Addition of angular momentum - Finding the second tower states

In summary, Homework Equations state that: $$J_|j, j_z>=\hbar \sqrt{j(j+1)-j_z(j_z-1)}|j,j_z-1>$$ $$S_|s, s_z>=\hbar \sqrt{s(s+1)-s_z(s_z-1)}|s,s_z-1>$$ $$L_|j, j_z>=\hbar \sqrt{l(l+1)-l_z(l_z-1)}|l,l_z-1>$$
  • #1
DataGG
Gold Member
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22

Homework Statement



I'm supposed to calculate all the states for a system with ##l=1## and ##s=1/2##. Let's say ##\vec{J} = \vec{L} + \vec{S}##. I want to find the Klebsch-Gordon coefficients.

I know that said system has 2 towers, one with ##j=3/2## and the other with ##j=1/2##. I've calculated all the states for ##j=3/2## but now I'm having some problems with ##j=1/2##.

So, for the second tower, we've two states: ##|j,j_z>=|1/2, 1/2>## and ##|j,j_z>=|1/2, -1/2>##

How am I supposed to find ##|j,j_z>=|1/2, 1/2>##? If I do that, finding ##|j,j_z>=|1/2, -1/2>## should be easily done by applying the ##J _## operator.

Homework Equations



$$J _ |j, j_z>=\hbar \sqrt{j(j+1)-j_z(j_z-1)}|j,j_z-1>$$
$$S _ |s, s_z>=\hbar \sqrt{s(s+1)-s_z(s_z-1)}|s,s_z-1>$$
$$L _ |j, j_z>=\hbar \sqrt{l(l+1)-l_z(l_z-1)}|l,l_z-1>$$

The Attempt at a Solution



Well, I've done well for the tower with ##j=3/2##. Now with this second tower, I don't know where to begin from. I think this is because for ##j = j_z = 3/2##, we know that ##j_z = l_z + s_z## which means ##l_z =1 ## and ##s_z = 1/2##. There's no other way.

For the case with ##j=1/2##, we've two options. ## l_z=0, s_z=1/2##, which is to say ##|l_z,s_z>=|0, 1/2> ## and ##l_z=1, s_z=-1/2##, which is to say ##|l_z,s_z>=|1, -1/2>##. Should I sum those states somehow?
 
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  • #2
You already know that one of the jz = 1/2 states is part of the j=3/2 representation. Since there are only two of those, which is the remaining one?
 
  • #3
Orodruin said:
You already know that one of the jz = 1/2 states is part of the j=3/2 representation. Since there are only two of those, which is the remaining one?

I'm not sure I understand what you're saying.. I know that, for ##j_z = 1/2## there's two states. One being for ##j=3/2## and the other for ##j=1/2##. That is:

##|j, j_z> = |3/2, 1/2>## and ##|j, j_z> = |1/2, 1/2>##. Now I need to write this last state using ##l_z## and ##s_z##, in order to find the Klebsch-Gordon coefficients.
 
  • #4
Yes, I am fully aware of that. What I am saying is that you know what the state with ##j_z = 1/2## and ##j = 3/2## is, since you have already computed the states with ##j = 3/2##. The state you are searching for must be orthogonal to this.
 
  • #5
Orodruin said:
Yes, I am fully aware of that. What I am saying is that you know what the state with ##j_z = 1/2## and ##j = 3/2## is, since you have already computed the states with ##j = 3/2##. The state you are searching for must be orthogonal to this.

Oh! I forgot that. I don't know why they need to be orthogonal though, but I guess that's a discussion for another thread. I'll see if I can solve it having that in mind!

Thank you Orodruin!
 
  • #6
DataGG said:
I don't know why they need to be orthogonal though

What can you say about the operator ##\hat J = \hat L + \hat S##?
 

1. What is the addition of angular momentum?

The addition of angular momentum is a mathematical concept used to determine the total angular momentum of a system composed of two or more objects with individual angular momenta. It is analogous to the addition of linear momentum in classical mechanics.

2. How is the second tower state calculated?

The second tower state is calculated by finding the product of the individual angular momenta of the objects in the system and then adding them together using the rules of vector addition. This results in a new angular momentum vector that represents the combined angular momentum of the system.

3. What is the significance of the second tower state?

The second tower state represents the maximum possible angular momentum that can be obtained by a system with given individual angular momenta. It is an important concept in quantum mechanics and is used to determine the allowed energy levels and states of a system.

4. How is the second tower state different from the first tower state?

The first tower state is the sum of the individual angular momenta of the objects in the system without considering the rules of vector addition. The second tower state takes into account the direction and orientation of the individual angular momenta, resulting in a more accurate representation of the total angular momentum of the system.

5. Can the second tower state ever be greater than the first tower state?

No, the second tower state can never be greater than the first tower state. This is because the rules of vector addition dictate that the sum of two vectors can never be greater than the magnitude of either vector alone. Therefore, the second tower state is always equal to or less than the first tower state.

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