Physics 513, Quantum Field Theory Homework 4 Due Tuesday, 30th September 2003

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1 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 1 Physics 513, Quantum Fiel Theory Homework Due Tuesay, 30th September 003 Jacob Lewis Bourjaily 1. We have efine the coherent state by the relation { } 3 k η k a k {η k } N exp (π 3 0. Ek For my own personal convenience throughout this solution, I will let 3 k η k a k A (π 3. Ek a Lemma: [ a p, e A] ηp e A. Ep proof: First we note that from simple Taylor expansion (which is justifie here, e A 1 + A + A + A ! Clearly a p commutes with 1 so we may write, [ ap, e A] [a p, A] + 1 [a p, A ] + 1 3! [a p, A 3 ] +..., [a p, A] + 1 ([a p, A]A + A[a p, A] + 1 ( [ap, A]A + A[a p, A]A + A[a p, A]A +..., 3! [a p, A] (1 + A + A + A3 + A +..., 3!! [a p, A]e A. Note that the step labelle * is unjustifie. To allow the use of * we must show that [a p, A] is an invariant scalar an therefore commutes with all the A s. This is shown by irect calculation. [a p, A] 3 k η k (π 3 [a p, a k ], Ek 3 k η k (π 3 (π 3 δ (3 ( p k, Ek. Ep This proves what was require for *. Ep is clearly a scalar because η an E p are real numbers only. But by emonstrating the value of [a p, A] we can complete the proof of the require lemma. Clearly, [ ap, e A] [a p, A]e A Ep e A. It is clear from the efinition of the commutator that a p e A [ a p, e A] + e A a p. Therefore it is intuitively obvious, an also proven that a p {η k } N a p e A 0, a p {η k } N ([ a p, e A] + e A a p 0, N Ep 0 + N e A a p 0, Ep a p {η k }. (1.1

2 JACOB LEWIS BOURJAILY b We are to compute the normalization constant N so that {η k } {η k } 1. I will procee by irect calculation. 1 {η k } {η k }, 3 k η N k a Ek k (π 0 e 3 {ηk }, N 0 e because we know that a k {η k } 3 k (π 3 η k Ek {ηk } η k Ek {η k }. So clearly 1 N e N e 1 3 k (π 3 η k E k, 3 k η (π 3 k E k. c We will fin the expectation value of the fiel φ(x by irect calculation as before. 3 p 1 ( φ(x {η k } φ(x {η k } {η k } ap (π 3 e i p x + a pe i p x {η k }, Ep 3 p 1 (π 3 Ep {η k} a p e i p x {η k } + {η k } a }{{} pe i p x {η k } }{{}, act with a p to the right act with a p to the left ( 3 p 1 η p (π 3 e i p x + e, i p x Ep Ep Ep 3 p (π 3 cos( p x. E p We will compute the expecte particle number irectly. 3 p N {η k } N {η k } {η k } (π 3 a pa p {η k }, 3 ( p (π 3 {η k } a p a p {η k }, 3 p ηp (π 3. E p e To compute the mean square ispersion, let us recall the theorem of elementary probability theory that ( N N N. We have alreay calculate N so it is trivial to note that N 3 k 3 p ηk (π 6. E k E p Let us then calculate N. N {η k } N 3 k 3 p {η k } {η k } (π 6 a k a ka pa p {η k }, 3 k 3 p η k (π 6 {η k } a k a p {η k }, E k E p 3 k 3 p η k η ( p (π 6 (π 3 δ (3 ( k p + {η k } a pa k {η k }, E k E p 3 k η k 3 k 3 p ηk (π 3 + E k (π 6. E k E p It is therefore quite easy to see that ( N N N 3 k ηk (π 3. E k

3 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 3. We are given the Lorentz commutation relations, [J µν, J ρσ ] i(g νρ J µσ g µρ J νσ g νσ J µρ + g µσ J νρ. a Given the generators of rotations an boosts efine by, L i 1 ɛijk J jk K i J 0i, we are to explicitly calculate all the commutation relations. We are given trivially that [L i, L j ] iɛ ijk L k. Let us begin with the K s. By irect calculation, [K i, K j ] [J 0i, J 0j ] i(g 0i J 0j g 00 J ij g ij J 00 + g 0j J i0, ij ij ; iɛ ijk L k. Likewise, we can irectly compute the commutator between the L an K s. This also will follow by irect calculation. [L i, K j ] 1 ɛlk [J ilk, J 0j ], 1 ɛilk i(g l0 J ij g i0 J lj g lj J i0 + g ij J l0, iɛ ijk J 0k ; iɛ ijk K k. We were also to show that the operators J i + 1 (Li + ik i J i 1 (Li ik i, coul be seen to satisfy the commutation relations of angular momentum. compute, First let us [J +, J ] 1 [ (L i + ik i, (L j ik i ], 1 ( [L i, L j ] + i[k i, L j ] i[l i, K j ] + [K i, K j ], 0. In the last line it was clear that I use the commutator [L i, K j ] erive above. The next two calculations are very similar an there is a lot of justification algebra in each step. There is essentially no way for me to inclue all of the etails of every step, but each can be verifie (e.g. i[k i, L j ] i[l j, K i ] ( iiɛ jik K k ɛ ijk K k...etc. They are as follows: [J+, i J+] j 1 [ (L i + ik i, (L j + ik j ], 1 ( [L i, L j ] + i[k i, L j ] + i[l i, K j ] + i[l i, K i ] [K i, K j ], 1 ( iɛ ijk L k ɛ ijk K k ɛ ijk K k + iɛ ijk L k, iɛ ijk 1 (Lk + ik k iɛ ijk J k +. Likewise, [J, i J ] j 1 [ (L i ik i, (L j ik j ], 1 ( [L i, L j ] i[k i, L j ] i[l i, K j ] + i[l i, K i ] [K i, K j ], 1 ( iɛ ijk L k + ɛ ijk K k + ɛ ijk K k + iɛ ijk L k, iɛ ijk 1 (Lk ik k iɛ ijk J k.

4 JACOB LEWIS BOURJAILY b Let us consier first the (0, 1 representation. For this representation we will nee to satisfy J i + 1 (Li + ik i 0 J i 1 (Li ik k σi. This is obtaine by taking L i σi an Ki iσi. The transformation law then of the (0, 1 representation is µν Φ (0, 1 iωµνj e Φ (0, 1, e i(θi L i +β j K j Φ (0, 1, e iθi σ i + βj K j Φ (0, 1. The calculation for the ( 1, 0 representation is very similar. Taking Li σi an Ki σi we get J+ i 1 (Li + ik i σi J i 1 (Li ik k 0. Then the transformation law of the representation is µν Φ ( 1,0 iωµνj e Φ ( 1,0, e i(θi L i +β j K j Φ ( 1,0, e iθi σ i βj K j Φ ( 1,0. Comparing these transformation laws with Peskin an Schroeer s (3.37, we see that ψ L Φ ( 1,0 ψ R Φ (0, a We are given that T a is a representation of some Lie group. This means that [T a, T b ] if abc T c by efinition. Allow me to take the complex conjugate of both sies. Note that [T a, T b ] [( T a, ( T b ] in general an recall that f abc are real. [T a, T b ] (if abc T c, [T a, T b ] if abc T c, [( T a, ( T b ] if abc ( T c. So by the efinition of a representation, it is clear that ( T a is also a representation of the algebra. b As before, we are given that T a is a representation of some Lie group. We will take the Hermitian ajoint of both sies. [T a, T b ] (if abc T c, (T a T b (T b T a if abc T c, T b T a T at b if abc T c, [T b, T a] if abc T c, [T a, T b ] if abc T c. So by the efinition of a representation, it is clear that T a is a representation of the algebra. c We efine the spinor representation of SU( by T a σa so that T 1 1 ( 0 1 T ( 0 i T i ( We will consier the matrix S iσ. Clearly S is unitary because (iσ (iσ 1. Now, one coul procee by irect calculation to emonstrate that ST 1 S 1 ( 0 1 T ST S 1 ( 0 i T i 0 ST 3 S 1 ( 1 0 T This clearly emonstrates that the representation T a is equivalent to that of T a.,

5 PHYSICS 513: QUANTUM FIELD THEORY HOMEWORK 5 From our efinitions of our representation of SO(3, 1 using J i + an J i, it is clear that (J i + J i. This coul be expresse as if ( 1, 0 (0, 1, or, rather L R. So what we must ask ourselves is, oes there exist a unitary matrix S such that SLS L but SKS K? If there i exist such a unitary transformation, then we coul conclue that L an R are equivalent representations. However, this is not possible in our SO(3, 1 representation because both L an K are represente strictly by the Pauli spin matrices so that ik L σ. It is therefore clear that there cannot exist a transformation that will change the sign of K yet leave L alone. So the representations are inequivalent.