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 observable results of measurements must be independent of the coordinate system (modulo the change of coordinates). As a particular example, consider a system consisting of a single free particle, and consider what happens when the coordinate system (or particle) is translated by a small amount δx along the x axis. The Hamiltonian for a free particle will be unchanged by the translation:

H(x') = H(x+δx) = H(x).

The translation can be represented by a linear operator (see Q.M. text) such that:
Dψ(x) = ψ(x + δx)

where D is the translation (displacement) operator.
If the displacement is infinitisimal, then we can expand ψ to get
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)

To demonstrate how the symmetry under translational leads to conservation of momentum, consider the what happens when we apply the translation operator, D, to the expression

H(x)ψ(x) = φ(x).

Applying the translation operator yields:
DH(x)ψ(x) = Dφ(x) = φ(x+δx) = H(x+δx)ψ(x+δx) = H(x)ψ(x+δx) = H(x) Dψ(x).

where I've used the invariance of H from above.
Subtracting the rightmost expression from the leftmost expression yields:
(DH(x) - H(x)D)ψ(x) = 0

and since ψ(x) is arbitrary, this can only hold in general if D and H commute,
[D,H] = 0.

By writing D in terms of the momentum operator, as done above, we show that the momentum operator must commute with H, which is the requirement for the momentum to be conserved.
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.
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.
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'> = <Dψ|Dψ> = <ψ|D^{†}D|ψ> = <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