What is the Number of Spin s States for Two Identical Particles?

[Xyius]

Number of Spin "s" States

Homework Statement

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 let's 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.

 

[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.

 

[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 let's 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..

<br /> \begin{pmatrix}<br /> 3 \\ 1<br /> \end{pmatrix}<br /> +<br /> \begin{pmatrix}<br /> 3 \\ 2<br /> \end{pmatrix}<br /> +<br /> \begin{pmatrix}<br /> 3 \\ 3<br /> \end{pmatrix}<br /> =<br /> 3+3+1=7<br />

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&gt;,|1,1&gt;,|-1,-1&gt;

So all together would be 10 symmetric states.

Is this thinking correct?

 

[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.

 

[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&gt;,|s-1,s-1&gt;,\dots.|-s,-s&gt; 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 $(2s+1)+\frac{1}{2}(2s+1)!$

 

[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.

 

[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..

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

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!)

 

[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.

 

[Xyius]

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