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 −iωpt+ip·x † iωpt−ip·x φ(t, x) = 3 p ape + ape (1) (2π) 2ωp (i) By taking an explicit hermitian conjugation, we find our result that φ† = 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 † † semi-definite then ωp is real for all values of p. Also (ˆa ) =a ˆ is needed.
(ii) Since π = φ˙ = ∂tφ classically, we try this on the operator (1) to find Z d3p rω π(t, x) = −i p a e−iωpt+ip·x − a† eiωpt−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.
−iωpt+ip·x −ipx The second way to simplify notation is to write e = e so that we choose p0 = +ωp.
−iωpt+ip·x ˆ† † −iωpt+ip·x The simplest way however is to write Abp =a ˆpe and Ap =a ˆpe . You can quickly † iHt −iHt check these satisfy the same commutation relations asa ˆp anda ˆp. In fact Abp = e aˆpe =a ˆp(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 † i 3 3 h † † 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 ωq [φ(t, x), π(t, y)] = 3 p 3 (−i) (2π) 2ωp (2π) 2 h i −iωpt+ip·x † iωpt−ip·x −iωqt+iq·y † iωqt−iq·y (ape + ape ), (aqe − aqe ) (5) Z 3 3 d p d q r ωp = (−i) 3 3 (2π) (2π) 4ωq h i h i −i(ωp−ωq)t+ip·x−iq·y † i(ωp−ωq)t−ip·x+iq·y † −e ap, aq + e ap, aq (6) Z 3 3 r d p d q ωp −i(ωp−ωq)t+ip·x−iq·y 3 3 = (−i) 3 3 −e (2π) δ (p − q) (2π) (2π) 4ωq +ei(ωp−ωq)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 †i 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 φˆ(t, x), φˆ(t, y) = [ˆπ(t, x), πˆ(t, y)] = 0 . (11) we see that they are of similar form Z 3 h ˆ ˆ i h d p 1 +ip·x † −ip·x φ(t = 0, x), φ(t = 0, y) = 3 p (ˆape +a ˆpe ), (2π) 2ωp Z 3 d q 1 +iq·y † −iq·y i 3 p (ˆaqe +a ˆqe ) (12) (2π) 2ωq h Z d3p rω [ˆπ(t = 0, x), πˆ(t = 0, y)] = (−i) p (ˆa e+ip·x − aˆ† e−ip·x), (2π)3 2 p p Z d3q rω i (−i) q (ˆa e+iq·y − aˆ† e−iq·y) (2π)3 2 q q (13)
Thus all we need to show that Z d3pd3q h i C = (ˆa e+ip·x + saˆ† e−ip·x), (ˆa e−iq·y + saˆ† e+iq·y) , (14) s s/2 p p q q (ωpωq)
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 = [ˆa , aˆ ] e+ip·x+iq·y + s aˆ , aˆ† e+ip·x−iq·y s s/2 p q p q (ωpωq) h † i −ip·x+iq·y h † † i −ip·x−iq·y +s aˆp, aˆq e + aˆp, aˆq e (15) Z 3 3 d pd q +ip·(x−y) −ip·(x−y) = s sδ(p − q)e − sδ(p − q)e (16) (ωp) Z 3 d p +ip·(x−y) −ip·(x−y) = s s e − e (17) (ωp) 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 +∞ and −∞ as we integrate out at high energy scales, at large |p|. 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 |p| “nicely”. The functions needed will always respect our space-time symmetries, so here will always be functions of |p|. 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 ωp = p + m then ωp = ω−p and hence
Z Z 3 3 iq·x 3 d p 1 ip·x † −ip·x iq·x d x φ(t = 0, x)e = d x 3 p ape + ape e (18) (2π) 2ωp Z 3 3 d p 1 i(p+q)·x † i(q−p)·x = d x 3 p ape + ape (19) (2π) 2ωp Z 3 d p 1 3 (3) † 3 (3) = 3 p ap(2π) δ (q + p) + ap(2π) δ (q − p) (20) (2π) 2ωp
1 † = p a−q + aq (21) 2ωq
Z Z d3p rω d3x π(t = 0, x)eiq·x = −i d3x p a eip·x − a† e−ip·x eiq·x (22) (2π)3 2 p p Z d3p rω = −i d3x p a ei(p+q)·x − a† ei(q−p)·x (23) (2π)3 2 p p Z d3p rω = −i p a (2π)3δ(3)(q + p) − a† (2π)3δ(3)(q − p) (24) (2π)3 2 p p rω = −i q a − a† (25) 2 −q q
† Thus solving for a−q and aq gives
Z r ! 3 −iq·x ωq 1 aq = d x e φ(0, x) + ip π(0, x) (26) 2 2ωq Z r ! † 3 iq·x ωq 1 aq = d x e φ(0, x) − ip π(0, x) (27) 2 2ωq (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 † i 3 −ip·x 3 iq·y ap, aq = d x e d y e (29) "r r # ωp 1 ωq 1 × φ(0, x) + ip π(0, x), φ(0, y) − ip π(0, y) (30) 2 2ωp 2 2ωq Z r ω r ω = d3x d3y e−ip·x+iq·y −i p [φ(0, x), π(0, y)] + i q [π(0, x), φ(0, y)] (31) 4ωq 4ωp Z r ω r ω = d3x d3y e−ip·x+iq·y p δ(3)(x − y) + q δ(3)(y − x) (32) 4ωq 4ωp Z r ω r ω = d3x ei(q−p)·x p + q (33) 4ωq 4ωp = (2π)3δ(3)(p − q) (34)
(v) Taking the derivative we have
Z 3 d p ip ip·x † −ip·x ∇φ(x) = 3 p ape − ape (35) (2π) 2ωp
so4 Z P = − d3x π(x)∇φ(x) (36)
Z 3 3 r 3 d p d q ωp ip·x † −ip·x q iq·x † −iq·x = − d x 3 3 ape − ape p aqe − aqe (37) (2π) (2π) 2 2ωq Z 3 3 √ 3 d p d q ωpq = − d x 3 3 √ (38) (2π) (2π) 2 ωq n i(p+q)·x i(p−q)·x † −i(p−q)·x † −i(p+q)·x † † o × e apaq − e apaq − e apaq + e apaq (39) 3 3 √ Z d p d q ωpq = − 3 3 √ (40) (2π) (2π) 2 ωq n 3 (3) † † 3 (3) † † o × (2π) δ (p + q) apaq + apaq − (2π) δ (p − q) apaq + apaq (41) Z d3p p = − −a a − a† a† − a a† − a† a (42) (2π)3 2 p −p p −p p p p p
† † Now p apa−p + apa−p is an odd function under p → −p and so integrates to zero. Thus
Z d3p p p = a a† + a† a (43) (2π)3 2 p p p p Z d3p p = 2a† a + (2π)3δ(3)(0) (44) (2π)3 2 p p Z d3p = p a† a (45) (2π)3 p p
since again pδ(3)(0) is an odd function (albeit it poorly defined!).
4NOTE THIS IS FOR t = 0 should add in more exponentials for non-tirvial times! QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 5
† 3 The interpretation is that apap d p is the operator giving the number of quanta in a small volume d3p centred at momentum p. This will indeed contribute p to the total momenta. We see the same
type of term for the energy, the Hamiltonian operator H, but we get a factor of ωp not p in that case.
(vi) From (2) we have
Z Z Z d3p rω d3x Π2 = d3x − i p a e−iωpt+ip·x − a† eiωpt−ip·x (2π)3 2 p p Z d3q rω . − i q a e−iωqt+iq·x − a† eiωqt−iq·x (46) (2π)3 2 q q Z Z d3p Z d3q rω ω = − d3x p q a e−iωpt+ip·x − a† eiωpt−ip·x (2π)3 (2π)3 4 p p −iωqt+iq·x † iωqt−iq·x aqe − aqe (47)
Using that Z d3xeip·x = (2π)3δ(p) (48)
we apply the R d3x and use the resulting delta function of δ3(p ± q) to eliminate the q integral. This gives us
Z Z d3p ω d3x Π2 = − p a a e−2iωpt − a a† − a† a + a† a† e+2iωpt (49) (2π)3 2 p −p p p p p p −p
From (2) we have
Z Z Z 3 3 2 3 d p 1 −iωpt+ip·x † iωpt−ip·x d x (∇φ) = d x 3 p ipape − ipape (2π) 2ωp Z 3 d q 1 −iωqt+iq·x † iωqt−iq·x . 3 p iqaqe − iqaqe (50) (2π) 2ωq Z Z 3 Z 3 3 d p d q p.q −iωpt+ip·x † iωpt−ip·x = d x 3 3 p iape − iape (2π) (2π) 4ωpωq −iωqt+iq·x † iωqt−iq·x . iaqe − iaqe (51)
Again we apply the R d3x and use the resulting delta function of δ3(p±q) to eliminate the q integral to find Z Z 3 2 3 2 d p p −2iωpt † † † † 2iωpt d x (∇φ) = 3 apa−pe + apap + apap + apa−pe (52) (2π) 2ωp Finally, in the same manner we find that
Z Z 3 3 2 d p 1 −2iωpt † † † † 2iωpt d x (φ) = 3 apa−pe + apap + apap + apa−pe (53) (2π) 2ωp QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 6 Putting this together we find 1 Z H = d3x Π2 + (∇φ)2 + m2φ2 (54) 2 Z 3 2 2 2 1 d p ωp + p + m † † = 3 apap + apap 2 (2π) 2ωp Z 3 −ω2 + p2 + m2 1 d p p −2iωpt † † 2iωpt 3 apa−pe + apa−pe (55) 2 (2π) 2ωp 1 Z d3p Z d3p 1 = ω a a† + a† a = ω a† a + (2π)3δ3(0) (56) 2 (2π)3 p p p p p (2π)3 p p p 2
p 2 2 with ωp = + p + m
(vii)
Z d3p 1 [H, a† a ] = ω a† a + , a† a (57) k k (2π)3 p p p 2 k k Z d3p = ω a† a , a† a (58) (2π)3 p p p k k Z d3p h i = ω a† a , a† a = 0 (59) (2π)3 p p p k k
Note that this result is not true for any interesting i.e. interacting theory. The interactions mix the modes of different momenta leading to a lack of conservation of particle number. Only a continuous symmetry can guarantee conserved numbers and those are usually linked to total numbers of various particles, not the individual quanta of one particle at one momentum.
2. Time evolution of annihilation operator
(i) From (56) and using the commutation relation
h † i 3 h † † i ap, aq = (2π) δ(p − q) , [ap, aq] = ap, aq = 0 , (60)
we have that Z d3q 1 [H, a ] = ω a† a + , a (61) p (2π)3 q q q 2 p Z d3q = ω a† a a − a a† a (62) (2π)3 q q q p p q q Z d3q = ω a† a a − (a† a a + (2π)3δ(p − q)a ) (63) (2π)3 q q q p q q p −p Z 3 = − d qωqδ(p − q))ap = −ωpap (64)
as required.
(ii) We want to prove that n n (itH) ap = ap [it(H − ωp)] (65) QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 7
using [H, ap] = −ωpap. First note that trivially the relation holds for n = 1. Now if we assume the relation holds for n = m then
m+1 m (itH) ap = (itH)(itH) ap (66) m = itHap [it(H − ωp)] (67) m = it(apH − ωpap)[it(H − ωp)] (68) m+1 = ap [it(H − ωp)] . (69)
So the relation holds for n = m + 1 if it holds for n = m. Thus, by induction, it holds for all n ≥ 0.
(iii) Start by expanding the first exponential
∞ ! X 1 eiHta e−iHt = (iHt)n a e−iHt (70) p n! p n=0 ! X 1 = (iHt)na e−iHt (71) n! p n ! X 1 = a [it(H − ω )]n e−iHt (72) n! p p n ! X 1 = a [it(H − ω )]n e−iHt (73) p n! p n it(H−ωp) −iHt = ape e (74)
it(H−ωp−H) = ape (75) (76)
Thus we see that5 iHt −iHt −itωp e ape = ape (77)
where we have used the obvious facts that [H,H] = 0 and [ωp,H] = 0. † The corresponding equation for ap follows from hermitian conjugation