The Single Particle Path Integral and Its Calculations

Lai Zhong Yuan

Summary Of Contents

• Introduction and Motivation

• Some Examples in Calculating Path Integrals– The Free Particle– The Harmonic Oscillator

• Perturbation Expansions

Introduction and Motivation

• In Q.M. – a probability amplitude associated with every method whereby an event in Nature can take place.

• Also associate an amplitude with the overall event by adding together the amplitudes of each alternative method.

• Can be seen via the famous double-slit experiment

The Double Slit Experiment

•1φ Amplitude for reaching the

detector while passing through hole 1

2φ Amplitude for reaching the detector while passing through hole 2

21 φφφ +=Total amplitude of reaching detector

since2Amplitudey ProbabilityProbabilit =

Total Probability of Electron Reaching Detector

221 φφ +=P

But – many ways to analyze concept of amplitude.

In this example – associate amplitude with each possible motion of electron from A to B .

Hence – total amplitude = sum of a contribution from each of the paths

A Thought Experiment

Each alternative path has own amplitude –Complete amplitude is sum of all possible paths.

Now suppose: More and more holes are drilled – till nothing more is left of the screens…

At each screen, path of electron specified by heights at which the electron passes positions

ED xx ,ED yy ,

A Thought Experiment II

A Thought Experiment III

• To each pair of heights corresponds an amplitude.

• Need to take integral of the amplitudes over all possible values of .

• Also, the time at which the electron passes each point in space – .

• Thus – a path is defined through • Total amplitude = sum over the amplitude

for source – detector travel following all possible paths.

ED xx ,

)(ty)(),( tytx

• But how does one define a sum over a path?

• One way is to approximate it by line segments + let segments 0

• Unitarity of Q.M.: Amplitude of the whole path is product of the amplitudes of each infinitesimal path

• Amplitude to propagate from to in time is given in terms of the unitary operator such that

iq fqT

iiHT

f qeq −=Amplitude

• Dividing time into infinitesimal segments each of length , one has

Completeness of means that

itiHtiH

fiiHT

f qeeqqeq δδ −−− = L

T NNTt /=δ

q

1=∫ qqdq

• And thus

• Taking a look at one individual factor

itiHtiH

NtiH

NNtiH

fj

N

j

iiHT

f

qeqqeq

qeqqeqdq

qeq

δδ

δδ

−−

−−

−−−

−

=

−

⎟⎠⎞

⎜⎝⎛ Π= ∫

112

211

1

1

L

L

jtiH

j qeq δ−+1

• for free particle case, along with the completeness of gives

,0)( =qVp

∫

∫

∫

−−

+−

−+

−+

+=

=

=

)()2/(

1)2/(

)2/ˆ(1

)2/ˆ(1

12

2

22

2

2

2

jj qqipmpti

jjmpti

jmpti

jjmpti

j

eedp

qppqedp

qppeqdpqeq

δ

δ

δδ

π

π

π

• The integral over yields, noting that it is Gaussian and employing the relation

21

21

2

]/))[(2/(21

2/])([21

)2/(1

2

2

tqqmti

tqqimj

mptiHj

jj

jj

et

mi

et

miqeq

δδ

δδ

δπ

δπ

−

−−+

+

+

⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −=

p

aedx

ax π22

21

=−

+∞

∞−∫

• Substituting the integrated expression into the expression for the product of yields

• Taking the continuum limit one replaces and

jtiH

j qeq δ−+1

∫ ∑Π⎟⎠⎞

⎜⎝⎛ −=

−

= + −−

=

−1

02

1 ]/)[()2/(1

0

22 N

j jj tqqmti

j

N

j

N

iiHT

f edqt

miqeqδδ

δπ

0→tδ22

1 ]/)[( qtqq jj &→−+ δ ∫∑ →−

=

TN

jdtt

0

1

0δ

∫∫−

=∞→Π⎟

⎠⎞

⎜⎝⎛ −= j

N

j

N

Ndq

tmitDq

1

0

22lim)(δ

π

• One has then the path integral representation:

• To obtain , one simply integrates over all possible paths such that

and • Hamiltonian for a particle in a potential

∫ ∫=−T

qmdti

iiHT

f etDqqeq 02

21

)(&

iiHT

f qeq −

)(tq

iqq =)0( fqTq =)()ˆ(qV

)]ˆ(21[ 2

0)(qVqmdti

iiHT

f

T

etDqqeq−− ∫= ∫

&

• But , and in general

• As , and restoring the Planck’s constant, the final expression is then

),()(21 2 qqLqVqm && =−

),(0)(

qqLdti

iiHT

f

T

etDqqeq&∫= ∫−

)(),(0

qSqqdtLT

=∫ &

)()(

qSi

i

HTi

f etDqqeq hh ∫=−

Some Examples in Calculating Path Integrals

I. The Free Particle

• The Lagrangian for a free particle is noted to be

• The full amplitude (or kernel) is then

),( qqL &

2),(

2qmqqL&

& =

⎟⎟⎠

⎞⎜⎜⎝

⎛−Π×

⎟⎠⎞

⎜⎝⎛=

∑∫=

−

−

=

∞→

N

jjjj

N

j

N

Nif

xxt

imdq

timqqK

0

21

1

0

2

)(2

exp

2lim),(

δ

δπ

h

h

• The full expression can be obtained by substituting the Lagrangian into the action

and performing the time integration.

• Alternatively, one can also apply the action principle along with the Hamilton equations for the free particle.

)(),(0

qSqqdtLT

=∫ &

• Taking note of the identity

one first considers the first integral over in the sum of terms in the exponent.

1x

( ) ( )

( ) ( ) ⎥⎦⎤

⎢⎣⎡ −

+−

+=

−−−−∞

∞−∫2exp

22

yxba

aba

ab

ebeuadu yubuxa

π

π

1q

• Including two of the terms one has

according to the integral identity displayed in the previous slide

21

2⎟⎠⎞

⎜⎝⎛

tim

δπ h

( )

( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡ −=

⎥⎦⎤

⎢⎣⎡ −+−⎟

⎠⎞

⎜⎝⎛ ∫

202

212

2011

22exp

22

)(22

exp2

qqt

imti

m

qqt

imqqt

imdqti

m

δδπ

δδδπ

hh

hhh

• It can be seen that the effect of integration on the is the replacements

( ) ( ) ( )203

202

223

2

l)exponentia in the androot square in the(both 32 :nintegratio

qqqqqq

ttq

−→−+−

→ δδiq

( ) ( ) ( )204

203

234

3

l)exponentia in the androot square in the(both 43 :nintegratio

qqqqqq

ttq

−→−+−

→ δδ

• Grouping and integrating N times, one finally has

• The final expression of the integral is

( )( ) ( )22

01squared Distance

1

ifN qqqq

TtNt

−=−→

=+→

+

δδ

( ) ⎥⎦⎤

⎢⎣⎡ −= 2

2exp

2),( ifjf qq

Tim

TimqqK

hhπ

II. The Harmonic Oscillator

• The harmonic oscillator Lagrangianspecial case of the quadratic Lagrangian

• For the harmonic oscillator (replacing :)

222

21

21 xmxmL ω−= &

xtextcxxtbxmL )()(21)(

21 22 −−+= &&

)( // abif txq →

• The action of the Lagrangian is of course

• In general, quadratic Lagrangians (such as har. osc.) can be solved by introducing

where , i.e., where the action is an extremum (S unchanged in the 1st order if is modified slightly.)

)()()( τττ yxx +=

y trajectorclassical)( =τx

)(τx

∫= LdtxS ))(( τ

• Difference between classical path and a possible alternative pathis .

• Evidently

• Classical path is constant

)(tx

)(tx)(ty

0)()( == ba tyty

• Setting a new “integration variable”

• the Lagrangian can be Taylor-expanded around as

)()()()()()(

txtxtytxtxty

&&& −=−=

xx

xx

yxLyy

xxLy

xL

yxLy

xLtxxLtxxL

&

&

&&

&&

&&

&&

,

22

222

2

2

221

);,();,(

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂+

∂∂∂+

∂∂+

∂∂+

∂∂+=

)(),( txtx &

• The Taylor expansion terminates after the second term since is a quadratic functional.

• Substituting the Lagrangian for the harmonic oscillator yields for the action:

);,( τxxL &

( )∫

∫∫∫

−+

⎟⎠⎞

⎜⎝⎛

∂∂+

∂∂+==

f

i

f

i

f

i

t

t

xx

t

t

t

t

yymdt

yxLy

xLdttxxLtxxdtLS

222

,

21

);,();,(

ω&

&&

&&&

• Performing integration by parts and usingfrom the definition of the classical

path yields the expression

Which gives for the amplitude

as the path integral only depends on time

0=xx

Sδδ

( )∫ −+= f

i

t

tcl yymdtSS 222

21 ω&

),0;,0(~),;,(exp),;,( if

iiffiiff ttK

txtxiStxtxK ⎥

⎦

⎤⎢⎣

⎡=

h

• Employing the usual definitions of the path integral one writes

• To deal with the multitude of integrals, one resorts to a method introduced by Gelfandand Yaglom

( )⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −−×

⎟⎠⎞

⎜⎝⎛=

∑

∫

=−

∞→

N

jjjj

N

NNab

ytyyt

mi

timdydyttK

0

2221

2

1

21

2exp

2lim),0;,0(~

ωδδ

δπ

h

hK

Gelfand-Yaglom Method• The expression

can be written in terms of matrix elements such that

and thus

( )⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −−∑

=−

N

jjjj ytyy

tmi

0

2221 2

12

exp ωδδh

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

Ny

yM1

η

( ) σηηωδ TN

jjj ytyymi −=⎥⎤

⎢⎡ −−∑ −

2221

1δj t ⎦⎣=0 22h

• Here is the matrix defined by

and thus the amplitude can be written

σ

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

−−

−−−

−

=

Nc

cc

tiit

m

O

Oh

OO

Oh

2

1

2

211

21121

12

2δ

δσ

( )∫ −⎟⎠⎞

⎜⎝⎛=

∞→σηηη

δπTN

N

Nd

timK exp

2lim~ 2

h

• is of the form with real and Hermitian diagonalizable by a unitary matrix

where is the diagonal matrix of eigenvalues of . For real eigenvectors is also real; one can set . Since

it is true that

σ~iσ σ~

σπ

σπξη

α α

ξξσσηη

det

2/

1

NNNN D

T

eded === ∏∫∫=

−−

UU Dσσ +=

Dσσ U

ηξ U=

1det =U

• The amplitude is then written as

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛

⋅⋅=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

∞→

+

∞→

σδδπ

σπ

δπ

det211

2lim

det2lim),0;,0(~

2/11

NN

NN

Nif

mtiti

m

timttK

hh

h

• The factor

• The determinant can be calculated by considering truncated matrices from

NN

N

N

p

c

c

mt

mti

≡≡

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=⎟⎠⎞

⎜⎝⎛

'det

)(

211

2112

detdet21

2

σ

δσδO

O

OO

Oh

jj ×N'σ

• By expanding in minors, it can be seen that they obey the recursion formula

, where

and ,

• Rewriting yields

1' +jσ

1

2

1)(2 −+ −⎟⎟

⎠

⎞⎜⎜⎝

⎛−= jjj pp

mtp ωδ

Nj ,,1K=

mtp /)(2 21 ωδ−=

10 =p

mp

tppp jjjj ω

δ−=

+− −+2

11

)(2

• If one writes for , then in the limit of , satisfies

with initial values

∞→→−−=⎟⎠⎞

⎜⎝⎛ −=

→=

Nmt

tppt

dtd

tp

as 11)(2)0(

0)0(2

01

0

ωδδ

δϕδϕ

jtpt δϕ =)( tjtt a δ+=0→tδ )(tϕ

ϕωϕmdt

d −=2

2

• Thus

is obtained by solving the differential equation

with initial conditions

1),( ,0),( =∂∂=

= attiii tt

tfttf

)(det2lim),( f

N

Nif tm

tittf ϕσδε =⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

∞→

h

0),(),(

2

2

=+∂

∂if

if ttft

ttfm ω

• And thus, one arrives at the amplitude for the harmonic oscillator

⎥⎦⎤

⎢⎣⎡

=

),;,(exp),(2

),;,(

iiffcif

iiff

txtxSittfi

m

txtxK

hhπ

Perturbation Expansions

• On general quadratic actions are relatively easy to solve using the path integral method.

• For general class of potentials perturbation expansion of path integral.

• Start with a general expression for the potential specialize to scattering problem.

• a particle moving in a potential has the kernel

where is as previously defined. • If the potential is small, then the potential

part is:

),( txV

)(tDx

)(),(2

exp),( 2 tDxdttxVxmiifKf

i

t

tVf

i∫ ∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡ −= &

h

Khhh

22

),(!2

1),(1),(exp ⎥⎦⎤

⎢⎣⎡⎟

⎠⎞

⎜⎝⎛+−=⎥⎦

⎤⎢⎣⎡− ∫∫∫

f

i

f

i

f

i

t

t

t

t

t

ttxVidttxVidttxVi

• Substituting into the original expression:

where:K+++= ),(),(),(),( )2()1(

0 ifKifKifKifKV

[ ]

[ ]

[ ] )(''),'(

)(2

exp21),(

)(),(2

exp),(

)(2

exp),(

2

2)2(

2)1(

2

0

tDxdsssxV

dssxVdtxmiifK

tdsDxssxVdtxmiiifK

tDxdtxmiifK

f

i

f

i

b

a

b

a

b

a

f

i

t

t

t

t

f

i

t

t

t

t

f

i

t

t

f

i

t

t

∫

∫∫ ∫

∫∫ ∫

∫ ∫

×

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

&

hh

&

hh

&

h

• Interchanging integration order and writing

• is the sum over all paths of the free-particle amplitude; potential term can be additionally interpreted.

[ ] )(),(2

exp)(

where

)(),(

2

)1(

tDxssxVdtxmisF

dssFiifK

f

i

t

t

t

t

f

i

f

i

∫ ∫

∫

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

−=

&

h

h

)(sF

• However –weighted at time s by

• Before and after s – free particle.

• Unitarity of propagators sum over all paths between i to c & cto f can be written

[ ]ssxV ),(

),(),( 00 icKcfK

• Thus, for , can be written as

• Substituting into the expression for

• Can be interpreted as a free particle propagating from i to c, is scattered, and finally propagates freely to f

cts = )()( ctFsF =

∫∞

∞−= ccc dxicKtxVcfKtF ),(),(),()( 00

),()1( ifK

∫−= f

i

t

t ccdtdxicKcVcfKiifK ),()(),(),( 00)1(

h

K),,(),,(),,( 210 ifKifKifK

• Thus, can be written as ),( ifKV

∫ ∫

∫∫

∫

⎥⎦⎤

⎢⎣⎡ +−−

=

+⎟⎠⎞

⎜⎝⎛

+−

=

Khh

Kh

h

d

ac

c

V

didKdVdfKiicKcVcfKiifK

ddidKdVdcKcVcfKi

dicKcVcfKiifK

ifK

τ

ττ

τ

),()(),(),()(),(

),(

),()(),()(),(

),()(),(),(

),(

0000

0

000

2

000

• Thus, one can also write

• An integral equation describing ifis known.

∫−= cVV dicKcVcfKiifKifK τ),()(),(),(),( 00h

VK 0K

• Operating on a wavefunction yields

• Substituting the series expansion of

Kh

+−

=

∫ ∫

∫ic

i

dxifdicKcVcfKi

dxififKb

)(),()(),(

)(),()(

00

0

τ

ψ

),( ifKV

∫= iV dxififKb )(),()(ψ

VK

• The first term gives the unperturbed wavefunction at time . Calling this term one has

which yields

ft φ

idxififKf )(),()( 0∫=φ

∫ ∫

∫

++

−=

Kh

h

dc

c

ddddVdcKcVcfKi

dccVcfKibb

ττφ

τφφψ

)()(),()(),(

)()(),()()(

002

0

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