The Dirac Equation. H. A. Tanaka

Transcription

1 The Dirac Equation H. A. Tanaka

2 Relativistic Wave Equations: In non-relativistic quantum mechanics, we have the Schrödinger Equation: H = i t H = p2 2m 2 = i 2m 2 p t i Inspired by this, Klein and Gordon (and actually Schrödinger) tried: E 2 = p 2 c 2 + m 2 c 4 = c 2 ( 1 2 c 2 t µ =(, 1, 2, 3 )= ct, x, m 2 c 2 ) = = m2 c 2 y, 2 z ( 2 2 t 2 2 µ µ + m 2 c 2 ) = Manifestly Lorentz Invariant

3 Issues with KG and Dirac: Within the context of quantum mechanics, this had some issues: As it turns out, this allows negative probability densities: Dirac traced this to the fact that we had second-order time derivative factor the E/p relation to get linear relations and obtained: and found that: 2 < p µ p µ m 2 c 2 =) ( apple p apple + mc)( p mc) apple = apple { µ, } = µ + µ =2g µ Dirac found that these relationships could be held by matrices, and that the corresponding wave function must be a vector. µ p µ mc =) (i~ µ mc) =

4 The Dirac Equation in its many forms: (i mc) = a a µ µ (i µ µ mc) = a a µ µ = a a 1 1 µ =(, 1, 2, 3 )= a 2 2 ct, x, a 3 3 y, z i ( ) mc = i ct 1 x 2 y 3 z mc =

5 Now the gamma Matrices: µ =(, 1, = = = i i i i 3 = , 3 ) = = 1 1 = = =( 1, , 3 )= 1 1, i i, 1 1 Note that this is a particular representation of the matrices Any set of matrices satisfying the anti-commutation relations works { µ, } = µ + µ =2g µ There are an infinite number of possibilities: this particular one (Björken-Drell) is just one example

6 In full glory: i ct z x + i y ct x i y z z x i y ct x + i y z ct mc mc mc mc A = p mc p p p mc A B p µ i µ = B Consider applying another matrix to this equation p + mc p p p + mc p mc p p p mc = p2 m 2 c 2 (p ) 2 p 2 m 2 c 2 (p ) 2 From problem 4.2c ( a)( b) =a b + i (a b) ( p) 2 = p p p 2 p 2 m 2 c 2 p 2 p 2 m 2 c 2 A B = But this is just the KG equation four times Wavefunctions that satisfy the Dirac equation also satisfy KG

7 Solutions to the Dirac Equation: Consider a particle at rest: (x) e ik x = e i ( E c t p x) k µ = 1 (E/c, p x,p y,p z ) Particle has no spatial dependence, only time dependence. (i ct mc) = 1 1 t t A B = imc2 A B Note that the equation breaks up into two independent parts: t A = i mc2 A t B = i mc2 B A(t) =e i( mc2 )t A() B(t) =e i( mc2 )t B()

8 Dirac s Dilemma: ψ B appears to have negative energy mc 2 i( A(t) =e )t A() B(t) =e i( mc 2 )t Why don t all particles fall down into these states (and down to - )? Dirac s excuse: all electron states in the universe up to a certain level (say E=) are filled. Pauli exclusion prevents collapse of states down to E = - We can excite particles out of the sea into free states This leaves a hole that looks like a particle with opposite properties (positive charge, opposite spin, etc.) e e B() e hole kf = Dirac originally proposed that this might be the proton

9 Excuse to Triumph 1932: Anderson finds positrons in cosmic rays Exactly like electrons but positively charged: Fits what Dirac was looking for Dirac predicts the existence of anti-matter and it is found

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11 Solutions to the Dirac Equation at Rest: 1(t) =e 1 imc2 t/ spin up 2(t) =e imc2 t/ spin down 1 positive energy solutions (particle) 3(t) =e +imc2 t/ 1 spin down 4(t) =e +imc2 t/ 1 spin up Note that all particles have the same mass negative energy solutions (anti-particle)

12 Pedagogical Sore Point All the discussion we had thus far about difficulties with relativistic equations, (negative probabilities, negative energies) is of historic interest Scientifically, the framework for dealing with quantum mechanics and special relativity (i.e. quantum field theory) had not been developed The old tools of NR quantum mechanics had reached their limit and new ones were necessary. In particular, the idea of a wavefunction had to be revisited Until this was done, there were many difficulties! Once QFT was developed, all of these problems go away. Both KG and Dirac Equations are valid in QFT No negative probabilities, no negative energies Nonetheless, the history and its course are rather interesting.

13 Plane Wave Solutions to the Dirac Equation: Consider a solution of the form: column vector (x) =e ik x u(k) Column vector of 4 elements with spacetime dependence space-time dependence and place it in the Dirac equation: (i µ µ mc)(x) = ( µ k µ mc)e ik x u(k) = ( µ k µ mc)u(k) =

14 What does this equation look like: µ k µ = k 1 k 1 2 k 2 3 k 3 we found this is: k k k k Note 2x2 notation u(k) u A u B So that the Dirac equation reads: k mc k k k mc ua u B = ( k mc)u A k u B = k u A ( k + mc)u B = u A = k ( k mc) u B k ( k + mc) u A = u B

15 Determing u u A = k ( k mc) u B k ( k + mc) u k A = u B ( k + mc) k ( k mc) =1 this means we can identify ħk p 2 k 2 =( k ) 2 m 2 c 2 p 2 =(E/c) 2 m 2 c 2 Use positive solutions We can now construct the column vector u: u 1 = N 1 u 2 = N 1 (p x ip y )c/(e + mc 2 ) p z c/(e + mc 2 ) p z c/(e + mc 2 ) (p x + ip y )c/(e + mc 2 ) u 3 = N p z c/(e mc2 ) (p x + ip y )c/(e mc 2 ) 1 (p, p) (± k, k) electrons positrons p = Use negative solutions u 4 = N p z p x ip y p x + ip y p z (p x ip y )c/(e mc 2 ) p z c/(e mc 2 ) 1

16 Normalization of the Wavefunction: We need to choose a standard normalization of the wavefunctions Note that multiples of the solutions are still solution The normalization convention simply fixes this arbitrary choice: u u =2E/c u = u 1 u 2 u (u T ) u 3 u =(u 1,u 2,u 3 u 4 ) u 4 for u1 u 1 u 1 = N 2 1+ p 2 (E+mc 2 ) 2 =2E/c N = (E + mc 2 )/c

17 Lorentz Properties: The Dirac equation works in all reference frames. What exactly does this mean? i, ħ, m and c are constants that don t change with reference frames. µ and ψ will change with reference frames, however. µ is a derivative that will be taken with respect to the space-time coordinates in the new reference frame. We ll call this µ how does ψ change? i µ µ mc = Lorentz Covariant ψ = Sψ where ψ is the spinor in the new reference frame Putting this together, we have the following transformation of the equation when evaluating it in a new reference frame i µ µ mc = i µ µ mc = What properties does S need to make this work? i µ µ (S) mc (S) =

18 The Properties of S Since we know how to relate space time coordinates in one reference with another (i.e. Lorentz transformation), we can do the same for the derivatives Using the chain rule, we get: µ x µ = x x µ x where we view x as a function x (i.e. the original coordinates as a function of the transformed or primed coordinates). Note the summation over ν if the primed coordinates moving along the x axis with velocity βc: x = (x + x 1 ) x 1 = (x 1 + x ) x 2 = x 2 x 3 = x 3 ( =,µ= ) ( =,µ= 1) x x = x x 1 etc. =

19 Transforming the Dirac Equation: i µ µ mc = i i µ µ mc = µ µ (S) mc (S) = i mc = i µ x x µ (S) mc (S) = S is constant in space time, so we can move it to the left of the derivatives i µ S x x µ mc (S) = Now slap S -1 from both sides S 1! i~ mc S = Since these equations must be the same, S must satisfy = S 1 µ S x x µ

20 Example: The parity operator For the parity operator, we want to invert the spatial coordinates while keeping the time coordinate unchanged: P = 1 x x 1 =1 = 1 1 x x 1 1 x 2 x 3 1 = 1 = 1 x 2 x 3 We then have = S 1 S 1 = S 1 1 S 2 = S 1 2 S 3 = S 1 3 S Recalling = S 1 = 1 1 ( ) 2 =1 { µ, µ S x x µ } =2g µ i = We find that γ satisfies our needs = = i = i i = i = i S P =

21 Next time Read I would encourage you to work out the examples in 7.6 yourself explicitly so that you start to gain some fluency with the Feynman rules Lots of notation, lots of stuff going on.... please stop by if you have questions!