# Spin-Statistics Theorem

bahamagreen
From Wikipedia "Spin-Statistics Theorem" General section:
Two indistinguishable particles, occupying two separate points, have only one state, not two. This means that if we exchange the positions of the particles, we do not get a new state, but rather the same physical state. In fact, one cannot tell which particle is in which position.
A physical state is described by a wavefunction, or – more generally – by a vector, which is also called a "state"; if interactions with other particles are ignored, then two different wavefunctions are physically equivalent if their absolute value is equal. So, while the physical state does not change under the exchange of the particles' positions, the wavefunction may get a minus sign.
Bosons are particles whose wavefunction is symmetric under such an exchange, so if we swap the particles the wavefunction does not change. Fermions are particles whose wavefunction is antisymmetric, so under such a swap the wavefunction gets a minus sign, meaning that the amplitude for two identical fermions to occupy the same state must be zero. This is the Pauli exclusion principle: two identical fermions cannot occupy the same state. This rule does not hold for bosons.

So,
Two indistinguishable particles... therefore one cannot tell which particle is is in which position.

Two particles have one state, not two.

Physical state described by a wavefunction (therefore the two particles are described by one wavefunction)

Two different wavefunctions are physically equivalent if absolute value is equal
(so the change in sign is ignored, I'm assuming this is a change in sign of the value of the amplitude)

Bosons; the swap does not change the wavefunction
Two indistinguishable particles... one wavefunction... the swap is the same as rotating the wave function "180 degrees" to realign with the two particles' original positions? Where the center of rotation is the midpoint between them?

There is no mention here of adding two things together, only a comparison of two things...

Fermions; the swap changes the wavefunction sign
So it would take two "180 degree" rotations to realign the points back to their original positions.
But how does change in sign of the wavefunction yield the amplitude zero? Are two wavefunctions or two wavefunction amplitudes from two locations being summed to zero? What happened to "one state, one wavefunction"? Is a single wavefunction amplitude value being derived from two individual wavefunction locations with opposite signs?

In the example given, we start with two identical fermions in the same state, no? Then swapping them gives a null result... how is this swap, or rotation, any different from shifting/rotating the position from which they are observed? If the amplitudes were of opposite sign before the swap, how are they in the same state before, but not after, with only an exchange of sign?

Ultimately, if two fermions are antisymmetric because of sign, what about a third fermion? Will it be antisymmetric with either or both of the others? If the state is redefined for three fermions, how can all three be antisymmetric with each other? What kind of rotations are implied? How are an arbitrary number of fermions all antisymmetric with each other?

Any help on clarifying this appreciated. As simple as possible, also appreciated.

Jolb
I think the reason you're getting confused is because you're trying to explain this all verbally. It makes much more sense if you look at the math.

A two-particle state is a direct product of two single-particle states. Each particle inhabits its own subspace. So the state |ψ(r1, r2)>=|ψ1(r1)>|ψ2(r2)> says that particle 1 has probability amplitude given by ψ1 at location r1, and particle 2 has probability amplitude given by ψ2 at location r2.

Remember that the wavefunction itself is not observable. The probability distribution of finding the particles is the only observable thing. Thus the state ψ(r) is physically the same as the state -ψ(r) since <ψ(r)|ψ(r)>=ψ*(r)ψ(r) is exactly equal to <-ψ(r)|-ψ(r)>=[-ψ*(r)][-ψ(r)].

So far I haven't said anything about indistinguishability or symmetry. Here is where they come in.

We define the exchange operator P as follows: P|ψ1(r1)>|ψ2(r2)> = |ψ1(r2)>|ψ2(r1)>. Basically the two particles have changed roles. If the particles are indistinguishable, then a change of roles can have no physical effect. There are then two possibilities: P|ψ>=|ψ> or P|ψ>=-|ψ>, since as I said before, |ψ> and -|ψ> are physically equivalent.

Now here is a fact that I will state without proof:
Bosons satisfy P|ψ>=|ψ> and Fermions satisfy P|ψ>=-|ψ>.
These are the symmetry requirements for bosons and fermions. The former is the symmetry requirement for bosons and the latter is the antisymmetry requirement for fermions. The spin-statistics theorem derives this fact but it requires much more advanced physics than what your question asks.

In order to satisfy these symmetry requirements, we symmetrize bosons' wavefunctions and antisymmetrize fermions' wavefunctions. For example, it is easy to see that the state |ψ1(r1)>|ψ2(r2)> is not necessarily symmetric or antisymmetric, but the state 2-1/2(|ψ1(r1)>|ψ2(r2)>+|ψ1(r2)>|ψ2(r1)>) is symmetric while the state 2-1/2(|ψ1(r1)>|ψ2(r2)>-|ψ1(r2)>|ψ2(r1)>) is antisymmetric. Thus only bosons can take the former state, and only fermions can take the latter state.

For three or more particle states, the symmetry requirements are satisfied for an exchange of any two particles. For example, in a three particle state, there are three operators P1,2, P2,3, and P3,1, and each one follows symmetry or antisymmetry. For example, for a fermion state |ψ(r1, r2, r3)>, one of the antisymmetry requirements would be that
P3,1|ψ(r1, r2, r3)>=-|ψ(r3, r2, r1)>.

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bahamagreen
Could you describe what these symbols... | > < mean in the notation?

Jolb
Sure. They're just a sloppy looking version of bra-ket notation, and they usually look more pretty like $\left | \Psi \right \rangle$ . (I was just too lazy to tex all of that.)

The symbols |, <, and > have no real meaning by themselves--they're more similar to a vector symbol (like the arrow on top of this guy: $\vec{v}$) or parentheses. A vector symbol by itself is meaningless, and parentheses by themselves are meaningless--same thing for |, >, and <. They only take on meaning when they're enclosing different entities. Mathematicians might write the same ideas differently, but the notation is handy for doing quantum physics, so even if it looks a little strange, it is useful.

|ψ> is called a "ket", and <ψ| is called a "bra". They represent an abstract quantum state, which mathematically is described as a unit vector in Hilbert space. (Technically, they sometimes also have components called "spinors" which are not the same as vectors.) Mathematically, a Hilbert space is an infinite-dimensional vector space with an L2 norm. To calculate the inner product between two objects in the space, you matrix multiply one object with the adjoint (the complex conjugate of the transpose) of the other object. The L2 norm of an object is the object's inner product with itself.

Basically, a ket |ψ> is notation for a unit vector (where the entries are complex numbers) and the bra <ψ| is notation that means the adjoint of the ket |ψ>. So |ψ> is a one-column matrix and <ψ| is a one-row matrix, where the elements are complex conjugates of one another. If you put them next to each other <ψ||ψ> the implicit operation is matrix multiplication and we drop double |'s, so <ψ||ψ>=<ψ|ψ>.

If we choose an orthonormal basis for the Hilbert space, such as the position basis, {|x>}, then the spatial wavefunction is given by
ψ(x) = <x|ψ>

The easiest way for me to think about this is to ignore that the states are infinite dimensional and simply think of the states as finite vectors. For example you might represent the ket |ψ> as the column vector [a, b, c, ...]T and its adjoint <ψ| as the row vector [a*, b*, c*, ...], where the T indicates a transpose and the stars indicate complex conjugates. Then the norm of |ψ> is
<ψ|ψ> = [a*, b*, c*, ...][a, b, c, ...]T = a*a + b*b + c*c + ...
by matrix multiplication.

This is all very important stuff for understanding the math of quantum mechanics, and it takes a while to get used to all of this. This is sort-of a crash course. So if you're not familiar with it, you're not alone. It takes a bit of study to remember all of this.

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