By Mark Srednicki

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**Extra resources for Quantum Field Theory: Problem Solutions**

**Example text**

17) This must be an orthogonal matrix of the form 1+θ ′′ . Using eq. 6), we find θ a θ ′b [T a , T b ] = iθ ′′c T c . Since the real parameters θ a and θ b can be chosen arbitrarily, this can only be true if [T a , T b ] = if abc T c , where the coefficients f abc are real. 3) a) From eq. 18) j aµ = i∂ µ ϕi (T a )ij ϕj . 19) θ a j µa = which yields b) Q = d3y j 0a (y), and j 0a = i∂ 0 ϕj (T a )jk ϕk = −iϕ˙ j (T a )jk ϕk = −iΠj (T a )jk ϕk . Q is time independent, so we can take y 0 = x0 . Then [ϕi (x), j 0a (y)] = −i[ϕi (x), Πj (y)](T a )jk ϕk (y) = δ3 (x−y)δij (T a )jk ϕk (y).

8), we get j µ = −v(1 + ρ/v)2 ∂ µ χ. In free-field theory, we then find k|j µ (x)|0 = ivkµ e−ikx , so f = v. b) We can compute corrections by treating j µ (x) as a vertex, and drawing Feynman diagrams with one external χ line. The single tree diagram just attaches that line to the vertex with vertex factor ivkµ , yielding f = v. Loop corrections will modify this to f = v(1 + O(λ)). 1) Aµν = 21 (B µν − B νµ ), T = gµν B µν , S µν = 12 (B µν + B νµ ) − 14 gµν T . 2) 4[Ni , Nj ] = [Ji , Jj ] − i[Ji , Kj ] − i[Ki , Jj ] − [Ki , Kj ] = iεijk (Jk − iKk − iKk + Jk ) = 4iεijk Nk .

C) Combining the results of parts (a) and (b), the symmetry is O(N ). d) A Dirac field is equivalent to two Weyl fields, hence the symmetry is U(2N ). e) Combining the results of parts (a) and (d), the symmetry is O(2N ). 1) Eq. 13) follows immediately because all components of χ anticommute with all components of ξ † . To get eq. 14), we write {Ψα , Ψβ } = = {χc , ξ a } {χc , χ†a˙ } {ξ †c˙ , ξ a } {ξ †c˙ , χ†a˙ } 0 σc0a˙ ˙ σ ¯ 0ca 0 δ3 (x−y) = γ 0 δ3 (x−y) . where we used eqs. 8) to get the second line.