Quantum Field Theory Homework 4: Coherent States and Lorentz Commutation Relations, Assignments of Quantum Mechanics

Download Quantum Field Theory Homework 4: Coherent States and Lorentz Commutation Relations and more Assignments Quantum Mechanics in PDF only on Docsity! PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 4 1 Physics 513, Quantum Field Theory Homework 4 Due Tuesday, 30th September 2003 Jacob Lewis Bourjaily 1. We have defined the coherent state by the relation |{ηk}〉 ≡ N exp {∫ d3k (2π)3 ηka † k√ 2Ek } |0〉. For my own personal convenience throughout this solution, I will let A ≡ ∫ d3k (2π)3 ηka † k√ 2Ek . a) Lemma: [ ap, e A] = ηp√ 2Ep eA. proof: First we note that from simple Taylor expansion (which is justified here), eA = 1 +A+ A 2 2 + A3 3! + . . . Clearly ap commutes with 1 so we may write, [ ap, e A] = [ap,A] + 12 [ap,A 2] + 1 3! [ap,A3] + . . . , = [ap,A] + 12 ([ap,A]A+A[ap,A]) + 1 3! ( [ap,A]A2 +A[ap,A]A+A[ap,A]A ) + . . . , ∗= [ap,A] ( 1 +A+ A 2 2 + A3 3! + A4 4! + . . . ) , = [ap,A]eA. Note that the step labelled ‘*’ is unjustified. To allow the use of ‘*’ we must show that [ap,A] is an invariant scalar and therefore commutes with all the A’s. This is shown by direct calculation. [ap,A] = ∫ d3k (2π)3 ηk√ 2Ek [ap, a † k], = ∫ d3k (2π)3 ηk√ 2Ek (2π)3δ(3)(~p− ~k), = ηp√ 2Ep . This proves what was required for ‘*.’ ηp√ 2Ep is clearly a scalar because η and Ep are real numbers only. But by demonstrating the value of [ap,A] we can complete the proof of the required lemma. Clearly, [ ap, e A] = [ap,A]eA = ηp√ 2Ep eA. ‘óπ²ρ ’²́δ²ι δ²Äιξαι It is clear from the definition of the commutator that apeA = [ ap, e A] + eAap. Therefore it is intuitively obvious, and also proven that ap|{ηk}〉 = NapeA|0〉, = N ([ap, eA ] + eAap ) |0〉, = N ηp√ 2Ep |0〉+N eAap|0〉, ∴ ap|{ηk}〉 = ηp√ 2Ep ap|{ηk}〉. (1.1) ‘óπ²ρ ’²́δ²ι δ²Äιξαι 2 JACOB LEWIS BOURJAILY b) We are to compute the normalization constant N so that 〈{ηk}|{ηk}〉 = 1. I will proceed by direct calculation. 1 = 〈{ηk}|{ηk}〉, = N ∗〈0|e ∫ d3k (2π)3 ηkak√ 2Ek |{ηk}〉, = N ∗〈0|e ∫ d3k (2π)3 ηk√ 2Ek |{ηk}〉 because we know that ak|{ηk}〉 = ηk√2Ek |{ηk}〉. So clearly 1 = |N |2e ∫ d3k (2π)3 η2k 2Ek , ∴ N = e− 12 ∫ d3k (2π)3 η2k 2Ek . c) We will find the expectation value of the field φ(x) by direct calculation as before. φ(x) = 〈{ηk}|φ(x)|{ηk}〉 = 〈{ηk}| ∫ d3p (2π)3 1√ 2Ep ( ape i~p·~x + a†pe −i~p·~x)|{ηk}〉, = ∫ d3p (2π)3 1√ 2Ep   〈{ηk}|apei~p·~x|{ηk}〉︸ ︷︷ ︸ act with ap to the right + 〈{ηk}|a†pe−i~p·~x|{ηk}〉︸ ︷︷ ︸ act with a†p to the left  , = ∫ d3p (2π)3 1√ 2Ep ( ηp√ 2Ep ei~p·~x + ηp√ 2Ep e−i~p·~x ) , = ∫ d3p (2π)3 ηp Ep cos(~p · ~x). d) We will compute the expected particle number directly. N = 〈{ηk}|N |{ηk}〉 = 〈{ηk}| ∫ d3p (2π)3 a†pap|{ηk}〉, = ∫ d3p (2π)3 ( 〈{ηk}|a†p←−−−− ap|{ηk−−−−→}〉 ) , = ∫ d3p (2π)3 η2p 2Ep . e) To compute the mean square dispersion, let us recall the theorem of elementary probability theory that 〈(∆N)2〉 = N2 −N2. We have already calculated N so it is trivial to note that N 2 = ∫ d3kd3p (2π)6 η2kη 2 p 4EkEp . Let us then calculate N2. N2 = 〈{ηk}|N2|{ηk}〉 = 〈{ηk}| ∫ d3kd3p (2π)6 a†kaka † pap|{ηk}〉, = ∫ d3kd3p (2π)6 ηkηp 2 √ EkEp 〈{ηk}|aka†p|{ηk}〉, = ∫ d3kd3p (2π)6 ηkηp 2 √ EkEp ( (2π)3δ(3)(~k − ~p) + 〈{ηk}|a†pak|{ηk}〉 ) , = ∫ d3k (2π)3 η2k 2Ek + ∫ d3kd3p (2π)6 η2kη 2 p 4EkEp . It is therefore quite easy to see that 〈(∆N)2〉 = N2 −N2 = ∫ d3k (2π)3 η2k 2Ek .