2-particle Wavefunction and the Pauli Exclusion Principle

[PeterDonis]

I understand what a Hilbert space is, but I always thought that a valid basis would be a spin up state and a spin down state

Those states form a basis for one particular Hilbert space, the Hilbert space of spin states for a single spin-1/2 particle.

or if we are talking about energy, the ground state, the first excited state, the second excited state, etc

Those states would form a basis for another particular Hilbert space, the Hilbert space of a single particle in a potential well. Strictly speaking, this particle would have to have zero spin, since you are only including energy in the specification of the basis states.

The Hilbert space you have been implicitly referring to in this discussion is neither of the above. In fact it's not entirely clear which one it is, since we have talked about several different possible scenarios, but we'll start with the simplest possible one that is relevant: the Hilbert space of a quantum system consisting of two identical spin-1/2 particles, which are both in the same energy level of some potential well. At this point we are ignoring any other characteristics of the particle or the potential well: no electric charge, no nucleus of the atom, etc. We are also ignoring any possibility of either particle jumping to some other energy level (for example, by absorbing a quantum of radiation).

This Hilbert space is an antisymmetric tensor product of two copies of the Hilbert space of spin states for a single spin-1/2 particle. Unfortunately, we don't have any way of writing down what that means without labeling the particles, but as I've said before, it's important to realize that these labels are artifacts of our math and do not correspond to anything in the actual system. If we label the particles as "particle 1" and "particle 2", then a simple tensor product (which is not quite what we're looking for, but will give us a start at getting there) of the Hilbert spaces for each particle would be a Hilbert space with a basis consisting of four mutually orthogonal vectors which we could write as $\vert \uparrow \rangle_1 \vert \uparrow \rangle_2$, $\vert \uparrow \rangle_1 \vert \downarrow \rangle_2$, $\vert \downarrow \rangle_1 \vert \uparrow \rangle_2$, $\vert \downarrow \rangle_1 \vert \downarrow \rangle_2$, where the arrows denote the "up" and "down" basis states for the single-particle Hilbert space and the subscripts denote which particle (1 or 2). In other words, the simple tensor product Hilbert spaces consists of all possible states you can form by multiplying together two single-particle states.

If you want some more information about tensor products of Hilbert spaces, see here:

https://en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces

But the Hilbert space we actually want is the antisymmetric tensor product; "antisymmetric" because we are dealing with fermions. This Hilbert space can be constructed in a number of ways, but the simplest is to take all possible states you can form from the following formula:

$$
\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left( \vert \psi_a \rangle_1 \vert \psi_b \rangle_2 - \vert \psi_b \rangle_1 \vert \psi_a \rangle_2 \right)
$$

where $\vert \psi_a \rangle$ and $\vert \psi_a \rangle$ are states belonging to the single-particle Hilbert space of spin states. So what we need is a valid set of basis states for this Hilbert space. One way to find them is to expand $\vert \psi_a \rangle$ and $\vert \psi_a \rangle$ in terms of the single-particle basis states and substitute into the above formula:

$$
\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left[ \left( a_{\uparrow} \vert \uparrow \rangle_1 + a_{\downarrow} \vert \downarrow \rangle_1 \right) \left( b_{\uparrow} \vert \uparrow \rangle_2 + b_{\downarrow} \vert \downarrow \rangle_2 \right) - \left( b_{\uparrow} \vert \uparrow \rangle_1 + b_{\downarrow} \vert \downarrow \rangle_1 \right) \left( a_{\uparrow} \vert \uparrow \rangle_2 +a _{\downarrow} \vert \downarrow \rangle_2 \right) \right]
$$

Then we can just expand out the products and collect terms:

$$
\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left[ \left( a_{\uparrow} b_{\uparrow} - b_{\uparrow} a_{\uparrow} \right) \vert \uparrow \rangle_1 \vert \uparrow \rangle_2 + \left( a_{\uparrow} b_{\downarrow} - b_{\uparrow} a_{\downarrow} \right) \vert \uparrow \rangle_1 \vert \downarrow \rangle_2 + \left( a_{\downarrow} b_{\uparrow} - b_{\downarrow} a_{\uparrow}\right) \vert \downarrow \rangle_1 \vert \uparrow \rangle_2 + \left( a_{\downarrow} b_{\downarrow} - b_{\downarrow} a_{\downarrow} \right) \vert \downarrow \rangle_1 \vert \downarrow \rangle_2 \right]
$$

You can see that two of these four terms vanish, and the other two have the same factor in front, just with a sign change, so we end up with (note that the factor involving the $a$ and $b$ coefficients must be $1$ if the state is normalized):

$$
\vert \psi \rangle_A = \frac{1}{\sqrt{2}} \left( \vert \uparrow \rangle_1 \vert \downarrow \rangle_2 - \vert \downarrow \rangle_1 \vert \uparrow \rangle_2 \right)
$$

In other words, the antisymmetric tensor product Hilbert space has just one basis state! And therefore it has just one state, period (since the only way to change the above basis state would be to multiply it by some complex number, and that doesn't change the state physically).

What does this mean, physically? It means that, if you want to put two spin-1/2 particles into the same energy level of a potential well, there is only one way to do it. What is that way? The obvious way: give them opposite spins. Note that that is what the above state is: a state in which the two spin-1/2 particles have opposite spins. It does not specify the spins of either particle individually: it just says they have opposite spins. In other words, this state is an entangled state: neither particle has a definite state individually, only the system consisting of the two particles does.

Note also that, since the two particles must have opposite spins, there is no state in this Hilbert space that corresponds to your example, where the two spins only differed by a small amount. What the above is telling us is that this is not physically possible for two spin-1/2 fermions in the same energy level of a potential well.

I'll stop here and let you digest the above. But you should understand it thoroughly before we try to go on to more complicated examples.

 

[Isaac0427]

Ah, this makes perfect sense now. And, I can see how this would apply to any situation where there is a discrete set of basics states, like the first two energy levels (for bound states) of a potential well.

So, I see that the spin up state and the spin down state are orthogonal. It would appear that any other two orthogonal states would provide the same results (i.e. having $| \uparrow \rangle \longrightarrow \frac{1}{\sqrt{2}}\left ( | \uparrow \rangle + | \downarrow \rangle \right )$ and $| \downarrow \rangle \longrightarrow \frac{1}{\sqrt{2}} \left ( | \uparrow \rangle - |\downarrow \rangle \right )$). Is this correct?

Thank you very much for all the help.

 

[PeterDonis]

It would appear that any other two orthogonal states would provide the same results

Yes, you can pick any two orthogonal single particle spin states and use them as a basis for the single particle Hilbert space, and things will work out the same way.

 

[vanhees71]

Well, I do see where I got confused. Griffiths did say that you could pretend that the particles are distinguishable for practical matters. I’m assuming then that it is not really the limit of the identical fermion equation, it is just another way look at the situation that yields the same practical result.

Hm, Griffiths seems to confuse students often with his quantum mechanics textbook. It seems not as good as his electrodynamics book then? I don't know his book, so I can't tell.

First of all one should indeed distinguish two things first: According to all what we know in Nature there are to kinds of indistinguishable particles only, namely bosons and fermions, for which many-body states are completely symmetric for bosons (antisymmetric for fermions) under exchange of two particles. There are esceptions for certain "quasiparticles" in interesting effectively two-dimensional quantum many-body systems, but let's leave out this somewhat exotic topic here. As far as the elementary particles are concerned there are indeed only bosons or fermions (at least as far as we know and as we more than successfully describe by the Standard Model).

Then there is spin, and according to the mathematics of the rotation group SO(3) and its covering group, the SU(2), there are half-integer and integer spin realizations. At a first glance, this seems to have nothing to do with whether a given particle with spin is a boson or fermion, but in relativistic quantum field theory, making some quite intuitive assumptions (locality, microcausality, boundedness of the Hamiltonian from below and the existence of a stable ground state) implies that for the proper realization of the proper orthochronous Poincare group for both massless and massive particles you are forced to quantize integer-spin (half-integer-spin) fields as bosons (fermions) to fulfill all these assymptions, which in turn lead to a physically meaningful particle interpretation (in the sense of asymptotic free states), a unitary S-matrix, and the linked-cluster principle. Indeed, all empirical evidence shows that half-integer-spin particles are always fermions and integer-spin particles are always bosons.

For the details see Weinberg, Quantum Theory of Fields, vol. I, which is a tough book, but it really provides the complete insight why QFT is the way it is!