Srednicki Chapter 6. QFT Problems & Solutions. A. George. May 24, dp j 2π eip j(q j+1 q j ) e ihδt (6.1.1) dq k. j=0. k=1

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1 Srednicki Chapter 6 QFT Probles & Solutions A. George May 24, 202 Srednicki 6.. a) Find an explicit forula for Dq in equation 6.9. Your forula should be of the for Dq = C j= dq j, where C is a constant that you should copute. We ll go back to equation 6.7: dp j eip jq j+ q j ) e ihδt 6..) ow we let a = q j+ q j. Since we re trying to get to equation 6.9, we ll ake the assuptions used in that equation see first new paragraph of page 45) now. In particular, we ll let: ow define Bq) = B q) a δt. H = Ap 2 + B q)p + Cq) 6..2) dp j eip ja e i[ap2 +B q)p+cq)]δt dp j e i[ap2 +Bq)p+Cq)]δt 6..3) 6..4) ow we re ready to do the integral. We ll take the p liits at ± since Srednicki tells us to use an arbitrary value, and these are good arbitrary values. Let s also define A, B, C δta, B, C), and supress the q-dependence for convenience. Then we have to do n integrals which look like this: i 3/2 ) πe ic B2 /4A) i2ap + B) dpe iap2 +Bp+C) = 2 A erfi 2 A 6..5) This iaginary error function goes to i at infinity and i at negative infinity, provided that A is positive. Then: dp j e iap2 +Bp+C) = i 5/2 e ic B2 /4A) π 2 for A > 0 A This gives: dp j e iap2 +Bp+C) = eic B2 /4A) 4πiA for A > 0

2 And then: dp j e iap2 +Bp+C) = ow we need to do this integral + ties. Hence, dp j e iap2 +Bp+C) = e 2iC B 2 /4A) 4πiA for A > ) ) +)/2 e 2iC B2 /4A) 4πiA ow we ll put back in the factor of δt in the A,B,C ters, as well as the q-dependence in the B and C ters. ) +)/2 dp j e iap2 +Bq)p+Cq)) /4A) = e 2iδtCq) Bq)2 δt4πia ow we put in the ters needed to recover 6..4). dp je iap2 +Bq)p+Cq)) = ) +)/2 /4A) e 2iδtCq) Bq)2 4πiA ow we have: dp je iap2 +Bq)p+Cq)) = ) +)/2 /4A) i e 2iδtCq) Bq)2 4πA dp je iap2 +Bq)p+Cq)) = ) +)/2 i ) +)/2 e 2iδtCq) Bq)2 /4A) 4πA ow we re ready to use equation 6..4), reebering to put A δta: ) +)/2 e /4A)) + iδtcq) Bq)2 6..7) 4iπδtA But we are being rather casual with this last ter. In fact, the q s have subscripts the js of equation 6.7) and the exponent is just a casual shorthand for a product over all the js. Let s revert to using this notation: ) +)/2 e iδtcq j) Bq j ) 2 /4A) 4iπδtA This can obviously be rewritten as: ) +)/2 e iδt Cq j) Bq j ) 2 /4A) 4πiδtA 6..8) 2

3 Obviously this first ter is the C alluded to in the proble stateent. ow let s think about the Lagrangian. The Lagrangian is given by: L = p q Ap 2 B q)p Cq) L = p q B q)) Ap 2 Cq) We use equation 6.0, which shows: p = q B )/2A). Then, L = 2Ap 2 Ap 2 Cq) L = Ap 2 Cq) ) 2 q B q) L = A Cq) 2A ow, recall that we defined Bq) = B q) q. Then, L = A ) 2 Bq) Cq) 2A L = Bq)2 4A ow we ll insert this into equation 6..8): Cq) Dqe iδt L ow let δt 0, which gives: Dqe i t t dtl which is Srednicki equation 6.9. Hence, we have derived the forula, and forally defined the path integral along the way. We derived the path etric soe tie ago, so let e state the result again here fro equation 6..7)): Dq = ) +)/2 dq j 6..9) 4πiδtA j= ote: Srednicki s solution to this proble leaves a lot to be desired. ot only does he assue a specific for for H without any justification, but he also stops at his analog of equation 6..8), without proving that 6..8) is equivalent to Srednicki equation 6.9. b) In the case of a free particle, VQ) = 0, evaluate the path integral of equation 6.9 explicitly. Hint: Integrate over q, then q 2, etc., and look for a pattern. Express your final answer in ters of q, t, q, t, and. Restore h by diensional analysis. 3

4 We ll go back to 6..8): Dqe iδt Cq k) Bq k ) 2 /4A) In this case, A =, B 2 = 0, C = 0. We can calculate B = q. Then: Dqe i 2 δt q k 2 It s a little bit difficult to deal with the derivative, so let s write in long for. Dqe i q j+ q j ) 2 2 δt Using equation 6..9): ) +)/2 2 dq j e i 2 4πiδt j= Taking advantage of the hint, we ll do the q integral: dq e icq 2 q ) 2 e icq q 0 ) 2 where C = /2δt). q j+ q j ) 2 δt We ll expand these products: dq e ic2q2 2q q 2 +q 0 )+q 2 2 +q2 0 )) This is of the for of equation 6..6). ow: dq e icq 2 q ) 2 e icq q 0 ) π 2 ic = 2iC e 2 q 2 q 0 ) 2 and then: dq e icq 2 q ) 2 e icq q 0 ) 2 = dq e icq 2 q ) 2 e icq q 0 ) 2 = π ic 2iC e 2 q 2 q 0 ) 2 iπδt ic e 2 q 2 q 0 ) 2 ow for the q 2 integral. I won t show all the calculus since it s the sae as before, the result is: dq 2 e icq 3 q 2 ) 2 e ic 2 q 2 q 0 ) 4iπδt 2 ic = 3 e 3 q 3 q 0 ) 2 At soe point we see the pattern. We increase the integer in the denoinator of the expononent, and weigh the factor in the coefficient accordingly. The result is: dq e icq 3 q 2 ) 2 e ic 2 q 2 q 0 ) 2iπδt 2 ic = + ) e + q + q 0 ) 2 4

5 Back to equation 6..9), plugging in our solutions for all the integrals: ) ) +)/2 q, t q, t jiδt = iδt j + ) j= e ic j+ q j+ q 0 ) 2 ote that the product acts only on the coefficient in front of the exponential, not on the exponential itself this is because the entire q n exponential was used to construct the q n exponential, and so forth). As for the prefactors, everything cancels except the + ter and the ter, but that is obviously uniportant). Then: ) +)/2 iδt + iδt ) /2 e ic + q + q 0 ) 2 which siplifies to: q, t q, t = + )iδt e i +)2δt q + q 0 ) 2 and then: it t ) e i 2 q q ) 2 t t where we noted that q 0 = q and q + = q. ow to restore the factors of h. The exponent needs to be diensionless, but it currently has units of Et. So we divide by h. q, t q, t = it t ) e i q q ) 2 2 h t t What units should the prefactor have? If we set t = t, then we obviously have q q = δq q ). This will get rid of one length unit, so it has diensions of inverse length. In the prefactor given, we should therefore divide by h. This will give units of s 2 ) /2 = s, which becoes inverse length when ultiplied by c and we are not told to restore the cs. Then, i ht t ) e i 2 h q q ) 2 t t 6..0) c) Copute q e iht t ) q by inserting a coplete set of oentu eigenstates, and perforing the integral over the oentu. Copare with your result fro part b). q e iht t ) q Inserting the coplete set of oentu eigenstates: dp q e iht t ) p p q The bra-ket on the right is easy to evaluate. The Hailtonian for a free particle is given by H = P2. So: 2 dp q e it t ) P2 2 p e ipq 5

6 The oentu operator acting on the oentu eigenstate will give the oentu eigenvalue, which we ll call p. Then, dp q e it t ) p2 2 p e ipq ow the operator between the bra and the ket is just a constant, so we can take it out: dpe it t ) p2 2 q p e ipq dpe p2 it t ) 2 e ipq q ) [ dpe i p 2 t t ] ) 2 pq q ) This is of the for of equation 6..6), but this tie A is negative. The result is that in equation 6..5), we get 2i rather than 2i fro the iaginary error function. As a result, equation 6..6) has the i in the nuerator rather than the denoinator. Using this result, we have: it t ) e i 2 q q ) 2 t t which is the sae as our result fro part b, equation 6..0), once we add in the factors of h as before. 6