Quantum Field Theory III

Transcription

1 Quantum Field Theory III Prof. Erick Weinberg April 5, Lecture 6 Let s write down the superfield (without worrying about factors of i or Φ = A(y + θψ(y + θθf (y = A(x + θσ θ A + θθ θ θ A + θψ + θθ( ψσ θ + θθf (1 Recall that the superfield is conveniently defined to satisfy DΦ = 0. The Lagrangian is written as L = K(Φ, Φ + W (Φ θθ + W (Φ θθ θ θ θ θ ( = dθ d θ K + dθ W + d θ W In the Wess-Zumino model the kinetic term and superpotential is Let s consider the transformation W = 1 mφ λφ3, K = Φ Φ (3 θ e iα θ, Φ e iα/3 Φ, Φ e iα/3 Φ, d θ e iα d θ (4 Then if we take m = 0 then the Lagrangian is invariant under this transformation. The component fields transform as A e iα/3 A, ψ e iα/3 ψ, F e 4iα/3 F (5 This symmetry is called R-symmetry, and if we require this symmetry then terms like d θ Φ will not occur in renormalization because it has to obey this symmetry. Now let s consider vector superfields with the constraint V = V V = c(x + iθχ i θ χ + i θθ(m + in i θ θ(m in θσ m θvm + iθθ θ λ i θ θθλ + 1 θθ θ θd (6 where c, M, N, D are real scalar fields, v m is a vector field, and χ and λ are Majorana spinor fields. Now v m is a candidate for gauge field. Because a SUSY transformation is written as ηq + η Q the condition V = V is maintained, so these fields remain real. We want to write down the transformation of the gauge field. We claim that V V + Φ + Φ (7 1

2 is a legitimate transformation. The component fields transform as c c + A + A, χ χ i ψ, M + in M + in if, v m v m i m (A A, (8 λ λ 1 σ m m ψ, D D + 1 (A + A (9 Now we can choose the real part of A and ψ and F to make c, χ, and M + in vanish. This gauge is called the Wess-Zumino gauge. However we still have the freedom in A A which is the imaginary part of A. So in this gauge the field is written as V = θσ θv m + θθ θ λ + θ θθλ + θθ θ θd (10 So if we just change the field by A A then we will have an ordinary gauge transformation on v m. Let s write a superfield as W α = 1 4 D β D βd α V (11 Then we immediately know that this is a chiral field because D β W α = 0. Now we want to know how it transforms under gauge transformation. We find that W α W α 1 4 D DD α Φ = W α D β D βd α Φ (1 Note that DΦ = 0 so the last term is just like an anticommutator acting on Φ, so we have W α W α D β( iσ m α β mφ = W α (13 So this field is gauge invariant, and that is reason we wrote it down in the first place. Its component form is W α = iλ α (y + θ α D(y i (σm σ n β α θ β ( m v n (y n v m (y + θθσ m α α m λ α (y (14 The Lagrangian for V will then be L V = 1 4 ( d θ W α W α θθ + = 1 D 1 4 vmn v mn iλσ m m λ d θ W α W α θ θ We can also add some mass term, but it will not be gauge invariant, and we need to use something like the Higgs mechanism. We will not do it here. Now let s consider the transformation of the matter fields (15 Φ e iqλ Φ, Φ Φ e iqλ, V V + i(λ Λ (16 We claim that the term Φ e qv Φ is gauge invariant, indeed Φ e qv Φ Φ e iqλ e q(v +iλ iλ e iqλ Φ = Φ e qv Φ (17 And this does not depend on the Wess-Zumino gauge at all. Now the super potential is a polynomial in Φ, so a term like Φ i Φ j Φ k will be invariant if q i + q j + q k = 0. A natural invariant Lagrangian is L = 1 ( W α W α θθ + 4 W β W β θ θ + Φ j eq jv j Φ j θθ θ θ + W (Φ j θθ + W (Φ j θ θ + κv θθ θ θ (18

3 The last term is called the Fayet-Iliopoulos term, and it is invariant because the D term is invariant under pure gauge transformation. Now let s look at the second term Φ e qv Φ = Φ Φ + qφ V Φ + q Φ V Φ (19 The first term is just the ordinary kinetic term. The second term will give us the interaction like qφ V Φ va A + v( A A + v m ψ σ m ψ + A λψ + A λ ψ + D A (0 The third term is more restrictive because the only term surviving V will be just v m, so it will give a term like q Φ V Φ v m v m A (1 This is the last term in the ordinary coupling of gauge field to a charged scalar. So the only new interaction is that between A, λ, and ψ. Now let s do supersymmetric QED. We have the gauge field A µ and the Dirac field ψ which has Majorana spinors. Let s count the number of real degree of freedom. The A µ has 4 1 = 3 real fields, and it has degrees of freedom, and ψ has 8 real fields, and 4 degrees of freedom. Now we need to put these into superfields, so we need to introduce λ called the photino, which is a Majorana field with physical degrees of freedom. We also need an auxiliary field with no physical degree of freedom. Now for the Dirac field because we have Majorana fields we need two superfields Φ + and Φ. They have bosonic partners A + and A which are called selectrons, and there are 4 real fields with 4 physical degrees of freedom. We also have the auxiliary fields F + and F with no physical degree of freedom. Now recall the Lagrangian of the theory (18. It will become L = 1 ( 4 (W α W α θθ + h.c. + Φ + eev Φ + + Φ θθ θ θ eev Φ + m Φ + Φ θθ + Φ θθ θ θ + Φ θ θ + κ V θθ θ θ ( The gauge transformation will read We claim that there is R-symmetry here Φ ± e ieλ Φ ± (3 θ e iα θ, V V, Φ ± e iq ±α Φ ± (4 and if we require q + +q = then the Lagrangian is R-invariant. For example we could choose q + = q = 1, then we will have λ e iα λ, A ± e iα A ±, ψ ± ψ ±, v m v m (5 Under this transformation the physical fields do not transform, and the supersymmetric fields gets transformed. This symmetry requires that all the supersymmetric particles have to appear or annihilate in pairs. Phenomelogically it means that if supersymmetric particles were created in early universe they will decay into the lightest particle, and the annihilation rate of these lightest particles will be smaller and smaller when the universe expands and finally freeze in. So they are a candidate for dark matter particles. We can write out the Lagrangian in component form L = 1 D + F + + F + e ( D A + + A + ( F j + h.c. + κd +... (6 A j=± j 3

4 From here we can write out the equations of motion for the auxiliary fields D + e ( A + + A + κ = 0, F± + ma = 0 (7 So the effective potential for A will become V = 1 κ + (m + 1 eκ A + + (m 1 eκ A + e 8 ( A + A Suppose κ = 0 then the minimum of this potential will be at A + = A = 0 and there is no spontaneous symmetry breaking for the gauge field of U(1. Now if κ > m /e then there will be spontaneous symmetry breaking of U(1. That is Abelian gauge theory. Now let s consider non-abelian case where the transformation law is (8 V = T a ijv a, W α = 1 4 D De V D α e V (9 e V e iλ e V e iλ (30 If we go through the same computation as above for the effective potential of A then we will get V = F i + 1 ga D a, Fj =, A j D a = g a A i T a ija j (31 There is no κ term now because V is not invariant under global gauge transformation. However the rest is just like what we had above. However we don t have supersymmetry. It has to be spontaneously broken. Let s just assume for now that the potential is V = A j j (3 If there is only one A field then V = 0 if / A = 0 and we have an unbroken vacuum. But we can t find a function of complex variable that its derivative is nonzero everywhere, so there is always unbroken supersymmetric vacuum if there is only one A. So now let s consider the O Raifeartaigh model where where λ, m, g are real and positive. Then the potential will becomes W = λφ 0 + mφ 1 Φ + gφ 0 Φ 1 (33 V = A 0 + A 1 + A = (34 λ + ga 1 + ma + ga 0 A 1 + m A 1 Assuming all parameters are nonzero, then let s find the minimum of V. We have V/ A = 0 which is A = (g/ma 0 A 1. Let s set A 1 = x where x > 0, then the potential is just V = (λ gx + m x (35 4

5 Then minimizing this is to find where V/ x = 0, which is where m g(λ gx = (m gλ + g x (36 If m > gλ then the minimum is at x = 0 so that A 1 = A = 0 and A 0 is anything we want. If m < gλ then the minimum is at gλ m x = g (37 then we have A 1 = ±i gλ m g, A = ig gλ m m g A 0 (38 and A 0 again can be arbitrary. This is no problem as we can have family of degenerate vacuum. The important thing for broken supersymmetry is that / A is nowhere vanishing. Let s come back to the SUSY QED V = 1 [ e( A + A + κ] ( + m A + + A (39 8 So we have a constraint on κ if we want a broken supersymmetry. Let s call M ij = W/ A i A j, then at minimum of V we should have 0 = A i A j ( W A i A j A j (40 But we know that / A never vanish, so the mass matrix M ij has to have zero eigenvalue, so we have massless fermion. This is analogous to the Goldstone boson, and we call this a Goldstino. Just like in Higgs mechanism the Goldstone boson combines with the gauge boson to give a massive vector particle, the Goldstino in supergravity combines with the spin 3/ gravitino to give massive gravitinos. 5