Phys624 Formalism for interactions Homework 6. Homework 6 Solutions Restriction on interaction Lagrangian. 6.1.

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1 Homework 6 Solutions 6. - Restriction on interaction Lagrangian Hermiticity

2 6.. - Lorentz invariance We borrow the following results from homework 4. Under a Lorentz transformation, the bilinears transform as follows, ψ Λ ψ () ψ ψλ () and the fact that Λ commutes with γ. This can be derived using the fact that γ anticommutes with all γ µ. Part (i) It is now easy to calculate the transformation property, ψγ ψ ψλ γ Λ ψ (3) = ψγ Λ Λ ψ (4) = ψγ ψ () Thus, it transforms the same way as a scalar under continuous Lorentz transformation. As shown elsewhere in this homework, this operator is odd under parity, so it is a pseudo-scalar operator. Part (ii) Given L int =( ψγ ψ)( ψγ ψ) (6) Since each of the terms is separately Lorentz (pseudo-scalar), the interaction term is Lorentz invariant as well. Part (iii) Repeating the calculation above, Λ ν µ ψγ µ γ ψ ψλ Λ ν µγ Λ ψ (7) = ψλ Λ ν µ γµ Λ γ ψ (8) = ψγ ν γ ψ (9) where we have used Λ Λ ν µγ µ Λ = γ ν, proved in an earlier homework. Thus, this spinor transforms like a vector under continuous Lorentz transformations. Again, this is seen to behave opposite to a vector under parity, so it is an axial-vector or a pseudo-vector.

3 Part (iv) Given, L int =( ψγ µ γ ψ)( ψγ µ γ ψ) (0) Since each of the terms above transforms honestly with the Lorentz index, the interaction made from contracted Lorentz indices is a Lorentz invariant Renormalizability The basis of estimating mass dimensions is the fact that the action is dimensionless in natural units ( = c = ). This can be seen from the time-evolution operator ( e i R dth ) or from the path-integral ( e is ). Part (i) The action for the kinetic term is S = d 4 x ψ / ψ () The γ-matrices and the Lorentz indices are unimportant for the dimension calculation. We get 4 inverse powers of mass from the volume integration, one power of mass from the derivative. Therefore, [ψ] 4 + = 0 () [ψ] = 3 (3) Part (ii) The action for the mass term, Thus, S = d 4 xm ψψ (4) [ψ] 4 + [m] = 0 () [m] = (6) Part (iii) We need the dimension of the scalar field. The action for the Yukawa term (with the kinetic term for the scalar included), S = d 4 x λφ ψψ + µφ µ φ (7) 3

4 Thus, [φ] 4 + = 0 (8) [φ] = (9) [φ ψψ] 4 + [λ] = 0 (0) [λ] = 0 () Part (iv) For the four-fermion interaction S = d 4 xκ( ψψ)( ψψ) () 4[ψ] 4 + [κ] = 0 (3) [κ] = (4) Since the co-efficient has negative mass-dimensions, it turns out that this interaction is nonrenormalizable Parity invariance Part (i) Under parity, ψ ψ p = η p γ 0 ψ () where η p =. Therefore, ψγ ψ ψ pγ 0 γ ψ p (6) = η p ψ γ 0 γ 0 γ γ 0 ψ (7) = ψ γ γ 0 ψ (8) = ψ γ 0 γ ψ (9) = ψγ ψ (30) Therefore, ψγ ψ is odd under parity. Part (ii) Repeating the same calculation, ψγ µ γ ψ ψ pγ 0 γ µ γ ψ p (3) = η p ψ γ 0 γ 0 γ µ γ γ 0 ψ (3) = ψ γ µ γ γ 0 ψ (33) = ψ γ µ γ 0 γ ψ (34) 4

5 The commutation property of γ µ and γ 0 can be characterized as γ µ γ 0 =( ) µ γ 0 γ µ (3) where ( ) µ = for µ = 0 and ( ) µ = for µ =,, 3. ψ γ µ γ 0 γ ψ = ( ) µ ψ γ 0 γ µ γ ψ (36) Therefore, the parity eigenvalue for ψγ µ γ ψ is ( ) µ. Part (iii) ψγ µ γ ψ ( ) µ ψγ µ γ ψ (37) The term in the Lagrangian transforms as follows under parity, ( ) ( ) ( ) ( ) hφ ψ γ ψ + h φ ψ +γ ψ hφ ψ +γ ψ + h φ ψ γ ψ (38) since the term with γ is odd under parity. Therefore, for parity to be a good symmetry, h = h, or in other words h should be real.

6 6. - Wick s theorem (Ex.3 L & P) In going from the scalar to fermions, we have to be careful about one fact. In all the definitions of time-ordering and normal-ordering, whenever we have to move a fermionic operator across another one, we introduce a negative sign. This is a matter of definition of the time-ordering and normal-ordering. In this problem we will see that this definition is useful in that it allows us to use Wick s theorem and all of subsequent Feynman diagram technology and makes the definitions consistent with anti-commutations. Consider the product of two fermion operators. In analogy with the scalar case, we can decompose the fields into the part with creation operator (ψ ) and the part with annihilation operator (ψ + ). ψ(x)ψ (x )=ψ + (x)ψ +(x )+ψ + (x)ψ (x )+ψ (x)ψ (x )+ψ (x)ψ +(x ) (39) = ψ + (x)ψ + (x ) ψ (x )ψ + (x)+{ψ + (x),ψ (x )} + ψ (x)ψ (x )+ψ (x)ψ + (x ) (40) where we have introduced the anti-commutator to make all the other terms normal-ordered (i.e. to have all the annihilation operators to the right of all the creation operators). Note that the anti-commutator of the two operators is a c-number. Therefore, where 0 is the vacuum state. {ψ + (x),ψ (x )} = 0 {ψ + (x),ψ (x )} 0 (4) = 0 ψ + (x)ψ (x ) 0 (4) = 0 (ψ + (x)+ψ (x))(ψ + (x )+ψ (x )) 0 (43) = 0 ψ(x)ψ (x ) 0 (44) ψ(x)ψ (x ) =: ψ(x)ψ (x ): + 0 ψ(x)ψ (x ) 0 (4) where we have used the following definition of the normal-ordered product. : ψ(x)ψ (x ): = ψ + (x)ψ + (x ) ψ (x )ψ + (x)+ψ (x)ψ (x )+ψ (x)ψ + (x ) (46) Notice that the normal-ordered product has a negative sign in the second term, when we move a fermion operator around the other. Now, the time-ordered product for two fermions, Therefore, T (ψ(x)ψ (x )) = θ(t t )ψ(x)ψ (x ) θ(t t)ψ (x )ψ(x) (47) T (ψ(x)ψ (x )) = θ(t t ) [: ψ(x)ψ (x ): + 0 ψ(x)ψ (x ) 0 ] θ(t t) [: ψ (x )ψ(x): + 0 ψ (x )ψ(x) 0 ] (48) = θ(t t ) [: ψ(x)ψ (x ): + 0 ψ(x)ψ (x ) 0 ] θ(t t)[ : ψ(x)ψ (x ): + 0 ψ (x )ψ(x) 0 ] (49) =: ψ(x)ψ (x ): + 0 θ(t t )ψ(x)ψ (x ) θ(t t)ψ (x )ψ(x) 0 (0) =: ψ(x)ψ (x ): + 0 T(ψ(x)ψ (x )) 0 () which is the statement of Wick s theorem. 6