Physics 249 Lecture 39, Dec 10th 2012
Reading: Chapter 12
HW 11: due Friday Dec 14th
2) A relativistic wave equation:
Start with the energy momentum relationship
E2 - p2 = m2
Quantizing
results in the Klein-Gordon equation
-
¶2
f +Ñ2f = m 2f
¶t 2
𝜕
𝜕
defining a four component vector or derivatives 𝜕 = (± 𝜕𝑡 , 𝜕𝑥 )
𝑖
𝜕 2 𝜙 = 𝑚2 𝜙
With a solutions:
f (x,t) = Ne ip×x-iEt
f (x,t) = Ne-ip×x+iEt
This is the solution for a spin zero (scalar particle). For instance a more complex solution
is found for the spin ½ electrons. However, it will let us investigate the concept of
antiparticles.
Consider multiplying the Klein Gordon equation by -i* and subtracting the complex
conjugate of the equation multiplied by -i. The mass terms don’t involve operators and
cancel out. Bringing out a time and space derivative
One can define a continuity equation as the change in time of the probability density in a
given volume is equal to the current of probability out of the volume:
This looks exactly like the equation above and allows us to interpret it.
The first term above is probability density  and the second the probability current j out
of the volume.
Calculating for  using the first wave function solution.  = 2E|N|2
Note the factor of 2E. The factor of 2E makes the probability density element d3x
Lorentz invariant. If you consider a particle in a Lorentz boosted from space contracts
since d3x d3x(1-v2)1/2. However, energy, the time like component dilates and the
probability goes up making the probability density element constant.
Now consider the wave functions with a negative sign in the exponential. These
solutions have negative probability density!
This issue can be interpreted and resolved by noting two things.
First, fundamentally the two solutions to the wave equations are independent.
y(x,t) = Ne ip× x-iEtand y (x, t) = Ne-ip×x+iEt .
It makes sense to identify them with independent particles both with energy E.
If you had one particle with probability density  = 2E|N|2 , E positive, and one
antiparticle with probability density  = -2E|N|2 , E also positive, then in if you consider
the total probability density of a system with one of each of these particles you can
annihilate them without changing the probability density. These equations predict that a
second solution to the wave equation should exist that corresponds to a particle that can
annihilate with the original equation. In fact we observe exactly this!
In electricity and magnetism the fundamental property that is associated with the particles
and antiparticle is the charge. The continuity equation implies that charge in conserved.
A more complex solution can be pursued by introducing the a general function u(p) into
the solution. If we do this we find that there are actually 4 independent solutions (6 if
you include the 2 we already found). Investigating the solutions using the same
techniques we find that 2 correspond to particles and two to antiparticles. Further
investigating the solutions by calculating the angular momentum we find that the two sets
of two solutions are subdivided into solutions with intrinsic angular momentum up ½ and
intrinsic angular momentum down -½. The property of spin can be understood directly
from the allowed wave equation solutions.
2) The photon
Considering Maxwell’s equations in differential form. There are several points relevant
to our discussion, which I will introduce (or review).
Both E and B can be expressed in terms of potentials. Without any time changing fields
the simple potential V for E and the vector potential A for B.
E: 𝐸⃑ = ∇𝑉
⃑ = ∇x𝐴
B: 𝐵
The potentials can be calculated directly from the charges and currents using the
following:
∇ ∙ 𝐸⃑ = 4𝜋𝜌
4𝜋
⃑ =
∇𝑥𝐵
𝐽
𝑐
equations for the fields in terms of the charges and currents by substituting the equations
for the fields in terms of the potentials into them. Then second derivatives of the
potentials are found in terms of the charges and currents. Inserting the ideas that time
changing fields can cause the other type of field and defining two four component vectors
𝐴 = (𝑉, 𝐴) and 𝐽 = (𝜌, 𝐽) then above equations gives you.
(−
1 𝜕2
4𝜋
+ ∇2 ) 𝐴 = −
𝐽
2
2
𝑐 𝜕𝑡
𝑐
In free space J=0 and this is the equation for a traveling wave, written in the form of the
potential of the traveling wave rather then the electric and magnetic fields.
(−
1 𝜕2
+ ∇2 ) 𝐴 = 0
𝑐 2 𝜕𝑡 2
With different constants and m=0 this is exactly the same as the Klein Gordon equation.
Since matter particles and photons are both governed by the same wave equation you
expect them to exhibit the same physics properties. For instance, similar wave and
particle like properties. However, you do expect specific differences due to factors like
the fact that the photon is massless and electrically neutral.
We would like to understand the effect of the potential caused by J on a particle. In
specifically we would like to understand the change in momentum or energy. The force
in proportional to E or B times the charge of the particle and thus proportional to the
derivative of the potential times the charge of the particle. Now remembering that force
in proportional to the differential change in momentum we can integrate the force to find
the change in momentum.
Δ𝑝 = ∫ 𝐹 = 𝑒 ∫ 𝐸, 𝐵 = 𝑒 ∫ 𝜕𝐴 = 𝑒𝐴
Thus 𝑝 → 𝑝 + 𝑒𝐴
Quantizing 𝑖ℏ𝜕 → 𝑖ℏ𝜕 + 𝑒𝐴
Then the effect of adding the potential to the wave equation is:
𝜕 2 𝜙 = 𝑚2 𝜙 → (𝜕 − 𝑖𝑒𝐴)(𝜕 − 𝑖𝑒𝐴)𝜙 = 𝜕 2 𝜙 − 𝑖𝑒(𝜕𝐴 + 𝐴𝜕) − 𝑒 2 𝐴2 = 𝑚2 𝜙
The A2 term can be identified as the energy of the potential and the other terms describes
how the potential changes momentums.
We can calculate the probability of the potential to cause a specific change in momentum
between any two momentum and energy states of a particle as:
𝑃 = 𝜙𝑓∗ 𝑉𝜙𝑖 = 𝜙𝑓∗ 𝑉𝜙𝑖 = −𝑖𝑒𝜙𝑓∗ (𝜕𝐴 + 𝐴𝜕)𝜙𝑖
However, we still have to figure out what to put in for A in this equation. This is where
observation such as the fact that the information of a change in the electromagnetic field
travels at the speed of light or that posing that the photon carries this information lead to
the right function form allowed physicists to understand the correct answer. Finally, they
idea that light can be thought of a traveling potential wave is consistent with the idea that
it can carry the information of the potential. If you put in the equation for the photon
potential you get the correct probability distribution!
This answer correctly described known problems. For instance Rutherford scattering
where the observed distribution of angles is directly related to the initial kinematics.
However, it also allowed us to solve unknown problems like what the distribution of
angles where when an e+ and e- annihilated and then pair produced.