Number of Angular Momentum States (3 particles)

[schlegelii]

Homework Statement

You have three particles with angular momentum. How many states are there for each of the total angular momentum values (3, 2, 1, 0)?

Hint: Start by coupling two particles with angular momentum 1 and find how many states each of the possible angular momentum values have. Then couple the third particle to the previous two states.

Relevant Equations

$j = \vert j_1 + j_2 \vert , \vert j_1 + j_2 - 1\vert, ..., \vert j_1 - j_2 \vert$
$m = -j , -j+1, ..., j$

I can solve the two particle system easily enough:

Using $j_1 = 1$ and $j_2 = 1$, the possible total angular momentum values are $j = 2, 1, 0$. With $m = -j , -j+1, ..., j$,
$j = 2: m = 2, 1, 0, -1, -2$ (5 states)
$j = 1: m = 1, 0, -1$ (3 states)
$j = 0: m = 0$ (1 state)

I can imagine the second part as if there were only two particles:
For $j_3 = 1$ and $j(combined) = 2$,
$j = 3: m = 3, 2, 1, 0, -1, -2, -3$ (7 states)
$j = 2: m = 2, 1, 0, -1, -2$ (5 states)
$j = 1: m = 1, 0, -1$ (3 states)

For $j_3 = 1$ and $j(combined) = 1$, (same as above)
$j = 2: m = 2, 1, 0, -1, -2$ (5 states)
$j = 1: m = 1, 0, -1$ (3 states)
$j = 0: m = 0$ (1 state)

For $j_3 = 1$ and $j(combined) = 0$,
$j = 1: m = 1, 0, -1$ (3 states)

But I cannot figure out what to do with number of states for the three particle system. Do I multiply? Add?

I want to end up with something like:
$j(total) = 3$ gives ___ states for each of the possible $j(total)$ states but I'm really confused how to get there.

 

[PeroK]

But I cannot figure out what to do with number of states for the three particle system. Do I multiply? Add?

What you have for two particles corresponds to systems with AM 0, 1 or 2.

When you add a third particle, you can add to anyone of these.

 

[schlegelii]

What you have for two particles corresponds to systems with AM 0, 1 or 2.

When you add a third particle, you can add to anyone of these.

My initial thought was to multiply the number of states you got from the AM 0, 1, 2 systems by the number of states you got by combining them, like:
There are 5 $j = 2$ states from combining $j_1$ and $j_2$, so
$j = 3$ = 7 states * 5 states = 35 states
$j = 2$ = 5 states * 5 states = 25 states
$j = 1$ = 3 states * 5 states = 15 states

Then doing the same for the other two #jcombined# states.

I think this would give 35 AM 3 states, 40 AM 2 states, 27 AM 1 states, and 3 AM 0 states. This seems really high, so I'm not sure that it is right?

 

[DrClaude]

My initial thought was to multiply the number of states you got from the AM 0, 1, 2 systems by the number of states you got by combining them, like:
There are 5 $j = 2$ states from combining $j_1$ and $j_2$, so
$j = 3$ = 7 states * 5 states = 35 states
$j = 2$ = 5 states * 5 states = 25 states
$j = 1$ = 3 states * 5 states = 15 states

Then doing the same for the other two #jcombined# states.

I think this would give 35 AM 3 states, 40 AM 2 states, 27 AM 1 states, and 3 AM 0 states. This seems really high, so I'm not sure that it is right?

You are overcounting.

Think of what you are doing in the first step, when you consider only two particles: you don't multiply by the degeneracy of the particle's angular momentum. It's the same for the combining the total $j$ of particles 1 and 2 with $j_3$.

 

[PeroK]

My initial thought was to multiply the number of states you got from the AM 0, 1, 2 systems by the number of states you got by combining them, like:
There are 5 $j = 2$ states from combining $j_1$ and $j_2$, so
$j = 3$ = 7 states * 5 states = 35 states
$j = 2$ = 5 states * 5 states = 25 states
$j = 1$ = 3 states * 5 states = 15 states

Then doing the same for the other two #jcombined# states.

I think this would give 35 AM 3 states, 40 AM 2 states, 27 AM 1 states, and 3 AM 0 states. This seems really high, so I'm not sure that it is right?

Only the values of AM are relevant. It doesn't matter how you get that value. There may be other aspects to these particles that distinguish them, but the AM state is defined purely by its numerical values.

A total AM of $j = 2$ is the same total AM whether that's $1 + 1 + 0, \ 1 + 0 + 1$ or $0 + 1 + 1$.

Likewise the state $j = 2, m = 1$ fully defined the AM state. You don't have separate AM states with the same AM values.

 

[DrClaude]

By the way, and this is a question of nomenclature, I wouldn't say that there are 5 $j=2$ states (after adding particles 1 and 2), but one state that is 5-fold degenerate. I would call $j=2$, $m=+2$ a level, not a state. Check to make sure what nomenclature is used by your professor / textbook.

 

[PeroK]

By the way, and this is a question of nomenclature, I wouldn't say that there are 5 $j=2$ states (after adding particles 1 and 2), but one state that is 5-fold degenerate. I would call $j=2$, $m=+2$ a level, not a state. Check to make sure what nomenclature is used by your professor / textbook.

$|j \ m \rangle$ are the orthonormal basis states/vectors, surely?

Wouldn't you possibly say that $j = 2$ is a level (like an energy level) with a five-fold degeneracy?

 

[schlegelii]

You are overcounting.

Think of what you are doing in the first step, when you consider only two particles: you don't multiply by the degeneracy of the particle's angular momentum. It's the same for the combining the total $j$ of particles 1 and 2 with $j_3$.

Only the values of AM are relevant. It doesn't matter how you get that value. There may be other aspects to these particles that distinguish them, but the AM state is defined purely by its numerical values.

A total AM of $j = 2$ is the same total AM whether that's $1 + 1 + 0, \ 1 + 0 + 1$ or $0 + 1 + 1$.

Likewise the state $j = 2, m = 1$ fully defined the AM state. You don't have separate AM states with the same AM values.

Sorry, I'm still somehow confused by this.
So do I just have 7 states for AM 3, 5 states for AM 2, 3 states for AM 1, and 1 state for AM 0?
Or do I add to get 7 for AM 3, 10 for AM 2, 9 for AM 1, and 1 for AM 0? (By adding the totals I got for the states at the end?)

 

[schlegelii]

By the way, and this is a question of nomenclature, I wouldn't say that there are 5 $j=2$ states (after adding particles 1 and 2), but one state that is 5-fold degenerate. I would call $j=2$, $m=+2$ a level, not a state. Check to make sure what nomenclature is used by your professor / textbook.

Good to know! My professor definitely refers to them as states, although my textbook does not.

 

[PeroK]

Sorry, I'm still somehow confused by this.
So do I just have 7 states for AM 3, 5 states for AM 2, 3 states for AM 1, and 1 state for AM 0?
Or do I add to get 7 for AM 3, 10 for AM 2, 9 for AM 1, and 1 for AM 0? (By adding the totals I got for the states at the end?)

If you add two $j=1$ systems you get $j = 0, 1$ or $2$

When you add a third $j=1$ system, you get $j = 0, 1, 2$ or $3$.

In the end, the maximum number of states you can possibly have is $1+3+5+7$. The theory of AM explicitly shows you have only three (independent) states for $j=1$, namely $m=−1,0,1$. You can't have any more!

And, in fact, you can't have any fewer either. You just apply the raising or lowering operator and you must get the next/previous state.

 

[schlegelii]

In the end, the maximum number of states you can possibly have is $1+3+5+7$. The theory of AM explicitly shows you have only three (independent) states for $j=1$, namely $m=−1,0,1$. You can't have any more!

And, in fact, you can't have any fewer either. You just apply the raising or lowering operator and you must get the next/previous state.

This makes a lot of sense to me! Thanks for the help.

 

[DrClaude]

$|j \ m \rangle$ are the orthonormal basis states/vectors, surely?

Wouldn't you possibly say that $j = 2$ is a level (like an energy level) with a five-fold degeneracy?

Indeed, $j = 2$ would be level, not a state. Let's take state to mean a basis state.

If you add two $j=1$ systems you get $j = 0, 1$ or $2$

When you add a third $j=1$ system, you get $j = 0, 1, 2$ or $3$.

In the end, the maximum number of states you can possibly have is $1+3+5+7$.

I find this ambiguous. What do you mean by $1+3+5+7$? Because surely this is no the right number of basis states.

 

[PeroK]

Indeed, $j = 2$ would be level, not a state. Let's take state to mean a basis state.I find this ambiguous. What do you mean by $1+3+5+7$? Because surely this is no the right number of basis states.

Yes, you're right. I forgot the extra degeneracy.

 

[andresB]

I wouldn't say that there are 5 $j=2$ states (after adding particles 1 and 2), but one state that is 5-fold degenerate.

I find that nomenclature to be strange, crearly they are diffeerent vectors in the Hilbert space.

 

[DrClaude]

I find that nomenclature to be strange, crearly they are diffeerent vectors in the Hilbert space.

As I replied above, I messed up and indeed you could call the individual $|j,m\rangle$ states.

I had the nomenclature for atomic physics in mind when I wrote that, which is why I mentioned levels. I apologize to all for the ensuing confusion!

 

[schlegelii]

Yes, you're right. I forgot the extra degeneracy.

Can you explain this a little further? I thought I got it but now I'm confused again.

 

[PeroK]

Can you explain this a little further? I thought I got it but now I'm confused again.

@DrClaude is right. There must be 27 basis states, as the composite system is the tensor product of three 3D spaces. I think I led you astray somewhat.

There can only be the 16 combinations for $j, m$. That means there must be a total degeneracy of 11, based on the same $j, m$ resulting from physically different combinations.

I must admit, I don't know an easy way to calculate where these degeneracies fit in. I found this for combining three electrons, which may be useful:

https://physics.stackexchange.com/q...rticles-what-are-the-corresponding-sta/209711

There must be a way to do something similar in your case.

 

[schlegelii]

I must admit, I don't know an easy way to calculate where these degeneracies fit in. I found this for combining three electrons, which may be useful:

https://physics.stackexchange.com/q...rticles-what-are-the-corresponding-sta/209711

There must be a way to do something similar in your case.

Wow, that three electron case is surprisingly complicated.

I agree that there should be 27 states. @DrClaude suggested earlier that I should think about what I did in the first step. If you just add the number of states you get after the second step you get 7+5+3+5+3+1+3, which is 27. This is the right number and seems like a reasonable way to do it, I'm just not sure I can justify why it would work.

 

[DrClaude]

Wow, that three electron case is surprisingly complicated.

I agree that there should be 27 states. @DrClaude suggested earlier that I should think about what I did in the first step. If you just add the number of states you get after the second step you get 7+5+3+5+3+1+3, which is 27. This is the right number and seems like a reasonable way to do it, I'm just not sure I can justify why it would work.

When you combine $j_1$ and $j_2$ into $j_{12}$, you have three possible values, $j_{12} = 2, 1, 0$.

Then when you combine with $j_3$, the possibilities depend on the value of $j_{12}$:
$$
\begin{align*}
j_{12} = 2 &\Rightarrow j_{123} = 3, 2, 1 \\
j_{12} = 1 &\Rightarrow j_{123} = 2, 1, 0 \\
j_{12} = 0 &\Rightarrow j_{123} = 1
\end{align*}
$$
so you see that there are three ways to get a total angular momentum of $j_{123} = 1$, and so on.

 

[PeroK]

Wow, that three electron case is surprisingly complicated.

I agree that there should be 27 states. @DrClaude suggested earlier that I should think about what I did in the first step. If you just add the number of states you get after the second step you get 7+5+3+5+3+1+3, which is 27. This is the right number and seems like a reasonable way to do it, I'm just not sure I can justify why it would work.

I had another look at this. I worked out the 3 electron case for myself. But, in doing so I came to the conclusion that it is difficult to short-cut the process, rather than working the whole thing out.

I wonder whether the question is simply to give the 16 "levels", represented by $|j \ m \rangle$, without specifying where the 11 degeneracies are?

The fundamental issue is that the simple combined states are not eigenstates of the $J^2$ operator. They are, of course, the obvious eigenstates of the $J_z$ operator. For example, in the spin 1/2 case:

$J^2 |\chi_+ \chi_+ \chi_- \rangle = \frac{\hbar^2}{4}[7|\chi_+ \chi_+ \chi_- \rangle + 4|\chi_+ \chi_- \chi_+ \rangle + 4|\chi_- \chi_+ \chi_+ \rangle]$

The others follow from sysmmetry.

And, of course, these all have $m = \frac 1 2$.

So, we have a 3D subspace spanned by the three basis states with two ups and one down. It turns out that there is one $j = \frac 3 2$ eigenstate here and two $j = \frac 1 2$ eigenstates here.

And, the same for the case with one up and two downs.

(I can post more of what I did if you are interested.)

The link I posted shows another way using more symmetry considerations. But, that's not easy/obvious either. Also, I suspect several of the posts on that link might be wrong.

Back to your case. I may be wrong but I think that identifying the degeneracies may be difficult without significant analysis: either using the $J^2$ operator, looking at symmetries or (ultimately) applying group theory.

I'd be interested in what the answer is and if, indeed, there is a quick way to work out where the degeneracies are for the three $J=1$ case.

 

[Vanadium 50]

I am away from my copy of Edmonds, so this is all from memory.

There is something called a 6-j symbol which is the analog of the Clebsch-Gordan coefficient that will do this automatically. There is also something called a Racah W-coefficient that I have never had occasion to use, which is supposed to be simpler. You can also use Young Tableau, which is a very slick technique which I relearn every couple of years.

But I think the easiest way to answer the question - especially if I just need the number of states and not the states themselves - is to count.

It's already been established there are (3)(3)(3) = 27 total states.

The state +1,+1,+1 has J = 3. m= 3, and so mush have the other six m projections for a total number of states of 7. I need 20 more. This is called the "fully antisymmetric state". (Because all particles have the same spin. )

There's also a j=0, m=0 fully symmetric state. Now we have 19 to go.

The j=2 states have 5 projections, and the j = 1 have 3. If I have two j=2 and three j=3, I have my 19. There's no other combination that works. So, if I did this right, (1+1+1) = one J=3, two J=2, three j=1 and a singlet.

Oh, and let us not call this the "three electron case". Different problem.

 

[PeroK]

I am away from my copy of Edmonds, so this is all from memory.

There is something called a 6-j symbol which is the analog of the Clebsch-Gordan coefficient that will do this automatically. There is also something called a Racah W-coefficient that I have never had occasion to use, which is supposed to be simpler. You can also use Young Tableau, which is a very slick technique which I relearn every couple of years.

But I think the easiest way to answer the question - especially if I just need the number of states and not the states themselves - is to count.

It's already been established there are (3)(3)(3) = 27 total states.

The state +1,+1,+1 has J = 3. m= 3, and so mush have the other six m projections for a total number of states of 7. I need 20 more. This is called the "fully antisymmetric state". (Because all particles have the same spin. )

There's also a j=0, m=0 fully symmetric state. Now we have 19 to go.

The j=2 states have 5 projections, and the j = 1 have 3. If I have two j=2 and three j=3, I have my 19. There's no other combination that works. So, if I did this right, (1+1+1) = one J=3, two J=2, three j=1 and a singlet.

Oh, and let us not call this the "three electron case". Different problem.

Basically, the argument is:

$j = 3$ has $7$ projections
$j = 2$ has $5$ projections
$j = 1$ has $3$ projections
$j =0$ has $1$ projection

If we assume that every projection has equal degeneracy (within a $j$ level), then we only have one solution:

$j = 3$ has $7$ projections, each with degeneracy $1$
$j = 2$ has $5$ projections, each with degeneracy $2$
$j = 1$ has $3$ projections, each with degeneracy $3$
$j =0$ has $1$ projection, with degeneracy $1$

Although, we need the added assumption that $j = 0$ has degeneracy $1$. Otherwise, there are other solutions, of course. Perhaps it's only $j =1, 2$ that have the flexibility to have degeneracy of more than $1$?

 

[Vanadium 50]

If we assume that every projection has equal degeneracy

If that were not true there would be a preferred direction in space.

Although, we need the added assumption that j = 0 has degeneracy 1.

We do. But if there were two orthogonal j=0 solutions, when I hit them with the raising operator three times I get two orthogonal j = 3 solutions, and an even number of j=2 and j=1, so the total number of states is even, not 27. So that's out. If I have three orthogonal j = 0 solutions, I need three orthogonal j = 3 solutions, and a number of j = 1 and j = 2 solutions divisible by three. But that only works if I have zero j=1 and j=2 solutions, so my ladder operator can't get to j=3. So that's out too. Four or more and I have too many states. So there's only one.

There is likely a general theorem on this. I can tell you I have never run across a counterexample.

Edit: nonsense struck through.

 

[PeroK]

If that were not true there would be a preferred direction in space.

Yes, of course.

 

[PeroK]

We do. But if there were two orthogonal j=0 solutions, when I hit them with the raising operator three times I get two orthogonal j = 3 solutions, and an even number of j=2 and j=1, so the total number of states is even, not 27. So that's out. If I have three orthogonal j = 0 solutions, I need three orthogonal j = 3 solutions, and a number of j = 1 and j = 2 solutions divisible by three. But that only works if I have zero j=1 and j=2 solutions, so my ladder operator can't get to j=3. So that's out too. Four or more and I have too many states. So there's only one.

There is likely a general theorem on this. I can tell you I have never run across a counterexample.

The raising and lowering operators are not unitary. And, in any case, as we move up and down the $j$ levels, orthogonal states must be getting mapped to non-orthogonal states at some point. Anyway, looking at four electrons, we have 16 states to divide up between:

$j = 2$ with $5$ levels
$j = 1$ with $3$ levels
$j = 0$ with $1$ level

We must have at least one of each, which leaves 7 degeneracies. That's either a 5 and two 1's or two 3's and a 1. I'd bet on the latter:

$j = 2$ with $5$ levels, degeneracy $1$
$j = 1$ with $3$ levels, degeneracy $3$
$j = 0$ with $1$ level, degeneracy $2$

If I have time sometime I might take a look at that and see what actually comes out.

 

[Vanadium 50]

You're right, on that.

The fully antisymmetric state is unique, by inspection. Every other bit of my argument is invalid. (But I still suspect a theorem)

 

[PeroK]

You're right, on that.

The fully antisymmetric state is unique, by inspection. Every other bit of my argument is invalid. (But I still suspect a theorem)

I think an interesting side issue is how the $m$ values could combine to give the total $j$ value. E.g.

$j=1$ corresponds to $J^2 = 2\hbar^2$, and $m = 0, \pm 1$ correspond to $J_z^2 = 0, \hbar^2$.

Naively, two of $J_x, J_y, J_z$ are $\pm 1$ and one is $0$.

$j=2$ corresponds to $J^2 = 6\hbar^2$, and $m = 0, \pm 1, \pm 2$ correspond to $J_z^2 = 0, \hbar^2, 4\hbar^2$.

Naively, two of $J_x, J_y, J_z$ are $\pm 1$ and one is $\pm 2$. That makes the state $|2, 0 \rangle$ impossible. That that is not the case seems to me to be an argument against hidden variables.

$j = \frac 3 2$ is interesting as $J^2 = \frac{15\hbar^2}{4}$ and $J_z^2 = \frac{\hbar^2}{4}, \ \frac{9\hbar^2}{4}$

And there is no way to make $J^2$ from allowable values of $J_x^2, J_y^2, J_z^2$.

I guess Bohmian mechanics must have an answer to that.

 

[PeroK]

Looking at four electrons, we have 16 states to divide up between:

$j = 2$ with $5$ levels
$j = 1$ with $3$ levels
$j = 0$ with $1$ level

We must have at least one of each, which leaves 7 degeneracies. That's either a 5 and two 1's or two 3's and a 1. I'd bet on the latter:

$j = 2$ with $5$ levels, degeneracy $1$
$j = 1$ with $3$ levels, degeneracy $3$
$j = 0$ with $1$ level, degeneracy $2$

If I have time sometime I might take a look at that and see what actually comes out.

Yes, that's what comes out. There are two $|0 \ 0 \rangle$ states:

 

[Vanadium 50]

That's either a 5 and two 1's or two 3's and a 1. I'd bet on the latter

Good bet. You can't have two fives.

Suppose there were two. One includes |++++>. The other one also has to add up to m=2, but be orthogonal. Since that's not possible, you only have one five.

Can you write down the explicit composition of both singlets?

 

[PeroK]

You can't have two fives.

Suppose there were two. One includes |++++>. The other one also has to add up to m=2, but be orthogonal. Since that's not possible, you only have one five.

Can you write down the explicit composition of both singlets?

There are six states with two up and two down electrons. The two dimensional singlet subspace is defined by:

$a|++-- \rangle + b|+-+- \rangle -(a+b)|+--+ \rangle -(a+b)|-++- \rangle + b|-+-+ \rangle + a|--++ \rangle$

For example, taking $a = 0$ or $b = 0$ gives the basis:

$|+-+- \rangle -|+--+ \rangle -|-++- \rangle + |-+-+ \rangle$,
$|++-- \rangle -|+--+ \rangle -|-++- \rangle + |--++ \rangle$

Actually, they are not orthogonal. $a = 0, \ b = 1$ and $a = 1, b = -1$ is better.

 

[Vanadium 50]

Thanks. I was thinking about when one has only one singlet.

 

[PeroK]

You're right, on that.

The fully antisymmetric state is unique, by inspection. Every other bit of my argument is invalid. (But I still suspect a theorem)

Here's a symmetry argument for the three spin-1 case.

First, we analyse by $j$ values (as before):

$j = 3$ has $7$ levels
$j = 2$ has $5$ levels
$j = 1$ has $3$ levels
$j = 0$ has $1$ level

There are numerous ways to meet the condition of degeneracies, assuming a common degeneracy across each level.

Then, we analyse by $m$ value:

$m = 3$ is $1$ dimensional
$m = 2$ is $3$ dimensional (cycles of $110$)
$m =1$ is $6$ dimensional (cycles of $100$ and $11-1$
$m = 0$ is $7$ dimensional (cycles of $10-1$ and $000$)

Now, $J^2$ is symmetric and preserves the $m$-value, hence preserves the above dimensionality, in terms of $|j m \rangle$ eigenstates. That might be your theorem:

There is only one $m = 3$ state, hence $j = 3$ has degeneracy $1$ - i.e. no degeneracy (as previously established).

$m = 2$ is $3$ dimensional. One dimension is for $j = 3$, leaving two for $j = 2$. Hence $j = 2$ has degeneracy $2$.

Looking at $m = 1$ implies $j = 1$ has degeneracy $3$.

And $m = 0$ implies $j =0$ has degeneracy $1$.

Looking at the negative values gives the same answer. So, we have, as expected:

$j = 3$ has $7$ levels with degeneracy $1$
$j = 2$ has $5$ levels with degeneracy $2$
$j = 1$ has $3$ levels with degeneracy $3$
$j = 0$ has $1$ level degeneracy $1$

The same argument also works for the established result in the four-electron case.

 

[Vanadium 50]

So the theorem is that for addition of 2 or 3 integral angular momenta j, there is exactly one singlet. For more angular momenta the number of singlets is larger: ten j = 1 have 603 singlets.

For four j = 1, the number of singlets is 2j + 1. For five, it's (5j² = 5j + 2)/2.