magnetic dipole-dipole interaction, where you have spin dependence directly in the original Hamiltonian regardless of exchange symmetry.
And it can get even muddier. If one is not interested in all possible states, but only the states a particle is likely to occupy under some conditions of interest, one can often write an effective Hamiltonian which is different from our original Hamiltonian but produces the same energies for the particular states of interest. And such an effective Hamiltonian might even include a spin-dependent term even though the original Hamiltonian does not. If the effective Hamiltonian has a spin-dependent term but the original does not, is there a "spin interaction"? Again, this is somewhat of a semantic debate, and I don't generally care about semantic debates. As long as one defines one's terms and stays consistent, I'm pretty flexible. But as above, this would still often be called an exchange interaction if in fact it originated from the exchange symmetry restrictions I discussed above.
But none of this means that pedrone is right about the spin dependence of the interaction being electromagnetic. It isn't for nuclear scattering.
