Homework 2 Solution

Homework 2 Solution - Quantum Field Theory HW 2 Solution...

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Quantum Field Theory – HW 2 Solution Key HW 2 Solution Key 1. In class, we showed that the general solution to the Klein-Gordon equation is: φ ( x ) = i d 3 p (2 π ) 3 1 r 2 E v p p a v p e i ( v p · vx - E v p t ) + a * v p e - i ( v p · - E v p t ) P (a) What is the energy for this Feld? ±or both this and for subsequent parts, I expect you to reduce the equation to a single 3-dimensional integral over v p or v q or similar. Sol. Based on the expression above, we get: ˙ φ ( x ) = i i d 3 p (2 π ) 3 E p r 2 E v p p a v p e i ( v p · - E v p t ) a * v p e - i ( v p · - E v p t ) P i φ ( x ) = i i d 3 p (2 π ) 3 p i r 2 E v p p a v p e i ( v p · - E v p t ) a * v p e - i ( v p · - E v p t ) P both of which we’ll need for the ensuing calculations. The energy density can be found by multiplying out the T 00 component of the stress energy tensor: ρ ( x ) = 1 2 ˙ φ 2 + 1 2 ( φ ) 2 + 1 2 m 2 φ 2 Multiplying (and collecting) the whole shebang is going to yield a total of 12 terms: ρ ( x ) = 1 2 i i d 3 pd 3 q (2 π ) 6 1 r 4 E p E q × b E p E q p a p a q e - i ( p μ + q μ ) x μ a p a * q e - i ( p μ - q μ ) x μ a * p a * q e i ( p μ - q μ ) x μ + a * p a * q e i ( p μ - q μ ) x μ P v p · v q p a p a q e - i ( p μ + q μ ) x μ a p a * q e - i ( p μ - q μ ) x μ a * p a * q e i ( p μ - q μ ) x μ + a * p a * q e i ( p μ - q μ ) x μ P + m 2 p a p a q e - i ( p μ + q μ ) x μ + a p a * q e - i ( p μ - q μ ) x μ + a * p a * q e i ( p μ - q μ ) x μ + a * p a * q e i ( p μ - q μ ) x μ PB These are ugly, but fortunately, they can be grouped. ±irst, note that we are actually computing: E = i d 3 ( x ) Now, let’s consider the a p a q terms: RHS = 1 2 i i i d 3 x d 3 pd 3 q (2 π ) 6 1 r 4 E p E q ( E p E q v p · v q + m 2 ) a p a q e - i ( p μ + q μ ) x μ = 1 2 i i d 3 pd 3 q (2 π ) 3 1 r 4 E p E q δ ( v p + v q ) ( E p E q v p · v q + m 2 ) a p a q e - i ( p 0 + q 0 ) t = 1 2 i d 3 p (2 π ) 3 1 R 4 E 2 p ( E 2 p + | v p | 2 + m 2 ) a v p a - v p e - i ( p 0 + q 0 ) t = 0 This vanishes since the term in the parentheses is necessarily zero by our dispersion relation: E 2 p = | v p | 2 + m 2
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The a * p a * q term vanishes identically – and necessarily – since the two terms are complex conjugates 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. E = 1 2 i i i d 3 x d 3 pd 3 q (2 π ) 6 1 r 4 E p E q ( E p E q + v p · v q + m 2 ) a p a * q e - i ( p μ - q μ ) x μ + C.C.
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Homework 2 Solution - Quantum Field Theory HW 2 Solution...

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