Physics 22A: HW3 solutions October 22, 202. a) It will help to start things off by doing soe gaussian integrals. Let x be a real vector of length, and let s copute dxe 2 xt Ax, where A is soe real atrix. First write the exponent as as A ij x i x j. In the upper indices this is syetric in i j, so only the syetric part of A will contribute and we can set A A T. This eans we can diagonalize A, A ij λ i δ ij (no su). So we get I dxe xt Ax [ i ] dy i e 2 λ iyi 2 i 2π (2π) λ i det A where y is an eigenbasis of A, obtained fro x by an orthogonal coordinate transforation y Ox. I ll write e Ax2 for e xt Ax. What about integrals with a linear ter, e 2 Ax2 Bx? These aren t uch harder, just coplete the square and shift the integration variable x x + A B to get dxe 2 Ax2 Bx () dxe 2 A(x+A B) 2 + 2 BA B e 2 BA B I (2) ow on to the proble. We want to evaluate the oentu integrals in the path integral expression giving the aplitude for the particle to travel fro point q at tie t to point q at tie t : q, t q, t DqDpe is (3) q(t )q, q(t )q dq k k j0 dp j pipi e i( 2π 2 p i δq i +V (q))δt δt where δq j q j+ q j. Coparing, we see that A iδt/ and B iδq j, which we ll write i q j δt. There are + of these integrals, and we get a fro the dp/2π. This gives the (2π) + result ( ) + q, t q, t dq k e iδt( 2 δ q2 +V (q)) (6) k (4) (5)
b) ow setting V to zero, we want to calculate the integrals over the qs, k Consider just the integral involving q : where a i/δt. Then I q2 ( ) + dq k e i 2δt j0 (q j+ q j ) 2 (7) I q dq e a 2 (q q 0 ) 2 e a 2 (q 2 q ) 2 (8) e 2 a(q2 0 +q2 2 ) dq e aq2 aq (q 0 +q 2 ) (9) B a(q 0 + q 2 ), A 2a (0) I q 2 (2πa )e 4 a(q 0+q 2 ) 2 e 2 a(q2 0 +q2 2 ) () 2 (2πa )e 4 a(q 2 q 0 ) 2 (2) (3) dq 2 I q e a 2 (q 3 q 2 ) 2 (4) dq 2 2 (2πa )e a 4 a(q 2 q 0 ) 2 e a 2 (q 3 q 2 ) 2 (5) 2 2 (2πa ) 3 (2πa )e a 6 (q 3 q 0 ) 2 (6) At this point it s probably easiest to just guess the general for and show it by induction. If the k th integral ter generates a factor e a 2k (q k q 0 ) 2 (ties stuff that doesn t atter) then I qk dq k e a 2 (q k+ q k ) 2 e a 2k (q k q 0 ) 2 (8) k k + (2πa )e a 2(k+) (q k+ q 0 ) 2 (9) We re intereted in the product of all these guys: (7) (20) I q e a 2(+) (q + q 0 ) 2 k k + (2πa ) (2) k 2πa e a 2(+) (q + q 0 ) 2 (22) + 2
Bringing in the noralization fro part a) gives the final result ( ) + q, t q, t I q (23) 2πi( + )δt e i 2(+)δt (q q ) 2 (24) 2πi(t t ) e i 2(t t ) (q q ) 2 (25) c) This is easy because for the free theory inserting a coplete set of p diagonalizes the entire hailtonian in one fell swoop. The coputation proceeds straightforwardly fro the definitions and the achinery fro the first part. q, t q, t q e iht e iht q (26) dp 2π q e ih(t t ) p p q (27) dp q ) 2π eip(q e i(t t p ) 2 2 (28) (q q ) 2 2πi(t t ) ei 2(t t ) (29) 2. Consider taylor expanding L(q, q) about the classical value: ( ) L L L L cl + q + q q q + ( ) 2 L q cl 2 q 2 q2 + 2 2 L q q q q + 2 L q 2 q2 + O( 3 ) q cl (3) The first variation vanishes; this is the stateent that we are expanding around a classical solution (saddle point of the action). ow specialize to the free particle lagrangian, L 2 q2. The first two ters in the second variation vanish in this case, and there are no ters of order δ 3, but this isn t necessary for the general conclusions of this proble. Let s think about q, t q, t. We re integrating over all paths taking us between the two points, which is the sae as integrating over all variations q about the classical path. As for the boundary conditions, we re fixing q on the endpoints, which is equivalent to fixing q 0 on the endpoints since we re taking q cl to hit our desired endpoints at the appropriate ties. (30) 3
q, t q, t Dqe il (32) q(t )q, q(t )q e il cl D qe i( 2 +... ) (33) q(t )0, q(t )0 (34) In the free theory this is e il cl q(t )0, q(t )0 D qe i q 2 2 (35) (36) The only dependence on the endpoints appears in the prefactor L cl, the action of the classical trajectory fro (q, t ) to (q, t ). The second factor in this expression represents all the quantu physics correcting the classical action for particle propagation. 3. q, t is defined to be the instantaneous eigenstate of the operator Q(t) in the heisenberg picture, Q(t) q, t q q, t, which evolves according to Q(t) e iht Q(0)e iht. Coputing Q(t) q, t e iht Q(0)e iht q, t, we see that the state q, t e iht q will satisfy Q(t) q, t q q, t. The active transforation of the operators Q is copatible with the passive transforation of the basis states q. 4. Srednicki 7.. Factor the denoinator in (7.2): de G(t t e ie(t t ) ) 2π (E (ω iɛ))(e + (ω iɛ)) To evaluate this via the ethod of contour integration, proote E to a coplex variable; then the function G has siple poles as E ±(ω iɛ), i.e. just below the real axis at +ω and just above at ω. When t > t, evaluate the integral by closing the contour in the lower half-plane, where the integrand reains finite if we take the arc to infinity. This picks up the pole at +ω, and cauchy s integral forula gives G(t t ) ( 2πi) e iω(t t ) 2ω(2π) i 2ω e iω(t t ) where the extra inus sign in the integral forula coes fro choosing a clockwise contour. For t < t, close the contour counterclockwise in the upper half-plane and pick up the ω pole to find G(t t ) i ) 2ω e+iω(t t (39) which shows the clai. 4 (37) (38)
Consider the effect of a different choice of iɛs. For exaple, we could have taken the poles to lie below the real axis. Then whenever t < t and we close the contour in the upper half-plane the green s function vanishes; this is the retarded green s function, appropriate for describing the effect of a disturbance at tie t on objects at a later tie t. Siilarly choosing the poles to lie above the real axis gives the advanced green s function. 5. Srednicki 7.2. The only tricky part is evaluating 2 t t t, but this isn t bad. Consider acting with t on t t when t < t and t > t, this gives + in the first case and in the latter, so t t t sign(t t ). The derivative of the sign function is twice the dirac delta, which follows fro writing the sign function as θ(t t ) θ(t t). So 2 t 2δ(t t ), and siple algebra gives the result. 5