Physics 2170: Introduction to Particle Physics

1
Physics 2170 Introduction to Particle Physics

• Lecture 16
• Greg Landsberg
• March 17, 2009

2
Accounting for Spin

• In non-relativistic QM, particles are described
  by Schroedingers equation which is the same
  independent of particle spin
• QFT is different
• Spin 0 Klein-Gordon Equation
• Spin ½ Dirac Equation
• Spin 1 Proca Equation
• Spin 2 no consistent QFT exists
• Not really important, once the Feynman rules are
  established, except for spin ½ particles
• Feynman rules for spin ½ particles intimately
  related to the very structure of the Dirac
  equation

3
Schroedingers Equation

• Can derive from classical energy-momentum
  relation
• Now, recall that in QM operators of momentum and
  energy are given by
• Hence
• This is Schroedingers equation
• Klein-Gordon equation can be obtained in the same
  fashion by replacing non-relativistic energy
  momentum relation with the relativistic one (and
  reintroducing )

4
Klein-Gordon Equation

• Using Einsteins relation for a free particle
  , which can be rewritten in a
  covariant fashionone obtains
• Now lets introduce covariant derivative
• In this notation,
  and Gordon-Klein equation becomes

5
Klein-Gordon Equation Problems

• Was actually found by Schroedinger before his
  non-relativistic equation
• However, didnt work for the hydrogen atom, and
  hence was abandoned
• This is because its an equation for spin-0, not
  spin ½ particles
• Also incompatible with statistical interpretation
  of Y(r)2 as the probability of finding
  particle at point r
• This was traced to the fact that G-K equation is
  second-order in t, unlike Ss one
• Diracs strategy was to find first order equation
  still compatible with the relativistic
  energy/momentum relation

6
Factorizing Einsteins Relation

• Imagine we only had energy component
• Then E2 m2 0 or (E m)(E m) 0 and the
  second order equation factorizes into two first
  order ones E m E -m
• Unfortunately, this doesnt work directly if
  momentum is included
• Indeed, lets look for a factorization of the
  type
• This is equivalent to
• So, the goal is to find eight coefficients bk and
  gl that satisfy the above equation

7
Solution

• The right-hand side of the equation, after
  opening the parentheses, becomes
• bkgkpkpl m(bk gk) m2
• We do not want any terms linear in p, thus bk
  gk
• Finally, to satisfy the equation, we need
  gkglpkpl pmpm
• Here is the problem while one can take g0 1
  and gi I, but its impossible to cancel all
  the cross terms
• So, it seems that the factorization approach
  failed however Dirac found a solution the above
  equation can be satisfied if the solution is
  sought not via complex numbers, but via complex
  matrices

8
Dirac Matrices

• We need a set of four matrices that satisfy the
  following conditions (g0)2 1, (gi)2 -1, and
  gmgn gngm 0 for any m ? n
• Since matrices do not commute, it may be possible
  to satisfy all three conditions
• A more economic way to write down the above
  conditions, is to use anticommutator, A,B AB
  BA gm,gn 2gmn, where gmn is the Minkowski
  tensor (1,-1,-1,-1)
• The minimum dimension of the matrix for which the
  solution is possible is four
• There are several equivalent sets of 4 matrices
  that satisfy the above equation in the 4x4 matrix
  space

9
Bjorken-Drell Convention

• We will use the following set of gamma matrices
  (introduced by Bjorken-Drell)where si are the
  Pauli matrices 1 is a 2 x 2 unity matrix, and
  0 is a 2 x 2 matrix of zeroes
• The equation now becomes
• We can now choose either of the two terms to be
  zero (it turns out that it doesnt matter which
  one)

10
Dirac Equation

• We finally do one more substitution pm ? i?m,
  which finally gives the Dirac equation
• The solution Y is a 4-column (bi-spinor or
  Dirac spinor)

11
Solutions of the Dirac Equation

• Consider Y that does not depend on the position
• This represents a particle with zero momentum,
  given pm ? i?m
• The Dirac equation for such a particle becomes
  ig0 ?Y/?t mY orwhere

12
Dirac Equation Solutions

• Thus we got two equations
• with the following solutions
• yA(t) exp(-imt)yA(0), yB(t) exp(imt)yB(0)
• The exp(-imt) exp(-iEt) is the characteristic
  time-dependence of a quantum state
• What about the other solution? It corresponds to
  a particle with negative energy E -m
• Just as we discussed early in this course, this
  solution describes an antiparticle, while the
  first one describes a particle!
• Dirac equation itself dictates the existence of
  anti-fermions, which are intrinsically coupled to
  fermions!

13
Solutions

• The four independent solutions are
• They represent electron with spin up, electron
  with spin down, positron with spin down, and
  positron with spin up
• Note that spin direction for the antiparticle is
  reversed negative-energy electron with spin up
  is a positive-energy positron with spin down

14
General Solutions

• Lets now look for general solution with non-zero
  momentum
• As in classical QM, lets look for plane-wave
  solutions of the type y(x) ae-ikxu(k)
• Here kx kmxm k0t - kr
• Just like for a regular exponent, ?my - ikmy
  (easy to check explicitly)
• Putting this in the Dirac equation, we
  getgmkmae-ikxu m ae-ikxu or (gmkm m)u 0

15
General Solutions (contd)

• Note that the above equation is purely algebraic
  (no derivatives)
• Thus,