feynman path integral - binghamton...
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Feynman Path Integral: Formulation of Quantum Dynamics Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: September 22, 2016)
What is the path integral in quantum mechanics?
The path integral formulation of quantum mechanics is a description of quantum theory which generalizes the action principle of classical mechanics. It replaces the classical notion of a single, unique trajectory for a system with a sum, or functional integral, over an infinity of possible trajectories to compute a quantum amplitude. The basic idea of the path integral formulation can be traced back to Norbert Wiener, who introduced the Wiener integral for solving problems in diffusion and Brownian motion. This idea was extended to the use of the Lagrangian in quantum mechanics by P. A. M. Dirac in his 1933 paper. The complete method was developed in 1948 by Richard Feynman. This formulation has proven crucial to the subsequent development of theoretical physics, because it is manifestly symmetric between time and space. Unlike previous methods, the path-integral allows a physicist to easily change coordinates between very different canonical descriptions of the same quantum system.
((The idea of the path integral by Richard P. Feynman))
R.P. Feynman, The development of the space-time view of quantum electrodynamics (Nobel Lecture, December 11, 1965).
http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html
Feynman explained how to get the idea of the path integral in his talk of the Nobel Lecture. The detail is as follows. The sentence is a little revised because of typo. ________________________________________________________________
I went to a beer party in the Nassau Tavern in Princeton. There was a gentleman, newly arrived from Europe (Herbert Jehle) who came and sat next to me. Europeans are much more serious than we are in America because they think that a good place to discuss intellectual matters is a beer party. So, he sat by me and asked, «what are you doing» and so on, and I said, «I’m drinking beer.» Then I realized that he wanted to know what work I was doing and I told him I was struggling with this problem, and I simply turned to him and said, ((listen, do you know any way of doing quantum mechanics, starting with action - where the action integral comes into the quantum mechanics?» «No», he said, «but Dirac has a paper in which the Lagrangian, at least, comes into quantum mechanics. I will show it to you tomorrow»
Next day we went to the Princeton Library, they have little rooms on the side to discuss things, and he showed me this paper. What Dirac said was the following: There is in quantum mechanics a very important quantity which carries the wave function from one time to another, besides the differential equation but equivalent to it, a kind of a kernel, which we might call ),'( xxK ,which carries the wave function )(x known at time t, to the wave function )'(x at time, t . Dirac points out that this function K was analogous to the quantity in classical mechanics that you would calculate if you took the exponential of /i , multiplied by the Lagrangian ),( xxL imagining that these
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two positions x, x’ corresponded t and t . In other words, ),'( xxK is analogous to
)],'
(exp[ xxx
Li
,
)],'
(exp[),'( xxx
LixxK
. (1)
Professor Jehle showed me this, I read it, he explained it to me, and I said, «what does he mean, they are analogous; what does that mean, analogous? What is the use of that?» He said, «you Americans ! You always want to find a use for everything!» I said, that I thought that Dirac must mean that they were equal. «No», he explained, «he doesn’t mean they are equal.» «Well», I said, «let’s see what happens if we make them equal.»
So I simply put them equal, taking the simplest example where the Lagrangian is
)(2
1 2 xVxM , but soon found I had to put a constant of proportionality A in, suitably
adjusted. When I substituted )/exp( Li for K to get
dxtxxxx
LiAtx ),()],'
(exp[),'(
, (2)
and just calculated things out by Taylor series expansion, out came the Schrödinger equation. So, I turned to Professor Jehle, not really understanding, and said, «well, you see Professor Dirac meant that they were proportional.» Professor Jehle’s eyes were bugging out-he had taken out a little notebook and was rapidly copying it down from the blackboard, and said, «no, no, this is an important discovery. You Americans are always trying to find out how something can be used. That’s a good way to discover things!» So, I thought I was finding out what Dirac meant, but, as a matter of fact, had made the discovery that what Dirac thought was analogous, was, in fact, equal. I had then, at least, the connection between the Lagrangian and quantum mechanics, but still with wave functions and infinitesimal times.
It must have been a day or so later when I was lying in bed thinking about these things, that I imagined what would happen if I wanted to calculate the wave function at a finite interval later. I would put one of these factors )/exp( Li in here, and that would give me the wave functions the next moment, t and then I could substitute that back into (2) to get another factor of )/exp( Li and give me the wave function the next
moment, t + 2, and so on and so on. In that way I found myself thinking of a large number of integrals, one after the other in sequence. In the integrand was the product of the exponentials, which, of course, was the exponential of the sum of terms like /L . Now, L is the Lagrangian and is like the time interval dt, so that if you took a sum of such terms, that’s exactly like an integral. That’s like Riemann’s formula for the integral
Ldt , you just take the value at each point and add them together. We are to take the
limit as 0 , of course. Therefore, the connection between the wave function of one instant and the wave function of another instant a finite time later could be obtained by
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an infinite number of integrals, (because goes to zero, of course) of exponential )/( iS where S is the action expression (3),
dtxxLS ),( . (3)
At last, I had succeeded in representing quantum mechanics directly in terms of the action S. This led later on to the idea of the amplitude for a path; that for each possible way that the particle can go from one point to another in space-time, there’s an amplitude. That amplitude is an exponential of /i times the action for the path. Amplitudes from various paths superpose by addition. This then is another, a third way, of describing quantum mechanics, which looks quite different than that of Schrödinger or Heisenberg, but which is equivalent to them. 1. Introduction
The time evolution of the quantum state in the Schrodinger picture is given by
)'()',(ˆ)( tttUt ,
or
)'('')',(ˆ')'()',(ˆ)( txxttUxdxtttUxtx ,
in the x representation, where K(x, t; x’, t’) is referred to the propagator (kernel) and given by
')]'(ˆexp[')',(ˆ)',';,( xttHi
xxttUxtxtxK
.
Note that here we assume that the Hamiltonian H is independent of time t. Then we get the form
)'(')',';,(')( txtxtxKdxtx .
For the free particle, the propagator is described by
])'(2
)'(exp[
)'(2)',';,(
2
tt
xxim
tti
mtxtxK
.
(which will be derived later) ((Note)) Propagator as a transition amplitude
4
',',
')'ˆexp()ˆexp(
')]'(ˆexp[)',';,(
txtx
xtHi
tHi
x
xttHi
xtxtxK
Here we define
xtHi
tx )ˆexp(,
, )ˆexp(, tHi
xtx
.
We note that
axtEi
atHi
xatx a )exp()ˆexp(,
,
where
aEaH aˆ .
((Heisenberg picture)) The physical meaning of the ket tx, :
The operator in the Heisenberg's picture is given by
)ˆexp(ˆ)ˆexp(ˆ tHi
xtHi
xH
,
txxxtHi
x
xxtHi
xxtHi
xtHi
tHi
xtHi
txxH
,)ˆexp(
)ˆexp(
ˆ)ˆexp(
)ˆexp()ˆexp(ˆ)ˆexp(,ˆ
This means that tx, is the eigenket of the Heisenberg operator Hx with the eigenvalue x.
We note that
HS tHi
t )ˆexp()(
.
5
Then we get
HHS txtHi
xtx ,)ˆexp()(
.
This implies that
Htxtx ,, ,
Sxx .
where S means Schrödinger picture and H means Heisenberg picture. 2. Propagator
We are now ready to evaluate the transition amplitude for a finite time interval
')]ˆexp[)].....ˆexp[)]ˆexp[
',',)',';,(
xtHi
tHi
tHi
x
txtxtxtxK
where
t t t '
N (in the limit of N )
tt0 tDt
where t0 = t' in this figure. We next insert complete sets of position states (closure relation)
')ˆexp()ˆexp(
)ˆexp(......)ˆexp(
)ˆexp(....
',',)',';,(
112
2321
11221
xtHi
xxtHi
x
xtHi
xxtHi
x
xtHi
xdxdxdxdx
txtxtxtxK
NN
NNN
This expression says that the amplitude is the integral of the amplitude of all N-legged paths.
6
((Note))
),(),,(),,(),,(),,.....(
),,(),,(),,(),,(),','(
11223344
44332211
txtxtxtxtx
txtxtxtxtx
NNNNNNNN
with
t' t0 t1 t2 t3 t4 ....... tN 3 tN 2 tN 1 t tN
We define
,' 0xx 0' tt ,
,Nxx Ntt .
We need to calculate the propagator for one sub-interval
1)ˆexp[ ii xtHi
x
,
where i = 1, 2, …, N, and
)ˆ(2
ˆˆ2
xVm
pH ,
Then we have
7
))(()](2
[1
))((])()2
(1[
))((])ˆ()2
ˆ(1[
))(())ˆ(2
ˆ(1
))((ˆ1
)ˆexp()ˆexp(
211
2
2111
2
211
2
21
2
21
11
tOxxVm
pt
ippxdp
tOxxtVi
pxm
pt
ippxdp
tOxxtVi
pxm
pt
ippxdp
tOxxVm
pt
ippxdp
tOxHti
ppxdp
xHti
ppxdpxtHi
x
iii
iiii
iiiii
iiii
iiiiiii
iiiii
iiiii
iiiiiii
where pi (i = 1, 2, 3, …, N),
or
])},()(
{exp[2
1
}]),()(
{exp[2
1
]),(exp[)](exp[2
1
]),(1)[(exp[2
1
))](2
ˆ(1[)ˆexp(
11
11
11
11
1
2
11
txpEt
xxp
idp
txpEtt
xxp
idp
xptEi
xxpi
dp
xptEi
xxpi
dp
xVm
pt
ixppxdpxtH
ix
iiii
ii
iiii
ii
iiiiii
iiiiii
ii
iiiiiii
where
E( pi , xi1 ) pi
2
2m V(xi1) .
Then we have
N
ii
iiii
NNN
N
txVm
p
t
xxp
i
dpdpdpdpdxdxdx
txtxtxtxK
11
21
121121
]))}(2
()(
{exp[
22.......
22.......lim
',',)',';,(
8
We note that
])(2
exp[2
}2
)({exp[2
}}2
)({exp[
2
21
2
1
21
t
xxtim
ti
m
tm
pixxp
idpt
m
p
t
xxp
idp
ii
iiii
iiiii
i
_____________________________________________________________________ ((Mathematica))
Clear"Global`"; f1 1
2 —Exp
—p x
p2
2 m —t;
Integratef1, p, , Simplify, — 0, m 0, Imt 0 &
m x22 t —
2 t —
m ______________________________________________________________________ Then we have
N
ii
ii
NN
N
xVt
xxmt
iti
mdxdxdxtxtxK
11
21
2/121
)}]()(2
{exp[
)2
(.......lim)',';,(
Notice that as N and therefore t 0 , the argument of the exponent becomes the standard definition of a Riemann integral
t
t
N
ii
ii
tN
xxdtLi
xVt
xxmt
i
'11
21
0
),()}()(2
{lim
,
where L is the Lagrangian (which is described by the difference between the kinetic energy and the potential energy)
9
t
x
O
)()(2
1),( 2 xVxmxxL .
It is convenient to express the remaining infinite number of position integrals using the shorthand notation
2/121 )
2(.......lim)]([ N
NN ti
mdxdxdxtxD
.
Thus we have
)]}([exp{)]([',',)',';,( txSi
txDtxtxtxtxK
,
where
),()](['t
t
xxdtLtxS .
The unit of S is [erg sec].
10
When two points at (ti, xi) and (tf, xf) are fixed as shown the figure below, for convenience, we use
),()]([ f
i
t
t
xxdtLtxS .
t1 t2
x1
x2
x
t
This expression is known as Feynman’s path integral (configuration space path integral). )]([ txS is the value of the action evaluated for a particular path taken by the particle. If one want to know the quantum mechanical amplitude for a point particle at x’, at time t’ to reach a position x, at time t, one integrate over all possible paths connecting the points with a weight factor given by the classical action for each path. This formulation is completely equivalent to the usual formulation of quantum mechanics.
The expression for ',',)',';,( txtxtxtxK may be written, in some loose sense, as
)exp()exp()exp(
])0,(
exp[',',
21
)(00
npathpathpath
pathallNN
iSiSiS
NiSttxxttxx
11
where the sum is to be taken over an innumerably infinite sets of paths.
t
x
xclt
(a) Classical case.
Suppose that 0 (classical case), the weight factor ]/exp[ iS oscillates very violently. So there is a tendency for cancellation among various contribution from neighboring paths. The classical path (in the limit of 0 ) is the path of least action, for which the action is an extremum. The constructive interference occurs in a very narrow strip containing the classical path. This is nothing but the derivation of Euler-Lagrange equation from the classical action. Thus the classical trajectory dominates the path integral in the small ħ limit.
In the classical approximation ( )S
',', 00 ttxxttxx NN "smooth function" )exp(
cliS. (1)
But at an atomic level, S may be compared with ħ, and then all trajectory must be added in
',', 00 ttxxttxx NN in detail. No particular trajectory is of overwhelming importance,
and of course Eq.(1) is not necessarily a good approximation.
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(b) Quantum case. What about the case for the finite value of /S (corresponding to the quantum case)? The
phase ]/exp[ iS does not vary very much as we deviate slightly from the classical path. As a result, as long as we stay near the classical path, constructive interference between neighboring paths is possible. The path integral is an infinite-slit experiment. Because one cannot specify which path the particle choose, even when one know what the initial and final positions are. The trajectory can deviate from the classical trajectory if the difference in the action is roughly within ħ. ((Note)) R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (extended edition) Dover 2005. 3. Free particle propagator In this case, there is no potential energy.
N
i
iiNN
N t
xxmt
i
ti
mdxdxdxtxtxK
1
212/121 }])(
2{exp[)
2(.......lim)',';,(
,
or
K(x, t;x' , t' ) limN
dx1 dx2 ....... dxN 1(
m
2it)N / 2 exp[
m
2it{(x1 x ' )2 (x2 x1)
2 (x3 x2 )2 .... (x xN1 )2}]
We need to calculate the integrals,
])'(4
exp[)(4
1
])()'{(2
exp[)2
(
22
2/1
212
211
2/21
xxti
m
ti
m
xxxxti
mdx
ti
mf
])(2
exp[)2
( 223
2/111 xx
ti
m
ti
mfg
f2 g1dx2
1
6(
m
it)1/ 2 exp[
m
6it(x3 x' ) 2 ]
g2 f2(
m
2it)1 / 2 exp[
m
2it(x4 x3 )2 ]
13
f3 g2dx3
1
8(
m
it)1/ 2 exp[
m
8it(x4 x' )2 ]
..........................................................................................................................................
K(x, t;x' , t' ) lim
N(
m
2Nit)1/ 2 exp[
m(x x' )2
2iNt],
or
K(x, t;x' , t' ) (
m
2i(t t' ))1/ 2 exp[
m(x x' )2
2i( t t' )] ,
where we uset t' Nt in the last part. ((Mathematica))
14
Free particle propagator;e=Dt
Clear"Global`";
exp_ : exp . Complexre_, im_ Complexre, im;
h1 x1 x'2 x2 x12 Expand;
f0x1_ :2 —
m
22
Exp m
2 — h1;
f1 Integratef0x1, x1, , Simplify, Im m
— 0 &
m x2x2
4 — m —
2
g1 f12 —
m
12
Exp m
2 — x3 x22 Simplify;
f2
g1 x2 Simplify, Im m
— 0 &
m x3x2
6 —
6 —
m
g2 f22 —
m
12
Exp m
2 — x4 x32 Simplify;
f3
g2 x3 Simplify, Im m
— 0 &
m x4x2
8 — m —
2 2
g3 f32 —
m
12
Exp m
2 — x5 x42 Simplify;
f4
g3 x4 Simplify, Im m
— 0 &
m x5x2
10 —
10 —
m
15
4. Gaussian path integral
The simplest path integral corresponds to the vase where the dynamical variables appear at the most up to quadratic order in the Lagrangian (the free particle, simple harmonics are examples of such systems). The the probability amplitude associated with the transition from the points ),( ii tx to ),( ff tx is the sum over all paths with the action as a phase angle, namely,
),(]exp[),;,( ifcliiff ttFSi
txtxK
,
where clS is the classical action associated with each path,
),,( txxdtLS clcl
t
t
cl
f
i
,
with the Lagrangian ),,( txxL described by the Gaussian form,
)()()()()()(),,( 22 tfxtextdxtcxxtbxtatxxL If the Lagrangian has no explicit time dependence, then we get
)(),( ifif ttFttF .
For simplicity, we use this theorem without proof.
)(]exp[),;,( ifcliiff ttFSi
txtxK
.
((Proof)) R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals. Let )(txcl be the classical path between the specified end points. This is the path which is an
extremun for the action S. We can represent )(tx in terms of )(txcl and a new function )(ty ;
)()()( tytxtx cl ,
where 0)()( fi tyty . At each t, the variables )(tx and )(ty differ by the constant . )(txcl (Of
course, this is a different constant for each value of t). Thus, clearly,
ii dydx ,
for each specific point ti in the subdivision of time. In general, we may say that
16
)()( tDytDx .
The integral for the action can be written as
f
i
t
t
if dtttxtxLttS ])),(),([],[ ,
with
)()()()()()(),,( 22 tfxtextdxtcxxtbxtatxxL . We expand ),,( txxL in a Taylor expansion around clcl xx , . This series terminates after the
second term because of the Gaussian form of Lagrangian. Then
clclclcl xxxxclcl yx
Lyy
xx
Ly
x
Ly
x
Ly
x
LtxxLtxxL
,
22
222
2
2
|)(2
1||),,(),,(
.
From here we obtain the action
f
i
f
i
clcl
f
i
clcl
f
i
clcl
t
t
t
t
xxifcl
t
t
xx
t
t
xxifclif
ytcyytbytadt
yx
Ly
x
LdtttS
yx
Lyy
xx
Ly
x
Ldt
yx
Ly
x
LdtttSttS
])()()([
)||(],[
|)(2
1
)||(],[],[
22
,2
2
222
2
2
The integration by parts and use of the Lagrange equation makes the second term on the right-hand side vanish. So we are left with
f
i
t
t
ifclif ytcyytbytadtttSttS ])()()([],[],[ 22 .
Then we can write
17
f
i
t
t
ifcl dtytcyytbytattStxS ])()()([],[)]([ 22 .
The integral over paths does not depend on the classical path, so the kernel can be written as
]exp[),(
)(}])()()([exp{]exp[),;,( 220
0
clif
t
t
y
y
cliiff
Si
ttF
tDydtytcyytbytai
Si
txtxKf
i
where
f
i
t
t
y
y
if tDydtytcyytbytai
ttF )(}])()()([exp{),( 220
0
.
It is defined hat ),( if ttF is the integral over all paths from 0y back to 0y during the
interval )( if tt .
((Note)) If the Lagrangian is given by the simple form
22 )()()(),,( xtcxxtbxtatxxL
then ),( if ttF can be expressed by
),0;,0(),( iiffif txtxKttF .
4. Evaluation of ),( if ttF for the free particle
We now calculate )',( ttttF if for the free particles, where the Lagrangian is given by the
form,
2
2
1),,( xmtxxL .
Then we have
f
i
t
t
y
y
if tDydtymi
ttF )(}2
exp{),( 20
0
.
18
Replacing the variable y by x, we get
f
i
t
t
x
x
if tDxdtxmi
ttF )(}2
exp{),( 20
0
In this case, formally ),( if ttF is equal to the propagator ),0;,0( iiff txtxK ,
}])(....)()({2
exp[)2
(
.......lim
),0;,0(),(
21
221
223
212
21
2/
121
NNNN
NN
iiffif
xxxxxxxxti
m
ti
m
dxdxdx
txtxKttF
where we put 0' xxi , and 0 xx f , and N
tt
N
ttt if '
.
How can we solve the integral? We use the eigenvalue problem (Gottfried and Yan). Before that we discuss the eigenvalue problem related to this problem.
Suppose that the matrix A (n x n) under the basis }{ ib , is given by
nnnnn
n
n
n
AAAA
AAAA
AAAA
AAAA
A
...
.......
.......
.......
...
...
...
ˆ
321
3333231
2232221
1131211
.
where n = N-1. We also a ket under the basis of }{ ib is given by
nx
x
x
x
.
.
.3
2
1
.
Now we calculate the average A as
19
jijji
T
nnnnnn
n
n
n
n
xAx
x
x
x
x
AAAA
AAAA
AAAA
AAAA
xxxxA
,1
*
*
3
2
1
321
3333231
2232221
1131211
*321
.
.
.
...
.......
.......
.......
...
...
...
...ˆ
AXX
where
nx
x
x
x
.
.
.3
2
1
X , nT xxxx ...321X .
We introduce the new basis }{ ia such that
iii aaaA ˆ , (Eigen value problem)
where ia is the eigenket of A with the eigenvalue ai. Suppose that can be expressed under
the basis of }{ ia ,
ii
i a .
Then we have
20
iiii
nn
n
a
a
a
a
a
A
*
3
2
1
3
2
1
*321
.
.
.
...000
.......
.......
.......
0...00
0...00
0...00
...ˆ
where
n
.
.
.3
2
1
η , nT ...321η .
We introduce the unitary operator U ;
ii bUa ˆ ,
where
ijij bUbab ˆ .
The eigenvalue problem can be rewritten as
iii bUabUA ˆˆˆ ,
or
iii babUAU ˆˆˆ ,
or
ijiij abUAUb ˆˆˆ .
21
We note that
ii
ijj
j bxa ,
where
i
ijii
iiji
ijij xUxbabax )( ,
or
XUXU Tˆˆ .
Note that each element of matrix U is real; TUU ˆˆ . Then we have
UX ˆ ,
since 1ˆˆ UU . We now evaluate the following integral;
}])(....)()({2
exp[)2
(
.......lim),(
21
221
223
212
21
2/
121
NNNN
NN
if
xxxxxxxxti
m
ti
m
dxdxdxttF
We note that
)....(2)....(2
)(....)()(
1232212
12
22
1
21
221
223
212
21
NNN
NNN
xxxxxxxxx
xxxxxxxxf
Using the matrix, f can be rewritten as
)ˆˆˆ(ˆˆˆˆ)ˆ(ˆ UAUUAUUAUXAXf T , with
22
1
2
1
.
.
.
.
Nx
x
x
X ,
21.....
12.....
.......
..12100
..01210
..00121
0.00012
A .
We solve the eigenvalue problems to determine the eigenvalues and the unitary operator, such that
n
UAU
..0000
.......
.......
0..000
0..000
0..000
0..000
ˆˆˆ4
3
2
1
,
where i is the eigenvalue of A. Then we have
1
1
2
1
2
1
4
3
2
1
11111
.
.
.
.
..0000
.......
.......
0..000
0..000
0..000
0..000
..
N
iii
Nn
Nf
The Jacobian determinant is obtained as
1det),...,,(
),...,,(
121
121
Uxxx
N
N
.
Then we have the integral
23
)'(2
2
)(det2
.....
1)
2()
2(
2...
22)
2(
}]...{2
exp[)2
(.......lim
)',(),(
121
2/)1(2/
121
2/
211
222
211
2/121
tti
m
tiN
m
Ati
m
ti
m
ti
m
m
ti
m
ti
m
ti
ti
m
ti
m
ti
mddd
ttFttF
N
NN
N
N
NNN
NN
if
where
NA N 1321 ...det
((Mathematica)) Example (N = 6). The matrix A (5 x 5): 51N
21000
12100
01210
00121
00012
A
The eigenvalues of A:
321 , 32 , 23 , 14 , 325 .
The unitary operator
U
24
1
1
2N
iiif .
5
4
3
2
1
0000
0000
0000
0000
0000
ˆˆˆ
UAU .
6det NA .
__________________________________________________________________________
25
Clear "Global` " ;
A1
2 1 0 0 01 2 1 0 0
0 1 2 1 00 0 1 2 10 0 0 1 2
;
eq1 Eigensystem A1
2 3 , 3, 2, 1, 2 3 , 1, 3 , 2, 3 , 1 ,
1, 1, 0, 1, 1 , 1, 0, 1, 0, 1 ,
1, 1, 0, 1, 1 , 1, 3 , 2, 3 , 1
1 Normalize eq1 2, 1 Simplify
1
2 3,
12
,1
3,
12
,1
2 3
2 Normalize eq1 2, 2 Simplify
12
,12
, 0,12
,12
26
3 Normalize eq1 2, 3 Simplify
1
3, 0,
1
3, 0,
1
3
4 Normalize eq1 2, 4 Simplify
12
,12
, 0,12
,12
5 Normalize eq1 2, 5 Simplify
1
2 3,
12
,1
3,
12
,1
2 3
UT 1, 2, 3, 4, 5 ; U Transpose UT ;
UH UT;
U MatrixForm
12 3
12
13
12
12 3
12
12
0 12
12
13
0 13
0 13
12
12
0 12
12
12 3
12
13
12
12 3
27
UH.U Simplify; UH.U MatrixForm
1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1
12345
;
s1 UH.A1.U Simplify; s1 MatrixForm
2 3 0 0 0 00 3 0 0 00 0 2 0 00 0 0 1 0
0 0 0 0 2 3
f1 Transpose .s1. FullSimplify
2 3 12 3 22 2 32 42 2 3 52
Det A1
6
K1 eq1 1, 1 eq1 1, 2 eq1 1, 3 eq1 1, 4
eq1 1, 5 Simplify
6
6. Equivalence with Schrodinger equation
The Schrödinger equation is given by
)(ˆ)( tHtt
i
.
For an infinitesimal time interval , we can write
28
)0(ˆ)0()( Hi
,
from the definition of the derivative, or
)0()](2
[
)0(ˆ)0()(
2
22
xxVxm
i
Hxi
xx
or
)0,()](2
[
)0(ˆ)0,(),(
2
22
xxVxm
i
Hxi
xx
in the x representation.
We now show that the path integral also predicts this behavior for the wave function. To this end, we start with
)0(')0,';,(')( xxxKdxx ,
or
)0,'()0,';,('),( xxxKdxx ,
where
)}]2
'(
)'(
2
1{exp[
2
)]2
',
'(exp[
2)0,';,(
2
2 xxV
xxm
i
i
m
xxxxL
i
i
mxxK
Then we get
)0,'()}]
2
'(
)'(
2
1{exp['
2),(
2
2
xxx
Vxx
mi
dxi
mx
.
29
We now define
xx' . Then we have
)0,()]2
(2
exp[2
),(2
xxV
iimd
i
mx
.
The dominant contribution comes from the small limit of . Using the Taylor expansion in the limit of 0
)()2
( xVxV ,
2
22 )0,(
2
)0,()0,()0,(
x
x
x
xxx
,
we get
)0,(]2
)(1[
)]()0,(
2
)0,()0,()[
2exp(
2
])0,(
2
)0,()0,()][(1)[
2exp(
2),(
2
22
2
222
2
222
xxm
ixV
i
xVi
x
x
x
xx
imd
i
m
x
x
x
xxxV
iimd
i
mx
Thus we have
)0,()](2
[)0,(),(2
22
xxVxm
ixx
,
which is the same as that derived from the Schrödinger equation. The path integral formalism leads to the Schrodinger equation for infinitesimal intervals. Since any finite interval can be thought of a series of successive infinitesimal intervals the equivalence would still be true. ((Note))
2/12 2)
2exp(
m
iimd
,
2/122 2
)2
exp(
m
i
m
iimd
.
30
((Mathematica))
Clear"Global`";
Integraln_ :
Integraten Exp m 2
2 — , , ,
Simplify, Im m
— 0 &;
K1 Tablen, Integraln, n, 0, 4;
K1 TableForm
0 2
m —
1 0
2 2
m —
32
3 0
4 3 2
m —
52
7. Motion of free particle; Feynman path integral
The Lagrangian of the free particle is given by
2
2x
mL .
Lagrange equation for the classical path;
0)()(
x
L
x
L
dt
d
,
or
ax , or
batx .
31
This line passes through (t', x'), (t, x);
)'('
'' 11 tt
tt
xxxx
,
t',x'
t,x
x1
t1
Then we have
)'('
'' 11 tt
tt
xxxx
,
and
22
1
11 )
'
'(
2
1)(
2
1)(
tt
xxm
dt
dxmtL
.
which is independent of t1. Consequently, we have
'
)'(
2)
'
'(
2)
'
'(
2)(
2
' '
12
12
'
11 tt
xxmdt
tt
xxmdt
tt
xxmdttLS
t
t
t
t
t
t
cl
,
and
32
])'(2
)'(exp[]exp[)',',,(
2
tti
xxmAS
iAtxtxK cl
.
((Approach from the classical limit))
To find A, we use the fact that as 0' tt , K must tend to )'( xx
]2
)'(exp[
2
1lim
])'(
exp[)(
1lim)'(
2
2
0
2
2
2/120
xx
xxxx
.
where
2
,
]2
)'(exp[
2
1),',(
2
2
xx
xxf
. (Gaussian distribution).
So we get
m
tti )'(2
,
)'(2)(
12/12 tti
mA
,
or
])'(2
)'(exp[
)'(2)',',,(
2
tti
xxm
tti
mtxtxK
.
Note that
)'(2)'(
tti
mttF articlefreep
.
((Evaluation of /S )) From the above discussion, S can be evaluated as
33
xh
xp
xmv
t
xxm
t
xmS
2222
)()(
2
2
,
or
xkxp
xmvS
2
1
2
1
2 .
where p is the momentum,
kp . Suppose that m is the mass of electron and the velocity v is equal to c/137. We make a plot of
S(radian) as a function of x (cm).
((Mathematica)) NIST Physics constant : cgs units
Clear"Global`";
rule1 c 2.99792 1010, — 1.054571628 1027,
me 9.10938215 1028;
K1 me c
2 —. rule1
1.2948 1010
K1 137
9.4511 107
34
10-8 10-6 10-4 0.01 1 100Dx cm
1
100
104
106
108
1010SÑ rad
Fig.
S(radian) as a function of x (cm), where v = c/137. m is the mass of electron.
8. Evaluation of S for the 1D system (example 8-1, Towmsend, 2nd edition)
We consider the Young’s double slit;
xxh
pxxt
xm
t
mxS
2
1
2
1
2)(2
2
.
The phase difference between two paths is evaluated as
xxkxpS
2
1
2
1
,
35
If
S is comparable to , the interference effect can be observed.Such a condition is satisfied
when
x . ((Note))
In classical physics, the phase difference is given by
x 2
.
9. Single slit experiment
Imagine the slit divided into many narrow zones, width y (= = a/N). Treat each as a secondary source of light contributing electric field amplitude E to the field at P.
We consider a linear array of N coherent point oscillators, which are each identical, even to their polarization. For the moment, we consider the oscillators to have no intrinsic phase difference. The rays shown are all almost parallel, meeting at some very distant point P. If the
36
spatial extent of the array is comparatively small, the separate wave amplitudes arriving at P will be essentially equal, having traveled nearly equal distances, that is
N
ErErErErE N
0002010 )()(...)()(
The sum of the interfering spherical wavelets yields an electric field at P, given by the real part of
]]...1[)(Re[
])(...)()(Re[))()()()(
0
)(0
)(0
)(0
113121
21
rrikrrikrriktkri
tkritkritkri
N
N
eeeerE
erEerEerEE
((Note)) When the distances r1 and r2 from sources 1 and 2 to the field point P are large compared with the separation , then these two rays from the sources to the point P are nearly parallel. The path difference r2 – r1 is essentially equal to sin.
Here we note that the phase difference between adjacent zone is
)(
)(
)(
)sin(sin)(
1
34
23
12
NN rrk
rrk
rrkN
akkrrk
where k is the wavenumber, 2
k . It follows that
)1()(
3)(
2)(
)(
1
14
13
12
Nrr
rrk
rrk
rrk
N
Thus the field at the point P may be written as
]]...1[)(Re[ )1(2)(0
1 Niiitkri eeeerEE
We now calculate the complex number given by
37
)2
sin(
)2
sin(
)(
)(
1
1
...1
2/)1(
2/2/2/
2/2/2/
)1(2
N
e
eee
eee
e
e
eeeZ
Ni
iii
iNiNiN
i
iN
Niii
If we define D to be the distance from the center of the line of oscillators to the point P, that is
)1(2
1)(
)1(2
1sin)1(
2
1
1
11
NrDk
krNkrkNkD
Then we have the form for E as
]~
Re[])
2sin(
)2
sin()(Re[ )(
0titkDi eE
N
erEE
The intensity distribution within the diffraction pattern due to N coherent, identical, distant point sources in a linear array is equal to
20 ~
2E
cSI
2
2
2
2
02
2
0
2
)2
(sin
)2
(sin
)2
(sin
)2
(sin
)2
(sin
mI
N
I
N
II
in the limit of N→∞, where
22
2)
2(sin
NN
2
00
0 )]([2
rEc
I
38
2
002
002
0 2)]([
2E
crNE
cNIIm
sinsin kaNkN
sink
where a = N. We make a plot of the relative intensity I/Im as a function of .
2
2
2
)2
(sin
mI
I
Note that
2
2
)2
(sin
2
2
d
-4 -2 0 2 4b2p
0.01
0.02
0.03
0.04
0.05
0.06IIm
39
The numerator undergoes rapid fluctuations, while the denominator varies relatively slowly. The combined expression gives rise to a series of sharp principal peaks separated by small subsidiary maxima. The principal minimum occur in directions in direction m such that
mmmk
a
mka
m
22
21
sin
sin22
10. Phasor diagram (i) The system with two paths
The phasor diagram can be used for the calculation of the double slilts (Young) interference.
We consider the sum of the vectors given by OS and ST . The magnitudes of these vectors is the
same. The angle between OS and ST is (the phase difference).
40
Fig. Phasor diagram for the double slit.
In this figure, RQTQSQO . 2/ STMSOM . Then we have
2cos2
2cos22
AOSOMOT .
The resultant intensity is proportional to 2OT ,
)cos1(22
cos4 2222
AAOTI .
Note that the radius R is related to OS (= A) through a relation
2sin2
RA .
When A = 1 (in the present case), we have the intensity I as
2cos4 2 I
The intensity has a maximum (I = 4) at n 2 and a minimum (I = 0) at
)2/1(2 n .
41
(ii) The system with 6 paths.
Fig. The resultant amplitude of N = 6 equally spaced sources with net successive phase
difference φ. = N φ = 6 φ. (iii) The system with 36 paths (comparable to single slit)
Fig. The resultant amplitude of N = 36 equally spaced sources with net successive phase
difference φ.
42
(iv) Single slit in the limit of N→∞
We now consider the system with a very large N. We may imagine dividing the slit into N narrow strips. In the limit of large N, there is an infinite number of infinitesimally narrow strips. Then the curve trail of phasors become an arc of a circle, with arc length equal to the length E0. The center C of this arc is found by constructing perpendiculars at O and T.
The radius of arc is given by
)(0 NRRE .
in the limit of large N, where R is the side of the isosceles triangular lattice with the vertex angle φ, and is given by
sinkaN , with the value being kept constant. Then the amplitude Ep of the resultant electric field at P is
equal to the chord OT , which is equal to
2
2sin
2sin2
2sin2 0
0
EE
REP .
Then the intensity I for the single slits with finite width a is given by
43
2)
2
2sin
(
mII .
where Im is the intensity in the straight-ahead direction where
The phase difference φ is given by
sin2sin2sin pa
ka . We make a plot of
I/Im as a function of , where p = a/ is changed as a parameter.
Fig. The relative intensity in single-slit diffraction for various values of the ratio p = a/. The
wider the slit is the narrower is the central diffraction maximum. 11. Gravity: Feynman path integral ((Calculation from the classical limit))
mgxxm
L 2
2 ,
)()(x
L
x
L
dt
d
,
gx ,
BAttg
x 12
11 2,
Initial conditions:
x1 = x and t1= t,
44
BAttg
x 2
2.
and '1 xx and '1 tt
BAttg
x ''2
' 2 ,
A and B are determined from the above two equations.
)'(2
')(2]2))('()['( 1111 tt
xttxttttgttx
,
)'(2
)'(2)2')('( 1
1
11 tt
xxtttttg
dt
dxx
.
Then we get the expression of the Lagrangian
12
1111 2),,( mgxx
mtxxL .
The Hamilton’s principle function is
)'(24
)]'()'(12)'(12)'([
),,()',',,(
2242
'
1111
tt
xxttgxxttgm
dttxxLtxtxSt
t
cl
)]',':,(exp[)',';,( txtxSi
AtxtxK cl
.
((Classical limit))
When 'xx and Ttt ' , we have
gmTxTmgtxtxScl '24
1)',':,( 32 ,
and
45
)'
exp(
)'
24exp()',';,(
1
32
mgTixA
mgTixTmg
iAtxtxK
where T is the time during which a particle passes through the system.
)24
exp( 321 Tmg
iAA
.
((Mathematica))
Clear"Global`"; eq1 x 1
2g t2 A t B;
eq2 x0 1
2g t02 A t0 B;
rule1 Solveeq1, eq2, A, B Flatten;
x1 1
2g t12 A t1 B . rule1 FullSimplify
t0 t1 g t t0 t t1 2 x 2 t t1 x0
2 t t0
v1 Dx1, t1 Simplify
g t t0 t t0 2 t1 2 x x02 t t0
L1 m
2v12 m g x1 FullSimplify
1
8 t t02m g t t0 t t0 2 t1 2 x x02
4 g t t0 t0 t1 g t t0 t t1 2 x 2 t t1 x0
K1 t0
tL1 t1 FullSimplify
m g2 t t04 12 x x02 12 g t t02 x x0
24 t t0
rule1 x x0, t t0 T; K2 K1 . rule1 Simplify
1
24g m T g T2 24 x0
46
12. Simple harmonics: Feynman path integral ((Classical limit))
220
2
2
1
2xmx
mL ,
)()(x
L
x
L
dt
d
,
xx 20 ,
)sin()cos( 10101 tBtAx .
Initial conditions:
x1 = x and t1= t,
x A cos0t Bsin0t , and
'1 xx and '1 tt ,
'sin'cos' 00 tBtAx .
A and B are determined from the above two equations.
)]'(sin[
)]'(sin[)](sin['
0
10101 tt
ttxttxx
,
)]'(sin[
)]'(cos[)](cos['
0
100100
1
11 tt
ttxttx
dt
dxx
.
Then we get the expression of the Lagrangian
)])'(cos[)](2cos['
)2'(cos['2[)]'([sin2
2
1
2),,(
102
102
100
2
20
21
20
21111
ttxttx
tttxxtt
m
xmxm
txxL
47
The Hamilton’s principle function is
]'2))'(cos()'[()'(sin(2
),,()',',,(
022
0
0
'
1111
xxttxxtt
m
dttxxLtxtxSt
t
cl
)]',':,(exp[)',';,( txtxSi
AtxtxK cl
,
or
}]'2))'(cos()'{()]'(sin[2
exp[)',';,( 022
0
0 xxttxxtt
imAtxtxK
.
In the limit of 0' tt , we have
])'()}'(sin{2
exp[)',',,( 2
0
0 xxtt
imAtxtxK
.
To find A, we use the fact that as 0' tt , K must tend to )'( xx ,
]2
)'(exp[
2
1lim
])'(
exp[)(
1lim)'(
2
2
0
2
2
2/120
xx
xxxx
.
where
2
,
]2
)'(exp[
2
1),',( 2
2
xx
xxf
(Gaussian distribution).
In other words
A 1
(2)1/ 2 , )}'(sin{2
1
0
02 tti
m
.
So we get
48
0
0 )}'(sin{2
m
tti
,
)}'(sin{2)(
1
0
02/12 tti
mA
or
]]'2)}'(cos{)'[()]'(sin[2
exp[
)}'(sin{2)',',,(
022
0
0
0
0
xxttxxtt
im
tti
mtxtxK
Note that
)'()}'(sin{2
)',0',,0(0
0 ttFtti
mtxtxK
and
)]'';,(exp[)'()',',,( txtxSi
ttFtxtxK cl
.
((Mathematica))
Clear"Global`";
expr_ : expr . Complexa_, b_ Complexa, bseq1 x A Cos0 t B Sin0 t;
seq2 x0 A Cos0 t0 B Sin0 t0;
srule1 Solveseq1, seq2, A, B Simplify Flatten;
x1 A Cost1 0 B Sint1 0 . srule1 Simplify;
v1 Dx1, t1 Simplify;
L1 m
2v12
m
202 x12 Simplify;
t0
tL1 t1 FullSimplify
1
2m 0 2 x x0 x2 x02 Cost t0 0 Csct t0 0
If the initial state of a harmonic oscillator is given by the displaced ground state wave function
(x,0) exp[
m0
2(x x0)
2].
49
When
x , with
0 m
.
Then we have
])(2
1exp[
1)0,( 2
04/1
,
and
)}]1(2
12{
2
1
2exp[
1
}]2
)2
({(2
1exp[)
2exp(
1
}]2
)sin()cos(){(2
1exp[)
2exp(
1
}])sin(
12
)cot(){(2
)sin(exp[)
2exp(
1
')0,'()0,';,(),(
00
00
00
0
0
2200
204/1
0
20
204/1
0
02
02
020
4/1
00
20
20
2004/1
titi
titititi
ti
ti
eeti
eeee
ti
titeti
t
itetti
dtKt
or
)]2sin4
1sin
2()cos(
2
1exp[
1
)}]2sin2cos1(2
1
)sin(cos2{2
1
2exp[
1),(
02
00002
004/1
002
0
00020
4/1
ttt
it
tit
titti
t
Finally we have
50
]))cos((exp[1
|),(| 2002/1
2 tt
((Mathematica))
Clear"Global`";
exp_ : exp . Complexre_, im_ Complexre, im;
KSH_, t_, 1_ :1
2 Sin0 t Exp
2 Sin0 t 2 12 Cos0 t 2 1;
0_ : 14 Exp 02
2;
f1
KSH, t, 1 01 1 FullSimplify, ImCott 0 1, 0 t 0 &
202 Cott 0 2 0 Csct 0
2 Cott 0 Csct 014 1 Cott 0
Amp1 f1 f1 FullSimplify
0 Cost 02
rule1 0 1, 0 1; H1 Amp1 . rule1;
PlotEvaluateTableH1, t, 0, 20, 2, , 3, 3,
PlotStyle TableThick, Hue0.1 i, i, 0, 10, AxesLabel "", "Amplitude"
-3 -2 -1 1 2 3x
0.1
0.2
0.3
0.4
0.5
Amplitude
13. Gaussian wave packet propagation (quantum mechanics)
)'('')',(ˆ')'()',(ˆ)( txxttUxdxtttUxtx ,
'))'(ˆexp[')',(ˆ)',';,( xttHi
xxttUxtxtxK
,
or
)'(')',';,(')( txtxtxKdxtx .
51
K(x, t; x’, t’) is referred to the propagator (kernel) For the free particle, the propagator is given by
])'(2
)'(exp[
)'(2)',';,(
2
tt
xxim
tti
mtxtxK
.
Let’s give a proof for this in the momentum space. ˆ H is the Hamiltonian of the free particle.
ikxekx
21
,
kEkH kˆ ,
with
Ek
2k2
2m,
k
Ek
,
)]'(2
)'(exp[2
1
'))'(2
exp[
'))'(ˆexp[
'))'(ˆexp[)',';,(
2
2
ttm
kixxikdk
xttm
kikkxdk
xttHi
kkxdk
xttHi
xtxtxK
Note that
])'(2
)'(}
)'(
)'({
2
)'(exp[)]'(
2)'(exp[
22
2
tti
xxm
tt
xxmk
m
ttitt
m
kixxik
,
and
i
kidk
)exp( 2 ,
52
or
])'(2
)'(exp[
)'(2)',';,(
2
tt
xxim
tti
mtxtxK
.
((Quantum mechanical treatment))
Probability amplitude that a particle initially at x’ propagates to x in the interval t-t’. This expression is generalized to that for the three dimension.
])'(2
|'|exp[]
)'(2[)',';,(
22/3
tt
im
tti
mttK
rrrr
.
We now consider the wave function:
)4
exp(2
1)0( 2
2
0
xx
xxiktx
.
where the probability
)2
exp(2
1)0( 2
22
xx
xtxP
,
has the form of Gaussian distribution with the standard deviation x . The Fourier transform is
given by
])(exp[2
])(exp[22
1
2
1
)4
exp()exp(2
1
2
1
)0()0(
20
24/1
20
2
2
2
0
kk
kk
xxikikxdx
txxkdxtk
xx
xx
x
xx
or
53
]2
)(exp[
2
1
]
2
12
)(exp[
2
1
1
2
1
])(2exp[2
)0(
2
20
2
20
20
22
kk
xx
xx
kk
kk
kktk
.
which is the Gaussian distribution with the standard deviation
xk
2
1.
Then we get the wave function at time t,
])
41(8
))4(8(
)4
1(4
)(exp[
21
1
2
1
])
21)(
21(4
)2
1}(2)4({
exp[
21
1
2
1
])
21(4
2)4(exp[
21
1
2
1
]24
2)4(exp[
21
12
22
1
]2
)'(
4
''exp['
22
1
)0(')0,';,(')(
42
2242
420
240
42
222
20
2
22
2
2
220
20
2
2
2
220
20
2
2
220
20
2
2
2
2
0
x
x
xx
x
x
x
x
xx
x
x
xx
x
x
x
x
xx
x
x
x
xx
x
x
xx
m
tm
kxtmxkmi
m
tm
tkx
m
it
m
it
m
itm
it
m
tikikxx
m
it
m
itm
tikikxx
m
it
itm
tikikxmx
t
imti
m
t
xximxxikdx
ti
m
xxtxKdxtx
Since
42
22
24
1
1
21
1
xx m
tm
it
54
the probability is obtained as
])
41(2
)(exp[
412
1)(
42
222
20
42
22
2
x
x
x
x m
tm
tkx
m
ttx
.
which has the form of Gaussian distribution with the standard deviation
42
22
41
x
xm
t
.
and has a peak at
m
tkx 0 .
The Fourier transform:
]2
2
14
)(exp[
2
12
1
])
21(4
2)4(exp[
21
1
2
1
2
1
)(2
1)()(
2
2
20
2
2
220
20
2
m
tikkk
m
itm
tikikxx
ikxdx
m
it
txdxetxxkdxtk
xx
x
x
xx
x
x
ikx
where
])
21(4
2)4(exp[
21
1
2
1)(
2
2
220
20
2x
x
xx
x
xm
itm
tikikxx
m
ittx
.
Then the probability is obtained as
55
]2
)(exp[
2
1)( 2
202
kk
kktk
.
where
xk
2
1 .
Therefore
22)0()( tktk .
0 1 2 3 4 5x
0.1
0.2
0.5
1.0
2.0
yx,t2
Fig. Plot of 2
)(tx as a function of x where the time t is changed as a parameter.
In summary, we have
])
41(2
)(exp[
412
1)(
42
222
20
42
22
2
x
x
x
x m
tm
tkx
m
ttx
.
and
56
]2
)(exp[
2
1)( 2
202
kk
kktk
.
14. Wave packet for simple harmonics (quamtum mechanics) ((L.I. Schiff p.67-68))
')0('')ˆexp(
)0()ˆexp()(
dxtxxtHi
x
ttHi
xtx
We define the kernel ),',( txxK as
nnnn
nn
xxtEi
xntEi
nx
xtHi
xtxxK
)'()()exp(
')exp(
')ˆexp(),',(
*
Note that
x , with
0 m
.
Then we have
)()'()'(')exp(),( * xxxdxtEi
tx nnn
n
.
We assume that
])(2
1exp[)( 22
4/1
2/1
axx ,
or
57
)(11
)( xnxn nn
,
or
])(2
1exp[
1)( 2
04/1
,
with
0 x0 . We need to calculate the integral defined by
])(2
1exp[)(
1
])(2
1exp[)(
])'(2
1exp[)'('
)'()'('
20
*4/1
204/1
2/1*2/1
224/1
2/1*
*
n
n
n
n
d
d
axxdx
xxdxI
Here
n () ( 2n n!) 1
2 e
2
2 Hn(). Then we get
I1
1 / 4 ( 2n n!) 1
2 d exp(2
2)Hn()exp[
1
2( 0 )2 ]) .
Here we use the generating function:
exp(2s s2 ) sn
n!n 0
Hn ().
Note that
58
0
200
2
0
20
22
0
22
)]2
1(exp[)(
!
])(2
1exp[)
2exp()(
!])(
2
1exp[)
2exp()2exp(
nn
n
nn
n
Hn
sd
Hn
sdssd
The left-hand side is
0
02
02/1
20
02/12
0022
!)
4exp(
)4
exp()]2
1(exp[)2exp(
n
nn
n
s
sssd
Thus we have
1/ 2 exp(0
2
4)
sn0n
n!n 0
ds n
n!Hn ()exp[(2 0
1
20
2 )]
n0
.
or
1/ 2 exp(0
2
4)0
n dHn () exp[( 2 0 1
20
2 )]
.
Then
I ( 2n n!) 1
2 exp(0
2
4)0
n ,
and
(x, t) exp(
i
Ent)
n (2n n!)
1
2 exp(0
2
4)0
nn(x) .
Since
En 0 (n
1
2) ,
or
exp(
i
Ent) exp(
i
20t in0t),
and
59
),(1
),( txt
,
we get
(,t) (2n n!) 1
2 exp(0
2
4)(0e
i0 t)n exp(
i
20t)n ()
n 0
,
or
0
02
0
20
4/1
0
200
201
4/1
)()2
(!
1)
224exp(
1
)()2
exp())(4
exp()!2(1
),(
0
2
0
nn
n
ti
nn
ntin
He
nt
i
Heti
ent
Using the generating function
1
n!(0e
i 0t
2)n Hn ()
n0
exp[1
40
2e2i n 0t 0e
i 0t],
we have the final form
)]sin(cos
)2sin2(cos4
1
224exp[
1
)4
1
224exp(
1),(
000
002
00
220
4/1
022
00
220
4/1
00
tit
titti
eeti
ttiti
OR
( , t)2
1
1/ 2 exp[0
2
2 2
1
20
2 cos20t 20 cos0t]
or
])cos(exp[1
),( 2002/1
2tt
( , t)
2 represents a wave packet that oscillates without change of shape about = 0 with
amplitude 0 and angular frequency 0.
60
15. Mathematica
,t 2
4 2 2 4
0.2
0.4
0.6
Fig. The time dependence of ])cos(exp[1
),( 2002/1
2tt
, where 0 =1. 0/2 T .
The peak shifts from = 0 at t = 0 to = 0 at t = T/4, = 0 at t = T/2, = - 0 at t =
3T/4, and = 0 at t = T.
61
62
16. Neutron interferometry
Suppose that the interferometry initially lies in a horizontal plane so that there are no gravitational effects. We then rotate the plane formed by the two paths by angle about the segment AC. The segment BD is now higher than the segment AC by l2sin.
63
Fig. Dependence of gravity-induced phase on angle of rotation . From R. Colella, A. W. Overhauser, and S. A. Werner, Phys.. Rev. Lett. 34 ( 1975) 1472.
From the previous discussion we have the propagator for the gravity
)exp(
)24
exp(
)exp(),;,(
01
032
00
mgTixA
mgTixTmg
i
Si
AtxtxK cl
For the path ABD and path ACD, we have
)}exp(1)]{(exp[)](exp[)](exp[ iACDSi
ACDSi
ABDSi
.
The phase difference between the path ABD and the path ACD is given by
sin1
sin1)()( 21
2
2 p
lglmTmgl
ACDSABDS
,
or
sinsin
2sin
21
221
221
2
lglmlglm.
where p is the momentum
mvp . T is related to l1 ( the distance over the horizontal line) as
22
1111 mlml
p
ml
v
lT .
The Probability is
)2
sin(cos4
222 .
64
Note that
221
2
2 lglm
,
with
mg=1.025 neV/cm. where g = 9.80446 m/s2 at Binghamton, NY (USA). The energy of neutron is evaluated as
222
2 2
22
nn Mk
ME
= 40.63 meV
for =1.419Å. ((Mathematica))
Clear "Global` " ;
rule1 1.054571628 10 27, mn 1.674927211 10 24,
qe 4.8032068 10 10, meV 1.602176487 10 15,
A0 10 8, neV 1.602176487 10 21, g 980.446,
1.419 A0 ;
mn gneV
. rule1
1.02497
E12
2 mn
2 2
meV. rule1
40.6266 17. Quantum-mechanical interference to detect a potential difference
65
A low-intensity beam of charged particles, each with charge q, is split into two parts. each part then enters a very long metallic tube shown above. Suppose that the length of the wave packet for each of the particles is sufficiently smaller than the length of the tube so that for a certain time interval, say from t0 to t, the wave packet for the particle is definitely within the tubes. During this time interval, a constant electric potential V1 is applied to the upper tube and a constant electric potential V2 is applied to the lower tube. The rest of the time there is no voltage applied to the tubes. Here we consider how the interference pattern depends on the voltages V1 and V2.
Without the applied potentials, the amplitude to arrive at a particular point on the detecting screen is
21 . The intnsity is proportional to
)( *212
*1
2
2
2
1
2
210 I .
With the potential, the Lagrangian in the path is modified as
qVLLL 00 .
Thus the wave functions are modified as
]exp[]exp[
)](exp[)](exp[
])(exp[])(exp[
22
11
02
201
1
12211
00
tiqV
tiqV
ttiqV
ttiqV
dtqVi
dtqVi t
t
t
t
66
where 0ttt . The intensity of the screen is proportional to
2
2121
2
22
11
2
])(
exp[
]exp[]exp[
tVViq
tiqV
tiqV
I
We assume that
)( 21 VVq .
Then we have
)( *212
*1
2
2
2
1
2
21 iii eeeI
When 021 , we have
2cos4)cos1(2 22
0
2
0
I
The intensity depends on the phase; I becomes maximum when n2 and minimum at
)2
1(2 n .
((Note)) Method with the use of gauge transformation
The proof for the expression can also be given using the concept of the Gauge trasnformation. The vector potential A and scalar potential are related to the magnetic field B and electric field E by
AB ,
tc
AE
1,
The gauge transformation is defined by
AA' ,
67
tc
1
' ,
where c is an arbitrary function. The new wave function is related to the old wave function through
)()exp()(' rc
iq
r .
Suppose that 0' . Then )()(' 0 rr is the wave function of free particle. Then we get
tc
1
, dtc .
The wave function )(r is given by
)()exp(
)()exp(
)()exp()(
0
0
0
r
r
r
dtViq
dtcc
iqc
iqr
When V is the electric potential and is independent of time t, we have
)()](exp[)( 00 r ttViq
r
This expression is exactly the same as that derived from the Feynman path integral. 18. Quantization of magnetic flux and Aharonov-Bohm effect
The classical Lagrangian L is defined by
Avv c
qqmL 2
2
1.
in the presence of a magnetic field. In the absence of the scalar potential ( 0 ), we get
AvAvv c
eL
c
qmL ccl
)0(2
2
1,
where the charge q = -e (e>0), A is the vector potential. The corresponding change in the action of some definite path segment going from ),( 11 nn tr to ),( 1 nn tr is then given by
68
n
n
t
t dt
ddt
c
ennSnnS
1
)1,()1,( )0()0( Ar
,
This integral can be written as
n
n
n
n
dc
e
dt
ddt
c
et
t
r
r
rAAr
11
,
where rd is the differential line element along the path segment.
Now we consider the Aharonov-Bohm (AB) effect. This effect can be usually explained in terms of the gauge transformation. Here instead we discuss the effect using the Feynman’s path integral. In the best known version, electrons are aimed so as to pass through two regions that are free of electromagnetic field, but which are separated from each other by a long cylindrical solenoid (which contains magnetic field line), arriving at a detector screen behind. At no stage do the electrons encounter any non-zero field B.
69
Fig. Schematic diagram of the Aharonov-Bohm experiment. Electron beams are split into two
paths that go to either a collection of lines of magnetic flux (achieved by means of a long solenoid). The beams are brought together at a screen, and the resulting quantum interference pattern depends upon the magnetic flux strength- despite the fact that the electrons only encounter a zero magnetic field. Path denoted by red (counterclockwise). Path denoted by blue (clockwise)
70
Incident Electron beams
Screen
Reflector
Reflector
Bout of page
C1
C2
slit-1slit-2P Q
Fig. Schematic diagram of the Aharonov-Bohm experiment. Incident electron beams go into
the two narrow slits (one beam denoted by blue arrow, and the other beam denoted by red arrow). The diffraction pattern is observed on the screen. The reflector plays a role of mirror for the optical experiment. The path1: slit-1 – C1 – S. The path 2: slit-2 – C2 – S.
Let B1 be the wave function when only slit 1 is open.
10,1,1 )](exp[)()(
PathB dc
ierArrr
, (1)
The line integral runs from the source through slit 1 to r (screen) through C1. Similarly, for the wave function when only slit 2 is open, we have
20,2,1 )](exp[)()(
PathB dc
ierArrr
, (2)
The line integral runs from the source through slit 2 to r (screen) through C2. Superimposing Eqs.(1) and (2), we obtain
20,210,1 )](exp[)()](exp[)()(
PathPathB dc
ied
c
ierArrrArrr
.
71
The relative phase of the two terms is
aArArrArrAr ddddPathPath
)()()()(21
,
by using the Stokes’ theorem, where the closed path consists of path1 and path2 along the same direction. The relative phase now can be expressed in terms of the flux of the magnetic field through the closed path,
c
ed
c
ed
c
ed
c
eaBaArA )( .
where the magnetic field B is given by
AB . The final form is obtained as
)]()exp()([)](exp[)( 0,20,12rrrArr id
c
iePathB
,
and is the magnetic flux inside the loop. It is required that
n2 . Then we get the quantization of the magnetic flux,
e
cnn
2 ,
where n is a positive integer, n = 0, 1,2,….. Note that
e
c2=4.135667510-7 Gauss cm2.
which is equal to 2 0 , where 0 is the magnetic quantum flux,
e
c
2
20
= 2.067833758(46) x 10-7 Gauss cm2. (NIST)
19. Example-1: Feynman path integral
We consider the Gaussian position-space wave packet at t = 0, which is given by
)2
exp(2
1)0(
2
2
x
tx (Gaussian wave packet at t = 0).
72
The Gaussian position-space wave packet evolves in time as
])(2
exp[1
2
1
]2
)(
2exp[
22
1
)0()0,;,((
2
2
2
20
2
20
0
000
mti
x
m
ti
t
xximxdx
ti
m
xxtxKdxtx
(1)
where
]2
)(exp[)
2()0,;,(
202/1
00 t
xxim
ti
mtxtxK
(free propagator) (2)
(Note) You need to show all the procedures to get the final form of tx ( .
(a) Prove the expression for tx ( given by Eq.(1).
(b) Evaluate the probability given by 2
(tx for finding the wave packet at the position x
and time t. (a) The Gaussian wave packet:
)2
exp(2
1)0(
2
2
x
tx . (Gaussian)
The free propagator:
]2
)(exp[)
2()0,;,(
202/1
00 t
xxim
ti
mtxtxK
.
Then we have
]2
)(
2exp[
22
1
)0()0,;,((
20
2
20
0
2/1
000
t
xximxdx
ti
m
xxtxKdxtx
Here we have
73
a
acb
a
bxa
cbxax
t
imxx
t
imxx
t
im
t
xxxximx
t
xximx
4
4)
2(
2)()
22
1(
2
)2(
22
)(
2
22
0
02
0
2
02
02
200
2
2
20
20
2
20
where
t
ima
22
12
,
t
imxb
,
t
imxc
2
2
.
Then we get the integral
)4
exp(
)2
(exp[)4
exp(
4)
2(exp[]exp[
2
200
2
22
0002
00
a
bc
a
a
bxadx
a
bc
a
bc
a
bxadxcbxaxdx
Note that when 02
1)Re(
2
a , the above integral can be calculated as
aydy
aa
bxadx
)exp(
1)
2(exp[ 22
00 ,
with the replacement of variable as )2
( 0 a
bxay and adxdy 0 . Thus we have
])
221
(42exp[
22122
1
)4
exp(22
1(
2
2
2
2
2
tim
timx
t
imx
timti
m
a
bc
ati
mtx
or
74
]22
exp[1
2
1
]
)(2
2exp[
1
2
1
])(2
2exp[
)2(2
2
1
])
221
(42exp[
)22
1(22
1(
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
mti
tmx
im
t
t
imx
mti
tm
ti
im
tmx
t
imx
mti
timtt
mx
t
imx
tit
imti
m
tim
timx
t
imx
tim
ti
mtx
or
])(2
exp[1
2
1
)](2
exp[1
2
1
)]1(2
exp[1
2
1
)22
exp(1
2
1(
2
2
2
2
2
2
2
22
2
2
222
2
mti
x
mti
mti
mti
t
imx
mti
mtit
imx
mti
mtit
imx
t
imx
mti
tx
Finally we get
])(2
exp[1
2
1(
2
2
2
mti
x
mti
tx
.
75
It is clear that at t = 0, tx ( is the original Gaussian wave packet.
(b)
])(2
exp[])(2
exp[1
2
1
])(2
exp[])(2
exp[1
2
1
)]()(2
exp[1
2
1(
2
224
2
2
224
22
4/1
2
224
2
224
2
2
224
22
2
2
2
224
2
2
mt
xmt
ie
mt
x
mt
mt
xm
tii
mt
x
mti
m
ti
mt
x
mti
tx
i
where
iem
t
m
ti2
2242 with )arctan(
22
22
m
t .
Then we have
])(
exp[1
2
1(
2
224
22
2
224
2
mt
x
mt
tx
The height of 2
(tx is
2
224
1
2
1
mt
.
The width is 2
2241
m
tx
.
20. Example-2: Feynman path integral
Suppose that the Gaussian wave packet is given by
76
)2
exp(2
1)0(
2
2
0 x
xiktx . (Gaussian)
Here we discuss how such a Gaussian wave packet propagates along the x axis as the time changes. The free propagator:
]2
)(exp[)
2()0,;,(
202/1
00 t
xxim
ti
mtxtxK
.
Then we have
]2
)(
2exp[
22
1
)0()0,;,((
00
20
2
20
0
2/1
000
xikt
xximxdx
ti
m
xxtxKdxtx
Here we have
a
acb
a
bxa
cbxax
t
imxx
t
mxkix
t
im
xikt
xxxximxxik
t
xximx
4
4)
2(
2)()
22
1(
2
)2(
22
)(
2
22
0
02
0
2
002
02
00
200
2
2
20
00
20
2
20
where
t
ima
22
12
, )( 0 t
mxkib
,
t
imxc
2
2
.
Then we get the integral
77
)4
exp(
)2
(exp[)4
exp(
4)
2(exp[]exp[
2
200
2
22
0002
00
a
bc
a
a
bxadx
a
bc
a
bc
a
bxadxcbxaxdx
Note that when 02
1)Re(
2
a , the above integral can be calculated as
aydy
aa
bxadx
)exp(
1])
2(exp[ 22
00 ,
with the replacement of variable as )2
( 0 a
bxay and adxdy 0 . Thus we have
])
22
1(42
exp[
22
122
1
)4
exp(22
1(
2
2
02
2
2
t
imt
imxik
t
imx
t
imti
m
a
bc
ati
mtx
or
78
]22
exp[1
2
1
]
)(2
2exp[
1
2
1
]
)(22
exp[)2(
22
1
])
22
1(42
exp[
)22
1(22
1(
2
2
022
2
2
2
2
02
2
2
2
2
02
2
2
2
02
2
m
ti
kt
mx
im
t
t
imx
m
ti
tm
ti
im
kt
mx
t
imx
m
ti
t
imt
kt
mx
t
imx
tit
imti
m
t
imt
imxik
t
imx
tim
ti
mtx
or
])(
2
1exp[)
2
)(
2exp(
1
2
1
))(
2
)(
2exp(
1
2
1(
2
224
202
2
224
420
2
4/1
2
224
2
224
22202
2
mt
m
tkx
mtt
m
tkxim
t
imx
e
mt
mtm
ti
tm
tkxim
t
imx
m
titx
i
where
iem
t
m
ti2
2242 with )arctan(
22
22
m
t .
Then we have
79
]
)(
)(exp[
1
2
1(
2
224
202
2
224
2
m
tm
tkx
m
ttx
.
This means the center of the Gaussian wave packet moves along the x axis at the constant velocity.
(i) The height of 2
(tx is
2
224
1
2
1
mt
.
(ii) The width is 2
2241
m
tx
.
(iii) The Group velocity is m
kvg
0 .
21. Summary: Feynman path integral
The probability amplitude associated with the transition from the point ),( ii tx to ),( ff tx is
the sum over all paths with the action as a phase angle, namely,
Amplitude = )exp( Si
pathsAll
,
where S is the action associated with each path. So we can write down
)exp()(
)exp(
,,),;,(
clif
t
tpathsAll
iiffiiff
Si
ttF
dtLi
txtxtxtxK
f
i
where Scl is the classical action associated with each path.
If the Lagrangian is given by the simple form
22 )()()(),,( xtcxxtbxtatxxL , then ),( if ttF can be expressed by
80
),0;,0(),( iiffif txtxKttF .
_______________________________________________________________________ REFERENCES R.P. Feynman, Nobel Lecture December 11, 1965: The development of the space-time view of
quantum electrodynamics http://www.nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html
R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals (extended edition) Dover 2005.
J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley, New York, 2011).
A. Das; Lecture Notes on Quantum Mechanics, 2nd edition (Hindustan Book Agency, 2012). John S. Townsend, A Modern Approach to Quantum Mechanics, second edition (University
Science Books, 2012). J. Mehra, The beat of different drum; The life and science of R. Feynman (Oxford University
Press, 1994). H. Rauch and S.A. Werner, Neutron Interferrometry: Lessons in Experimental Quantum
Mechanics (Oxford, 2000). K. Gottfried and T.-M. Yan, Quantum mechanics: Fundamentals, 2nd edition (Springer, 2003). M. Reuter, Classical and Quantum Dynamics: From Classical Paths to Path Integral, 3rd
edition (Springer, 2001). L.S. Schulman, Techniques and Applications of Path Integration (Wiley-Interscience, 1981). Colella, A. W. Overhauser, and S. A. Werner, Phys.. Rev. Lett. 34 ( 1975) 1472.”Observationof
Gravitationally Induced Quantum Interference.” _______________________________________________________________________ APPENDIX-I Mathematical formula-1
)4
exp()exp(2
2 ca
b
acbxaxdx
for 0]Re[ a __________________________________________________________________________ APPENDIX-II Action in the classical mechanics
We start to discuss the calculus of variations with an action given by the form
f
i
t
t
dtxxLS ],[ ,
81
where dt
dxx . The problem is to find has a stationary function )(xxcl so as to minimize the
value of the action S. The minimization process can be accomplished by introducing a
parameter .
ti tf
xi
xf
x
t
xcl t
x t
Fig.
ii xtx )( , ff xtx )( ,
)()()( ttxtx cl ,
where is a real number and
iicl xtx )( , ffcl xtx )( ,
0)( it , 0)( ft ,
82
dtdx
x )(0
,
f
i
t
t
clcl dtttxttxLxS ))()(),()((][ ,
has a minimum at = 0.
f
i
t
t
cl dttxLxS )]([][ 0 ,
0][
0
xL
,
dL
L 0|
,
f
i
f
i
f
i
f
i
f
i
t
t
t
t
tt
t
t
t
t
dttx
L
dt
d
x
L
dttx
L
dt
dt
x
Ldtt
x
L
dttx
Lt
x
LxS
)()]([
)()(|)}()(
)}()({][
(1) The Taylor expansion:
f
i
t
t
clcl dtttxttxLxS ))()(),()((][
((Fundamental lemma)) If
0)()( f
i
t
t
dtttM
83
for all arbitrary function (t) continuous through the second derivative, then M(t) must
identically vanish in the interval fi ttt .
________________________________________________________________________ From this fundamental lemma of variational and Eq.(1), we have Lagrange equation
0)(
x
L
dt
d
x
L
. (2)
L can have a stationary value only if the Lagrange equation is valid.
In summary,
2
1
),(x
x
dtxxLS ,
S = 0 0)(
x
L
dt
d
x
L
.