# I Physical eigenstates of systems of n particles of spins sᵢ?

Tags:
1. Aug 17, 2016

### tomdodd4598

I am relatively well versed when it comes to systems of spin, or doing the maths for them at least, but am unsure whether all of the {L2, Lz, (other required quantum numbers)} basis eigenstates for a general system of n particles of spins si, where si is the spin of the ith particle, can actually exist in nature. I am new to the concept and therefore don't know the full ins and outs of requiring to symmetrise or antisymmetrise wave functions depending on whether you're dealing with bosons or fermions, and I can only imagine this places restrictions on the spins the particles can have. It's also possible the n particles may contain both bosons and fermions, and in that case I'm even more clueless. I also understand whether the particles are distinguishable or not plays a major role, and whether, for example, this is assumed or not in the example below.

For example, suppose I had three particles, two of spin 1/2 and one of spin 1. The eigenstates of L2, and Lz, |s,m>, are |2,2>, |2,1>, |2,0>, |2,-1>, |2,-2>, |1,1>1, |1,0>1, |1,-1>1, |1,1>2, |1,0>2 and |1,-1>2 (an additional quantum number is needed to distinguish between the |1,m> states).
Of these, which ones could actually exist, or could some groups of them be realised in different scenarios?

2. Aug 17, 2016

### Staff: Mentor

If particles have different spins, then they are definitely not identical, and you can treat them independently.

In the case of the two spin-1/2, you get the "classic" singlet + triplet states. The singlet state combines with all three possible states for the spin-1 particle, giving
\begin{align*} |1,1\rangle_3 &= | 0, 0 \rangle_{1/2} \otimes |1,1\rangle_1 \\ |1,0\rangle_3 &= | 0, 0 \rangle_{1/2} \otimes |1,0\rangle_1 \\ |1,-1\rangle_3 &= | 0, 0 \rangle_{1/2} \otimes |1,-1\rangle_1 \\ \end{align*}
(the index indicates whether it is the 3-body state, the state of the two spin-1/2 particles, or the state of the spin-1).

For the triplet of the spin-1/2, each state of the triplet combines with the three states of the spin-1. According to the rules of addition of angular momenta, for $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$, the allowed values for $S$ are
$$S = S_1 + S_2, S_1 + S_2-1, \ldots, \left| S_1 - S_2 \right|$$
which in this case gives $S = 2, 1, 0$. So the three-body states will be $|2,2\rangle_3$, $|2,1\rangle_3$, $|2,0\rangle_3$, $|2,-1\rangle_3$, $|2,2-\rangle_3$, $|1,1\rangle_3$, $|1,0\rangle_3$, $|1,-1\rangle_3$, $|0,0\rangle_3$. (You were missing that last one in the OP.) These states can be expressed in terms of the spin-1/2 and spin-1 states using the proper Clebsch-Gordan coefficients.

3. Aug 17, 2016

### tomdodd4598

Ah, yes, I did miss the |0,0> state - thanks. I wrote a Mathematica script a month or so ago that can give me the set of orthogonal states, and this is what I get for two spin-1/2 and one spin-1, where, assuming particles 1 and 2 are the spin-1/2 particles and particle 3 is the spin-1 particle,
the first component of the vector is the probability amplitude for finding particle 1 with m=1/2, particle 2 with m=1/2 and particle 3 with m=1,
the second component is the p.a. for finding particle 1 with m=1/2, particle 2 with m=1/2 and particle 3 with m=0,
the third component is the p.a. for finding particle 1 with m=1/2, particle 2 with m=1/2 and particle 3 with m=-1,
the second component is the p.a. for finding particle 1 with m=1/2, particle 2 with m=-1/2 and particle 3 with m=1, etc:

|2,2> = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
|2,1> = [0, 1/√2, 0, 1/2, 0, 0, 1/2, 0, 0, 0, 0, 0]
|2,0> = [0, 0, 1/√6, 0, 1/√3, 0, 0, 1/√3, 0, 1/√6, 0, 0]
|2,-1> = [0, 0, 0, 0, 0, 1/2, 0, 0, 1/2, 0, 1/√2, 0]
|2,-2> = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
...
|0,0> = [0, 0, 1/√3, 0, -1/√6, 0, 0, -1/√6, 0, 1/√3, 0, 0]

So I've found all of the spin eigenstates, but if we now also think about spatial wave functions, if the spin-1/2 particles are identical, the wave function has to be anti-symmetric with respect to swapping the two particles. Doesn't that mean that the only possibilities are that the spatial part is symmetric and the spin part is anti-symmetric or that the spatial part is anti-symmetric and the spin part is symmetric?

4. Aug 18, 2016

### Staff: Mentor

That's correct.

5. Aug 18, 2016

### tomdodd4598

Ok, I see now. I'm assuming then, that, when two bosons are exchanged, the whole wave function needs to be symmetric (so both the spatial and spin need to be symmetric anti-symmetric with respect to swapping them), and if there are no indistinguishable particles, then there's no restriction of this sort. Thanks :)