# Number of Spin s States

1. Apr 10, 2014

### Xyius

Number of Spin "s" States

1. The problem statement, all variables and given/known data

For a system of two identical particles with spin s, determine the number of symmetric
and anti-symmetric spin states.

2. The attempt at a solution

This does not seem like a problem that is that difficult, but I am having some trouble with it.

I know that for two spin 1/2 particles, the four different possible states are..

1.$$|1/2,1/2>$$
2. $$|-1/2,-1/2>$$
3. $$\frac{1}{\sqrt{2}}[|1/2,-1/2>+|-1/2,1/2> ]$$
4. $$\frac{1}{\sqrt{2}}[|1/2,-1/2>-|-1/2,1/2> ]$$

Where states 1, 2, and 3 are symmetrical states, and state 4 is antisymmetrical.

Each particle also has a space function associated with it, $\Phi(\vec{x}_1)$ and $\Phi(\vec{x}_2)$. I know that If the space function is antisymmetric, the spin part must be symmetric (and vice versa).

So if the problem were asking for two particles with spin 1/2, then the total amount of symmetric spin states would be 3 and the total number of antisymmetric spin states would be 1.

So now lets generalize to two spin "s" particles.

The first thing I am confused about is that I have never dealt with any spin value besides 1/2. So say I have a spin 1 particle. Would the only spin values be +1 and -1 just like the 1/2 case? Or would it be +1,0,-1? I am pretty sure it would be +1,0,and -1. If not, then the possible states will look the same as the spin 1/2 case, except 1/2 is replaced be s. But I don't know if this is correct.

2. Apr 10, 2014

### tman12321

You are correct in that, for the case of a spin 1 particle, its quantum number along the z-axis say, m, can take the values +1, 0, and -1. In general the allowed m for a spin s particle are s, s-1, ... , 1, 0, -1, ... , -s+1, -s. Thus there are 2s+1 allowed values of m. This means that two spin s particles should have (2s+1)^2 total composite states, each of which are eigenstates of Sz, say. Your job is to find how many symmetric and antisymmetric linear combinations of these states there are.

3. Apr 11, 2014

### Xyius

Well one thing I know right away, is that any state kept where both particles are in the same state will be symmetric. In other words..

$$|s,s>,|s-1,s-1>, \dots |-s,-s>$$

Are all symmetric states because when I switch particle 1 and 2, I get the same state. So that tells me right away that I have at LEAST 2s+1 symmetric states. However, this does not include the linear combinations.

The linear combinations are where I am having some confusion.

So lets go back to a spin 1 particle. We have six possible combinations where the two spins aren't equal..

$$|-1,0>,|-1,1>,|0,1>$$
$$|0,-1>,|1,-1>,|1,0>$$

In order to make a symmetric or anti-symmetric linear combination, we must have.. (just one example)

$$|1,0>\pm |0,1>$$

Because when we switch particle 1 or 2, we have either the same state (symmetric) or the negative of the original state (anti-symmetric).

But all together we can have..

$$(|-1,0> \pm |0,-1>)+(|-1,1> \pm |1,-1>)+(|0,1> \pm |1,0>)$$

So would I need to count this expression as only 1 combination? Or do I count each term in this expression as one combination as wel as the whole expression itself? If it is the latter, would the total number of combinations be..

$$\begin{pmatrix} 3 \\ 1 \end{pmatrix} + \begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 3 \\ 3 \end{pmatrix} = 3+3+1=7$$

So the total anti-symmetric states in this particular case would be 7.

For the symmetric case, we have 7, plus the terms that have the same spin. So..

$$|0,0>,|1,1>,|-1,-1>$$

So all together would be 10 symmetric states.

Is this thinking correct?

4. Apr 11, 2014

### tman12321

For two spin 1 particles, there are only nine possible composite basis states. The sum of the number of symmetric and anti-symmetric states cannot be more than this. Your long expression (under "But all together we can have...") is not one combination, but six. To see this, realize that three +/- can vary independently. Add to this the three symmetric states you found (having the same m for both particles, say m1=m2), and you get nine states. Similarly you could have used the three sets of +/- combinations for a total of six. (These are easier to work with.) Add to this the three m1=m2 symmetric states again, and you find nine total states. You are very close to getting the answer. Now just generalize this to two spin s particles. Another minor detail is that the states should be normalized, so you are missing constants out front.

5. Apr 11, 2014

### Xyius

So the total number of antisymmetric states would be 3, and the total number of symmetric states would be 6?

I forgot to mention I was omitting the normalization constants for simplicity.

Okay so for two spin "s" particles, we have the following $m_1=m_2$ states..

$$|s,s>,|s-1,s-1>,\dots.|-s,-s>$$ So we have at least $2s+1$ symmetric states so far.

So for the s=1 particles, the total number of linear combinations was $(2(1)+1)!$. Generalizing, for a spin "s" particle, the total number of linear combinations is $(2s+1)!$

From the $3!$ linear combinations, half were symmetric, and half were antisymmetric. Generalizing, the total number of anti-symmetric states would be $\frac{1}{2}(2s+1)!$ and the total number of symmetric states would be $(2s+1)+\frac{1}{2}(2s+1)!$

6. Apr 11, 2014

### tman12321

No, there are no factorials. (What is your justification for the factorial?) For s>=3/2, we have (2s+1)! > (2s+1)^2 already, which cannot be. You have (2s+1)^2 total basis states. Again, the sum of the number of symmetric and anti-symmetric states cannot be more than this. (2s+1) of these are symmetric right off the bat, i.e. those with m1=m2. This leaves you with (2s+1)^2 - (2s+1) states, some of which are symmetric and some of which are anti-symmetric. Figure out how many there are of each.

7. Apr 11, 2014

### Xyius

I guess the reason why I was so quickly to put factorials in there is because I heard my professor mention a factorial relationship somewhere in the lectures. Now that I think about it though, it was for a different topic.

Let's try this again. This time I start with factorials but end up not having them in the final expression.

So for a particular value of s, there will be 2s+1 values it can take on. Since there are two particles, the total number of linear combinations that are possible is $\begin{pmatrix} 2s+1 \\ 2 \end{pmatrix}$ This simplifies to..

\begin{align} \begin{pmatrix} 2s+1 \\ 2 \end{pmatrix}=\frac{(2s+1)!}{(2s+1-2)!}=\frac{(2s+1)(2s)(2s-1)!}{(2s-1)!}=(2s+1)(2s) \end{align}

So for $s=\frac{3}{2}$, we have a total of 12 linear combinations, and 4 $m_1=m_2$ states making a total of 16, or $4^2$. For s=1 we have 6 linear combinations, with three $m_1=m_2$ states making a total of 9 states or $3^2$, and for $s=\frac{5}{2}$, we have 30 linear combinations, and 6 $m_1=m_2$ states, making a total of 36 states or $6^2$.

So after seeing these three examples, I think this answer is correct now. So just like in the s=1 case, half of these combinations will be symmetric and the other half will be anti-symmetric, in other words.

Total number of Symmetric combinations = $(2s+1)+s(2s+1)=(2s+1)(s+1)$
Total number of Anti-Symmetric Combinations = $s(2s+1)$

Adding these two together we get..

$(2s+1)(s+1)+s(2s+1)=(2s+1)(s+1+s)=(2s+1)(2s+1)=(2s+1)^2$

Which is the total number of basis states! (Hope this is correct!)

8. Apr 11, 2014

### tman12321

I think that you are mostly correct, i.e. I believe your final result is true. There are a few minor points however. Your use of the binomial coefficient is incorrect. 2s+1 choose 2 notation means (2s+1)!/[2!*(2s+1-2)!]. This gives the number of ways of choosing two elements from 2s+1 elements. However, you correctly calculated the number of permutations, i.e. (2s+1)!/(2s+1-2)!. Also, I believe it is easier to see that (2s+1)^2-(2s+1) = (2s+1)2s+(2s+1)-(2s+1) = (2s+1)2s. Half of these are symmetric and half are anti-symmetric. Just little details though.

9. Apr 11, 2014

### Xyius

I see your points! Thanks a lot for all your help!