Thirring Model. Brian Williams. April 13, 2010

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1 Thirring Model Brian Williams April 13, Introduction The Thirring model is a completely soluble, covariant (1+1)-dimensional quantum field theory of a two-component Dirac spinor We can write the generalized Thirring Lagrangian as, L S = i 2 ψγ µ µ ψ i 2 ( µ ψ)γ µ ψ 1 2 ( µa ν )B µν A ν µ B µν µ2 2 A µa µ B µνb µν +gj µ A µ + σ 2 j µj µ I ll call this the Sommerfield Lagrangian for it is the more general case of the original Thirring Lagrangian, L 0 = i ψγ µ µ ψ + σ 2 j µj µ that Charles Sommerfield studied in the 1960 s In L S we have the usual two-component Dirac spinor ψ but also a spin-1 field A µ and a totally antisymmetric tensor field B µν This model could be relevant for it fits naturally with supersymmetry In the Lagrangians I have implicitly defined the classical current, j µ ψγ µ ψ Important: To attain total covariance of the theory (and for other reasons we will ecounter) we actually must more carefully define j µ We will be using the good-man s metric g 00 = g 11 = 1 The 2 2 Dirac matrices γ µ obey the usual Clifford algebra, {γ µ, γ ν } = 2g µν We choose the representation via Pauli matrices, ( γ 0 0 i = σ 2 = i 0 ), γ 1 = iσ 1 = ( ) 0 i i 0 We have the familiar γ 5 = σ 3 = ( )

2 with the useful relation γ µ γ ν = g µν + ɛ µν γ 5 where ɛ 10 = ɛ 01 = 1 is the antisymmetric tensor Using this antisymmetric tensor we can write B = 1 2 ɛ µνb µν B µν = Bɛ µν which eplicitly shows that the field B µν has only one degree of freedom off-shell 2 EoM We can easily derive the equations of motions and of course, iγ µ µ ψ = γ µ (ga µ + σj µ )ψ ν B µν = gj µ µ 2 A µ B µν = µ A ν ν A µ We also have the familiar field commutation relations [A 1 (), B( )] = iδ( ) and {ψ α (), ψ β ( )} = δ αβ δ( ) 3 The Action Principle As with any quantum field theory, our goal is to compute vac-vac epectation values of time ordered products Call such a time ordered product (R) + As usual the vac-vac amplitude with no source is, 0 0 = ( DA µ D ψ ep i ) d 2 L So varying the epectation value gives the general relation, δ 0 (R) + 0 = i 0 d 2 (RδL) δ(r) + 0 The familiar procedure is to now introduce source terms which will allow us to gain insight on the vacuum The eternal fields we introduce are φ µ and J µ The the Lagrangian becomes, L L + φ µ j µ + A µ J µ 2

3 This modifies the equations of motion a bit, iγ µ µ ψ = γ µ (ga µ + σj µ + φ µ ) ν B µν = gj µ + J µ µ 2 A µ From here on, we assume that the time-ordered products do not have eplicit dependence on the eternal fields So, But also, δ δφ µ () 0 (R) + 0 = i 0 (j µ ()R) + 0 δ δj µ () 0 (R) + 0 = i 0 (A µ ()R) + 0 g 0 (R) + 0 = i 0 σ 0 (R) + 0 = i 0 2 d 2 (Rj µ ()A µ ()) + 0 d 2 (Rj µ ()j µ ()) + 0 This allows us to write out differential equationa for the vev s, g 0 (R) + 0 eσg = i d 2 δ δ δφ µ () δj µ () 0 (R) + 0 eσg σ 0 (R) + 0 eσg = i d 2 δ δ 2 δφ µ () δφ µ () 0 (R) + 0 eσg Where I have adopted the convention for the subscripts e, σ and g which means the eternal fields and couplings are turned on Simple integration yields, [ 0 (R) + 0 eσg = ep ig d 2 δ δ δφ µ () δj µ () i ] 2 σ d 2 δ δ δφ µ () δφ µ 0 (R) + 0 e () The main point is that we now have an epression for the vev with coupling as a function of the vev without coupling, which is a much easier quantity to calculate Green s functions for Specifically we have the definition of the n point fermionic Green s function G (n) cσge( 1,, n, 1,, n) G (n) cσge(, ) = in 0 (ψ( 1 ) ψ( n ) ψ( n) ψ( 1 )) + 0 eσg 0 0 eσg The subscripts maintain the same convention and the c just indicates that we re looking at the causal solutions Now using our epression from above we can write, { [ ep ig d 2 δ δ G (n) eσg(, δφ µ() δj ) = µ () i 2 σ ]} d 2 δ δ δφ µ() δφ µ () G (n) ce (, ) 0 0 e { [ ep ig d 2 δ δ δφ µ() δj µ () i 2 σ ]} d 2 δ δ δφ µ() δφ µ () 0 0 e Where we now eplicitly have the form of the coupled Green s functions in terms of the uncoupled Green s functions 3

4 4 Computing G (n) ec (, ) The n point Green s function can be written as the Slater determinant of 2 point Green s functions, G ec ( 1, 1 G (n) ec (, ) G ec( 1, n) ) = G ec ( n, 1 ) G ec( n, n) Each of the two point functions satisfy the non-homogenous Dirac equation, [ iγ µ µ γ µ φ µ ()]G ec (, ) = δ( ) Now turn φ off to get the regular Dirac equation, iγ µ µ G c (, ) = δ( ) Recall the 2-d representation of the Dirac delta function, δ( d 2 p ( ) = ) µ (2π) 2 eipµ So we have solution, G c (, ) = d 2 p γ µ p µ ( ) µ (2π) 2 p 2 iη eipµ = 1 γ µ ( ) µ 2π ( ) 2 + iη Thus, where we have introduced the propagator Now look at G c (, ) = iγ µ µ c ( ) c ( ) = i 4π log[( ) 2 + iη] [ iγ µ µ γ µ φ µ ()]γ 0 G ec (, ) = δ( )γ 0 ( g µ0 + ɛ µ0 γ 5 )( i µ φ µ )G ec (, ) = γ 0 δ( ) Take µ = 0, then the LHS is just ( i 0 φ 0 )G ec Taking µ = 1, the LHS is ( iγ 5 1 γ 5 φ 1 )G ec So we can write, (i 0 + iγ φ 0 + γ 5 φ 1 )G ec (, ) = γ 0 δ( ) I claim that the eternal field solution is, G ec (, ) = e i[f () F ( )] G c ( ) 4

5 where, First notice that, F () = i d 2 ξ G c (, ξ)γ 0 [φ 0 (ξ) + γ 5 φ 1 (ξ)] G(, ξ)γ 0 [φ 0 (ξ) + γ 5 φ 1 (ξ)] = γ µ µ c ( ξ)γ ν φ ν (ξ) So we can write F () more compactly as F () = γ µ µ d 2 ξ c ( ξ)γ ν φ ν (ξ) Now, [ iγ µ µ γ µ φ µ ()]G ec (, ) = { [γ µ µ F ()G c (, ) + δ( )] γ µ φ µ ()G c (, ) } ep[i(f F )] = {(γ µ µ ) 2 [ ] } d 2 ξ ( ξ)γ ν φ ν (ξ) γ µ φ µ () G c (, ) ep[i(f F )] + δ( ) = { } d 2 ξ δ( ξ)γ ν φ ν (ξ) γ µ φ µ () G c (, ) ep[i(f F )] + δ( ) = δ( ) So it checks out The determinant thus has the form, { G (n) ce (, ) = P Where, = P ( 1) P ep ( 1) P ep { i i i i } [F ( i ) F ( P (i) )] G c ( j P (j) ) j } d 2 ξ n (i) µ ( i, P (i) ; ξ)φµ (ξ) G c ( j P (j) ) n (i) µ ( i, P (i) ; ξ) = γν γ µ ν [ ( ξ) ( P (i) ξ)] j = ( µ + γ 5 µ )[ ( ξ) ( P (i) ξ)] 5

6 5 (Re)Defining the Current We look at the dependence of 0 0 e on the eternal fields First recall, δ δφ µ () 0 0 e = i 0 j µ () 0 Tentatively we wrote j µ () = ψ()γ µ ψ() So referring to our epression for G c (, ) and letting η 0 we have, 0 ψ( )γ µ ψ() 0 e = e 2π tr α ep[i(f () F ( ))] γν ( ) ν ( ) 2 = 1 2π tr α ep[i(f () F ( ))] γν ζ ν aζ 2 Where the trace is taken over all spinor indices and we have written ( ) µ aζ µ There is an obvious singularity as To remove this and also to preserve the charge symmetry ψ ψ we define, so that i 2 lim j µ () = 1 2 lim [ ψ( )γ µ ψ() γ µ ψ( ) ψ() ] 0 ψ( )γ µ ψ() γ µ ψ( ) ψ() 0 e 0 0 e = 1 4π lim tr αγ µ { ep[if () if ( )] ep[if ( ) if ()] } About = we have to O(a 2 ), γν ζ ν aζ 2 ep[if () if ( )] ep[if ( ) if ()] = (1 + ia λ [F () F ( )]ζ λ ) (1 ia λ [F () F ( )]ζ λ ) Thus, i 2 lim = 2ia λ F ()ζ λ 0 ψ( )γ µ ψ() γ µ ψ( ) ψ() 0 e = i ζ λ ζ ν 0 0 e 2π ζ 2 tr α[γ µ λ F ()γ ν ] = i π ζ λ ζ ν ζ 2 [gµσ gν τ ɛ µσ ɛ τ ν] σ λ d 2 ξ c ( ξ)φ τ (ξ) There is one more subtlety we need to take into account in defining j µ Note that the above epression is not invariant with respect to the path by which we approach which 6

7 takes the covariance out of the theory To retain covariance we average over two directions of approach, namely ζ µ and ζ µ Our final result is (this time with spinor indices!), j µ () = 1 [ ( 4 γ αβ lim ψα a ) ( a 0 2 ζ ψ β + a ) 2 ζ + ψ α ( a 2 ζ ) ( ψ β + a 2 ζ ) ( ψ β a ) ( 2 ζ ψα + a ) 2 ζ ψ β ( a 2 ζ ) ( ψα + a 2 ζ )] You may ask at this point if all of this touching up with j µ is even allowed Lorentz invariance is obvious, but canonical commutation rules (as we shall see) will not be so simple Also, the current as it is written is not invariant under the gauge transformation ψ() ψ()+iδλ()ψ() and ψ() ψ() iδλ() ψ(), which will have to be considered Now with this current, δ e δφ µ () 0 0 e = i d 2 y v µτ ( y)φ τ (y) where Integrating out, Or more succinctly as, Now look at, v µτ (z) = 1 2π (gµσ g ντ ɛ µσ ɛ ντ ) σ ν (z) [ i 0 0 e = 0 0 J ep 2 ] d 2 y d 2 y φ µ (y)v µν (y y )φ ν (y ) 0 0 e = 0 0 J ep[φvφ] δ δj µ () 0 0 = i 0 A µ() 0 J Using equations of motion in the absence of φ µ we have, ( + µ 2 )A µ () = J µ µ ν A ν and taking the inner product, [ ( + µ 2 ) 0 A µ () 0 J = J µ () 1 ] µ 2 µ ν J ν () 0 0 J = [g µν 1µ 2 µ ν ] J ν () 0 0 J Integrating we find, 0 0 J = ep [ i 2 ] [ ] i d 2 y d 2 y J µ (y)w µν (y y )J ν (y ) = ep 2 JwJ 7

8 Where, w µν (z) = [g µν 1µ 2 µ ν ] c (µ; z) And finally c (µ; z) = 1 (2π) 2 d 2 p e ipµ z µ p 2 + µ 2 iη which is the usual Klein-Gordon Green s function for a boson of mass µ in two dimensions 6 Interlude: Completing the Square We proceed to eponentiate quadratic functional derivatives of quadratic functionals in the fields Let us write ( ) φ χ = J and Finally, let C = ( ) σ g g 0 V = ( ) v 0 0 w ( ) nµ ( N = N µ ( i, P (i) ; ξ) = i, P (i) ; ξ) 0 So the desired amplitude is, [ i δ Ω 0 0 eσg = ep 2 δχ C δ ] (ep[in χ])(ep[i/2χvχ]) δχ We can complete the square with the ansatz, So that, χ = χ + N V 1 [ Ω = Ω ep i ] 2 N V 1 N where [ i Ω δ = ep 2 δχ C δ ] [ ] i δχ ep 2 χ Cχ After differentiating, integrating and more massaging we get the final result, [ Ω = Ω 0 ep i ] 2 N C(1 + VC) 1 N ep[iχ(1 + VC) 1 N ] ep[i/2χ(1 + VC) 1 Vχ] And where the amplitude of the free theory Ω σg (with no eternal fields) is Ω 0 = ep [ 12 ] tr log(1 + VC) 8

9 7 Properties of the Solution Converting the language from the last section we have, [ i 0 0 eσg = Ω 0 ep 2 φv φ + iφuj + i ] 2 JW J which is in our short notation fxg d 2 ξ d 2 ξ f µ (ξ)x µν (ξ ξ )g ν (ξ ) And where (I list here for reference only It is good to see the structure of these functions as they come from the act of Gaussian integration) V µν (z) = 1 [ 2 µ ν 2π 1 (α + λ) 2 c (z) + λ(µ g µν µ ν ) (1 + α)(1 + α + λ) c(µ ; z) g ] µν 1 + α δ(z) U µν (z) = g [ ] 2 µ ν 2πµ 2 1 (α + λ) 2 c (z) µ g µν µ ν 1 + α + λ c(µ ; z) W µν (z) = 1 [ 2λ µ ν µ 2 1 (α + λ) 2 c (z) + (1 + ] α)(µ g µν µ ν ) c (µ ; z) 1 + α + λ The coupling constants are adjusted, α = σ 2π λ = g2 2πµ 2 and we have an adjusted mass for the Bosonic Green s function, µ 2 = 1 + α + λ 1 + α µ 2 The free theory vac-vac amplitude is computed as, { 0 0 σg = ep 1 [ d 2 δ (2) (0) log[(1 α λ)(1 + α)] µ 2 µ 2 dr 2 c (r; 0) This gives a trivial result if you look at it closely We will reeamine this when look at a modified Lagrangian and it so happens that the Delta functions cancel and we are left ]} 9

10 with a non-trivial finite amplitude The two-point fermionic Green s function with sources off turns out to be, [ { 2(α + λ) G σgc (, 2 ) = ep 2πi 1 (α + λ) 2 [ c (0) c ( )] + }] λ (1 + α)(1 + α + λ) [ c(µ ; 0) c (µ ; )] G c ( ) As this two point function stand it is ultraviolet divergent for c (0) and c (µ ; 0) diverge at high energies There is also infrared difficulties with the fermionic Green s function but can be avoided by defining asymptotic fermionic states We must supply an ultraviolet cutoff in order to further make sense of the limiting process of defining the current 8 Ultraviolet Cutoff 9 Many-Fermion Green s Function 10 Bosonic Operators 11 Commutation Relations 12 Gauge Symmetries 13 Making Sense of the Current 14 New Lagrangian and Correcting the Vac-Vac Amplitude 10