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Lecture XI 32

Lecture XI: Statistical Mechanics and Semi-Classics

Connection of Path Integral to Classical Statistical Mechanics

Consider flexible string held under constant tension and confined to gutter potential

x

u

V(u)

Potential energy stored in spring due to line tension:

from segment x to x + dx, dVT = T

extension [(dx2 + du2)1/2 dx] T dx (xu)2/2

VT[xu]

dVT =1

2

L0

dx T (xu(x))2

External (gutter) potential: Vext[u] L0

dx V[u(x)]

According to Boltzmann principle, equilibrium partition function

Z= tr eF = Du(x) exp L0

dx

T

2(xu)

2 + V(u)

cf. quantum mechanical transmission amplitude Mapping:

Z=b.c.

Dq(t) exp

i

t0

dt

mq2

2 V(q)

Wick rotation t

i

imaginary (Euclidean) time path integralt

0

idt (tq)2

0

d(q)2,

t0

idtV(q) 0

dV(q)

Z=b.c.

Dqexp

1

0

dm

2(q)

2 + V(q)

N.B. change of relative sign!

(a) Classical partition function of one-dimensional systemcoincides with quantum mechanical amplitude

Z= dqq|eiHt/|qt=i

where time is imaginary, and plays the role of temperature

Lecture Notes October 2005

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Lecture XI 33

More generally, path integral for d-dimensional quantum system

corresponds to classical statistical mechanics of d + 1-dimensional system

(b) Quantum partition function

Z= tr(eH) =

dqq|eH|q

i.e. Zcan be interpreted through dynamical transition amplitude q|eiHt/|qevaluated at imaginary time t = i.

(c) In semi-classical limit ( 0), PI dominated by stationary configurations of actionS[p, q] = dt(pq H(p, q))

S = S[p + p,q+ q] S[p, q]=

dt [pq+ pq ppH qqH] + O(p2, q2,pq)

=

dt [p (q pH) + q( p qH)] + O(p2, q2,pq)

i.e. Hamiltons classical e.o.m.: q = pH, p = qH with b.c. q(0) = qI, q(t) = qF(Similarly, with Lagrangian formulation : S = 0

(dtq

q) L(q, q) = 0)

qqI

qF h

1/2

t

q

q(t)

Contributions to PI from fluctuations around classical paths?

Usually, exact evaluation of PI impossible resort to approximation schemes...

Saddle-point and Stationary Phase analysis

Consider integral over single variable

I =

dz ef(z)

Integral dominated by minima of f(z); suppose unique i.e. f(z0) = 0

Lecture Notes October 2005

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Lecture XI 34

Taylor expand around minimum: f(z) = f(z0) + (z z0) 0 f(z0) +12 (z z0)2f(z0) +

I ef(z0)

dz e(zz0)2f(z0)/2 =

2

f(z0)ef(z0)

Example : (s + 1) =

0

dzzsez =

0

dz ef(z), f(z) = z s ln z

f(z) = 1 s/z i.e. z0 = s, f(z0) = s/z20 = 1/si.e. (s + 1) 2se

(ss ln s)

Stirlings formula

If minima not on contour of integration deform contour through saddle-pointe.g. (s + 1), s complex

What if exponent complex? Fast phase fluctuations cancellationi.e. expand around region of slowest (i.e. stationary) phase and use identity

dz eiaz2/2 =

2

aei/4

Can we apply same approach to analyse the FPI?Yes: but we we must develop new technology;

basic tool of QFT the Gaussian functional integral!

Lecture Notes October 2005