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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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