Homework 2 Solution

Since we integrate d3 p over all p space this

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Unformatted text preview: erent reason. Let’s consider the aa terms: d3 pd3 q (2π )6 =− d3 x =− RHS d3 pd3 q (2π )3 =− µ Ep qi ap aq e−i(pµ +qµ )x 4Ep Eq Ep qi ap aq e−i(Ep +Eq )t δ (p + q ) 4Ep Eq d3 p Ep (−pi ) ap a−p e−2iEp t (2π )3 2 4E p This integral is odd in direction i. The ap a−p term is clearly even, as is everything that contains Ep . The only component that isn’t is pi , which is odd, and thus the entire contents of the integral is odd. Since we integrate d3 p over all p-space, this integral vanishes. We get the same result for a∗ a∗ since they are complex conjugates of one another. pq As with the energy term, we are left with: d3 pd3 q (2π )6 µ Ep qi ap a∗ e−i(pµ −qµ )x + C.C. q 4Ep Eq = d3 x = d3 pd3 q δ (p − q ) (2π )3 = Pi d3 p Ep pi ap a∗ + C.C. p (2π )3 2Ep Ep qi ap a∗ e−i(Ep −Eq )t + C.C. q 4Ep Eq d3 p p i ap a∗ p (2π )3 = exactly as expected. 2. This is from Tong, but I really like it: The Fourier decomposition of a real scalar field and its conjugate momentum in the Schroedinger picture is given by: d3 p (2π )3 ˆ φ(x) = π (x) ˆ =i 1 ap eip·x + ap e−ip·x ˆ ˆ† 2Ep d3 p (2π )3 Ep ap eip·x − ap e−ip·x ˆ ˆ† 2 Show that the the commutation relations: ˆ ˆ φ(x), φ(y ) = [ˆ (x), π (y )] = 0 π ˆ ˆ φ(x), π (y ) ˆ = iδ (x − y) imply that: [ˆp , aq ] = aˆ ap , aq ˆ ˆ† ap , aq = 0 ˆ† ˆ† = (2π )3 δ (p − q ) a point I glossed ove...
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