Read Martin&Shaw chapter 4.

Recall the QM equation for a particle:

i hbar dψ/dt = H ψ

If H = HIn a similar way, we can construct an operator that generates infinitisimal rotations about an axis, for instance, the z axis:

R_{z} = 1 + δφ d/dφ

The operator for the z component of angular momentum is
J_{z} = -i hbar [x d/dy - y d/dx] = -i hbar d/dφ

with the usual definition for φ in spherical coordinates.
We can use this to express the rotation operator in terms of the angular momentum operator:
R_{z} = 1 + i J_{z} δφ / hbar.

A finite rotation, Δφ is constructed by taking the limit as n goes to infinity of n infinitisimal rotations:
R_{z} = lim_{n to infinity}(1 + (i/hbar)J_{z}Δφ/n)^{n} = exp(iJ_{z}Δφ/hbar)

Again, it is a few additional steps to prove conservation of the z component of angular momentum.
Let
H(x)ψ(x) = φ(x).

Then
R_{n}H(x)ψ(x) = R_{n}φ(x) = φ(x') = H(x')ψ(x') = H(x)ψ(x') = H(x)R_{n}ψ(x),

where x' is the rotated coordinate, and I've used the invariance of the Hamiltonian, H(x') = H(x) [this holds generally for any central potential].
Subtracting the leftmost and rightmost terms yields
[R_{n}, H] = 0.

Writing R in terms of the angular momentum operator J (or L as done in the text), leads to the conclusion that J commutes with the Hamiltonian, and therefore the angular momentum is a conserved quantity.
Recall from quantum mechanics that the above is true for the angular momentum projected along one axis, normally taken to be the z axis, J_{z}.
The angular momenta along the other axes (J_{x} and J_{y}) do not commute with J_{z} or each other.
Therefore we can't measure all 3 simultaneously.
Instead, we can measure J_{z} and (the square of) the total angular momentum.

[J_{z}, H] = [J^{2}, H] = [J_{z}, J^{2}] = 0.

Hadrons are composite particles composed of quarks bound by the strong force. The quarks are fermions, posessing an intrinsic spin of S = 1/2, and can be bound together in states of orbital angular momentum L. The combination of orbital angular momentum L and spin S yields the total angular momentum of the composite particle, J. How L and S combine to yield J will be discussed in a later lecture.

Note that in general, L and S need not be separately conserved, only J. But we will find that often it is a good approximation to assume that L and S are separately conserved -- this is especially true in objects composed of one heavy and one or two light quarks. Please read section 4.2.3 on "Angular momentum in the quark model" to see how to apply this to some example states.

We'll come back to the rotation operator. Now let's move to a discrete symmetry, parity.

The parity operation, denoted P, is a reversal of sign on all coordinates.

Pψ(**r**) = ψ(-**r**)

This is clearly a discrete transformation.
Application of parity twice returns the initial state implying that PA wavefunction will have a defined parity if and only if it is an even or odd function. For example:

ψ = cos(x), Pψ = cos(-x) = cos(x) = ψ thus ψ is even and P=1

ψ = sin(x), Pψ = sin(-x) = -sin(x) = -ψ thus ψ is odd and P=-1

ψ = cos(x)+sin(x), Pψ = cos(x)-sin(x) which is neither ±ψ

and thus the last function has no defined parity.
ψ = sin(x), Pψ = sin(-x) = -sin(x) = -ψ thus ψ is odd and P=-1

ψ = cos(x)+sin(x), Pψ = cos(x)-sin(x) which is neither ±ψ

Copyright © Robert Harr 2005