# Collision between two particles with different spin

• Like Tony Stark
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.
Like Tony Stark
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?

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.

topsquark
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?

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.

DrClaude and topsquark

Replies
1
Views
838
Replies
1
Views
1K
Replies
3
Views
941
Replies
8
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
16
Views
714
Replies
2
Views
2K