ft1ls02p_08

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 8.323 Relativistic Quantum Field Theory I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 1. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department ����� �� �������� �� � ������ �� ���� � ����� ��� ��� �� �������� ��� � ��� ���� ������ ����������� ������� ����� ������ � — ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� ���� ���� �������� �� ������������ �������� Equation of motion: ( Initial condition: φ(x) = φin (x) . + m2 )φ(x) = j (x) . (2.1) (2.2) Eqs. (2.1) and (2.2) Solution: =⇒ unique solution for Heisenberg operator φ(x). � d4 y DR (x − y)j (y) , (2.3) φ(x) = φin (x) + i where DR (x − y) is the retarded propagator: ( x + m2 )DR (x − y) = −iδ (4) (x − y) where DR (x − y) = 0 if x0 < y0 (retarded) . (2.4) ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� ���� ���� –1– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 2. We know that DR (x − y) = θ(x0 − y0 ) 0 |[φin (x) , φin (y)]| 0 � � d3 p 1 � −ip·(x−y) 0 0 = θ (x − y ) e − eip·(x−y) √ (2π)3 2Ep p0 =Ep = p 2 +m2 (2.5) Note that DR (x − y) is defined by the free wave equation. It can be written in terms of [φin (x) , φin (y)] as above, or in terms of [φout (x) , φout (y)], but not in terms of [φ(x) , φ(y)]. θ(x0 − y0 ) in DR is hard to deal with, but for x0 ≡ t > t2 we can set θ(x0 − y0 ) = 1. Then � � � d3 p 1 � −ip·(x−y) 4 φ(x) = φin (x) + i d y j (y) e − eip·(x−y) . (2.6) (2π)3 2Ep Repeating, φ(x) = φin (x) + i Define ˜(p) ≡ so φ(x) = φin (x) + i � d3 p 1 � ˜(p)e−ip·x − ˜(−p)eip·x (2π)3 2Ep �� � � d3 p 1 i � ain (p ) + � ˜(p) e−ip·x = 3 ( 2π ) 2Ep 2Ep � � � i † + ai n ( p ) − � ˜(−p) eip·x 2Ep � � d3 p 1� � aout (p )e−ipx + h.c. . = 3 ( 2π ) 2Ep ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� ������ ��� � ��� ���� � d y j (y ) 4 ����� ������� ��������� �� �� ������� ���� ���� –2– � � d3 p 1 � −ip·(x−y) e − eip·(x−y) . (2π)3 2Ep � (2.6) d4 y eip·y j (y) , (2.7) � (2.8) –3– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 3. So i aout (p ) = ain (p ) + � ˜(p) 2Ep a†ut (p ) o = a† ( p ) in −� i (2.9) 2Ep ˜(−p) , where since j (x) is real, and ˜(−p) = ˜∗ (p) , p0 = (2.10) p 2 + m2 . (2.11) � Thus, only the mass shell component (p0 = p 2 + m2 ) of ˜(p) results in particle creation. This is just the classical phenomenon of resonance o ccurring in the quantum field theory setting. � ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –4– ������� �������������� ������� �� ��� ��� It is useful to construct a unitary transformation that relates in and out quantities. Remembering that DR (x − y) = θ(x0 − y0 ) 0 |[φin (x) , φin (y)]| 0 , recall also that [φin (x) , φin (y)] is a c-number, so 0 |[φin (x) , φin (y)]| 0 = [φin (x) , φin (y)]. So for x0 ≡ t > t2 , � φ(x) = φout (x) = φin (x) + i d4 y [φin (x) , φin (y)] j (y) . (2.12) If we define B≡ � d4 y j (y) φin (y) , (2.13) (2.14) then φout (x) = φin (x) + i [φin (x) , B ] . But [φin (x) , B ] is also a c-number, so we can write φout (x) = e−iB φin (x) eiB . ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� (2.15) –5– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 4. Since φout (x) = e−iB φin (x) eiB , (2.15) we know from the uniqueness of the Fourier expansion that aout (p ) = e−iB ain (p )eiB . (2.16) We can also verify that this equation is true by using i aout (p ) = ain (p ) + � ˜(p) 2Ep (2.9a) with � � † ain (p ) , ain (q ) = (2π)3 δ (3) (p − q ) . (2.17) ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –6– Define S ≡ eiB (the ������ S -Matrix) Mapping of states: aout (p ) |0out = 0 S −1 ain (p )S |0out = 0 =⇒ ain (p ) S |0out = 0 This implies, up to a phase, the S |0out = |0in . We can redefine the phase of |0out (or |0in ) so that S |0out = |0in . ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� ��� S ������� (2.18) (2.19) (2.20) –7– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 5. On one particle states, S |p out = S a†ut (p ) |0out o = S a†ut (p )S −1 S |0out � o �� � � �� � † |0in ai n ( p ) = |p in In general, we could show that � �� � S �p 1 . . . p N,out = �p 1 . . . p N,in . (2.21) (2.22) ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –8– ������ �������� �� S We know that S=e iB � =e i d4 y j (y) φin (y) . (2.23) It is useful to write S so that all the annihilation operators are on the right. Let � � � d3 p 1� 4 � i B = i d y j (y ) ain (p )e−ip·y + a† (p )eip·y = G + F , in (2π)3 2Ep (2.24) where � � d3 p d3 p 1 1 † � � F =i ˜(p)ain (p ) , G = i ˜(−p)ain (p ) , (2.25) (2π)3 2Ep (2π)3 2Ep where we recall that ˜(p) ≡ ����� ������� ��������� �� �� ������� � d4 y eip·y j (y) . (2.7) –9– ���� ���� ������ ��� � ��� ���� 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 6. So S = eiB = eF +G . (2.26) F and G do not commute, but [F , G] is a c-number and therefore commutes with both F and G. Whenever F and G commute with [F , G], eF +G = eF eG e− 2 [F , G] . 1 (2.27) ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –10– ����� ����� e To prove this identity, define F +G =e F 1 G − 2 [F , G] ee H1 (λ) ≡ eλ(F +G) , H2 (λ) = eλF eλG e− 2 λ 1 2 [F,G] . (2.28) Clearly H1 (0) = H2 (0) = I (identity operator), and dH1 (λ) = (F + G) H1 (λ) . dλ (2.29) So if we can show that H2 (λ) obeys the same differential equation as above, then it follows that H2 (λ) = H1 (λ). You’ll get to show this on your next problem set. This is actually a special case of the Baker-Campbell-Hausdor formula, which has the general form eF eG = eF +G+ 2 [F , G]+...(iterated 1 commutators) . (2.30) We’ll prove this, to o, on a problem set so on. ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –11– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 7. ��������� �� ��� ���� ��������� So (2.26) F and G do not commute, but [F , G] is a c-number and therefore commutes with both F and G. Whenever F and G commute with [F , G], eF +G = eF eG e− 2 [F , G] . 1 S = eiB = eF +G . (2.27) Recalling � 1 d3 p � F =i ˜(p)a† (p ) , in 3 ( 2π ) 2Ep one sees that [F , G] = − � = ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� � G=i 1 d3 p � ˜(−p)ain (p ) , (2.25) 3 ( 2π ) 2Ep � d3 p 1 d3 q 1 � � ˜(p) ˜(−q) [a† (p ), ain (q )] 3 3 ( 2π ) � � i n �� 2Ep (2π) 2Eq −(2π )3 δ (3) (p −q ) (2.31) d3 p ( 2π )3 1 |˜(p)|2 . 2Ep –12– So S=e iB 1 = exp − 2 � � d3 p 1 |˜(p)|2 (2π)3 2Ep � eF eG . (2.32) ���� ���� ����� ������� ��������� �� �� ������� ������ ��� � ��� ���� –13– 8.323 Lecture Notes 2: Particle Pro duction by a Classical Source, Part I I (incomplete), p. 8. �� ��� �� �� ��� ����������� The probability that no particles are pro duced by the source is given by P (no particle pro duction) = | 0out |0in |2 . (2.33) Logic: physical state is |0in , independent of time (in the Heisenberg picture). |0out = state with no particles for t > t2 . So, | 0out |0in |2 is the probability that the physical state of the system would be measured to have 0 particles at t > t2 . To express the answer in terms of the S -matrix, recall |0out = S −1 |0in =⇒ 0out | = 0in | S. (2.34) So 0out |0in = 0in |S | 0in � � � (2.35) � � F G� � 1 d3 p 1 2 0in �e e � 0in . = exp − |˜(p)| 2 (2π)3 2Ep ������ ��� � ��� ���� ����� ������� ��������� �� �� ������� ���� ���� –14– ...
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