Homework 2 Solution

# This leaves only the aa terms im only going to write

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Unformatted text preview: p0 +q0 )t 4Ep Eq 2 −Ep + |p|2 + m2 ap a−p e−i(p0 +q0 )t =0 This vanishes since the term in the parentheses is necessarily zero by our dispersion relation: 2 Ep = |p|2 + m2 The a∗ a∗ term vanishes identically – and necessarily – since the two terms are complex conjugates pq to one another. This leaves only the aa∗ terms. I’m only going to write out the terms with the negative exponents, and leave C.C. to correspond to the complex conjugate. d3 pd3 q (2π )6 µ 1 Ep Eq + p · q + m2 ap a∗ e−i(pµ −qµ )x + C.C. q 4Ep Eq = 1 2 d3 x = 1 2 d3 pd3 q (2π )3 = 1 2 d3 p (2π )3 = E 1 2 d3 p 1 2 2Ep ap a∗ + C.C q (2π )3 2Ep 1 δ (p − q ) Ep Eq + p · q + m2 ap a∗ e−i(Ep −Eq )t + C.C. q 4Ep Eq 1 2 4Ep 2 Ep + |p|2 + m2 ap a∗ + C.C q The Complex conjugate simply doubles the result. Multiplying it up we get: d3 p ∗ Ep ap ap (2π )3 E= (1) (b) What is the momentum for this ﬁeld? Sol. This result is much simpler, but involves an additional trick. First, we note the general relation: ˙ d3 xφ(x)φ,i (x) Pi = − that we derived from the T i0 component of the stress-energy tensor from our Noether relation. Plugging in, we get 4 terms: Pi = − d3 x d3 pd3 q (2π )6 µ µ µ µ Ep qi ap aq e−i(pµ +qµ )x − ap a∗ e−i(pµ −qµ )x − a∗ a∗ ei(pµ −qµ )x + a∗ a∗ ei(pµ −qµ )x q pq pq 4Ep Eq Here, the aa and a∗ a∗ terms vanish for a di...
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## This document was uploaded on 01/21/2014.

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