Problems

A particle of spin 3/2, at rest in the laboratory, disintegrates into two particles, one of spin ½ and one of spin 0.

(a)  What values are possible for the relative orbital angular momentum of the two particles?  Show that there is only one possible value if the parity of the relative orbital state is fixed.

(b)  Assume the decaying particle is initially in the eigenstate of Sz with eigenvalue   Is it possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state?

Solution:

(a)  The initial value of j is j=3/2.  We need the final value to be j=3/2, since angular momentum of an isolated system is conserved.  Let us call the initial particle particle a, the spin ½ paticle particle b, and the spin 0 particle particle c.  Then the total spin of the two particles in the final state is S=Sb+Sc=Sb.  Therefore the spin quantum number is s=½.  The possible values for the orbital angular momentum quantum number are l=1 and l=2. (j=l+s, l-s; j=l+½, l-½, implies l=1 or l=2.)  The parity of the orbital state is (-1)l.  If the parity is odd, we have l=1, if the parity is even, we have l=2.

(b)  Let P(+) be the probabilities of finding the spin ½ particle in the |+> state.  Do states with different l give us different P(+)?  Assume that the final state has l=1.  It may be written as |j1,j2;j,m>=|1,½;3/2,m>. To find P(+) we need to write it as a linear combination of |j1,j2;m1,m2>=|1,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2.  Using a table of Clebsch-Gordan coefficients we find

Now assume that the final state has l=2. It may be written as |j1,j2;j,m>=|2,½;3/2,m>.
We may write it as a linear combination of |j1,j2;m1,m2>=|2,½;ml,ms>.
m can take on the values 3/2, ½, -½, -3/2.  We find

Yes, it is possible to determine the parity of the final state by measuring the probabilities of finding the spin ½ particle either in the |+> or |-> state.

Given are 3 coupled spins (s1=s2= s3=½). Starting with find 7 other linearly independent normalized states |S,Sz> that are eigenstates of S2=(S1+S2+S3)2 and Sz=S1z+S2z+S3z.  You should obtain one quartet for s=3/2 and two independent doublets for s=½.  You can assume that basis states such as are normalized.  If you wish you may use without proof the relation
Solution:
In this problem we are asked to add more than two angular momenta.  Combining S1 and S2 to form S12=S1+S2 we can form states with s12=1 (triplet states) and a state with s12=0 (singlet state).
S=S12+S3. For S12=0 we have S=S3, s=½.  The eigenstates of S2 and Sz are

and

These two states are the first doublet.

For s12=1 we can have s=3/2 or s=½.  The eigenstate of S2 and Sz are

These two states are the second doublet.

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These four states are the quartet.

If you do not have a table of Clebsch-Gordan coefficients you can proceed in the following way.

because this is the only way to make m=-3/2.

J+=S1++S2++S3+.

Similarly because this is the only way to make m=3/2.

To find we look for a linear combination which is normalized and orthogonal to

 and .

(We cannot find this state by operating with J+ or J- on one of the above states, since such an operatiom changes m only.)

a+b+c=0, -a+b=0, a2+b2+c2=1.

a=b, 2b2+c2=1, 2b+c=0, 

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Assume the scattering of p mesons by nucleons takes place chiefly through the intermediate state of the meson nucleon system with a total isospin of I=3/2, but not through the state with I=½.  Compute the relationship of the differential effective cross section of the following three reactions for the same relative energies, scattering angles, and orientations of the spins.
(i) , (ii) , (iii) .
Solution:
Charge independence of the nuclear force lead to the introduction of a new conserved quantum number, isospin.  The isospin operator I operates in isospin space.  The three components of I are I1, I2, and I3. They satisfy the commutation relations .  Therefore common eigenstates of I2 and I3 can be found.  They are denote by |I,I3> in isospin space.

I2|I,I3>=I(I+1)|I,I3>, I3|I,I3> = I3|I, I3>.

Directions in isospin space have nothing to do with directions in ordinary space.

For the proton we have |I,I3>=|½,½>, and for the neutron we have |I,I3>=|½,-½>.  We have q=e(I3+½) for nucleons.

The p meson exists in 3 states of roughly the same mass, p+, p0, and p-.  Consequently it is assigned isospin quantum numbers I=1, I3=0,1,-1.  The strong interaction is charge independent, i.e. it does not depend on I3. It can, however, depend on I.  Charge conservation implies that I3 is conserved, and the strong interaction conserves I.

In the reaction the initial and final states have I3=3/2 and are therefore described in isospin space by |I,I3>=|3/2,3/2>.

The states in have I3=-½ and can be a linear superposition of |I,I3>=|3/2,-½> and |½,-½>. 
Let Ia denote the isospin of the p- and Ib the isospin of the proton.  Using a table of Clebsh-Gordan coefficients we find

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The states in have I3=-½ and can be a linear superposition of |I,I3>=|3/2,-½> and |½,-½>.  The initial state is the same as above, and the final state is

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The cross section is proportional to he matrix element connecting the scattered initial state and the particular final state observed.    U is an operator that cares about isospin.  Let U=U1 if it operates on an initial state with I=½ and U=U3 if it operates on an initial state with I=3/2.  H does not change the isospin of a pure state, therefore the matrix elements between initial and final states with different I are zero.

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If  then