Collision between two particles with different spin

In summary, the conversation discusses the Hilbert space for each particle and the system, as well as the CSCO for considering the total Hamiltonian. It also mentions the initial and final states of the system and how to calculate the transition probability. The conversation also touches on energy levels and bound vs unbound states.
  • #1
Like Tony Stark
179
6
Homework Statement
Consider the following 1D scattering problem concerning two different spin particles: particle 1 (projectile) with spin ##s_{1}=\frac{1}{2}## in the state ##\ket{\frac{1}{2} \frac{1}{2}}## and particle 2 (target fixed at ##x=0##) with spin ##s_{2}=1## in the state ##\ket{1 0}##.

The interaction Hamiltonian is ##V=-\frac{\lambda}{h^2} \delta(x) \vec{s_1} \cdot \vec{s_2}##, with ##\lambda>0##.

Particle 1 approaches particle 2 from the left, with energy ##E##.

1) Determine the Hilbert space for this problem
2) Define the two CSCO (the canonical and the addition of angular momentum one) considering the total Hamiltonian
3) Determine the energies requiered to have bound states and find these states
4) Determine the energies requiered to have unbound states and find these states
5) Determine the spin of the particles after the collision
6) Compute the different transition probabilities of the projectile
Relevant Equations
##H_{total}=H_1 \otimes ... \otimes H_N##
1) The Hilbert space for each particle and the system are:
##H_1={\ket{\frac{1}{2} \frac{1}{2}}; \ket{\frac{1}{2} -\frac{1}{2}}}##

##H_2={\ket{1 1}; \ket{1 0}; \ket{1 -1}}##

##H=H_1 \otimes H_2##

2) I'm not sure what "considering the total Hamiltonian" means, but I think that the two CSCO are:

Canonical: ##{(S_1)^2, (S_2)^2, S_{1z}, S_{2z}}##
Addition: ##{(S_1)^2, (S_2)^2, S^2, S_{z}}##3)4) As for these ones, I don't know how to proceed. I'd use partial wave analysis but the thing is that I don't know what to do with the spin part.

5)6) The initial state of the system is:
##\ket{1/2; 0}##, which is a state from the total Hilbert space.

Once I know the final state of the system, I'll be able to write the final state of the particle 1 in terms of the kets from ##H_1##. Then, the transition probability will be computed evaluating the square of the inner product between each ket from ##H_1## and the final state of particle 1.

But I don't know how to calculate the final state of the system. Should I rewrite the initial state in terms of the kets from other basis?
 
Physics news on Phys.org
  • #2
The trick is to rewrite ##\vec{s}_1 \cdot \vec{s}_2## in terms of the CSCO after addition. Start by considering
$$
S^2 = (\vec{s}_1+ \vec{s}_2)^2 = (\vec{s}_1+ \vec{s}_2) \cdot (\vec{s}_1+ \vec{s}_2)
$$
Like Tony Stark said:
5)6) The initial state of the system is:
|1/2;0⟩, which is a state from the total Hilbert space.
The initial state is not an eigenstate of the Hamiltonian.
 
  • Like
Likes topsquark
  • #3
DrClaude said:
The trick is to rewrite ##\vec{s}_1 \cdot \vec{s}_2## in terms of the CSCO after addition. Start by considering
$$
S^2 = (\vec{s}_1+ \vec{s}_2)^2 = (\vec{s}_1+ \vec{s}_2) \cdot (\vec{s}_1+ \vec{s}_2)
$$
Yes, I know that ##\vec{S_1} \cdot \vec{S_2}=\frac{1}{2} [S^2-(S_1)^2-(S_2)^2]##. That means that the energy levels are:

$$E=-\frac{\lambda}{2h^2} \delta(x) [s(s+1)-s_1(s_1+1)-s_2(s_2+1)]$$

$$E=-\frac{\lambda}{2h^2} \delta(x) [s(s+1)-\frac{11}{4}]$$

with ##s=\frac{1}{2}, \frac{3}{2}##, ##s_1=\frac{1}{2}## and ##s_2=1##.

And those are the energy levels for the states ##\ket{S M}## (addition basis). Are these the bound states? And what about the unbound ones?
 
  • #4
Like Tony Stark said:
And those are the energy levels for the states ##\ket{S M}## (addition basis). Are these the bound states? And what about the unbound ones?
In this case you can define a bound state to be one that the interaction energy is greater than the sum of each individual particle's energy and likewise an unbound state has the sum of individual energies greater than the interaction energy.
 
  • Like
Likes DrClaude and topsquark

1. What is spin in particle collisions?

Spin is an intrinsic property of particles that describes their angular momentum. In particle collisions, spin plays a crucial role in determining the outcome of the collision and the resulting particles.

2. How does the spin of two particles affect their collision?

The spin of two particles can either be aligned (parallel) or anti-aligned (anti-parallel) when they collide. This affects the overall angular momentum of the system and can result in different types of interactions, such as elastic or inelastic collisions.

3. Can particles with different spin values collide?

Yes, particles with different spin values can collide. However, the outcome of the collision will depend on the relative spin orientations of the particles and the conservation of angular momentum.

4. What happens when particles with opposite spin collide?

When particles with opposite spin collide, they can either scatter or annihilate each other, depending on the specific particles involved and the energy of the collision. This is due to the conservation of spin angular momentum.

5. How is spin conservation applied in particle collisions?

Spin conservation is one of the fundamental principles in particle collisions. It states that the total spin angular momentum of the system before and after the collision must remain the same. This is used to predict the outcomes of collisions and to study the properties of particles.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
805
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
890
Replies
8
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Replies
0
Views
149
Replies
16
Views
556
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
1K
Back
Top