Physics 772 Peskin and Schroeder Problem 3.4.! R R (!,! ) = 1 ı!!

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1 Physics 77 Peskin and Schroeder Problem 3.4 Problem 3.4 a) We start with the equation ım = 0. Define R L (!,! ) = ı!!!! R R (!,! ) = ı!! +!! Remember we showed in class (and it is shown in the text) that if L transforms as a left-handed Weyl fermion, then L transforms as a right-handed fermion. Furthermore, remember that it was shown in the text and in the notes that µ =µ. Since µ is the lower-right block of µ, and since is given in block-diagonal form by R L in the upper-left block and R R in the lower right block, it follows that R µr R L = µ. Thus, We thus µ( ) µ@ = µ R R R L ( ) µ@ = R R R ım! ı (R R R L ımr R So this equation is Lorentz invariant. We have We thus have = R R ım ) = ım = ım ı = ım µ = ım µ = ım ( m ) = ım We may = This gives us (@ + m ) = 0

2 b) We have S = S = d 4 x[ + ım ( T )] d 4 x[ ı(@ µ ) µ ım ( T )] Where we have used the fact that =. Partially integrating the first term gives S = d 4 x[ı µ@ µ ım ( T )] = S Varying with respect to Moving the ( a)ı µ µ b a gives us the Euler-Lagrange equation ım ( b ba( a all the way to the left gives a)+( ım =0 a) ab b) = 0 Similarly, varying with respect to a yields ı(@ µ b ) µ ba ( a) = ım ( b ba( a)+( a) ab b) = ım b ba ( a) We then get ı( µ) µ ı( µ = ım( ) T = ım which is the conjugate of the Majorana equation. c) We may write the Dirac Lagrangian as L = (ı/@ m) = ı L + ı R m( L R + R L) Substituting in L =, R = i,wefind L = + ı( ı T = + ı ım( = + + ım( T ) m( (i )+( ı T T ) ) ) ) Note that if we take = we get (up to a factor of ) the same Lagrangian we had in part b).

3 d) We showed in class that the Lagrangian was invariant under the symmetry! e ı.sincethissymmetrymultiplies L and R by the same phase, we find the global symmetry! e ı! e ı Plugging into the Lagrangian, it is clear that the phases cancel in each term. For the theory of part b), we µ J µ µ ( µ ) = (@ µ ) µ = ) = m( ) + m = m<( ) The divergence is proportional to the mass, because the mass term breaks the symmetry in ( T and both transform with the same phase under the symmetry). For the theory of part c), we µ J µ µ ( µ µ ( µ ) = ) )) The Dirac equation can be written in terms of Weyl spinors as This can be rewritten as which gives us L m R = 0 R m L = 0 = = µ J µ = m<( ) Now one can write So = a ab b b = ab a = b ba a µ J µ = 0 This is because the current we have written is the current for the symmetry of the Lagrangian we found above. Finally, want to construct a theory of N two-component fermions with O(N) symmetry. We thus are not interested in a symmetry involving complex phases, 3

4 but we do want to consider the possibility of an odd number of fields. So we can take the theory from part b) as our starting point. Consider the Lagrangian L = NX i= i i + ım ( T i i i i ) Nothing in the Lagrangian acts on the i index, so it s easy to see that, if M is a matrix in the fundamental representation of O(N) (som T M =, and M is real), then this Lagrangian is invariant under i! M i j j. e) So we quantize the Majorana theory of parts a) and b) by taking { a (x), b (y)} = ab (3) (x y) Now for the Lagrangian in part b), the kinetic term (the term involving derivatives) is the same as for the Weyl theory of the left-handed spinor. So we may define a (x) =ı a(x). { a (x), b (y)} = ı ab (3) (x y) As we remember from computing the Hamiltonian for the Dirac theory (or any theory where the kinetic term is linear), the time derivative term drops from the Hamiltonian, and we are only left with the other terms. So we get H = = = = d 3 x[ ı!!@ ım ( T )] d 3 x ı T!!@ m( T )] d 3 x ı [ (@ 0 m ) T 0 m ) m( T )] d 3 x ı ]= d 3 0 ] where in the second line we took the hermitian conjugate, and then interchanged the two fields. Remember that satisfies the Klein-Gordon equation, we know it satisfies the dispersion relation! = k + m. So as usual, we expand in solutions of the wave equation (here, the Majorana equation), with annihilation operators multiplying the solutions with space-time dependence e ıp x.wethus get H = d N!a N a N up to a normal-ordering constant. Equivalently, we can note from part c) that the Majorana Lagrangian is the same as the Dirac Lagrangian with the identification R (x) =ı L (x). This is the same as saying C (x)c = (x). This identification reduces implies the identification b! p = a! p ; in other words, identifying a Majorana particle with its anti-particle. This means that the Hamiltonian for the Majorana Lagrangian can be written in terms of the oscillators we had defined for the Dirac theory in the same way, but with the number of oscillators reduced by a factor of two due to the identification. 4

5 Physics 77 Peskin and Schroeder Problem 3.5 Problem 3.5 a) We start with L µ + + F F Under = ı T = F F = we find L µ ( ı T )+@ µ ( ı T µ + F ) +( F ) + F ( )+( ) F = ı@ µ µ + ı(@ µ µ ı µ( F ) +( F + + F ( )+(ı@ µ )F µ = ı@ µ µ + ı(@ µ µ ı +( ) ' ı T (@ ) ı (@ )+ı ı T ' 0 where the ' is used after integration by parts, where the total derivative terms have been dropped. So the change in the Lagrangian is indeed a total divergence. b) We have apple L = m F + ım T +(c.c.)

6 Under the variation given above we have ( L) = m( ı T )F + m ( )+ ım T ( F ) + ım( F ) T +(c.c.) = ım T F + ım T F + ım + ım( T F ) +(c.c.) We now have = ım T F + ım T F + ım T µ + ım T F ım µ +(c.c.) = ım T F + ım T µ ım T F + ım µ +(c.c.) = ım a ab b F + a µ ım ba µ ım a ab b F + ım a µ b µ +(c.c.) = 0 L µ + + F F + m F + m F + ım T ım The Euler-Lagrange equations for F and F are then Substituting in gives us F = m F = m L µ m + + ım ( T ) In this equation we see that the fermion has a mass term of the form given in problem 3.4, and the mass is the same as that of the scalar. c) We now have L µ µ i + i i + Fi F ] + F i + W [ ] i j + c.c. j The top line of the Lagrangian is just the Lagrangian we considered in part a), and we know that it is invariant under supersymmetry up to a total divergence.

7 The variation of the second line is given W [ ] L = F i + F i j + j + W [ ] (( i) T j + T 3 W [ ] k T k j)+c.c. = ( i W [ ] + F i ( ı T j)+ j + W [ ] (( F i i ) T j + T j = W [ i + F i ( ı T j)+ j W [ j F i T j + W [ ] (( µ ) j@ µ i + T j = ı ] µ i 3 W [ ] ( a ab k)( c i k + W [ ] ( µ j@ µ i + T i µ j j = ı W [ ] i@ µ j 3 W [ ] ( a b k c k + W [ ] j a µ ab b j + a j µ ba b )@ µ 3 W [ ] ( ı T k) T k ( F j j )) + 3 W [ ] ( ı T k) T k µ j )+c.c. i + c.c. cd d j ) d j )( bc ad ac bd ) where in the last line we performed a partial integration of the first term, and have explicitly expanded out the Pauli matrices in the second term. Note b that the second term vanishes; one can see this by interchanging k and c i, while switching the b and c and k and i indices (and similarly for the second function pair). The remaining terms cancel, giving us L = 0 If W = g 3 /3, then the Euler-Lagrange equations for F and F are yielding F = g F = g L µ + g ( ) + ıg T The Euler-Lagrange equations for and are then given +g + ıg = 0 ıg = 0 ıg j j 3

8 Physics 77 Peskin and Schroeder Problem 3.6 Problem 3.6 a) Peskin and Schroeder already give the 6 Lorentz structures, so we need to normalize them properly. The list is:, 0,ı i, 5,ı 0 i 0 5, 5, i,ı i j. One can see that the trace of the product of any two di erent structures will vanish by explicit computation. b) We may write (ū A u )(ū 3 B u 4 ) = ū a u b ū c 3u d 4M abcd M abcd = Define the 6 matrices (G cb )by(g cb ) ad = G abcd,wherea, b are the matrix indices and c and d are the labels of the 6 matrices. We then find A ab B cd A ab B cd = X r s (G rs ) ad c b r,s = X rs (G rs ) ad (H rs ) cb where the matrices H rs are defined by (H rs ) cb = c r b s. Since the 6 Lorentz structures we defined above define a complete set of 4x4 matrices (any 4x4 matrix is a linear combination of the matrices above), we can express the matrices H rs and G rs instead in that basis and write A ab B cd = X CCD AB C ad D cb C,D (ū A u )(ū 3 B u 4 ) = X C,D for some coe cients CAB CD. We may then write C AB CD(ū C u 4 )(ū 3 D u ) ( A ab)( B cd) = CCD( AB ad)( D cb) ( A ab)( B cd)( C da)( D bc) = CCD( AB ad)( D cb)( C da)( C bc) ( A ab)( D bc)( B cd)( C da) = CCD( AB ad)( C da)( D cb)( D bc) Summing over indices a, b, c and d gives us Tr[ADBC] = CCDTr[C AB ]Tr[D ] CCD AB = 6 Tr[ADBC]

9 c) For (ū u )(ū 3 u 4 ), we essentially have = =. So the completeness relation tells us (ū u )(ū 3 u 4 ) = X (ū C u 4 )(ū C 3 u ) 4 For (ū µ u )(ū 3 µ u 4 ), note that the metric sign in the case of spatial indices (µ =,, 3) has the same e ect as the ı factor in our basis of matrices. Since, up to this ı, wehave A = B, we only have a non-zero CCD AB if C = D (to see this, just commute the A over to the B... for all of our structures ( A ) =, C so we are left with ±Tr[ D ] ). Moreover, we are summing over the µ. So for example, we see that the C = terms will cancel, since each one commutes with µ for two choices of µ and anti-commutes for two choices. This choices which commute give an opposite sign in the trace from those which anti-commute, so the sum over all terms vanishes. We need only remember that commutes with all µ and 5 anti-commutes with all of them. anti-commutes with three of them and commutes with one, while 5 commutes with three and anti-commutes with one. It follows that C (ū µ u )(ū 3 µ u 4 ) = X C (C)(ū C u 4 )(ū 3 C u ) where (C) = for C =, (C) = for C = 5, (C) = for C = and (C) = for C = 5.