Field Theory 263: Problem Set 1

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1 Field Theory 263: Problem Set Michael Good Jan 27, 2006 Problem : Show that Pauli s equation, [ 2m ( i e A( r)) c 2 e 2mc σ B( r)] αβ ψ β ( r) = i t ψ α( r) can be more compactly written as where π = i e c A( r). 2m [( π( r) σ)( π( r) σ)] αβψ β ( r) = i t ψ α( r) Show that the equation in a rotated frame,(r i = R ijr j etc.) 2m [( π ( r ) σ)( π ( r ) σ)] αβ ψ β( r ) = i t ψ α( r ) is indeed given by the rotated form of the original equation, i.e., R αβ ( 2m [( π( r) σ)( π( r) σ)] γβψ β ( r) i t ψ γ( r)) = 0 This demonstrates the rotational covariance of the Pauli equation. b. Show that the following Maxwell s equation ν F µν = e c J µ is Lorentz covariant, i.e., the equation in another Lorentz frame νf µν = e c J µ is equivalent to the Lorentz transform of the original equation Λ µ ν( α F βα e c J β ) = 0 Answer:

2 Using σ i σ j = δ ij + iɛ ijk σ k or also from the fundamental product rule of geometric algebra, we obtain the familiar relation, With we have (σ A)(σ B) = A B + iσ (A B) 2m [( π( r) σ)( π( r) σ)] αβψ β ( r) = i t ψ α( r) 2m [π2 + iσ (π π)] αβ ψ β = i t ψ α where the vector arrows and their respective functions have been supressed. Plugging in for π 2m [(p e c A)2 + iσ (p e c A) (p e c A)] αβψ β = i t ψ α 2m [(p e c A)2 i e c σ (p A + A p)] αβψ β = i t ψ α The cross product terms can be evaluated, as is also done in Ryder p.54 by, Subtracting, [p i, A j ] = i i A j Multiplying by ɛ ijk, [p i, A j ] [p j, A i ] = i ( i A j j A i ) (p i A j p j A i )ɛ ijk + (A i p j A j p i )ɛ ijk = i ( i A j j A i )ɛ ijk Summing over i and j give the k component of i B, therefore, we can use I now have p A + A p = i A = i B 2m [(p e c A)2 i e c σ ( i B)] αβψ β = i t ψ α which, unsupressing vector signs and functional dependence, is [ 2m ( i e A( r)) c 2 e 2mc σ B( r)] αβ ψ β ( r) = i t ψ α( r) To demonstrate the rotational covariance of the Pauli equation, we need to show that this: 2

3 R αβ ( 2m [( π( r) σ)( π( r) σ)] γβψ β ( r) i t ψ γ( r)) = 0 equals this: 2m [( π ( r ) σ)( π ( r ) σ)] αβ ψ β( r ) = i t ψ α( r ) In order to simplify my life, I m going to drop the functional dependence notation, the vector hats, and for a trial run, even the indices. This will better demonstrate the overall idea. Watch out for the indicies, as they can make life harder than it really is. The solution sketch goes like this: R{[π σ)(π σ)]ψ} = t ψ R{[π σ)(π σ)]r ψ } = t ψ R{[π σ)r R(π σ)]r ψ } = t ψ (π σr)(π σr)ψ = t ψ (π σ)(π σ)ψ = t ψ Now, in order to make things grossly explicit, I m going to put in indicies and keep track in glory detail. First I need to change ψ β ψ β = δ βλ ψ λ = R βρ R ρλψ λ = R βρ ψ ρ So we have, including the indicies and simply rotating the right side term, R αγ { 2m [(π iσ i )(π k σ k )] γβ ψ β } = i t ψ α Substituting my ψ β, the left hand term becomes R αγ 2m [(π iσ i )(π k σ k )] γβ R βρ ψ ρ Inserting an identity matrix right smack in the middle, we have 2m R αγ(π i σ i ) γδ R δλ R λɛ(π k σ k ) ɛβ R βρ ψ ρ = i t ψ α If I use the property, σ i R ik = Rσ k R, I conclude that the left hand side becomes 2m (π iσ j R ji ) αλ (π k σ i R ik ) λρ ψ ρ 3

4 Using, π j = R jiπ i, this becomes 2m (π jσ j ) αλ (π iσ i ) λρ ψ ρ Lets simplify those indicies and our equation is now 2m [(π jσ j )(π iσ i )] αρ ψ ρ = i t ψ α As the ρ s are of course just dummies, we have 2m [( π ( r ) σ)( π ( r ) σ)] αβ ψ β( r ) = i t ψ α( r ) which demonstrates the rotational covariance of the Pauli equation. Part b. If we evaluate Λ µ β ( αf βα e c J β ) = 0 we have Λ µ β αf βα Λ µ e β c J β = 0 Λ µ β αf βα e c J µ = 0 Λ µ β αf βα = e c J µ Using, Λ ν α α = ν, and applying an inverse transformation, yields, Λ α ν ν = Λ α ν Λ ν α α Λ α ν ν = α Plugging this into α, Λ ν α ν = α Λ µ β (Λν α ν)f βα = e c J µ νλ µ β Λν αf βα = e c J µ νf µν = e c J µ 4

5 therefore this Maxwell equation is Lorentz covariant. Problem 2: The Klein Paradox. Consider a one dimensional potential barriar problem for the KG equation for which H classical = p 2 + m 2 + V { V0 x > 0 V = 0 x < 0 For stationary state solutions φ(x, t) = e iet f(x) we have f(x) = Ae ikx + Be ikx k = E 2 m 2 x < 0 f(x) = Ce ikx K = (E V 0 ) 2 m 2 x > 0 We expect that for V 0 > E m = kinetic energy (the non-classically allowed case) K is imaginary so that f(x) is purely damped. Show that this expectation is NOT borne out for V 0 > E + m. Draw the energy spectrum for both regions x < 0 and x > 0; try to explain the paradox. Answer: The expectation that K is imaginary is not borne out when V 0 > E + m because K 2 is now positive. This can be seen by just plugging into V 0 > E + m K 2 = (E V 0 ) 2 m 2 where the first term will now be larger than m 2. We would normally expect the wavefunction to be even more strongly damped, since V 0 > E + m, but because K 2 > 0 we have a region which is oscillatory like the region x < 0. Basically, a particle can be confined in the region x < 0 and tunnel through the region where x > 0 and V 0 > E m, and then behave like it was in an attractive potential in region x > 0 and V > E + m. Despite the fact that V > E + m seemingly implies it would behave like a particle in a very strong repulsive potential. The free-particle solutions to the Dirac equation exhibit an energy spectrum from m to and from m to. The condition to have negative energy oscillatory solutions with an applied small positive potential is: 5

6 < E < m + V From this we can see that E doesn t have to be negative. One way to explain the Klein-Gordon paradox is that we have a negative energy solution despite E > 0. We would just increase V enough so an oscillatory negative energy solution can have the same positive energy as an oscillatory positive energy solution where x < 0. Problem 4: a. Show that cċ ɛċȧ aȧ ɛ ab φ b = 2 φ c. b. Show that if φ obeys ( 2 + m 2 )φ a = 0, so does χ. Answer: Substituting a ḃ = (σ µ) a ḃ µ and ɛ ab = ɛȧḃ = iσ 2 we will have cċ ɛċȧ aȧ ɛ ab φ b = (σ µ ) cċ µ (iσ 2 )ċȧ (σ ν ) aȧ ν (iσ 2 ) ab φ b = (σ µ ) cċ (iσ 2 )ċȧ (σ ν ) aȧ (iσ 2 ) ab µ ν φ b = (σ µ ) cċ (iσ 2 )ċȧ (σν T )ȧa (iσ 2 ) ab µ ν φ b = (σ µ ) cċ (σ ν )ċb µ ν φ b 6

7 Where I used σ 2 σ T i σ 2 = σ i in the last step. Now applying the anticommunitive property of the matrices, Then I ve got {σ µ, σ ν } = 2δ µ ν Therefore (σ µ σ ν ) b c = 2 {σ µ, σ ν } b c = 2 2{δ µν} b c = {δ µν } b c (σ µ ) cċ (σ ν )ċb µ ν φ b = (δ µν ) b c µ ν φ b = µ µ φ c = 2 φ c which is what we set out to prove. cċ ɛċȧ aȧ ɛ ab φ b = 2 φ c. For part b, we know that φ satisfies ( 2 + m 2 )φ a = 0, so to prove ( 2 + m 2 ) χ c = 0 we shall start with and multiply by cċ ɛ cb to get b ḃ ɛḃȧ χȧ = mφ b cċ ɛ cb b ḃ ɛḃȧ χȧ = cċ ɛ cb mφ b Applying our property that we proved in part (a), we may write the left hand side as So we have cċ ɛ cb b ḃ ɛḃȧ χȧ = 2 χċ Using cċ ɛ cb φ b = m χċ, 2 χċ = cċ ɛ cb mφ b This is 2 χċ = m 2 χċ ( 2 + m 2 ) χċ = 0 7

8 Problem 5: a. Denote the set of 6 γ-matrices by P = (, γ 5, γ µ, γ 5 γ µ, σ µν ). Find their trace. Show that (, γ 5, γ µ ) are linearly independent. b. Simplify γ µ γ µ, γ µ γ ν γ µ, and γ µ γ ν γ ρ γ µ. c. Find T r(γ µ γ ν ), T r(γ γ γ ν γ l ), and T r(γ µ γ ν γ l γ σ ). Answer: T r() = 4 as these are four by four matrices. In a particular representation, say the standard Dirac-Pauli representation, we have for γ 5 : ( ) 0 I T r(γ 5 ) = T r = 0 I 0 but this is true as the trace of any odd number of gamma matrices is zero. T r(γ 5 ) = T r(γ γ γ 5 ) = T r(γ γ 5 γ ) = T r(γ γ γ 5 ) = T r(γ 5 ) T r(γ 5 ) = 0 T r(γ µ ) = T r(γ 5 γ 5 γ µ ) = T r(γ 5 γ µ γ 5 ) = T r(γ 5 γ 5 γ µ ) = T r(γ µ ) As it is equal to its negative, it must vanish. T r(γ µ ) = 0 T r(γ 5 γ µ ) = T r(γ µ γ 5 ) = T r(γ 5 γ µ ) Via anticommunitivity, for the first step and cyclic property of the trace for the second step. T r(γ 5 γ µ ) = 0 The four by four spin matrices, contain T r(σ µν ) = T r( i 2 [γ µ, γ ν ]) = i 2 T r(γ µγ ν γ ν γ µ ) Now for any n n matrices U and V 8

9 Thus T r(uv ) = T r(v U) T r(σ µν ) = 0 Linear independence of, γ 5 and γ µ means if then A + Bγ 5 + Cγ m u = 0 A = B = C = 0 I can show A = 0 by multiplying by and taking the trace: T r(a + Bγ 5 + Cγ µ ) = 0 T r(a) + T r(bγ 5 ) + T r(cγ µ ) = 0 and from above, the last two terms where shown to be zero, T r(a) = 0 A = 0 If A = 0 then Bγ 5 + Cγ µ = 0 must imply B and C are zero for linear independence. Multiplying by γ 5 and taking the trace again, B + Cγ µ γ 5 = 0 T r(b) = T r(cγ µ γ 5 ) T r(b) = 0 Thus we have left, which implies B = 0 Cγ µ = 0 C = 0 And we have shown that, γ 5 and γ µ are linaer independent. b. Here we have 9

10 γ µ γ µ = η µν γ µ γ ν = 2 η µν{γ µ, γ ν } = η µν η µν = 4 where I used {γ µ, γ ν } = 2η µν. So we ve got: γ µ γ µ = 4 It s not hard to simplify γ µ γ ν γ µ, just apply the anticommutivity property, and from the above box, γ µ γ ν γ µ = (2η µν γ ν γ µ )γ µ = 2γ ν γ ν γ µ γ µ γ µ γ ν γ µ = 2γ ν γ ν 4 γ µ γ ν γ µ = 2γ ν The same process goes for γ µ γ ν γ ρ γ µ : γ µ γ ν γ ρ γ µ = (2η µν γ ν γ µ )(γ ρ γ µ ) Using the above box 2γ ρ γ ν γ ν γ µ γ ρ γ µ = 2γ ρ γ ν γ ν ( 2γ ρ ) = 2(γ ρ γ ν + γ ν γ ρ ) applying the anticommuntivity property, γ µ γ ν γ ρ γ µ = 2{γ ρ, γ ν } = 4η ρν γ µ γ ν γ ρ γ µ = 4η νρ c. The trace can be found though anticommutation and cyclicity. T r(γ µ γ ν ) = T r(2η µν γ ν γ µ ) Cycle the last term and recall the T r() = 4: T r(γ µ γ ν ) = 2η µν 4 T r(γ µ γ ν ) Bring the last term to the other side and viola: T r(γ µ γ ν ) = 4η µν Next, any odd number gamma trace will be zero, for this three gamma trace we have, T r(γ γ γ ν γ ρ ) = T r(γ 5 γ 5 γ γ γ ν γ ρ ) 0

11 as γ 5 γ 5 =, and since {γ 5, γ µ } = 0 I can say Cycle through T r(γ γ γ ν γ ρ ) = T r(γ 5 γ γ γ ν γ ρ γ 5 ) T r(γ γ γ ν γ ρ ) = T r(γ 5 γ 5 γ γ γ ν γ ρ ) = T r(γ γ γ ν γ ρ ) As the trace equals the negative of itself, it must vanish: T r(γ γ γ ν γ ρ ) = 0 Lastly, we can evaluate T r(γ µ γ ν γ ρ γ σ ) through repeated use of the anticommunitivity relation. Apply it again, and again, T r(γ µ γ ν γ ρ γ σ ) = T r(2η µν γ ρ γ σ γ ν γ µ γ ρ γ σ ) T r(γ µ γ ν γ ρ γ σ ) = T r(2η µν γ ρ γ σ γ ν 2η µρ γ σ + γ ν γ ρ γ µ γ σ ) T r(γ µ γ ν γ ρ γ σ ) = T r(2η µν γ ρ γ σ γ ν 2η µρ γ σ + γ ν γ ρ 2η µσ γ ν γ ρ γ σ γ µ ) Bring the last term over to the otherside after cycling it once and we get 2 T r(γ µ γ ν γ ρ γ σ ) = 2 T r(η µν γ ρ γ σ γ ν η µρ γ σ + γ ν γ ρ η µσ ) Using our T r(γ µ γ ν ) = 2η µν from above, T r(γ µ γ ν γ ρ γ σ ) = 4(η µν η ρσ η µρ η νσ + η µσ η νρ )