Relativistic Quantum Mechanics

Four Vectors The covariant position-time four vector is defined as x0 =ct, x1 =x, x2 =y, x3 =z And is written compactly as xm

Lorentz Transformations Transforms from x,y,z into x’,y’,z’ i.e from S into a moving reference frame S’ which has an uniform velocity of v in the x-hat direction And back again

Now combining four vectors with Lorentz But we may more compactly write this as Rows are m index, And n is column

Einstein Summation Convention Repeated indices are to be summed as one (i.e. leave off the sigmas) OR When a sub/superscript (letter) appears twice on one side of the equation, summation with respect to that index is implied

Lorentz invariance

The Metric Tensor, g

Covariant xm=gmnxn Where the subcripted x is called the covariant four vector And the superscripted x is called the contravariant four vector I=xm xm Note: a ·b=ambm

a · b is different from a 3 dimensional dot product If a2>0, then am is timelike If a2<0, then am is spacelike If a2=0, then am is lightlike

From vectors to tensors A second rank tensor, smn carries two indices, has 42=16 components, and transforms like two factors of L : A third rank tensor, smnl carries 3 indices, has 43=64 components, and transforms like three factors of L :

What I’ve used these things for years? In this hierarchy, a vector is a tensor of rank 1, and scalar (invariant) is a tensor of rank 0

A couple of quick definitions Proper time is the time w.r.t. to the moving reference frame. The differential to the left indicates the proper time interval in the moving reference frame. It should be getting larger and larger as v approaches c. This is our old friend velocity.

Proper Velocity Very invariant!

In relativity, momentum is the product of mass and proper velocity

Klein-Gordon Equation The trouble is that this equation does not describe spin; although we have used kets for convenience, these kets cannot describe angular momentum transformations. Actually, this equation works well for pions…

Deriving Dirac Dirac’s basic strategy was to factor pmpm –m2c2 into a perfect square; now if p is zero, then Now either p0-mc or p0+mc guarantees that pmpm –m2c2 =0

We don’t want any terms linear in momentum: need perfect squares Need to get rid of the cross-terms… Dirac said let the gammas be matrices, not numbers

Recall for Spin, Pauli Spin Matrices Dirac noted that Note: 1 with an arrow is the identity matrix and therefore, these are 4 x 4 matrices

Now for the last bit Dirac Equation The wavefunction is NOT a four vector but is called a Dirac spinor or bi-spinor

Ugly little sucker, isn’t it? A Pretty Form:

A “simple” example problem Consider a particle at rest, p=0 So the Dirac equation becomes Where yA carries the upper components and yB carries the lower components

SO Recall is the time dependent solution of SE So one of these states represents the rest mass of “normal” matter. The other state represents a “negative” energy state. Interpreted as the rest mass of the positron so

Wavefunctions

What if there is momentum? Then substitute h-bar * k Then take the inner product with alpha See your text on pages 803 and 804 for details

Okay, what about potentials?

Let E-V+mc2 =2mc2 So if the energy is low (less than mc2), then the SE is used otherwise, we must use Dirac Equation