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
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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?
 
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  • #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.
 
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  • #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.
 
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FAQ: Collision between two particles with different spin

What is spin in the context of particle physics?

Spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Unlike classical angular momentum, spin does not correspond to any actual spinning motion but is a quantum property that has only discrete values.

How does spin affect the collision between two particles?

Spin affects the collision between particles by determining the possible outcomes and interaction dynamics. The conservation of angular momentum, including spin, must be respected in any particle collision. Additionally, the spin states of the particles can influence the probability of certain interaction processes due to quantum mechanical selection rules.

What are the possible spin states for particles during a collision?

The possible spin states for particles during a collision depend on the intrinsic spin values of the particles involved. For example, if two particles with spin-1/2 collide, the combined system can be in a singlet state (total spin 0) or a triplet state (total spin 1). The specific spin states are determined by the quantum mechanical addition of angular momenta.

How do different spin values influence the scattering cross-section?

Different spin values influence the scattering cross-section by altering the interaction potential and the symmetry properties of the wave functions involved. Particles with different spin configurations can interfere constructively or destructively, leading to variations in the scattering amplitude and, consequently, the cross-section.

Can particles with different spins annihilate each other during a collision?

Particles with different spins can annihilate each other if they are particle-antiparticle pairs, such as an electron and a positron. The specific outcome of the annihilation process depends on the initial spin states and the conservation laws governing the interaction, such as the conservation of angular momentum and parity.

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