¨

Q + („¦2 ’ ∆„¦2 )Q + fi q i = 0,

i

2

qi + ωi q + fi Q = 0.

¨ (5.34)

1

A. Caldiera, A. J. Leggett, Physical Review Letters 46 (1981) 211.

5.2. CONSTRUCTING GREEN FUNCTIONS 123

Using our initial value Green function, we solve for the qi in terms of Q(t)

fi2

t

fi qi = ’ sin ωi (t ’ „ )Q(„ )d„. (5.35)

ωi

’∞

The resulting motion of the qi feeds back into the equation for Q to give

t

¨

Q + („¦2 ’ ∆„¦2 )Q + F (t ’ „ )Q(„ ) d„ = 0, (5.36)

’∞

where

fi2

F (t) = sin ωi (t) (5.37)

ωi

i

is a memory function.

Caldeira and Leggett de¬ne a spectral function

fi2

π

δ(ω ’ ωi ),

J(ω) = (5.38)

2 ωi

i

in terms of which

2 ∞

F (t) = ’ J(ω) sin ω(t) dω. (5.39)

π 0

By taking di¬erent forms for J(ω) we can represent a wide range of environ-

™

ments. To obtain a friction force proportional to Q we need J ∝ ω. We will

actually set

Λ2

J(ω) = ·ω 2 , (5.40)

Λ + ω2

where Λ is a high-frequency cuto¬, introduced to make the integral over ω

well-behaved. With this choice

· ωΛ2eiωt

2 2

∞ ∞

dω = sgn (t)· Λ2 e’Λ|t| . (5.41)

’ J(ω) sin(ωt) dω = 2 + ω2

π 2πi Λ

0 ’∞

Therefore,

t t

·Λ2 e’Λ|t’„ | Q(„ ) d„

F (t ’ „ )Q(„ ) d„ = ’

’∞ ’∞

·¨

™

= ’·ΛQ(t) + · Q(t) ’ Q(t) + · · · . (5.42)

2Λ

124 CHAPTER 5. GREEN FUNCTIONS

Now,

fi2 ·Λ2

2 J(ω) 2

∞ ∞

2

’∆„¦ ≡ = dω = dω = ·Λ. (5.43)

2

Λ2 + ω 2

ωi π ω π

0 0

i

The counter-term thus cancels the O(Λ) frequency shift, and, ignoring terms

with negative powers of the cuto¬, we end up with viscously damped motion

¨ ™

Q + · Q + „¦2 Q = 0. (5.44)

The oscillators in the bath absorb energy but, unlike a pair of coupled oscil-

lators which trade energy rhythmically back and forth, the incommensurate

motion of the many qi prevents them from cooperating for long enough to

return any energy to Q(t).

5.2.3 Modi¬ed Green Functions

When the equation Ly = 0 has a non trivial-solution, there can be no unique

solution to Ly = f , but there still will be solutions provided f is orthogonal

to all solutions of L† y = 0.

Example: Consider

2

Ly ≡ ’‚x y = f (x), y (0) = y (1) = 0. (5.45)

The equation Ly = 0 has one non-trivial solution, y(x) = 1. The operator

L is self-adjoint, L† = L, and so there will be solutions to Ly = f provided

1, f = 01 f dx = 0.

We cannot de¬ne the the green function as a solution to

2

’‚x G(x, x ) = δ(x ’ x ), (5.46)

1

δ(x ’ x ) dx = 1 = 0, but we can seek a solution to

because 0

2

’‚x G(x, x ) = δ(x ’ x ) ’ 1 (5.47)

as the right-hand integrates to zero.

2

A general solution to ’‚x y = ’1 is

1

y = A + Bx + x2 , (5.48)

2

5.2. CONSTRUCTING GREEN FUNCTIONS 125

and the functions

1

yL = A + x2 ,

2

1

yR = C ’ x + x2 , (5.49)

2

obey the boundary conditions at the left and right ends of the interval, re-

spectively. Continuity at x = x demands that A = C ’ x , and we are left

with

C ’ x + 1 x2 , 0 < x < x

2

G(x, x ) = (5.50)

12

C ’ x + 2 x , x < x < 1,

There is no freedom left to impose the condition

G (x ’ µ, x ) ’ G (x + µ, x ) = 1, (5.51)

but it is automatically satis¬ed ! Indeed,

G (x ’ µ, x ) = x

G (x + µ, x ) = ’1 + x . (5.52)

We may select a di¬erent value of C for each x , and a convenient choice