What Are the Final |L,S> States After Ground State Pionic Atom Decays?

[dipstik]

Homework Statement

A pion is a spin-0 particle with a negative unit charge. A neutron is a spin-1/2
particle with no charge. A proton is a spin-1/2 particle with positive unit charge.
One can construct an unstable version of the hydrogen atom where the nucleus
contains both a proton and a neutron, but uses a pion instead of an electron. (No
electrons in this problem!) This is called a "pionic atom".

A.) In 1951, it was observed that the ground state of a spin-1 "pionic atom" could
spontaneously decay into two neutrons (and nothing else). (The spin-1 refers to
the original total spin of the entire atom, including the original nucleus.)
Given that total angular momentum must be conserved during this decay process, list
ALL of the allowable final values for |L,S>, where S is the total spin of the 2
resulting neutrons and L is the orbital angular momentum of the two resulting
neutrons (orbiting around each other). Ignore symmetry constraints.

B) Now use symmetry arguments to determine which of your answers from A) will
actually happen. Hint: the symmetry of two neutrons with an orbital angular
momentum is (-1)^L. Make sure you show your reasoning.

Homework Equations

J=L+S
|s1-s2|<=s1+s2<=s1+s2 with integer steps
Clebsch–Gordan Table for 1/2 x 1/2 (maybe)
fermions only exist in antisymmetric states: (-1)^L=-1 for fermions

The Attempt at a Solution

A) The initial total angular momentum J0=L0+S0, and it is given that the nucleus is spin 1 (S=1) and the pion is spin 0 and L=0 (ground state). This implies that J0=1. Because this is conserved we require that the final total angular momentum is Jf=1. This requires Lf+Sf=1. This is possible by |1,0> or |0,1> final |L,S> states (the |1/2,1/2> is not allowed because the neutrons are 1/2 which gives S=0,1 only)

B) Only the L=1 state obeys the (-1)^L=-1 fermionic condition. This implies that the |0,1> is not allowed.

 

[vela]

You're kind of mixing up notation. I'll use the convention where a bolded letter corresponds to an angular momentum vector, and the corresponding lowercase letter is the quantum number associated with the square of the vector. For example, L=(L_x, L_y, L_z) and the operator L² has eigenvalues of the form l(l+1)\hbar^2.

A) The initial total angular momentum J0=L0+S0, and it is given that the nucleus is spin 1 (S=1) and the pion is spin 0 and L=0 (ground state). This implies that J0=1.

The total angular momentum has three contributions: the spin I of the nucleus, the spin S of the pion, and the orbital angular momentum L, so you have J=I+S+L. As you noted, L=S=0, so J=I.

From what you wrote in the problem statement above, you weren't directly told the nuclear spin as you suggest here, but you were told the total angular momentum. You didn't need to do any work to deduce it.

Because this is conserved we require that the final total angular momentum is Jf=1. This requires Lf+Sf=1. This is possible by |1,0> or |0,1> final |L,S> states (the |1/2,1/2> is not allowed because the neutrons are 1/2 which gives S=0,1 only)

In the final state, you have J=L+S, where J is the total angular momentum, L is the orbital angular momentum of the neutrons, and S is the total spin of the neutrons. From conservation of angular momentum, you know j=1, and as you correctly noted, the neutron spins can combine to give you an s=0 singlet state or an s=1 triplet state. You know that j=1 has to lie between l+s and |l-s|. When s=0, you must have l=1. But when s=1, you have more possibilities, some of which you missed. What values of l allow j=1?

B) Only the L=1 state obeys the (-1)^L=-1 fermionic condition. This implies that the |0,1> is not allowed.

 

[dipstik]

in the case of |s,m> the triplet gives |1,1>, |1,0> and |1,-1>, but that does not seem possible in this case because l and s are greater than or equal to zero.

If I allow myself to be open to the possibility that j=1 is only within the range of allowed j values, then i could allow l to be 0 or 1 (even 2 and greater it seems if this is the logic) when s=1, but i would think that l+s must equal 1.

thanks for the help vela, but as is often the case, more information sometimes leads to more data for me to muddle around incorrectly. you did a good job, but i confuse myself often.