Quantum Field Theory: Problem Solutions by Mark Srednicki

By Mark Srednicki

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Thus [ϕ(x), j 0 (y)] = −i[ϕ(x), Π(y)]ϕ(y) = δ3 (x−y)ϕ(y), and integrating over d3y yields [ϕ(x), Q] = ϕ(x). Next, let F (α) ≡ e−iαQ ϕe+iαQ , and note that F ′ (α) = e−iαQ i[ϕ, Q]e+iαQ . Since [ϕ, Q] = ϕ, this becomes F ′ (α) = ie−iαQ ϕe+iαQ = iF (α). Therefore F (n) (α) = in F (α), (n) (0)αn /n! = and F (n) (0) = in F (0) = in ϕ. Thus, by Taylor expansion, F (α) = ∞ n=0 F ∞ ϕ n=0 in αn /n! = ϕe+iα . b) Since [H, Q] = 0, He−iαQ |θ = 0, so e−iαQ |θ must be a linear combination of vacua. Then, since e+iαQ ϕe−iαQ = e−iα ϕ, we have θ|e+iαQ ϕe−iαQ |θ = e−iα θ|ϕ|θ ; using eq.

Examining eq. 13), we see that the integrand in eq. 18) is simply the O(α) contribution to πρ(s). [The s → s + iǫ prescription is implicit in eq. ] So ∞ −1 = 1 − A + O(α2 ), this verifies we conclude that, to O(α), A = − 4m 2 ds ρ(s). Since Zϕ ∞ −1 Zϕ = 1 + 4m2 ds ρ(s) to O(α). Incidentally, you could try to carry out this integral over s with finite ε, then take the ε → 0 limit, and hence get the value of κA . 4. 2) We start with the Cauchy integral formula for second derivative, Π′′ (k2 ) = 2!

44) by xk y l and integrate over d3x and d3y. We get [D ki , Dlj ] = i d3x d3y xk y l Πa (x)∇ix δ3 (x−y)∇jy ϕa (y) − Πa (y)∇jy δ3 (x−y)∇ix ϕa (x) = −i d3x d3y y l ∇ix (xk Πa (x))∇jy ϕa (y) − xk ∇jy (y l Πa (y))∇ix ϕa (x) δ3 (x−y) = −i d3x xl ∇i (xk Πa )∇j ϕa − xk ∇j (xl Πa )∇i ϕa = −i d3x xl (δik Πa + xk ∇i Πa )∇j ϕa − xk (δjl Πa + xl ∇j Πa )∇i ϕa . 54) The δik and δjl terms will vanish when we contract with εmki εnlj ; thus we have [J m , J n ] = −iεmki εnlj d3x xk xl ∇i Πa ∇j ϕa − xk xl ∇j Πa ∇i ϕa = +iεmki εnlj d3x Πa ∇i (xk xl ∇j ϕa ) − ∇j (xk xl ∇i ϕa ) = +iεmki εnlj d3x Πa (δik xl + δil xk + xk xl ∇i )∇j − (i↔j) ϕa = +iεmki εnlj d3x Πa (δil xk ∇i − δjk xl ∇i )ϕa = −iεmki εnlj d3x (δil xk T 0j − δjk xl T 0i ) = −iεmki εnlj (δil Dkj − δjk D li ) = −i (εmki εnij )D kj − (εmji εnlj )D li = −i (δjm δkn − δjk δmn )D kj − (δin δlm − δil δmn )D li = −i Dnm − D mn = +iεmnp J p .

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