e i x k k e it \u03c9 k \u03c9 k 2 \u03c9 k 2 \u03c9 k 1 \u03c9 k 1 \u03c9 k e it \u03c9 k \u03c9 k \u03b4 3 k k 0 a k a k d

E i x k k e it ω k ω k 2 ω k 2 ω k 1 ω k 1 ω k

This preview shows page 4 - 7 out of 7 pages.

e - i x · ( k + k ) e it ( ω k + ω k ) = 2 ω k 2 ω k ( - 1 ω k + 1 ω k ) e it ( ω k + ω k ) δ 3 ( k + k ) = 0 [ a k , a k ] = d 3 x d 3 y 2 ω k (2 π ) 3 / 2 2 ω k (2 π ) 3 / 2 ( 1 ω k δ 3 ( x - y ) + 1 ω k δ 3 ( x - y ) ) e - ikx e ik y = d 3 x 2 ω k 2 ω k (2 π ) 3 ( 1 ω k + 1 ω k ) e i x · ( k - k ) e it ( ω k - ω k ) = 2 ω k 2 ω k ( 1 ω k + 1 ω k ) e it ( ω k - ω k ) δ 3 ( k - k ) = δ 3 ( k - k ) The last equality in the first commutation relation follows because the prefactor of the delta function vanishes for k = - k , and the delta function vanishes for all k ̸ = - k , so it vanishes for all k . The last equality in the second commutation relation follows because the prefactor of the delta function is 1 for k = k , and is irrelevant for all k ̸ = k because the delta function vanishes. Of course, [ a k , a k ] = 0 follows by hermitian conjugation. (c) On one hand, U (Λ) φ ( x ) U (Λ) = d 3 k (2 π ) 3 / 2 2 ω k ( U (Λ)( 2 ω k a k ) U (Λ) e ikx + U (Λ)( 2 ω k a k ) U (Λ) e - ikx ) since the integration measure d 3 k (2 π ) 3 2 ω k and inner product kx are Lorentz invariant. On the other hand, U (Λ) φ ( x ) U (Λ) = φ x ) = d 3 k (2 π ) 3 / 2 2 ω k ( ( 2 ω k a k ) e ik Λ x + ( 2 ω k a k ) e - ik Λ x ) 4
Image of page 4
so that d 3 k (2 π ) 3 / 2 2 ω k ( U (Λ)( 2 ω k a k ) U (Λ) e ikx + U (Λ)( 2 ω k a k ) U (Λ) e - ikx ) = d 3 k (2 π ) 3 / 2 2 ω k ( ( 2 ω k a k ) e ik x ) + ( 2 ω k a k ) e - ik x ) ) = d 3 k (2 π ) 3 / 2 2 ω k ( ( 2 ω k a k ) e i - 1 k ) x + ( 2 ω k a k ) e - i - 1 k ) x ) = d 3 k (2 π ) 3 / 2 2 ω k ( ( 2 ω Λ k a Λ k ) e ikx + ( 2 ω Λ k a Λ k ) e - ikx ) Since Fourier expansions (or this one with oddly-normalized modes) are unique, we conclude that U (Λ)( 2 ω k a k ) U (Λ) = 2 ω Λ k a Λ k Problem 3 d 3 k (2 π ) 3 e ikx ω k = 2 π 1 - 1 d (cos θ ) 0 dk k 2 (2 π ) 3 e ikr cos θ ω k e - k t = 1 (2 π ) 2 1 ir 0 dk k ω k e - k t ( e ikr - e - ikr ) = 1 (2 π ) 2 1 ir -∞ dk k ω k e - k t + ikr This can be evaluated with a contour integral in the complex k plane on the contour pictured below. The square root in ω k introduces branch points at ± ; we connect them with a branch cut along the imaginary axis through complex infinity, out of the way of our desired integration path. k μ - μ ∖∖∖∖∖∖∖∖ The integrand has no poles inside this contour, so the contour integral vanishes. 5
Image of page 5
0 = I dk k μ 2 + k 2 e - i μ 2 + k 2 t + ikr = -∞ dk k μ 2 + k 2 e - i μ 2 + k 2 t + ikr + + ε i + ε idk ik μ 2 + ( ik ) 2 e - i μ 2 +( ik ) 2 t + i ( ik ) r + i ∞- ε - ε idk ik ( - 1) μ 2 + ( ik ) 2 e - i μ 2 +( ik ) 2 t + i ( ik ) r + π/ 2 - ε 0 k 2 e i 2 φ μ 2 + k 2 e i 2 φ e - i μ 2
Image of page 6
Image of page 7

You've reached the end of your free preview.

Want to read all 7 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture