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Relativistic Quantum Mechanics

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1 Relativistic Quantum Mechanics

2 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

3 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

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

5 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

6 Lorentz invariance

7 The Metric Tensor, g

8 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

9 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

10 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 :

11 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

12 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.

13 Proper Velocity Very invariant!

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

15 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…

16 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

17 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

18 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

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

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

21 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

22 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

23 Wavefunctions

24 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

25 Okay, what about potentials?

26 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


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