# QM Spin of bound state

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Hey guys,

Homework question:

Two exotic particles of spins 1 and 3/2 form a bound state.

a) What are the possible values of the total spin of the bound state?

--

So, I know that:

$s_{1}=3/2$

$s_{2}=1$

$S = (s_{1}+s_{2}) , (s_{1}+s_{2}-1) ... |s_{1}-s_{2}| = 5/2, 3/2, 1/2$

Also, 'm', is defined according to what I've read as: m=-s, -s+1,... +s

So in this case, m would be:

$m_{1} = -3/2, -1/2, 1/2, 3/2$

$m_{2} = -1, 0, 1$

Which is where I am getting stuck. My m1 and m2 have different number of components. When I compose the possible states, how does this work out??

$|s m> = \sum_{m1+m2=m} C_{m_{1}m_{2}m}^{s_{1}s_{2}s} |s_{1} m_{1} > | s_{2} m_{2}>$

So I'm not sure how to proceed. I checked examples in the book, and there's one about two particles of spins 1 and 3/2, where they go at it from "the other way" (if you already know your m1 and m2). Here is what they say:

$|\frac{3}{2} \frac{1}{2}| 1 0> = \sqrt{\frac{3}{5}}|\frac{5}{2} \frac{1}{2}> + \sqrt{\frac{1}{15}}|\frac{3}{2} \frac{1}{2}> + \sqrt{\frac{1}{3}}|\frac{1}{2} \frac{1}{2}>$

If you put particles of spin 3/2 and 1 in the box, and you know that the first has m1=1/2 and the second has m2=0 (so m is necessarily 1/2) and you measure the total spin, s, you could get 5/2 (wit probability 3/5), or 3/2 (with probability 1/15), or 1/2 (with probability 1/3). Again, the sum of the probabilities is 1.

In this example, though, they seem to be picking the values they want to use of m1 and m2. My question asks for all values in general. How do I pair up m1 and m2 that don't have the same number of components??

I'm stuck. Help!

~moo

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Spin 1 and spin 3/2 can combine to give spin 5/2, 3/2 and 1/2 (correct)

They can align (5/2) anti-align(1/2) and have the spin-1 give zero projection (3/2)

m is bounded by ±S so the magnetic sublevels will take all values from -5/2 to 5/2.

(Clebsch-Gordan coefficients are buried somewhere in here

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Spin 1 and spin 3/2 can combine to give spin 5/2, 3/2 and 1/2 (correct)

They can align (5/2) anti-align(1/2) and have the spin-1 give zero projection (3/2)

m is bounded by ±S so the magnetic sublevels will take all values from -5/2 to 5/2.

(Clebsch-Gordan coefficients are buried somewhere in here

Yeah I have the coefficients in my book.

So I'm going to have a huge list of possible eigenstates and eigenvalues? Not that I might doing the work, I just thought it looked.. weird.

Merged post follows:

Consecutive posts merged

m is bounded by ±S so the magnetic sublevels will take all values from -5/2 to 5/2.

Wait, this just sunk in -- my book defines m=m1+m2. Wouldn't that make m=-3/2, -1, -1/2, 0, 1/2, 1, 3/2 ?

I'm so confuuuuused.

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Wait, this just sunk in -- my book defines m=m1+m2. Wouldn't that make m=-3/2, -1, -1/2, 0, 1/2, 1, 3/2 ?

No, for example, 3/2 + 1 = 5/2.

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MMm.. I don't know why I 'm so confused. I'll keep going and see if I can maybe frame my confusion better.

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m = m1 + m2. You have m1 = -3/2, -1/2, 1/2, 3/2 and m2 = -1, 0, 1. You can get to any combination with a half-integral value between -5/2 and 5/2 with those. If you look at a table of Clebsch-Gordan coefficients, it will list all of the combinations of states. You should be able to see that all of these magnetic sublevels are represented.

(Yeah, it's confusing. I didn't do this until Grad school)

m= 0 isn't possible, since m1 has no m=0 state. m = ±1 isn't possible, either — all of the m1 states are half-integral

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So, wait, m isn't the combination of m1 and m2 as combining the sets.. it sounds to me like m1 is int jumps between (+/-)s1, m2 is between (+/-) of s2 and m is between the (+/-) of the total spin s.

So to get m, I don't need to even think of m1 and m2, I just need to get s (by doing s1+s2 = 5/2). And then the m of the bound state is between (-/+)s =(-/+)5/2 in integer jumps.

So, m = -5/2, -3/2, -1/2, 1/2, 3/2, 5/2

Is that right?

Err, confusion grows. s isn't just 5/2.

$s= (s1+s2) , (s1+s2-1) ... |s1-s2|$

My combined s, then, is

$s= 5/2, 3/2, 1/2$

Do I have an m *per* value of S then?

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Hey guys,

Homework question:

Two exotic particles of spins 1 and 3/2 form a bound state.

a) What are the possible values of the total spin of the bound state?

--

So, I know that:

$s_{1}=3/2$

$s_{2}=1$

$S = (s_{1}+s_{2}) , (s_{1}+s_{2}-1) ... |s_{1}-s_{2}| = 5/2, 3/2, 1/2$

Also, 'm', is defined according to what I've read as: m=-s, -s+1,... +s

So in this case, m would be:

$m_{1} = -3/2, -1/2, 1/2, 3/2$

$m_{2} = -1, 0, 1$

Which is where I am getting stuck. My m1 and m2 have different number of components. When I compose the possible states, how does this work out??

$|s m> = \sum_{m1+m2=m} C_{m_{1}m_{2}m}^{s_{1}s_{2}s} |s_{1} m_{1} > | s_{2} m_{2}>$

So I'm not sure how to proceed. I checked examples in the book, and there's one about two particles of spins 1 and 3/2, where they go at it from "the other way" (if you already know your m1 and m2). Here is what they say:

$|\frac{3}{2} \frac{1}{2}| 1 0> = \sqrt{\frac{3}{5}}|\frac{5}{2} \frac{1}{2}> + \sqrt{\frac{1}{15}}|\frac{3}{2} \frac{1}{2}> + \sqrt{\frac{1}{3}}|\frac{1}{2} \frac{1}{2}>$

If you put particles of spin 3/2 and 1 in the box, and you know that the first has m1=1/2 and the second has m2=0 (so m is necessarily 1/2) and you measure the total spin, s, you could get 5/2 (wit probability 3/5), or 3/2 (with probability 1/15), or 1/2 (with probability 1/3). Again, the sum of the probabilities is 1.

In this example, though, they seem to be picking the values they want to use of m1 and m2. My question asks for all values in general. How do I pair up m1 and m2 that don't have the same number of components??

I'm stuck. Help!

~moo

The spin is one [to] one and a half, duh!

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The spin is one [to] one and a half, duh!

What???

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So, wait, m isn't the combination of m1 and m2 as combining the sets.. it sounds to me like m1 is int jumps between (+/-)s1, m2 is between (+/-) of s2 and m is between the (+/-) of the total spin s.

So to get m, I don't need to even think of m1 and m2, I just need to get s (by doing s1+s2 = 5/2). And then the m of the bound state is between (-/+)s =(-/+)5/2 in integer jumps.

So, m = -5/2, -3/2, -1/2, 1/2, 3/2, 5/2

Is that right?

Yes, though you can do it both ways. Adding m1 and m2 will give you the same values (3/2 + 1 = 5/2, 3/2 + 0 = 3/2, etc.)

Err, confusion grows. s isn't just 5/2.

$s= (s1+s2) , (s1+s2-1) ... |s1-s2|$

My combined s, then, is

$s= 5/2, 3/2, 1/2$

Do I have an m *per* value of S then?

For any of the spin states, the magnetic sublevels will be limited to ±S, so if you have the S=1/2 state, m = ±1/2

The ways to get the S=1/2 state are:

a combination of s=1/2 and s=0 (±1/2 are the only possibilities for m)

or S=3/2 and S=1, but not aligned (i.e. m1 and m2 do not have the same sign), so you have |3/2,3/2>|1,-1> or |3/2,1/2>|1,0> or |3/2,-1/2>|1,1>, all of which give you |1/2,1/2>, and then three more combinations representing |1/2,-1/2>

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or S=3/2 and S=1, but not aligned (i.e. m1 and m2 do not have the same sign), so you have |3/2,3/2>|1,-1> or |3/2,1/2>|1,0> or |3/2,-1/2>|1,1>, all of which give you |1/2,1/2>, and then three more combinations representing |1/2,-1/2>

If I translate what you wrote into the equation:

$|s m> = \sum_{m1+m2=m} C_{m_{1}m_{2}m}^{s_{1}s_{2}s} |s_{1} m_{1} > | s_{2} m_{2}>$

It will become:

$|\frac{1}{2} , -\frac{1}{2} > = A |\frac{3}{2}, \frac{3}{2}>|1, -1> + B |\frac{3}{2}, \frac{1}{2}>|1, 0> + C |\frac{3}{2}, -\frac{1}{2}>|1, 1>$

Where A,B and C are the constants from the table we can find in your link and my book. -- right?

My confusion is -- how do you know that m1=3/2 (the part for "A") goes with m2=-1, and that m1=1/2 (part "B") goes with m2=0 ?

Why not m1=3/2 to go with m2=0?

I'm sorry if the questions might seem repetitive or silly, I'm trying to wrap my head around this. I get the broad subject - spin as angular momentum $I\omega$. To be honest, I think it's a combination of making mathematical sense of spin/magnetic sublevels as well as understanding the bra/ket notation.

Thanks!!

Merged post follows:

Consecutive posts merged

My confusion is -- how do you know that m1=3/2 (the part for "A") goes with m2=-1, and that m1=1/2 (part "B") goes with m2=0 ?

Why not m1=3/2 to go with m2=0?

Okay, I think the coin just dropped.

m1=3/2 and m2 must equal -1, to make m=m1+m2=1/2

when m1=-1, m2=1/2 to again make m=1/2

Which is the condition in the left-hand side 'bra/ket'.

Chuh-ching.

Don't you love it when that happens?

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If I translate what you wrote into the equation:

$|s m> = \sum_{m1+m2=m} C_{m_{1}m_{2}m}^{s_{1}s_{2}s} |s_{1} m_{1} > | s_{2} m_{2}>$

It will become:

$|\frac{1}{2} , -\frac{1}{2} > = A |\frac{3}{2}, \frac{3}{2}>|1, -1> + B |\frac{3}{2}, \frac{1}{2}>|1, 0> + C |\frac{3}{2}, -\frac{1}{2}>|1, 1>$

Where A,B and C are the constants from the table we can find in your link and my book. -- right?

My confusion is -- how do you know that m1=3/2 (the part for "A") goes with m2=-1, and that m1=1/2 (part "B") goes with m2=0 ?

Why not m1=3/2 to go with m2=0?

Then you'll get the 3/2 state, not the 1/2 state

I'm sorry if the questions might seem repetitive or silly, I'm trying to wrap my head around this. I get the broad subject - spin as angular momentum $I\omega$. To be honest, I think it's a combination of making mathematical sense of spin/magnetic sublevels as well as understanding the bra/ket notation.

Thanks!!

Not at all. It took a lot of spectroscopy study for me to get used to all of this — having to figure out why the hyperfine structure of the alkali atoms had the values they did helped see that this has actual application.

Merged post follows:

Consecutive posts merged

Okay, I think the coin just dropped.

m1=3/2 and m2 must equal -1, to make m=m1+m2=1/2

when m1=-1, m2=1/2 to again make m=1/2

Which is the condition in the left-hand side 'bra/ket'.

Chuh-ching.

Don't you love it when that happens?

You bet.

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