Quantum Field Theory Badis Ydri Department of Physics, Faculty of Sciences, Annaba University, Annaba, Algeria. December 10, 2012 1
Concours Doctorat Physique Théorique Théorie Des Champs Quantique v1 2012-2013 1.5 h Problem 1 DeterminethetransformationruleunderLorentztransformationsof ψ, ψψ, ψγ µ ψ. Recall that ψ(x) ψ (x ) = S(Λ)ψ(x), where S(Λ) is the spinor representation of the Lorentz transformation Λ = exp( iω αβ J αβ /2) given by S(Λ) = exp( i 2 ω αβγ αβ ). Problem 2 Calculate the 2 point function < 0 T(ˆφ(x 1 )ˆφ(x 2 )) 0 > in φ four theory up to the 1st order in preturbation theory using the Gell-Mann Low formula and Wick s theorem. Express each order in perturbation theory in terms of Feynman diagrams. Recall that the Gell-Mann Low formula, the S matrix and the interaction Lagrangian for a phi-four theory are given by ( ) T(ˆφ(x 1 )ˆφ(x 2 )...) = S 1 T ˆφ in (x 1 )ˆφ in (x 2 )...S. S = e i d 4 yl int (y). L int = λ 4!ˆφ 4 in. 2
Concours Doctorat Physique Théorique Théorie Des Champs Quantique v2 2012-2013 1.5 h Problem 1 The Heisenberg field ˆφ(t, x), the Schrödinger field ˆφ( x) and the Diracfield ˆφ I (t, x) are related by ˆφ(t, x) = U 1 (t)ˆφ( x)u(t), ˆφ(t, x) = Ω 1 (t)ˆφ I (t, x)ω(t). The U and Ω are evolution operators satisfying the Schrödinger equation with Hamiltonians Ĥ = Ĥ0 + ˆV and ˆV I respectively where Show that U and Ω are related by ˆV I = e itĥ0ˆve itĥ0. U = e itĥ0 Ω. Show that ˆφ( x) and ˆφ I (t, x) are related by ˆφ I (t, x) = e itĥ0ˆφ( x)e itĥ0. Show that the Dirac field ˆφ I (t, x) satisfies the free Heisenberg equation, viz i tˆφi = [ˆφ I,Ĥ0]. Problem 2 The free Hamiltonian H 0 of a scalar phi-four is given by Ĥ 0 = 1 [ ] ˆP + ( p)ˆp( p)+e p 2 V ˆQ + ( p)ˆq( p), E p 2 = p2 +m 2. >0 The Schrödinger scalar field ˆφ( x) and its conjugate momentum field ˆπ( x) are given in terms of ˆQ( p) and ˆP( p) by ˆφ( x) = 1 V We give the canonical commutation relation ˆQ( p)e i p x, ˆπ( x) = 1 V [ˆQ( p), ˆP + ( p)] = ivδ p, q. ˆP( p)e i p x. Show that the Dirac field ˆφ I (t, x) is a free field satisfying the Klein-Gordon equation. 3
Concours Doctorat Physique Théorique Théorie Des Champs Quantique v3 2012-2013 1.5 h Problem 1 Show that the 2 point function for a free scalar field is given by < 0 T ˆφ(x)ˆφ(y) 0 d 4 >= (2π) 4 p 2 m 2 +iǫ e ip(x y). We give ˆφ(x) = ( d 3 p 1 â( p)e ipx +â( p) + e ). ipx (2π) 3 2E( p) [â( p),â + ( q)] = (2π) 3 δ 3 ( p q). Problem 2 Compute for a free scalar field (without recourse to Wick s theorem) the 4 point function D(x 1,x 2,x 3,x 4 ) =< 0 T ˆφ(x 1 )ˆφ(x 2 )ˆφ(x 3 )ˆφ(x 4 ) 0 >. Assume for simplicity that x 0 1 > x0 2 > x0 3 > x0 4. 4
Concours Doctorat Physique Théorique Solutions Théorie Des Champs Quantique 2012-2013 Solution 1.v1 The Dirac spinor ψ changes under Lorentz transformations as Since (γ µ ) + = γ 0 γ µ γ 0 we get (Γ µν ) + = γ 0 Γ µν γ 0. Therefore In other words As a consequence ψ(x) ψ (x ) = S(Λ)ψ(x). (0.1) S(Λ) = e i 2 ω αβγ αβ. (0.2) S(Λ) + = γ 0 S(Λ) 1 γ 0 (0.3) ψ(x) ψ (x ) = ψ(x)s(λ) 1. (0.4) ψψ ψ ψ = ψψ. (0.5) ψγ 5 ψ ψ γ 5 ψ = ψψ. (0.6) ψγ µ ψ ψ γ µ ψ = Λ µ ν ψγ ν ψ. (0.7) ψγ µ γ 5 ψ ψ γ µ γ 5 ψ = Λ µ ν ψγ ν γ 5 ψ. (0.8) We have used [γ 5,Γ µν ] = 0 and S 1 γ µ S = Λ µ νγ ν. Finally we compute ψγ µν ψ ψ Γ µν ψ = ψs 1 Γ µν Sψ = ψ i 4 [S 1 γ µ S,S 1 γ ν S]ψ = Λ µ αλ ν β ψγ αβ ψ. (0.9) Solution 2.v1 We have ( ) < 0 T(ˆφ(x 1 )ˆφ(x 2 )) 0 > = < 0 T ˆφ in (x 1 )ˆφ in (x 2 )S 0 > ( ) = < 0 T ˆφ in (x 1 )ˆφ in (x 2 )e i d 4 yl int (y) 0 > ( ) = < 0 T ˆφ in (x 1 )ˆφ in (x 2 ) 0 > +i d 4 y 1 < 0 T ( ) ˆφ in (x 1 )ˆφ in (x 2 )L int (y 1 ) 0 > +... 5 (0.10)
The 0th order term is the free propagator, viz ( ) < 0 T ˆφ in (x 1 )ˆφ in (x 2 ) 0 >= D F (x 1 x 2 ). (0.11) For a phi-four theory the interaction Lagrangian is L int = λ 4!ˆφ 4 in. (0.12) The 1st order term is thus ( ) i d 4 y 1 < 0 T ˆφ in (x 1 )ˆφ in (x 2 )L int (y 1 ) 0 > = i λ 4! ( ) d 4 y 1 < 0 T ˆφ in (x 1 )ˆφ in (x 2 )ˆφ 4 int (y 1) 0 >. We apply Wick s theorem. The result is sum over all contractions. We have 5.3 = 15 contractions in total However only two contractions are really distinct. These are: The disconnected diagram (free propagator times the figure of 8 diagram): We contract the two external points x 1 and x 2 together. The internal point y 1 can be contracted in 3.1 = 3 different ways. We have therefore three identical contributions coming from these three contractions. We get (0.13) 3 i( λ 4! )D F(x 1 x 2 ) d 4 zd F (0) 2 = 1 8 ( iλ) d 4 zd F (x 1 x 2 )D F (0) 2. (0.14) The connected diagram (tadpole): We contract one of the external points with one of the internal points. There are four different ways for doing this. The remaining external point must then be contracted with one of the remaining three internal points. There are three different ways for doing this. In total we have 4.3 = 12 contractions which lead to the same contribution. We have 12 i( λ 4! ) d 4 zd F (x 1 z)d F (x 2 z)d F (0) = 1 2 ( iλ) d 4 zd F (x 1 z)d F (x 2 z)d F (0). (0.15) Solution 1.v2 We want to check that U = e itĥ0 Ω. (0.16) We take the derivative with respect to time of both sides i t U = i t e itĥ0.ω+e itĥ0.i t Ω = Ĥ0U +e itĥ0.ˆv I Ω = Ĥ0U + ˆVe itĥ0 Ω = ĤU. (0.17) 6
Next we have ˆφ(t, x) = U 1 (t)ˆφ( x)u(t) = Ω 1 (t)ˆφ I (t, x)ω(t). (0.18) By using the result of the previous question we obtain immediately This gives immediately We start from A straightforward calculation leads to Next we start from We compute Ω 1 e itĥ0ˆφ( x)e itĥ0 Ω = Ω 1ˆφI (t, x)ω. (0.19) e itĥ0ˆφ( x)e itĥ0 = ˆφ I (t, x). (0.20) ˆφ(t, x) = Ω 1 (t)ˆφ I (t, x)ω(t). (0.21) i tˆφ(t, x) = Ω 1 i tˆφi Ω 1 +[ˆφ(t, x),ω 1 i t Ω] i tˆφ = i t U 1.ˆφU +U 1ˆφ.i t U 1 = U [Ĥ, ˆφ]U = Ω 1 i tˆφi Ω 1 +[ˆφ(t, x),ω 1ˆVI Ω] = Ω 1 i tˆφi Ω 1 +Ω 1 [ˆφ I, ˆV I ]Ω. (0.22) ˆφ(t, x) = U 1 (t)ˆφ( x)u(t). (0.23) = U 1 [Ĥ0, ˆφ]U U 1 [ˆV, ˆφ]U From (0.22) and (0.24) we conclude that Solution 2.v2 We start from = Ω 1 [Ĥ0,e itĥ0ˆφ( x)e itĥ0 ]Ω Ω 1 [ˆV I,e itĥ0ˆφ( x)e itĥ0 ]Ω = Ω 1 [Ĥ0,Ωˆφ(t, x)ω 1 ]Ω Ω 1 [ˆV I,Ωˆφ(t, x)ω 1 ]Ω = Ω 1 [Ĥ0, ˆφ I ]Ω Ω 1 [ˆV I, ˆφ I ]Ω. (0.24) i tˆφi = [ˆφ I,Ĥ0]. (0.25) i tˆφi = [ˆφ I,Ĥ0]. (0.26) 7
We compute immediately that Similarly starting from the result i tˆφi = e itĥ0 [ˆφ,Ĥ0]e itĥ0 = ie 1 itĥ0 e i p x itĥ0 ˆP( p)e V = i 1 e i p x ˆPI ( p) V = iˆπ I. (0.27) We compute i tˆπ I = [ˆπ I,Ĥ0], (0.28) i tˆπ I = e itĥ0 [ˆπ,Ĥ0]e itĥ0 = ie 1 itĥ0 e i p x E 2 itĥ0 p ˆQ( p)e V = i 1 e i p x E 2 V ˆQ p I ( p) = i 1 e i p x ( p 2 +m 2 )ˆQ I ( p) V = i 1 V ( 2 m 2 ) e i p x ˆQI ( p) From (0.27) and (0.29) we get the Klein-Gordon equation In other words ˆφ I is a free field. Solution 1.v3 First recall that = i( 2 m 2 )ˆφ I. (0.29) ( µ µ +m 2 )ˆφ I = 0. (0.30) < 0 T ˆφ(x)ˆφ(y) 0 >= θ(x 0 y 0 ) < 0 ˆφ(x)ˆφ(y) 0 > +θ(y 0 x 0 ) < 0 ˆφ(y)ˆφ(x) 0 >. (0.31) For x 0 > y 0 we compute < 0 T ˆφ(x)ˆφ(y) 0 > = < 0 ˆφ(x)ˆφ(y) 0 > d 3 p 1 = (2π) 3 2E( p) e ip(x y). (0.32) 8
Consider now the integral (ǫ > 0) dp 0 i dp 0 i I = 2π p 2 m 2 +iǫ e ip(x y) = 2π (p 0 )2 E 2 ( p)+iǫ e ip0 (x0 y0)+i p( x y). (0.33) There are two poles to the integrand at p 0 = ± E 2 ( p) iǫ. We enclose the pole in the lower half complex plane by a semicircle of infinite radius R. Clearly dz i y 0 )+i p( x y) Rde iθ i y = 0) e 2πz 2 E 2 ( p)+iǫ e iz(x0 2π R 2 e 2iθ E 2 ( p)+iǫ ersinθ(x0 ircosθ(x0 y 0 )+i p( x y) 0, R. (0.34) This is because sinθ < 0 and x 0 > y 0. Hence we can close the contour and write dz i I = 2πz 2 E 2 ( p)+iǫ e iz(x0 y0 )+i p( x y). (0.35) By using the residue theorem we have now (with z 0 = E 2 ( p) iǫ) I = (2πi).(z z 0 ) i 1 2π(z z 0 )(z +z 0 ) e iz 0(x0 y0) e i p( x y) = e ip(x y) 2E( p). (0.36) Now in the last equation p 0 = E( p). Thus for x 0 > y 0 we have < 0 T ˆφ(x)ˆφ(y) 0 d 3 p 1 > = (2π) 3 2E( p) e ip(x y) d 4 = (2π) 4 p 2 m 2 +iǫ e ip(x y). (0.37) For x 0 < y 0 the pole of interest is the one in the upper half complex plane and thus we must close the contour in the upper half complex plane. The rest of the calculation is similar. Solution 2. v3 We have for x 0 1 > x0 2 > x0 3 > x0 4. On the other hand D(x 1,x 2,x 3,x 4 ) = < 0 ˆφ(x 1 )ˆφ(x 2 )ˆφ(x 3 )ˆφ(x 4 ) 0 > = e ip 4x 4 < 0 â( p 1 )ˆφ(x 2 )ˆφ(x 3 )â + ( p 4 ) 0 >. (0.38) p 4 ˆφ(x 3 )â + ( p 4 ) 0 >= < 0 â( p 1 )ˆφ(x 2 ) = p 1 e ip 1x 1 1 2E( p4 ) e ip 4x 3 0 > + â + ( p 3 )â + ( p 4 ) 0 > e ip 3x 3. (0.39) p 3 1 2E( p1 ) < 0 eip 1x 2 + < 0 â( p 1 )â( p 2 )e ip 2x 2. (0.40) p 2 9
Hence { D(x 1,x 2,x 3,x 4 ) = e ip 1x 1 e ip 4x 4 e ip 2x 2 e ip 3x 3 < 0 â( p 1 )â( p 2 )â + ( p 3 )â + ( p 4 ) 0 > p 1 p 4 p 2 p } 3 1 1 + 2E( p1 ) 2E( p4 ) eip 1x 2 e ip 4x 3 = e ip 1x 1 e ip 4x 4 e ip 2x 2 e ip 3x 3 < 0 â( p 1 )â( p 2 )â + ( p 3 )â + ( p 4 ) 0 > p 1 p 4 p 2 p 3 + D F (x 1 x 2 )D F (x 3 x 4 ). (0.41) Use now < 0 â( p 1 )â( p 2 )â + ( p 3 )â + ( p 4 ) 0 >= (2π) 3 δ 3 ( p 1 p 4 )(2π) 3 δ 3 ( p 2 p 3 )+(2π) 3 δ 3 ( p 1 p 3 )(2π) 3 δ 3 ( p 2 p 4 ). (0.42) We get finally D(x 1,x 2,x 3,x 4 ) = D F (x 1 x 4 )D F (x 2 x 3 )+D F (x 1 x 3 )D F (x 2 x 4 )+D F (x 1 x 2 )D F (x 3 x 4 ). (0.43) 10