QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 1 Solutions 4: Free Quantum Field Theory 1. Heisenberg picture free real scalar field We have Z 3 d p 1 −i!pt+ip·x y i!pt−ip·x φ(t; x) = 3 p ape + ape (1) (2π) 2!p (i) By taking an explicit hermitian conjugation, we find our result that φy = φ. You need to note that all the parameters are real: t; x; p are obviously real by definition and if m2 is real and positive y y semi-definite then !p is real for all values of p. Also (^a ) =a ^ is needed. (ii) Since π = φ_ = @tφ classically, we try this on the operator (1) to find Z d3p r! π(t; x) = −i p a e−i!pt+ip·x − ay ei!pt−ip·x (2) (2π)3 2 p p (iii) In this question as in many others it is best to leave the expression in terms of commutators. That is exploit [A + B; C + D] = [A; C] + [A; D] + [B; C] + [B; D] (3) as much as possible. Note that here we do not have a sum over two terms A and B but a sum over an infinite number, an integral, but the principle is the same. −i!pt+ip·x −ipx The second way to simplify notation is to write e = e so that we choose p0 = +!p. −i!pt+ip·x ^y y −i!pt+ip·x The simplest way however is to write Abp =a ^pe and Ap =a ^pe . You can quickly y iHt −iHt check these satisfy the same commutation relations asa ^p anda ^p. In fact Abp = e a^pe =a ^p(t) for a free field, i.e. we have just applied time evolution to the free field Heisenberg picture operators. We have the usual commutation relations for the annihilation and creation operators with contin- uous labels (here the three-momenta) 1 h y i 3 3 h y y i ap; aq = (2π) δ (p − q) ; [ap; aq] = ap; aq = 0 : (4) The first field commutation relation is then Z 3 Z 3 r d p 1 d q !q [φ(t; x); π(t; y)] = 3 p 3 (−i) (2π) 2!p (2π) 2 h i −i!pt+ip·x y i!pt−ip·x −i!qt+iq·y y i!qt−iq·y (ape + ape ); (aqe − aqe ) (5) Z 3 3 d p d q r !p = (−i) 3 3 (2π) (2π) 4!q h i h i −i(!p−!q)t+ip·x−iq·y y i(!p−!q)t−ip·x+iq·y y −e ap; aq + e ap; aq (6) Z 3 3 r d p d q !p −i(!p−!q)t+ip·x−iq·y 3 3 = (−i) 3 3 −e (2π) δ (p − q) (2π) (2π) 4!q +ei(!p−!q)t−ip·x+iq·y(2π)3(−δ3(p − q)) (7) Z d3p 1 = i e+ip·(x−y) + e−ip·(x−y) (8) (2π)3 2 i = 2δ3(x − y) (9) 2 1 −3 h yi Note that this gives the annihilation and creation operators units of energy (in natural units). The usual ai; aj = iδij is for the case of discrete labels. QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 2 Thus we find the canonical equal-time commutation relation [φ(t; x); π(t; y)] = iδ3(x − y) : (10) For the commutators h i φ^(t; x); φ^(t; y) = [^π(t; x); π^(t; y)] = 0 : (11) we see that they are of similar form Z 3 h ^ ^ i h d p 1 +ip·x y −ip·x φ(t = 0; x); φ(t = 0; y) = 3 p (^ape +a ^pe ); (2π) 2!p Z 3 d q 1 +iq·y y −iq·y i 3 p (^aqe +a ^qe ) (12) (2π) 2!q h Z d3p r! [^π(t = 0; x); π^(t = 0; y)] = (−i) p (^a e+ip·x − a^y e−ip·x); (2π)3 2 p p Z d3q r! i (−i) q (^a e+iq·y − a^y e−iq·y) (2π)3 2 q q (13) Thus all we need to show that Z d3pd3q h i C = (^a e+ip·x + sa^y e−ip·x); (^a e−iq·y + sa^y e+iq·y) ; (14) s s=2 p p q q (!p!q) is zero where s = ±1 as the field (momenta) commutator is proportional to C+ (C−). Using (3) and (4) we find exactly as before that Z d3pd3q h i C = [^a ; a^ ] e+ip·x+iq·y + s a^ ; a^y e+ip·x−iq·y s s=2 p q p q (!p!q) h y i −ip·x+iq·y h y y i −ip·x−iq·y +s a^p; a^q e + a^p; a^q e (15) Z 3 3 d pd q +ip·(x−y) −ip·(x−y) = s sδ(p − q)e − sδ(p − q)e (16) (!p) Z 3 d p +ip·(x−y) −ip·(x−y) = s s e − e (17) (!p) If you now change integration variable in the second term to p0 = −p you will find it is of exactly the same form as the first term and these cancel giving zero as required. Aside - Trying to be Careful This is really rubbish isn't it! The integrals here look divergent as they scale as energy (=momen- tum=mass) to the power (3 − s). The oscillatory factor just means we are adding +1 and −∞ as we integrate out at high energy scales, at large jpj. The whole thing looks (and is) badly defined. QFT is not very reasonable about divergences. Riemann-Lebesgue lemma2 is the real answer here but unlikely to be helpful to Imperial Physics students who aren't given the relevant mathematical tools. If we set up the problem more carefully we could apply this lemma and you'd be happy. In practice I haven't set up the maths carefully. I think you will find most QFT text books are the same. We really should define everything carefully to make sure that such integrals are always defined properly. 2See https://bit.ly/2KOsAkA. QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 3 The quick and dirty way here is to remember that these integrals here are part of an operator expression coming from commutators even if it is a unit operator. To evaluate these expressions the operator must act on something. So you should let your expression act on a dummy function f(p). Now this dummy function has to be of the right type, some well behaved function. Basically that has to be something that falls off for large jpj \nicely". The functions needed will always respect our space-time symmetries, so here will always be functions of jpj. A suitable example might be something that falls off as exp(−p:p=(2σ2)) where you can choose sigma to be as big as you like (take it to infinity only after everything else is done). Now your integrals are of the right form for Riemann-Lebesgue lemma to apply. Alternatively, as each integral is now well behaved and finite, you can start to manipulate them. In our case we would need f(p) = f(−p) for our odd/even arguments to work but the space-time symmetries guarantee that any function we have in practice will have that symmetry. To be more precise we should set everything up carefully. This is what the Axiomatic QFT3 is very careful about. p 2 2 (iv) Given !p = p + m then !p = !−p and hence Z Z 3 3 iq·x 3 d p 1 ip·x y −ip·x iq·x d x φ(t = 0; x)e = d x 3 p ape + ape e (18) (2π) 2!p Z 3 3 d p 1 i(p+q)·x y i(q−p)·x = d x 3 p ape + ape (19) (2π) 2!p Z 3 d p 1 3 (3) y 3 (3) = 3 p ap(2π) δ (q + p) + ap(2π) δ (q − p) (20) (2π) 2!p 1 y = p a−q + aq (21) 2!q Z Z d3p r! d3x π(t = 0; x)eiq·x = −i d3x p a eip·x − ay e−ip·x eiq·x (22) (2π)3 2 p p Z d3p r! = −i d3x p a ei(p+q)·x − ay ei(q−p)·x (23) (2π)3 2 p p Z d3p r! = −i p a (2π)3δ(3)(q + p) − ay (2π)3δ(3)(q − p) (24) (2π)3 2 p p r! = −i q a − ay (25) 2 −q q y Thus solving for a−q and aq gives Z r ! 3 −iq·x !q 1 aq = d x e φ(0; x) + ip π(0; x) (26) 2 2!q Z r ! y 3 iq·x !q 1 aq = d x e φ(0; x) − ip π(0; x) (27) 2 2!q (28) 3See https://bit.ly/2wrg1ID. QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 4 Hence since [φ(x); φ(y)] = [π(x); π(y)] = 0 Z h y i 3 −ip·x 3 iq·y ap; aq = d x e d y e (29) "r r # !p 1 !q 1 × φ(0; x) + ip π(0; x); φ(0; y) − ip π(0; y) (30) 2 2!p 2 2!q Z r ! r ! = d3x d3y e−ip·x+iq·y −i p [φ(0; x); π(0; y)] + i q [π(0; x); φ(0; y)] (31) 4!q 4!p Z r ! r ! = d3x d3y e−ip·x+iq·y p δ(3)(x − y) + q δ(3)(y − x) (32) 4!q 4!p Z r ! r ! = d3x ei(q−p)·x p + q (33) 4!q 4!p = (2π)3δ(3)(p − q) (34) (v) Taking the derivative we have Z 3 d p ip ip·x y −ip·x rφ(x) = 3 p ape − ape (35) (2π) 2!p so4 Z P = − d3x π(x)rφ(x) (36) Z 3 3 r 3 d p d q !p ip·x y −ip·x q iq·x y −iq·x = − d x 3 3 ape − ape p aqe − aqe (37) (2π) (2π) 2 2!q Z 3 3 p 3 d p d q !pq = − d x 3 3 p (38) (2π) (2π) 2 !q n i(p+q)·x i(p−q)·x y −i(p−q)·x y −i(p+q)·x y y o × e apaq − e apaq − e apaq + e apaq (39) 3 3 p Z d p d q !pq = − 3 3 p (40) (2π) (2π) 2 !q n 3 (3) y y 3 (3) y y o × (2π) δ (p + q) apaq + apaq − (2π) δ (p − q) apaq + apaq (41) Z d3p p = − −a a − ay ay − a ay − ay a (42) (2π)3 2 p −p p −p p p p p y y Now p apa−p + apa−p is an odd function under p ! −p and so integrates to zero.