Homework 8 with Solutions - Advanced Quantum Mechanics I | PHYSICS 513, Assignments of Quantum Mechanics

Download Homework 8 with Solutions - Advanced Quantum Mechanics I | PHYSICS 513 and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 513, Quantum Field Theory Homework 8 Due Tuesday, 11th November 2003 Jacob Lewis Bourjaily Problem 4.1 We are to consider the problem of the creation of Klein Gordon particles by a classical source. This process can be described by the Hamiltonian H = Ho + ∫ d3x − j(x)φ(x), where Ho is the Klein-Gordon Hamiltonian, φ(x) is the Klein-Gordon filed, and j(x) is a c-number scalar function. Let us define the number λ by the relation λ = ∫ d3p (2π)3 1 2Ep |j̃ (p)|2. a) We are to show that the probability that the source creates no particles is given by P (0) = ∣∣∣∣〈0|T { exp [ i ∫ d4x j(x)φI(x) ]} |0〉 ∣∣∣∣ 2 . Without loss of understanding we will denote φ ≡ φI . Almost entirely trivially, we see that HI = − ∫ d3x j(x)φ(x). Therefore, P (0) = ∣∣∣∣〈0|T { exp [ −i ∫ dt′ HI(t′) ]} |0〉 ∣∣∣∣ 2 , = ∣∣∣∣〈0|T { exp [ i ∫ d4x j(x)φ(x) ]} |0〉 ∣∣∣∣ 2 . ‘óπ²ρ ’²́δ²ι δ²ιξαι b) We are to evaluate the expression for P (0) to the order j2 and show generally that P (0) = 1− λ +O(λ2). First, let us only consider the amplitude for the process. We can make the näıve expansion 〈0|T { exp [ i ∫ d4x j(x)φ(x) ]} |0〉 = 〈0|1|0〉+ i ∫ d4x j(x)〈0|φ(x)|0〉 − . . . . For every odd power of the expansion, there will be at least one field φIo that cannot be contracted from normal ordering and therefore will kill the entire term. So only even terms will contribute to the expansion. It should be clear that the amplitude will be of the form ∼ 1−O(j2)+O(j4)−. . .. Let us look at the O(j2) term. That term is given by 〈0|T { −½ (∫ d4x j(x)φ(x) )2} |0〉 = −1 2 ∫ d4xd4y j(x)j(y)〈0|T{φ(x)φ(y)}|0〉, = −1 2 ∫ d4xd4y j(x)j(y) DF (x− y), = −1 2 ∫ d4xd4y ∫ d4p (2π)4 i p2 −m2 + i²e −ip(x−y)j(x)j(y), = −1 2 ∫ d4p (2π)4 ∫ d4x j(x)e−ipx ︸ ︷︷ ︸ j̃ (p) ∫ d4y j(y)eipy ︸ ︷︷ ︸ j̃ ∗(p) i p2 −m2 + i² , = −1 2 ∫ d4p (2π)4 |j̃ (p)|2 i p2 −m2 + i² , = −1 2 ∫ d3p (2π)3 ∫ dp0 (2π) |j̃ (p)|2 i p2 −m2 + i² . 1 2 JACOB LEWIS BOURJAILY We know how to evaluate the integral ∫ dp0 (2π) |j̃ (p)|2 i p2 −m2 + i² = ∫ dp0 (2π) |j̃ (p)|2 i (p0)2 − E2p + i² , = ∫ dp0 (2π) |j̃ (p)|2 i (p0 − Ep)(p0 + Ep) . The function has a simple pole at p0 = −Ep with the residue i|j̃ (p)|2 p0 − Ep ∣∣∣∣ p0=−Ep = − i|j̃ (p)| 2 2Ep . We know from elementary complex analysis that the contour integral is 2πi times the residue at the pole. Therefore, −1 2 ∫ d3p (2π)3 ∫ dp0 (2π) |j̃ (p)|2 i p2 −m2 + i² = − 1 2 ∫ d3p (2π)3 1 2Ep |j̃ (p)|2, = −1 2 λ. Because we now know the amplitude to the first order of λ (or, rather, the second order of j), we have shown, as desired, that P (0) = |1− ½λ + . . . |2 ∼ 1− λ +O(λ2). ‘óπ²ρ ’²́δ²ι δ²ιξαι c) We must represent the term computed in part (b) as a Feynman diagram and show that the whole perturbation series for P (0) in terms of Feynman diagrams is precisely P (0) = e−λ. The term computed in part (b) can be represented by faf ≡ −λ. It has two points (neither originated by the source) and a time direction specified (not to be confused with charge or momentum). We can write the entire perturbation series as P (0) = ∣∣∣∣〈0|T { exp [ i ∫ d4x j(x)φ(x) ]} |0〉 ∣∣∣∣ 2 =  1 +faf+ faffaf + faffaffaf + faffaffaffaf + · · ·   2 . To get the series we must figure out the correct symmetry factors. If one begins with 2n vertices, then n of them must be chosen as ‘in’; there are 22n/2 = 2n ways to do this. After that, each one of the ‘in’ vertices must be paired with one of the ‘out’ vertices; you can do this n! ways. So the symmetry factor for the term with n uninteracting propagators is S(n) = 2n · n!. We may now compute the probability explicitly. P (0) =  1 +faf+ faffaf + faffaffaf + faffaffaffaf + · · ·   2 , = ( ∞∑ n=0 (−λ)n 2nn! )2 , = ( ∞∑ n=0 (−λ/2)n n! )2 , = ( e−λ/2 )2 , ∴ P (0) = e−λ. ‘óπ²ρ ’²́δ²ι δ²ιξαι