Srednicki Chapter 55

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1 Srednicki Chapter 55 QFT Problems & Solutions A. George August 3, 03 Srednicki 55.. Use equations and A i, A j ] = Π i, Π j ] = 0 (at equal times) to verify equations This is our third time solving this tedious problem, so let s note from the outset that all polarization vectors k have the same magnitude, and therefore ω = ω. Anyway, we begin by using equation 55.6: a ( k), a ( k )] = iε ( k) d 3 xe ikx 0 A(x), iε ( k ) d 3 x e ik x ] 0 A(x ) This is: a ( k), a ( k )] = ε ( k)ε ( k ) d 3 xd 3 x e ikx 0 A(x), e ik x ] 0 A(x ) We take the derivative as indicated at the bottom of page 34. We also use equation Then: a ( k), a ( k )] = ε ( k)ε k ) d 3 xd 3 x e ikx Π(x) iωe ikx A(x), e ik x Π(x ) iωe ik x A(x ) Only the cross terms survive. This gives: a ( k), a ( k )] = iωε ( k)ε ( k ) d 3 xd 3 x e i(kx+k x ) Π(x), A(x )] e i(kx+k x ) Π(x ), A(x)] } Now we take this second term and swap the two dummy variables: a ( k), a ( k )] = iωε ( k)ε ( } k ) d 3 xd 3 x e i(kx+k x ) e i(kx +k x) Π(x), A(x )] We now use equation The delta function sets x = x, which causes the term in braces to vanish. Thus: a ( k), a ( k )] = 0 We can take the Hermitiian conjugate of this to find: a ( k), a ( k )] = 0

2 The other commutator is: a ( k), a ( k )] = d 3 xd 3 x ε ( k) e ikx 0A(x), ε ( k ) e ik x ] 0A(x ) We write this dot product in index notation: a ( k), a ( k )] = ε i ( k)ε j ( k ) d 3 xd 3 x e ikx 0 A i (x), e ik x ] 0 A j (x ) Now we take the derivative, and recall that A = Π. Then: a ( k), a ( k )] = ε i ( k)ε j k ) d 3 xd 3 x e ikx Π i (x) + iωe ikx A i (x), e ik x Π j (x ) iωe ik x A j (x ) Next we recall A i (x), A j (x )] = Π i (x), Π j (x )] = 0. a ( k), a ( k )] = iωε i ( k)ε j ( k ) d 3 x d 3 x e i(k x kx) Π i (x), A j (x ) ] + A i (x), Π j (x ) ]} Now we have: a ( k), a ( k )] = iωε i ( k)ε j d 3 x d 3 x e i(k x kx) A j (x ), Π i (x) ] + A i (x), Π j (x ) ]} Now we use equation 55.0: a ( k), a ( k )] = ωε i ( k)ε j (π) 3 ei(k x kx) e i k ( x x) (δ ji k +e i k ( x x ) (δ ij k Now we can change the integral k k. Thus: a ( k), a ( k )] = ωε i ( k)ε j i k j k )} (π) 3 ei(k x kx) e i k ( x x ) (δ ij k Now in the second term, one of the dot products is ε ( k) k, where k is integrated over all possible values. The result, therefore, is zero: for every component of k in the same direction as ε ( k), the k will have the component in the opposite direction, resulting in a cancellation. Thus, the second term vanishes. Note that this has nothing to do with equation 55.3, as Srednicki claims in his solutions. Vectors k, k, and k are all distinct. j k i k i k j k ) )} This gives: a ( k), a ( k )] = ωε i ( k)ε j (π) 3 ei(k x kx) e i k ( x x ) δ ij

3 This dot product gives: a ( k), a ( k )] = ωε ( k) ε ( k ) Now we can use 3.7 and 55.4: a ( k), a ( k )] = ωδ (π) 3 ei(k x kx) e i k ( x x ) d 3 x d 3 x e i(k x kx) δ 3 ( x x ) Doing the x integral: a ( k), a ( k )] = ωδ 3 d 3 x e i(k k)x The temporal components of the exponential vanish, since ω = ω as discussed previously. Now we use 3.7 again on the spatial parts, which gives: a ( k), a ( k )] = ω(π) 3 δ δ 3 (k k) Srednicki 55.. Use equations 55., 55.4, 55.9, and to verify equation We begin with equation 55.9: H = Π iπ i + ja i j A i J i A i + H coul Now we have equation 55., and recall that Π = A. Then: H = (iω) ε i ( k)a ( k)e ikx + ε i k)a ( k)e ikx (iω) ε i ( k )a ( k )e ik x +ε i ( k )a ( k )e ik x (iω) ]+ (i) + ε i ( k )a ( k )(k j)e ik x + ε i ( k )a ( k )( k j)e ik x ε i ( k)a ( k)(k j )e ikx + ε i ( k)a ( k)( k j )e ikx ] ] J(x) A(x) + H coul Now we do some multiplication and factor terms from the first line with terms from the second and third lines. The result is: H = ε ( k) ε ( k )a ( k)a ( k )e i(k+k )x (ω + k k ) + ε ( k) ε ( k )a ( k)a ( k )e i(k+k )x (ω + k k )+ ε ( k) ε ( k )a ( k)a ( k )e i(k k)x ( k k ω ) + ε ( k) ε ( k )a ( k)a ( k )e i(k k )x ( k k } ω ) J(x) A(x) + H coul The next step is to move from the Hamiltonian density to the Hamiltonian, by integrating both sides. H = d 3 x ε ( k) ε ( k )a ( k)a ( k )e i(k+k )x (ω + k k ) 3

4 + ε ( k) ε ( k )a ( k)a ( k )e i(k+k )x (ω + k k ) ε ( k) ε ( k )a ( k)a ( k )e i(k k)x ( k k + ω ) ε ( k) ε ( k )a ( k)a ( k )e i(k k )x ( k k } + ω ) d 3 x J(x) A(x) + H coul Now the last two terms look good! In the first term, we use 3.7 to do the x-integral: H = (π)3 ε ( k) ε ( k )a ( k)a ( k )δ 3 ( k + k )e iωt (ω + k k ) + ε ( k) ε ( k )a ( k)a ( k )δ 3 ( k+ k )e iωt (ω + k k ) ε ( k) ε ( k )a ( k)a ( k )δ 3 ( k k)( k k +ω ) ε ( k) ε ( k )a ( k)a ( k )δ 3 ( k k )( k k } + ω ) d 3 x J(x) A(x) + H coul Now we can use the delta functions to perform the k integral: H = (π)3 ε (π) 3 ω ( k) ε ( k)a ( k)a ( k)e iωt (ω k ) + ε ( k) ε ( k)a ( k)a ( k)e iωt (ω k ) ε ( k) ε ( k)a ( k)a ( k)( k + ω ) ε ( k) ε ( k)a ( k)a ( k)( } k + ω ) d 3 x J(x) A(x) + H coul Now we recall ω = k, so the first two terms vanish, and we can third and fourth terms: H = (π)3 ε (π) 3 ( k) ε ( k)a ( k)a ( k)ω + ε ( k) ε ( k)a ( k)a k)ω Now we use equation 55.4: H = (π)3 δ (π) 3 a ( k)a ( k)ω + δ a ( k)a k)ω We can do the sum over : H = ω a ( k)a ( k) + a k)a ( k) =± We combine these first two terms with Then: H = ω a ( k)a k) + (π) 3 ωδ 3 (0) =± 4

5 which is: H = =± ω a ( k)a ( k) + =± (π) 3 ω δ 3 (0) Using 3.8: H = =± ω a ( k)a ( k) + =± d 3 k (π) 3 ω (π)3 ω δ 3 (0) We simplify, and take the sum: H = ω a ( k)a ( k) + =± d 3 k ωδ 3 (0) This gives: H = =± ω a ( k)a ( k) + E 0 (π) 3 δ 3 (0) As in chapter 3, we interpret V = (π) 3 δ 3 (0): H = ω a ( k)a ( k) + E 0 V =± which is equation