The Path Integral for Relativistic Worldlines

Transcription

1 The Path Integral for Relativistic Worldlines B. Koch with E. Muñoz and I. Reyes based on: Phys.Rev. D96 (2017) no.8, and arxiv: Afunalhue, La parte y el todo,

2 Content PI of the RPP, Status Local Symmetry: Velocity Rotations Constructing the PI of the RPP Conclusion 2

3 The Path Integral Propagator A B ha Bi Dx e S A,B 3 (after Wick rotation)

4 Non Relativistic Propagator ha Bi Dx e S A,B S = tb,b t A,A dt m 2 ( ~x) 2 = X i m 2 (~x i+1 ~x i ) 2 2 Two Nice Features Quadratic in field variable Can be connected (Chapman, Kolmogorov) 4

5 ha Bi = i (N i Non Relativistic Propagator d d x i e P i m 2 (~x i+1 ~x i ) 2 2 ) Feature 1: Quadratic Simple Gaussian integrals Feature 2: Chapman Kolmogorov ha Ci = d d x b ha BihB Ci Probability conservation & stepwise construction 5 of PI

6 Relativistic Propagator ha Bi Dx e S A,B S = d s dx µ d 2 Why intersting, why here Unsolved NON PERTURBATIVE problem Simplest system with general covariance Two Problems! 6

7 ha Bi =? i Relativistic Propagator 8 >< >: N i d d x i e P i r (x µ i+1 xµ i )2 2!9 >= >; Problem 1: Square root Horrible integrals & still wrong result Problem 2: No Chapman Kolmogorov ha Ci 6= d d x B ha BihB Ci No probability conservation & no stepwise construction 7 of PI

8 Relativistic Propagator Solutions in the Literature Hamiltonian formalism (classically equivalent) Evades P1 Redefine probability *2 Solves P1 & P2, but high price Restrict PI to spheres, or other approx. *3 Interesting, neither P1 nor P2 are solved Ignore the problems and do QFT right away Thats what we mostly do 8 *1

9 Relativistic PI: our Proposal A B ha Bi Dx e S A,B 9

10 Relativistic Propagator ha Bi Dx e S A,B s dx µ 2 with S = d d can be done, if one considers Three issues: Issue 1: Local reparametrizations (known) Issue 2: Local velocity rotations (trivial?) Issue 3: Measure without anomalies 10

11 Relativistic Propagator ha Bi Dx e S A,B S = d s dx µ d 2 using I1,I2,I3 works Functional *0 Geometric *00 Fadeev Popov method stepwise proof 11

12 Stepwise proof A ha Bi B = = ha Bi ha Bi Don t count again! Strategy Clarify geometry meaning of I1,2,3 Calculate ha Bi 1 using I2,3 Show with I1,2 ha Bi 1 contains ha Bi

13 Issue 1: Local reparametrizations (known) S = d s dx µ d 2 Invariant under! 0 ( ) We fix proper time such that = ( ) with dx µ d 2 =1 Geometric over counting: A B A C B 13

14 Issue 2: Local velocity rotations (trivial?) S = d s dx µ 2 = d d p v µ v µ Invariant under v µ! v 0µ = µ ( )v with v 0µ v 0 µ = v µ v µ Factor out of PI if S = S 0 and L = L 0! 14

15 Issue 3: Measure without anomalies When performing transformation v µ! v 0µ = µ ( )v define right measure invariant under this symmetry: Dx! Dx 0 = Dx Geometric example for two step propagator 15

16 Stepwise proof A ha Bi B = = ha Bi ha Bi Strategy Clarify geometry meaning of I1,2,3 Calculate ha Bi 1 using I2,3 Show with I1,2 ha Bi 1 contains ha Bi

17 Calculate ha Bi 1 using I2,3 A B ha Bi 1 = N d 2 x 1 ~x 1 1 e (S A,~x 1 +S ~x1,b) Change of integration coordinates x 1 = S 2M cos( ) (x 1,y 1 )! (S, ) 17 y 1 = x f 2 s S x f M 2 1 sin( ) x f = ~ B ~ A

18 Calculate ha Bi 1 using I2,3 A B S ~x 1 h0, ~x f i 1 = N 2,1 (t i,f ) 1 x f M ds 2 0 2(S/M) 2 x 2 f (1 + cos(2 )) d 1 8 p exp [ S] (S) 2 (x f M) 2 Change of integration coordinates x 1 = S 2M cos( ) (x 1,y 1 )! (S, ) y 1 = x f 2 s S x f M 2 1 sin( ) 18 x f = ~ B ~ A

19 Calculate ha Bi 1 using I2,3 0 x f S ~x 1 h0, ~x f i 1 = N 2,1 (t i,f ) 1 x f M ds 2 0 2(S/M) 2 x 2 f (1 + cos(2 )) d 1 8 p exp [ S] (S) 2 (x f M) 2 For! 0 one sees S = S 0 and L = L 0 I2 velocity rotation! Naive, measure depends on 1 1 ~x 1 0 ~x f ~x 1 19 : anomaly I3 cancels anomaly

20 Calculate ha Bi 1 using I2,3 0 x f S h0, ~x f i 1 = N 2,1 2 0 d 1 x f M ds 1 p exp [ S] (S) 2 (x f M) 2 For! 0 nothing changes h0, ~x f i 1 = N K 0 (x f M) 20

21 Calculate ha Bi 1 using I2,3 0 x f 0 S h0, ~x f i 1 = N 2,1 2 0 d 1 x f M ds 1 p exp [ S] (S) 2 (x f M) 2 For! 0 nothing changes h0, ~x f i 1 = N K 0 (x f M) 21

22 Calculate ha Bi 1 using I2,3 0 x f 0 S h0, ~x f i 1 = N 2,1 2 0 d 1 x f M ds 1 p exp [ S] (S) 2 (x f M) 2 For! 0 nothing changes h0, ~x f i 1 = N K 0 (x f M) 22

23 Stepwise proof A ha Bi B = = ha Bi ha Bi Strategy Clarify geometry meaning of I1,2,3 Calculate ha Bi 1 using I2,3 Show with I1,2 ha Bi 1 contains ha Bi

24 Show with I1,2 ha Bi 1 contains ha Bi 2...? A ha Bi B = = ha Bi ha Bi

25 Show with I1,2 ha Bi 1 contains ha Bi 2... ~x 1 h0 xf i 2 : 0 x f ~x 2 S 1,f! 0 For nothing changes ~x 2! ~x

26 Show with I1,2 ha Bi 1 contains ha Bi 2... ~x 1 ~x h0 x f i 2 : 0 x f S 1,f! 0 For nothing changes (I2) ~x 2! ~x 0 2 0! ~x 1! ~x 0 2 straight line: I1 reparametrizations 26

27 Show with I1,2 ha Bi 1 contains ha Bi 2... ~x 1 ~x 0 2 h0 xf i 2 : 0 x f S 1,f! 0 For nothing changes (I2) ~x 2! ~x 0 2 0! ~x 1! ~x 0 2 straight line: I1 reparametrizations h0 x f i 2 1,2 h0 x f i 1 27

28 Stepwise proof A ha Bi B = = ha Bi ha Bi Strategy Clarify geometry meaning of I1,2,3 Calculate ha Bi 1 using I2,3 Show with I1,2 ha Bi 1 contains ha Bi 2... h0 x f i = N h0 x f i 1 = N K 0 (Mx f ) 28

29 Concluding Comments Generalization to D dimensions PI of RPP action can be done, considering I1,I2,I3 Chapman Kolmogorov becomes trivial Future work 29

30 Thank You 0 0 x f S 30

31 Literature 0) B. K., E. Muñoz and I. Reyes; Phys.Rev. D96 (2017) no.8, ) B. K., E. Muñoz; arxiv: ) J. Polchinski, String Theory, Cambridge U. P., ISBN , page 145.; H. Kleinert, Path Integrals in Quantum Mechanics, World Scientific Publishing, ISBN , page M. Henneaux and C. Teitelboim, Annals Phys. 143, 127 (1982). 2) P. Jizba and H. Kleinert, Phys. Rev. E 78, (2008). 3) E. Prugovecki, Il Nuovo Cimento, 61 A, N.2, 85 (1981). H. Fukutaka and T. Kashiwa, Annals of Physics, 176, 301 (1987). Padmanabhan, T. Found Phys (1994) 24: