Read Martin&Shaw chapter 4.

Symmetries and conservation laws play an important role in particle physics. A symmetry is simply a statement that the results of an experiment must be independent of our choice of coordinate system, our definition of up and down, our the assignment of positive and negative charge, and the like. It results from an invariance of the equations of physics under a transformation of one of the above.

Some symmetries we look at are fundamental while others are only approximate. The fundamental symmetries have associated with them a conserved quantity. Conserved quantities play a vital role in the advance of our understanding of particle physics, but we must be constantly vigilant for signs of a violation of their conservation.

Symmetries can be either continuous (coordinate transformations) or discrete (charge conjugation). The conserved quantity of a continuous symmetry is additive (momentum or energy). The conserved quantity of a discrete symmetry is multiplicative (parity).

Recall the QM equation for a particle:

i hbar dψ/dt = H ψ

If H = HIf we are dealing with an isolated system, the physical law represented by this equation is independent of any coordinate system, and therefore the equation itself is independent of the particular choice of coordinate system. Hence if ψ(x) is a solution in one coordinate system, then ψ'(x') must be a solution in a primed coordinate system, translated by Δx from the unprimed system. In particular, this must be true for an infinitisimal translation δx.

ψ'(x') = ψ(x+δx) = ψ(x) + δx dψ(x)/dx = Dψ

with the translation operator D defined as D = 1 + δx d/dx.
We can write this in terms of the momentum operator p
D = 1 + (i/hbar) δx p = 1 + (i/hbar) pδx

We build up a finite translation by taking the limit as n goes to infinity of n infinitisimal tranlations:

D = lim_{n to infinity}(1 + (i/hbar)pΔx/n)^{n} = exp(ipΔx/hbar)

From here it is a few steps to demonstrate conservation of momentum.
In 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.
Then the rotation operator is
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.
As they should be, the operators D and R are unitary.
That is, they leave the normalization of ψ' unchanged since D^{*}D = exp(-ipΔx/hbar)exp(ipΔx/hbar) = 1:

<ψ'|ψ'> = <ψD*|Dψ> = <ψ|ψ> = 1

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

For any system bound by a central potential, V(r), the wave function can be decomposed into radial and angular parts, with the angular parts described by spherical harmonics:

ψ(r,θ,φ)=χ(r)Y_{l}^{m}(θ,φ)

The spherical harmonics are given by:
Sqrt{(2l+1)(l-m)!/4π(l+m)!} P_{l}^{m}(cosθ)exp(i mφ)

The parity operation on spherical coordinates changes r to -r, θ to π-θ, and φ to φ+π.
Thus:
exp(i mφ) goes to exp(i mφ + i mπ) = exp(i mπ)exp(i mφ) = (-1)^{m}exp(i mφ)

and
P_{l}^{m}(cosθ) goes to P_{l}^{m}(cos(π-θ)) = (-1)^{l+m}p_{l}^{m}(cosθ)

Assembling this yields:
P Y_{l}^{m} = (-1)^{l}Y_{l}^{m}

Copyright © Robert Harr 2005