Quantum Mechanics - Addition of Angular Momentum

In summary, the conversation discusses the concept of addition of angular momentum and the possible values for J and M. It also touches upon the difference between identical and non-identical particles in terms of the symmetry of the wave function. The conversation ends with a question about constructing states with the lowest energy.
  • #1
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Just when I think I've understood addition of angular momentum, I find a problem that completely questions everything I think I know. Okay so here's the problem: question 3/II/32D on page 68 http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken] .

For the possible values of J and M as I understand it, [tex]j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)[/tex] and therefore for two j=1 systems surely J=2,1, or 0 and M=-2,-1,0,1,2?

Thus we must find expressions for the states [tex]|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~[/tex]?

So letting [tex]|1~m_1\rangle |1~m_2\rangle=|m_1\rangle|m_2\rangle[/tex] where [tex]m_i=\pm 1[/tex], we start with the top state J=2:

[tex]|2~2\rangle=|1\rangle|1\rangle[/tex]

Then apply [tex]J_-[/tex] to get [tex]|2~ 1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle+|1\rangle|0\rangle)[/tex].

Orthog. combination gives us a J=1 state: [tex]|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)[/tex]. But if we try and apply [tex]J_+[/tex] to this in order to get [tex]|1 ~2\rangle[/tex] we just get zero, why? How do we get [tex]|1 ~2\rangle[/tex]?
 
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  • #2
Tangent87 said:
Just when I think I've understood addition of angular momentum, I find a problem that completely questions everything I think I know. Okay so here's the problem: question 3/II/32D on page 68 http://www.maths.cam.ac.uk/undergrad/pastpapers/2005/Part_2/list_II.pdf [Broken] .

For the possible values of J and M as I understand it, [tex]j_1+j_2\geq J\geq|j_1-j_2|,~j_1+j_2\geq M\geq-(j_1+j_2)[/tex] and therefore for two j=1 systems surely J=2,1, or 0 and M=-2,-1,0,1,2?
That's not quite correct. First you find the allowed values for J. Then for each value of J, M runs from -J to +J. So for this problem, since you're interested in J=1, you only have the states |1 1>, |1 0>, and |1 -1>. There are no other J=1 states (so it should make sense to you why you got 0 when you applied J+ below).
Thus we must find expressions for the states [tex]|1~ 2\rangle ,~|1 ~1\rangle ,~|1 ~0\rangle ,~|1 ~-1\rangle ,~and |1 ~-2\rangle ,~[/tex]?

So letting [tex]|1~m_1\rangle |1~m_2\rangle=|m_1\rangle|m_2\rangle[/tex] where [tex]m_i=\pm 1[/tex], we start with the top state J=2:

[tex]|2~2\rangle=|1\rangle|1\rangle[/tex]

Then apply [tex]J_-[/tex] to get [tex]|2~ 1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle+|1\rangle|0\rangle)[/tex].

Orthog. combination gives us a J=1 state: [tex]|1 ~1\rangle=\frac{1}{\sqrt{2}}(|0\rangle|1\rangle-|1\rangle|0\rangle)[/tex]. But if we try and apply [tex]J_+[/tex] to this in order to get [tex]|1 ~2\rangle[/tex] we just get zero, why? How do we get [tex]|1 ~2\rangle[/tex]?
 
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  • #3
vela said:
That's not quite correct. First you find the allowed values for J. Then for each value of J, M runs from -J to +J. So for this problem, since you're interested in J=1, you only have the states |1 1>, |1 0>, and |1 -1>. There are no other J=1 states (so it should make sense to you why you got 0 when you applied J+ below).

Ah I see, yes that makes sense. Also I don't understand the second paragraph of the question, I guess I need to understand what's the difference between identical and non-identical particles?
 
  • #4
Right. It has to do with the Pauli exclusion principle and the symmetry of the wave function.
 
  • #5
vela said:
Right. It has to do with the Pauli exclusion principle and the symmetry of the wave function.

Ok I have read this article http://en.wikipedia.org/wiki/Identical_particles and I think I'm starting to understand. So if the particles are identical the state must be symmetric since the antisymmetric state would be zero. And if the particles are non-identical the state can be either symmetric or anti-symmetric? Is that logic correct?
 
  • #6
Yes, that's right.
 
  • #7
vela said:
Yes, that's right.

So then how do we construct the states, we're given [tex]\psi_i(x)[/tex] and told that we have to construct states of lowest energy (i.e. [tex]E_1[/tex]) so do we only care about [tex]\psi_1^A(x)[/tex] and [tex]\psi_1^B(x)[/tex]?
 

1. What is angular momentum in quantum mechanics?

Angular momentum in quantum mechanics is a property of a quantum system that describes its rotational motion. It is related to the rotational symmetry of the system and is quantized, meaning it can only take on certain discrete values.

2. How is angular momentum added in quantum mechanics?

In quantum mechanics, angular momentum is added using the rules of vector addition. This means that the individual angular momenta of each particle or subsystem are combined to give the total angular momentum of the system.

3. What is the total angular momentum of a system with multiple particles?

The total angular momentum of a system with multiple particles is the sum of the individual angular momenta of each particle. This total angular momentum is a conserved quantity, meaning it remains constant over time unless an external torque is applied.

4. Can the total angular momentum of a system change?

The total angular momentum of a system can only change if an external torque is applied. In quantum mechanics, this external torque can come from interactions with other systems or from the application of external forces.

5. How is angular momentum quantized in quantum mechanics?

Angular momentum in quantum mechanics is quantized because it can only take on discrete values. This is due to the inherent nature of quantum systems, which can only exist in certain energy states. The quantization of angular momentum is described by the laws of quantum mechanics and is a fundamental aspect of the theory.

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