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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Feynman Integral and
Convolution Products
H. Vic Dannon
[email protected]
March, 2008.
Abstract
Feynman tried to express the Propagator of the
Schrödinger equation as an Infinite Convolution Product that
became known as the Feynman Path Integral.
Feynman was not aware of Polya-Laguerre’s Infinite Convolution
Products, and did not know that the Schrodinger Propagator
cannot be written as an Infinite Convolution Product.
Feynman’s construction is the same whether we use one iteration
or
infinitely
many
iterations,
the
limit
is
superfluous,
Schrodinger’s Propagator cannot be decomposed, and Feynman’s
Integral remains, as defined, Schrodinger’s Propagator.
The Feynman Integral, and approach to Quantum Mechanics, are
in fact, the Schrodinger Propagator, and the wave function
approach to Quantum Mechanics, renamed after Feynman.
Keywords: Feynman Path Integral, Convolution Product, Quantum
Mechanics, Schrodinger Propagator.
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
2000 Mathematics subject classification 81S40, 44A35, 81Q05
PACS 31.15.xk; 03.65.-w
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H. Vic Dannon
Contents
Introduction……………………………………………………………..5
1. Schrodinger’s Propagator of a Free Particle……………….8
1.1 The form of Schrodinger’s Propagator…………………………….8
1.2 The constant for Schrodinger’s Propagator…………………….10
1.3 Schrodinger’s Propagator………………………………………….12
1.4 Born’s Probability for the Free Particle………………………...12
2. Feynman Path Integral for a Free Particle……………….13
3. Schrodinger’s propagator for the harmonic Oscillator..17
3.1 The form of Schrodinger’s Propagator…………………………..17
3.2 The constant for Schrodinger’s Propagator…………………….21
3.3 Schrodinger’s Propagator………………………………………….21
3.4 Born’s Probability for the Harmonic Oscillator….…………….21
3.5 Feynman’s Integral for the Harmonic Oscillator….…………..22
4. Infinite Convolution Products………………………………...23
4.1 The Kernel of an Infinite Convolution Product………………..23
4.2 The Convolution-Kernel Integral Theorem…………………….28
4.3 The Kernel as a Probability Density Function…………………28
4.4 The Convolution Transform……………………………………....29
4.5 The Convolution-Transform Integral Theorem………………..32
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H. Vic Dannon
5. Schrodinger Propagator & Infinite Convolution
Product………………………………………………………………33
5.1 Schrodinger Propagator is not an Infinite Convolution
Product…………………………………………………………………...34
References……………………………………………………………...36
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Introduction
De-Broglie associated with a moving particle a wave, and
Schrodinger derived for a particle of mass m moving in the
direction of x , free of forces, the wave equation
∂ψ
i= ∂ 2 ψ
=
.
∂t
2m ∂x 2
Born suggested that ψ(x , t )
2
is the probability to find the particle
in the time interval [t, t + dt ] , and in the length interval
⎡ x , x + dx ⎤ .
⎣
⎦
If the particle is at xa , at time ta , and at xb , at time tb , its
Schrodinger’s wave function is given by
xa =∞
ψ(xb , tb ) =
∫
K (xb , tb ; xa , ta )ψ(xa , ta )dxa ,
xa =−∞
where
⎡
⎤1/2
m
i
⎥ exp
K (xb , tb ; xa , ta ) = ⎢
⎢⎣ 2πi=(tb − ta ) ⎥⎦
=
t =tb
∫
L (x min , x min , t )dt ,
t =ta
is Schrodinger’s Propagator, and x min (t ) is the least-energy-path
of the particle.
Feynman attempted to express K (xb , tb ; xa , ta ) by what is known
as an Infinite Convolution Product.
He showed that if the path is partitioned into the segments
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
a = x 0, x1, x 2 ...x N = b
so that
ta = t0 < t1 < t2 < ... < tN = tb ,
then
K (xb , tb ; x N −1, tN −1 ) ∗ K (x N −1, tN −1; x N −2, tN −2 ) ∗ ... ∗ K (x1, t1; xa , ta ) = K (xa , ta ; xb , tb ) ,
where the symbol ∗ denotes the Convolution Product.
This equality fails to yield an Infinite Convolution Product.
To obtain a genuine Infinite Convolution Product, the path
partition points must be determined by zeros of the Laplace
Transform of the Kernel K (xa , ta ; xb , tb ) , and the Kernel has to be
generated by
basic
Non-Schrodinger kernels that depend on
those zeros.
Feynman’s taking the limit N → ∞ makes no difference, because
the equality holds for any natural number N , and the limit
process does not obtain any new result.
Feynman’s Path Integral amounts to impossible expectations of
the Schrodinger Propagator.
In the following, we derive the Schrodinger
Propagator for a
particle in a force-free motion, and for a harmonic oscillator, in
order to understand Feynman’s failed attempt to express these
Propagators as Infinite Convolution Products.
Then, we use Polya-Laguerre theory of Infinite Convolution
Products to explain why Feynman’s Path Integral does not exist,
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
why Schrodinger’s Propagator need not be renamed after
Feynman, and why Feynman’s New approach to Quantum
Mechanics is Schrodinger’s Wave Mechanics.
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H. Vic Dannon
1.
Schrodinger’s
Propagator
of
a
Free
Particle
Consider a particle of mass m , at the point a , at time ta , moving
in a force-free zone, in the direction of x at speed x =
dx
, and
dt
arriving at the point b , at time tb .
Schrodinger’s wave function is given by
xa =∞
ψ(xb , tb ) =
∫
K (xb , tb ; xa , ta )ψ(xa , ta )dxa ,
xa =−∞
where K (xb , tb ; xa , ta ) is Schrodinger’s Propagator, obtained along
the particle’s least-energy-path.
We aim to derive the known formula stated in the introduction:
1.1 The Form of Schrodinger’s Operator
There is a constant A so that K (xb , tb ; xa , ta ) =
8
1
A
e
i
m
2= (tb −ta )
(xb −x a )2
.
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Proof: For a particle of mass m moving in the direction of x at
speed x =
dx
, in a force-free zone, Lagrange energy function is
dt
L free (x , x, t ) = 21 mx 2 .
The path of minimal energy, solves the Euler-Lagrange equation
d ∂L ∂L
−
= 0.
dt ∂x ∂x
Here,
d ∂L
d
= mx = mx ,
dt ∂x
dt
∂L
= 0,
∂x
and the Euler-Lagrange equation is
mx = 0 .
Thus, the least-energy-path
x min (t ) ,
has constant velocity
x min (t ) = const .
Since the particle is at the point a , at time ta , and at the point b ,
at time tb ,
x min (t ) =
xb − x a
tb − ta
Hence, the Lagrangian along the path of least energy is
1
2
L free (x min , x min , t ) = mx min
2
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
2
1 ⎛⎜ xb − xa ⎞⎟
= m⎜
⎟
2 ⎝⎜ tb − ta ⎠⎟⎟
Therefore,
t =tb
∫
t =ta
t =tb
L free (x min , x min , t )dt =
∫
t =ta
=
2
⎛ x − xa ⎞⎟
m ⎜⎜ b
⎟ dt
2 ⎜⎝ tb − ta ⎠⎟
1
1
2
2
m
( xb − x a )
tb − ta
Thus, Schrodinger’s Propagator is
i
=
t =tb
∫
e t =ta
1
K free (, a ) =
A
e
1
=
A
i
Lfree (x min,xmin,t )dt
m
(x −x )2
2=(tb −ta ) b a
.,
We proceed to determine the constant A .
1.2 The Constant for Schrodinger’s Propagator
1
⎡
⎤2
1
m
⎥
=⎢
A ⎢⎣ 2πi=(tb − ta ) ⎥⎦
Proof: Schrodinger’s wave function is
xa =∞
1
ψ(xb , tb ) =
Ax
∫
K free (xb , tb ; xa , ta )ψ(xa , ta )dxa
a =−∞
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Gauge Institute Journal, Volume 5, No 2, May 2009
i
xa =∞
=
1
Ax
e
∫
H. Vic Dannon
m
(x −x )2
2=(tb −ta ) b a
ψ(xa , ta )dxa ,
a =−∞
Letting
tb = ta + δt ,
x b = x a + δx ,
we have
xa =∞
1
ψ(xb , ta + δt ) =
Ax
∫
e
i
m
(x −x )2
2=(tb −ta ) b a
ψ(xb − δx , ta )dxa .
a =−∞
Hence, to first order,
xa =∞
∂ψ
1
ψ(xb , ta ) +
δt ≈
∂t
Ax
∫
e
i
a =−∞
m
(x −x )2
∂ψ ⎞⎟
2=(tb −ta ) b a ⎛⎜
δx ⎟dx .
⎜⎜ ψ(xb , ta ) −
∂x ⎠⎟ a
⎝
Equating coefficients of ψ(xb , ta ) on both sides, we conclude that
xa =∞
A=
∫
e
i
m
(x −x )2
2=(tb −ta ) b a
dxa
xa =−∞
xa =∞
∫
=
e
m
(x −x )2
2i=(tb −ta ) b a
−
dxa .
xa =−∞
Let ξ =
m
(x − xb ) . Then, d ξ =
2i=(tb −ta ) a
m
dx , and
2i=(tb −ta ) a
ξ =∞
A=
=
2
1
e −ξ d ξ
∫
m
2i=(tb −ta ) ξ =−∞
1
π .,
m
2i=(tb −ta )
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Substituting the constant A , we obtain Schrodinger’s Propagator,
and Born’s probability for the free particle:
1.3 Schrodinger’s Propagator
1
⎤2
⎡
m
⎥
K (xb , tb ; xa , ta ) = ⎢
⎢ 2πi=(t − t ) ⎥
b
a ⎦
⎣
e
i
m
(x −x )2
2=(tb −ta ) b a
1.4 Born’s Probability for the Free Particle
P (b, a ) =
m
.
2π=(tb − ta )
12
.
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
2.
Feynman
Path
Integral
for
a
Free
Particle
Feynman attempted to write K (xb , tb ; xa , ta ) as an Infinite
Convolution Product, over the least-energy-path of the free
particle.
We examine his derivation.
Divide the time interval [ta , tb ] into N equal sub-intervals of
length
Δt =
tb − ta
,
N
with Partition points
ta = t0 < t1 < t2 < ..... < tN −1 < tN = tb .
Consider
1
⎤2
⎡
m
⎥
K (x1, t1; x 0, t0 ) = ⎢
⎢⎣ 2πi=(t1 − t0 ) ⎥⎦
1
⎤2
⎡
m
⎥
K (x 2, t2 ; x1, t1 ) = ⎢
⎢⎣ 2πi=(t2 − t1) ⎥⎦
1
⎤2
⎡
m
⎥
K (x 3, t3; x 2, t2 ) = ⎢
⎢⎣ 2πi=(t3 − t2 ) ⎥⎦
13
e
e
⎛
⎞⎟
⎜⎜
m
⎟
2
(
−
)
x
x
⎜⎜
0 ⎟⎟⎟⎟
⎜⎜⎝ 2i=(t1−t0 ) 1
⎠
⎛
⎞⎟
⎜⎜
m
(x2 −x1)2 ⎟⎟⎟⎟
⎜⎜
⎜⎜⎝ 2i=(t2 −t1)
⎠⎟
e
,
,
⎛
⎞⎟
⎜⎜
m
(x 3 −x2 )2 ⎟⎟⎟⎟
⎜⎜
⎜⎜⎝ 2i=(t3 −t2 )
⎠⎟
,
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
…………………………………………………………….,
1
⎤2
⎡
m
⎥
K (x N −1, tN −1; x N −2 , tN −2 ) = ⎢
⎢ 2πi=(t
⎥
N −1 − tN −2 ) ⎦
⎣
1
⎤2
⎡
m
⎥
K (x N , tN ; x N −1, tN −1 ) = ⎢
⎢ 2πi=(t − t
⎥
)
N
N −1 ⎦
⎣
e
e
⎛
⎞⎟
⎜⎜
m
2 ⎟⎟⎟
⎜⎜
(
−
)
x
x
N −1 N −2 ⎟⎟⎟
⎜⎜ 2i=(t
⎜⎝
⎠
N −1−tN −2 )
⎛
⎞⎟
m
⎜⎜⎜
2
(xN −xN −1) ⎟⎟⎟⎟⎟
⎜⎜
⎜⎜⎝ 2i=(tN −tN −1)
⎠⎟
,
.
Then,
K (x 2, t2; x1, t1) ∗ K (x1, t1; x 0, t0 )
x1 =∞
∫
=
K (x 2, t2; x1, t1)K (x1, t1; x 0, t0 )dx 1
x1 =−∞
=
m
2πi=(t1 − t0 )
x1 =∞
∫ e
⎛
⎞ ⎛
⎞
m
m
(x 2 −x1 )2 ⎟⎟⎟ +⎜⎜
(x −x )2 ⎟⎟
⎜⎜
⎟⎠ ⎝⎜ 2i= (t1 −t0 ) 1 0 ⎠⎟
⎜⎝ 2i= (t2 −t1 )
dx1 .
x1 =−∞
By [Feyn1, p.357]
x =∞
∫
e
α(x 2 − ξ )2 + β (ξ −x1 )2
dξ =
x =−∞
−π
α+β
e
αβ
(x −x )2
α +β 2 1
Here,
−π
=
α+β
⎛
⎜⎜
−π2i= ⎜⎜
⎜
m ⎜⎜
⎜⎜
⎝t
1
1
1
2
− t1
+
1
t1 − t 0
14
⎞⎟2
⎟⎟
⎟⎟
⎟⎟ =
⎟⎟
⎠⎟
1
−π2i= ⎛⎜ (t2 − t1)(t1 − t0 ) ⎞⎟2
⎟⎟ .
⎜
m ⎜⎝
t2 − t 0
⎠⎟
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
m
m
2i=(t2 − t1 ) 2i=(t1 − t0 )
αβ
m
=
=
α+β
m
m
2i=(t2 − t0 )
+
2i=(t2 − t1 ) 2i=(t1 − t0 )
Substituting in K (x 2, t2; x1, t1) ∗ K (x1, t1; x 0, t0 ) ,
m
−π2i= ⎜⎛ (t2 − t1)(t1 − t0
=
⎜
2πi=(t1 − t0 )
m ⎜⎝
t2 − t 0
1
⎤2
⎡
m
⎥
=⎢
⎢⎣ 2πi=(t2 − t0 ) ⎥⎦
e
1
) ⎞2
⎟⎟
⎠⎟⎟
e
m
(x −x )2
2i= (t2 −t0 ) 2 0
⎛
⎞⎟
⎜⎜
m
(x2 −x 0 )2 ⎟⎟⎟⎟
⎜⎜
⎜⎜⎝ 2i=(t2 −t0 )
⎠⎟
= K (x 2, t2 ; x 0, t0 ) .
Therefore,
K (x 3, t3 ; x 2 , t2 ) ∗ K (x 2 , t2 ; x1, t1 ) ∗ K (x1, t1; x 0 , t0 )
= K (x 3 , t3 ; x 2 , t2 ) ∗ K (x 2 , t2 ; x 0 , t0 )
= K (x 3 , t3 ; x 0 , t0 )
Indunctively,
K (xb , tb ; x N −1, tN −1 ) ∗ K (x N −1, tN −1; x N −2, tN −2 ) ∗ ... ∗ K (x1, t1; xa , ta ) = K (xb , tb ; xa , ta )
Letting N → ∞ ,
lim K (xb , tb ; x N −1, tN −1 ) ∗ K (x N −1, tN −1; x N −2, tN −2 ) ∗ ... ∗ K (x1, t1; xa , ta )
N →∞
= K (xb , tb ; xa , ta ) .
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Feynman defined the limit, that is, the already known Schrodinger
Propagator, K (xb , tb ; xa , ta ) , as his Path Integral for the free
particle.
But since equality holds for any natural number N , the limit
process makes no difference, and serves no purpose.
Therefore, Feynman Path Integral is the Schrodinger Propagator
renamed after Feynman.
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
3.
Schrodinger’s
Propagator
for
the
Harmonic Oscillator
For a particle of mass m oscillating harmonically along x , at
angular frequency ω , Schrodinger wave function satisfies the
equation
i= ∂ 2 ψ
m 2 2
∂ψ
=
+
ω x ψ,
2m ∂x 2 2i=
∂t
and is given by
xa =∞
ψ(xb , tb ) =
∫
K harm (xb , tb ; xa , ta )ψ(xa , ta )dxa ,
xa =−∞
where K harm (xb , tb ; xa , ta ) is Schrodinger’s Propagator, obtained
along the particle’s least-energy-path.
We aim to derive the known formula [Feyn1,p.198] for it.
3.1 The Form of Schrodinger’s Propagator
There is a constant A so that
mω
1 2i= sin ω(tb −ta )((xb +xa )cos ω(tb −ta )−2xbxa )
Kharm (xb , tb ; xa , ta ) =
.
A
e
2
2
Proof: For a particle of mass m oscillating harmonically along x ,
at angular frequency ω , Lagrange energy function is
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
1
1
Lharm (x , x, t ) = mx 2 − m ω 2x 2 .
2
2
The path of minimal energy, solves the Euler-Lagrange equation
d ∂L ∂L
−
= 0.
dt ∂x ∂x
Here,
d ∂L
d
= mx = mx ,
dt ∂x
dt
∂L
= −m ω 2x ,
∂x
and the Euler-Lagrange equation is
mx + m ω 2x = 0 .
Thus, the least-energy-path is
x min (t ) = A cos ωt + B sin ωt ,
If the particle location is xa , at time ta , and xb , at time tb ,
A cos ωta + B sin ωta = xa
A cos ωtb + B sin ωtb = xb .
Solving for A , and B ,
A=
xa
xb
cos ωta
cos ωtb
sin ωta
sin ωtb
sin ωta
sin ωtb
=
18
xa sin ωtb − xb sin ωta
sin ω(tb − ta )
Gauge Institute Journal, Volume 5, No 2, May 2009
cos ωta
B =
cos ωtb
cos ωta
cos ωtb
xa
xb
sin ωta
sin ωtb
=
H. Vic Dannon
−xa cos ωtb + xb cos ωta
sin ω(tb − ta )
The Lagrangian along the path of least energy is
Lharm (x min , x min , t ) =
1
1
2
2
mx min
− m ω 2x min
2
2
=
2
1 ⎡
⎤
m ⎢ ( −Aω sin ωt + B ω cos ωt ) − ω 2(A cos ωt + B sin ωt )2 ⎥
2 ⎣
⎦
=
1
m ω 2 ⎡⎢ (B 2 − A2 )cos 2ωt − 2AB sin 2ωt ⎤⎥ .
⎣
⎦
2
Hence,
t =tb
∫
Lharm (x min , x min , t )dt =
t =ta
1
= m ω2
2
t =tb
∫
t =ta
⎡ (B 2 − A2 )cos 2ωt − 2AB sin 2ωt ⎤ dt
⎢⎣
⎥⎦
=
t =tb
1
m ω ⎡⎢ (B 2 − A2 )sin 2ωt + 2AB cos 2ωt ⎤⎥
⎣
⎦t =ta
4
=
1
m ω ⎡⎢ (B 2 − A2 )(sin 2ωtb − sin 2ωta )
⎣
4
+2AB(cos 2ωtb − cos 2ωta ) ⎤⎦
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Substituting
2
2
B −A =
2AB =
=
xb2 cos 2ωta + xa2 cos 2ωtb − 2xa xb cos w(tb + ta )
sin2 ω(tb − ta )
−xb2 sin 2ωta − xa2 sin 2ωtb + 2xa xb sin w(tb + ta )
sin2 ω(tb − ta )
1
mω
⎡ (x 2 + x 2 )sin 2ω(t − t )
a
b
a
4 sin2 ω(tb − ta ) ⎢⎣ b
−2xa xb cos w(tb + ta )[sin 2ωtb − sin 2ωta ]
+2xa xb sin w(tb + ta )[cos 2ωtb − cos 2ωta ] ⎤⎦
=
1
mω
⎡ (x 2 + x 2 )sin 2ω(t − t )
a
b
a
4 sin2 ω(tb − ta ) ⎢⎣ b
−2xa xb cos w(tb + ta )[2 sin ω(tb − ta )cos ω(tb + ta )]
+2xa xb sin w(tb + ta )[−2 sin ω(tb + ta )sin ω(tb − ta )] ⎤⎦
=
1
mω
⎡ (x 2 + x 2 )cos ω(t − t ) − 2x x ⎤
a
b
a
a b ⎥⎦
2 sin ω(tb − ta ) ⎢⎣ b
Therefore, Schrodinger’s Propagator is
i 2= sinmωω(t −t )((x +x )cos ω(tb −ta )−2xbxa )
a
e
1
Kharm (xb , tb ; xa , ta ) =
A
b
Similarly to the proof of 1.2, we obtain
20
2
b
2
a
.,
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
3.2 The Constant for Schrodinger’s Propagator
1
⎡
⎤2
1
mω
⎥
=⎢
A ⎢⎣ 2πi= sin ω(tb − ta ) ⎥⎦
Substituting the constant, we obtain the propagator, and the
probability for the harmonic oscillator.
3.3 Schrodinger’s Propagator
K harm (xb , tb ; xa , ta ) =
1
⎤2
⎡
m
⎥
=⎢
⎢⎣ 2πi= sin ω(tb − ta ) ⎥⎦
i 2= sinmωω(t −t )((x +x )cos ω(tb −ta )−2xbxa )
e
b
a
2
b
2
a
3.4 Conditional Probability for the Harmonic Oscillator
A particle of mass m , at the point a , at time ta , oscillating
harmonically along x , at angular velocity ω , with speed x =
dx
,
dt
will arrive at the point b , at time tb with the conditional
probability
P(b, a ) =
mω
.
2π= sin ω(tb − ta )
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
3.5 Feynman Integral for the Harmonic Oscillator
Similarly
to
section
2,
Feynman
attempted
to
express
Schrodinger’s Propagator for the harmonic oscillator
⎡
mω
⎢
⎢⎣ 2πi= sin ω(tb
⎤
⎥
− ta ) ⎥⎦
1
2
e
mω
(x b2 +x a2 )cos ω(tb −ta )−2xb xa )
(
2i= sin ω(tb −ta )
,
as an Infinite Convolution Product.
Again, applying the limit process to the convolution property for
Schrodinger Propagators serves no purpose, and Feynman Path
Integral for the harmonic oscillator, is the
Propagator renamed after Feynman.
22
Schrodinger
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
4.
Infinite Convolution Products
The evolution of Schrodinger’s wave function from time t1 to time
t2 , is given by the Convolution
ξ =∞
ψ(x , t2 ) =
∫
G (x , t ; ξ,t1 )ψ(ξ, t1 )d ξ ,
ξ =−∞
where the Schrodinger Propagator G (x , t2, ξ, t1 ) is obtained along
the path of least energy.
It is a particular case of the Convolution Transform
u =∞
F (t ) =
∫
G(t − u )f (u )du ,
u =−∞
that transforms the function f (u ) into the function F (t ) .
Then, the Kernel function G (t ) may be an Infinite Convolution
Product of basic kernel functions g1(t ) , g 2(t ) , g 2(t ) … Namely,
G(t ) = g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ ...
Is Schrodinger Propagator such kernel G (t ) ?
4.1
The Kernel of an Infinite Convolution Product.
We denote a complex number by
23
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
z = x + iy .
Following the notations of [Hirs1], [Hirs2], and [widd1].
b , and ak ≠ 0 , k = 1, 2, 3,...
are real numbers so that
1
a12
+
1
a22
+
1
+ ... < ∞ .
a32
Consider the basic kernel function
⎧
⎪
a1ea1 (t −b )−1, a1(t − b) − 1 < 0
⎪
g1(t ) = ⎨
⎪
0,
a1(t − b ) − 1 ≥ 0
⎪
⎩
⎧
1
⎪
a1 (t −b )−1
⎪
,
a
e
t
<
+b
⎪
1
⎪
a
1
.
=⎪
⎨
⎪
1
⎪
0,
t ≥
+b
⎪
⎪
a
⎪
⎩
1
The Two-Sided Laplace Transform of g1(t ) is
t =∞
L { g1(t )} =
∫
g1(t )e −ztdt
t =−∞
t=
=
1
+b
a1
∫
a1ea1 (t −b )−1e −ztdt
t =−∞
t=
=
24
a1
ea1b +1
1
+b
a1
∫
t =−∞
e(a1 −z )tdt
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
1
(a1 −iz )( +b )
1
a1
= a b +1
e
e 1 a1 − iz
a1
= e −biz
1
z
1−
a1
e
z
−
a1
.
For k = 2, 3, 4...n , consider the basic kernel functions
⎧⎪ a t −1
1
k
⎪
,
<
a
e
t
⎪⎪ 2
⎧⎪a eakt −1, a t − 1 < 0
ak
⎪
2
k
= ⎪⎨
.
gk (t ) = ⎨
⎪⎪ 0,
⎪
1
akt − 1 ≥ 0 ⎪
⎩
t ≥
⎪⎪ 0,
ak
⎪⎩
The Two-Sided Laplace Transform of gk (t ) is
t =∞
L { gk (t )} =
∫
gk (t )e −ztdt
t =−∞
t=
=
1
ak
∫
akeakt −1e −ztdt
t =−∞
t=
=
ak
e
1
ak
∫
e(ak −z )tdt
t =−∞
1
(ak −z )
ak 1
ak
=
e
e ak − z
25
Gauge Institute Journal, Volume 5, No 2, May 2009
=
H. Vic Dannon
1
z
1−
ak
e
z
−
ak
.
The Convolution Product of g1(t ) , and g 2(t ) is defined by
u1 =∞
g1(t ) ∗ g 2(t ) =
∫
g1(t − u1 )g 2(u1 )du1
u1 =−∞
By the Convolution Theorem, the Two-Sided Laplace Transform of
the Convolution Product g1(t ) ∗ g 2(t ) is
L { g1(t ) ∗ g 2(t )} = L { g1(t )} × L { g 2(t )}
= e −bz
1
z
1−
a1
e
z
−
a1
1
z
1−
a2
e
z
−
a2
.
Similarly, the Two-Sided Laplace Transform of the convolution
product
g1(t ) ∗ g 2(t ) ∗ ... ∗ gn (t )
is
L { g1(t ) ∗ g 2(t ) ∗ ... ∗ gn (t )} =
= L { g1(t )} × L { g 2(t )} × ... × L { gn (t )}
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Gauge Institute Journal, Volume 5, No 2, May 2009
=
1
H. Vic Dannon
1
1
z
ebz ⎛
⎞⎟ a
z
⎜⎜ 1 − ⎟e 1
⎜⎝
a1 ⎠⎟⎟
z
⎛
⎞
⎜⎜ 1 − z ⎟⎟e a2
a2 ⎠⎟⎟
⎝⎜
...
1
z
⎛
⎞
⎜⎜ 1 − z ⎟⎟e an
an ⎠⎟⎟
⎝⎜
The denominator functions
z
z
z
⎛
⎛
z ⎞ ⎛
z ⎞
z ⎞
En (z ) = ebz ⎜⎜ 1 − ⎟⎟⎟e a1 ⎜⎜ 1 − ⎟⎟⎟e a2 ...⎜⎜ 1 − ⎟⎟⎟e an
⎜⎝
a1 ⎠⎟ ⎝⎜
a2 ⎠⎟
an ⎠⎟
⎝⎜
are analytic for any complex number z , and converge uniformly
[Widd1, p. 173], to the infinite product
⎛
z
E (z ) = ebz ⎜⎜ 1 −
⎜⎝
a
z
z
z
⎞⎟ a ⎛
z ⎞ ⎛
z ⎞
⎟⎟e 1 ⎜⎜ 1 − ⎟⎟⎟e a2 ⎜⎜ 1 − ⎟⎟⎟e a3 ...
⎟ ⎝⎜
a2 ⎠⎟ ⎝⎜
a3 ⎠⎟
1⎠
on every compact set of the z plane.
Thus, E (z ) is analytic for any complex number z .
Therefore, for −∞ < t < ∞ , the Infinite Convolution Product
G (t ) = g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ ....
is the Inverse Two-Sided Laplace Transform
z =i ∞
⎧
1
⎪ 1 ⎫
⎪
⎪
L−1 ⎪
⎨
⎬=
⎪
⎪
⎪
⎩ E (z ) ⎪
⎭ 2πi
1
=
2π
∫
z =−i ∞
y =∞
∫
y =−∞
27
1 zt
e dω .
E (z )
1 iyt
e dy
E (iy )
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
Rather than evaluate directly the Infinite Convolution Product, as
Feynman
attempted,
we
evaluate
the
Two-Sided
Laplace
Transforms
L { g1(t )} , L { g 2(t )} , L { g 3(t )} …,
and invert their product.
This follows from the Convolution-Kernel Integral Theorem:
4.2 The Convolution-Kernel Integral Theorem
G (t ) = g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ ....
= L−1 { L { g1(t )} L { g 2(t )} L { g 3(t )} ...}
⎧
⎫
⎪
⎪
⎪
⎪
⎪
⎪
y =∞ ⎪
⎪
⎪
⎪
1
1
1
1
iyt
⎪
⎪
=
×
×
×
e
dy . ,
...
⎨
⎬
∫
iby
iy
iy
⎪
⎪
2π
e
⎛
⎪
y =−∞ ⎪
iy ⎞⎟ a1 ⎛⎜
iy ⎞⎟ a2
⎪
⎪
⎜
1 − ⎟⎟e
1 − ⎟⎟e
⎪
⎪
⎜
⎜
⎪
⎪
⎜
⎟
⎜
⎟
a
a
⎝
⎠
⎝
⎠
⎪
⎪
1
2
⎩
⎭
4.3 The Kernel G (t ) as a Probability Density Function
− a1 < x < a1 ,
In the strip
1
e
bz
×
1
⎛
⎞
⎜⎜ 1 − z ⎟⎟e
⎜⎝
a1 ⎠⎟⎟
z
a1
×
1
⎛
⎞
⎜⎜ 1 − z ⎟⎟e
a2 ⎠⎟⎟
⎝⎜
z
a2
×
28
1
⎛
⎞
⎜⎜ 1 − z ⎟⎟e
a3 ⎠⎟⎟
⎝⎜
z
a3
× ... =
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
= L { g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ ....}
t =∞
=
∫ { g1(t ) ∗ g2(t ) ∗ g3(t ) ∗ ....}e−ztdt
t =−∞
At z = 0 ,
t =∞
∫ { g1(t ) ∗ g2(t ) ∗ g3(t ) ∗ ...}dt
=
t =−∞
1
E (0)
=1
Also,
g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ ... ≥ 0 ,
because for any n ,
g1(t ) ∗ g 2(t ) ∗ ... ∗ gn (t ) ≥ 0 .
Thus, G (t ) is a probability density function.
It is shown (in [Hirs1], [Widd1]) that its Mean is
b,
and its Variance is
1
a12
4.4
+
1
a22
+
1
a32
+ ... .
The Convolution Transform
The convolution of G (t ) with an integrable function f (t ) ∈ L1 is
the Convolution Transform determined by the sequencea1, a2 , a 3 ,. ,
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
F (t ) = G(t ) ∗ f (t )
u =∞
∫
=
G(t − u )f (u )du
u =−∞
u =∞
=
f (t − u ) ⎡⎣ g1(u ) ∗ g 2(u ) ∗ g 3(u ) ∗ ... ⎤⎦ du .
∫
u =−∞
The following examples were derived from [Hirs1, p.79].
Laplace
ak
G (t )
1, 2, 3..
t −et
t
e −e ∗ e−e
Laplace
Stieltjes
ν =1
2
−5
2
−1
,
,
2
,
−3
2 2
,
3 5
τ =∞
1
Γ(1 − z )
ee
Iterated
Φ(ξ )
E (z )
∫
τ =0
⎛ τ =∞
⎞
⎜⎜ e −τ / θϕ(τ )d τ ⎟⎟ 1 e −ξθd θ
⎟
∫ ⎜⎜ ∫
⎠⎟ θ
θ=0 ⎝ τ =0
θ =∞
1
t
[Γ(1 − z )]2
1
t/2
1
Stieltjes
e t /2 − e −t /2
[Γ(z )Γ(1 − z )]2
1
1
z Γ(z )Γ(1 − z )
ν = 2
... − 2,
−1, 1, 2..
(e
t /2
Meijer
ν = 0
)
1
Stieltjes
ν >0
+e
−t /2 2
⎛ν
Γ⎜
⎜⎝ 2
< 1
∫
2
e K ν (e )
Γ
(
1 + ν − z
∫
⎞
− z⎟
⎟⎠
τ =0
∫
30
ϕ(τ )
ν
(ξ + τ )
dτ
dτ
ξτK 0(ξτ )ϕ(τ )d τ
τ =0
τ =∞
z +1
)(
2
τ =∞
1 − ν − z
Γ
2
ϕ(τ )
τ =0 ( ξ + τ )
2
⎡ ⎛ 1 − z ⎞⎤
⎟⎟ ⎥
⎢ Γ ⎜⎜
⎢⎣ ⎜⎝ 2 ⎠⎟ ⎥⎦
t
ϕ(τ )
dτ
ξ+τ
log ξ − log τ
ϕ(τ )d τ
ξ−τ
τ =∞
2z +1
Meijer
−1 < ν
τ =0
τ =0
τ =∞
⎞ ⎛ν
+ z ⎟Γ⎜
⎟⎠ ⎝⎜ 2
t
e K 0 (e )
t
∫
Γ(ν )
(e t /2 + e −t /2 )ν
t
∫
τ =∞
Iterated
Stieltjes
τ =∞
1
Γ(iz )Γ(1 − iz )
e t /2 + e−t /2
e−ξτϕ(τ )d τ
2
)
∫
τ =0
ξτK ν (ξτ )ϕ(τ )d τ
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
To see how the Convolution Transform F (t ) of
t
f (u ) with the
Kernel e te −e is the Laplace Transform Φ(ξ ) of ϕ(τ ) , write
u =∞
∫
F (t ) =
e t −ue−e
t −u
f (u )du .
u =−∞
Multiply both sides by e −t . Then,
u =∞
∫
e −t F (t ) =
e−e
t −u
f (u )e −udu .
u =−∞
Put τ = e −u . Then,
1
u = log ,
τ
d τ = −e −udu ,
and
τ =0
∫
e −t F (t ) =
τ =∞
t
1
e −e τ f (log )( −d τ )
τ
τ =∞
=
∫
τ =0
t
1
e −e τ f (log )d τ .
τ
Denote e t = ξ . Then,
1
F (log ξ ) =
ξ
Φ(ξ )
τ =∞
1
−ξτ
e
f
(log
)d τ .
∫
τ
τ =0
ϕ(τ )
Thus,
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
τ =∞
Φ(ξ ) =
∫
e −ξτϕ(τ )d τ .
τ =0
We obtain the Convolution Transform, similarly to the way we
obtain the Convolution-Kernel Integral Theorem, by applying the
Two-Sided Laplace Integral Theorem,
4.5 The Convolution-Transform Integral Theorem
F (t ) = ( g1(t ) ∗ g 2(t ) ∗ g 3(t ) ∗ .... ) ∗ f (t )
= L−1 {( L { g1(t )} L { g2(t )} L { g 3(t )} ... ) L { f (t )} }
⎧⎛
⎞⎟
⎪⎜
⎪
⎟
⎜
⎪⎜
y =∞ ⎪
⎟⎟
⎜⎜ 1
⎪
⎟
1
1
1
=
×
× ... ⎟⎟⎟ L { f (t )}}e iytdy ,
⎨⎪⎜⎜ iby ×
∫
iy
iy
⎪⎜⎜ e
2π
⎛
⎟⎟⎟
y =−∞ ⎪
iy ⎞⎟ a1 ⎛⎜
iy ⎞⎟ a2
⎪
⎜
⎜
⎟⎟
⎪⎜
⎜⎜ 1 − ⎟⎟e
⎜⎜ 1 − ⎟⎟e
⎪⎜
⎟
⎜
a
a
⎝
⎝
⎠
⎝
⎠
⎠
⎪
⎩
1
2
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
5.
Schrodinger Propagator and Infinite
Convolution Product
Feynman attempted to express the Schrodinger Propagator as an
Infinite Convolution Product over the least energy path of the
particle.
To that end he divided the time interval [ta , tb ] into N equal
subintervals, and by the properties of the Gaussian had
K (tN , tN −1 ) ∗ K (tN −1, tN −2 ) ∗ ... ∗ K (t1, t0 ) = K (b, a ) ,
where, for the force-free particle,
1
⎤2
⎡
m
⎥
K (tk , tk −1 ) = ⎢
⎢⎣ 2πi=(tk − tk −1 ) ⎥⎦
e
⎛
⎞⎟
⎜⎜
m
2
(xk −xk −1 ) ⎟⎟⎟⎟
⎜⎜
⎜⎝ 2i=(tk −tk −1 )
⎠⎟
.
As N → ∞ , the finite convolution product becomes infinite, but
its limit is the same K (b, a ) .
Indeed, Schrodinger’s Propagator is not an Infinite Convolution
Product.
An Infinite Convolution Product requires a sequence of real
numbers
a1, a2, a3,... with
1
a12
+
33
1
a22
+
1
a32
+ ... < ∞ ,
Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
that are roots of the analytic infinite product E (z ) .
But the Schrodinger Propagator K (b, a ) , is a kernel G (ξ ) with no
roots, and its Two-Sided Laplace Transform
1
, is such that
E (z )
E (z ) has no roots either. Therefore, G (ξ ) is not an Infinite
Convolution Product.
5.1
Schrodinger’s
propagator
is
not
an
Infinite
Convolution Product
Denote xb − xa ≡ ξ , and K (b, a ) = G (ξ )
Then,
1
= E (z ) has no roots.
L {G(ξ )}
Thus, the Schrodinger Propagator is not an Infinite Convolution
Product, and the Infinite Convolution Method cannot be applied to
the Feynman Integral.
Proof:
Denote
m
= −c .
2=(tb − ta )
Then,
1
⎤2
⎡
m
⎥
K (b, a ) = ⎢
⎢⎣ 2πi=(tb − ta ) ⎥⎦
e
⎛
⎞⎟
⎜⎜
m
(xb −xa )2 ⎟⎟⎟⎟
⎜⎜
⎜⎜⎝ 2i=(tb −ta )
⎠⎟
=
c
c −i ξ 2
πi
e
Therefore, the Two-Sided Laplace Transform of K (b, a ) is
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Gauge Institute Journal, Volume 5, No 2, May 2009
L { K (b, a )} =
=
H. Vic Dannon
ξ =∞
−c ξ2 −z ξ
c
∫ e i e dξ
πi ξ =−∞
c
πi
ξ =∞
−(c /i )ξ
∫ e
2
(ez ξ + e−z ξ )d ξ
ξ =0
By [Haan, Table 26, formula 14],
∞
∫ e−q x (e2px + e−2px )dx =
2 2
0
Identifying
q =
1 2 2
π e p /q .
q
c / i , p = z / 2 , we have
=e
i
z2
4c .
Hence,
z2
−i
1
E (z ) =
= e 4c has no zeros. ,
L { K (b, a )}
Feynman’s construction of his Path Integral is erroneous because
the Schrödinger Propagator K (b, a ) cannot be decomposed into an
Infinite Convolution Product.
The Feynman Integral is precisely the Schrodinger Propagator,
and Feynman’s New approach to Quantum Mechanics is
Schrodinger’s Wave Mechanics.
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Gauge Institute Journal, Volume 5, No 2, May 2009
H. Vic Dannon
References
[Feyn1] Feynman and Hibbs, “Quantum Mechanics and Path Integrals”
McGraw-Hill, 1965.
[Feyn2]. Feynman, R. P., “Space time Approach to Non-relativistic Quantum
Mechanics”, in “Selected Papers on Quantum Electrodynamics” edited by
Julian Schwinger, Dover, 1958. p. 321.
[Haan] De Haan, D. Bierens, “Nouvelles Tables D’Integrales Definies”,
Edition of 1867 – Corrected. Hafner Publishing.
[Hirs1] Hirschman,
I., and
Widder, D., The Convolution Transform,
Princeton University Press, 1955.
[Hirs2] Hirschman,
I., and
Widder, D., “The Laplace Transform, The
Stieltjes Transform, and their Generalizations”, in “Studies in Real and
Complex Analysis”. Edited by Hirschman, I.. Published by Mathematical
Association of America, 1965
[Widd1] Widder D., “An introduction to Transform Theory”, Academic Press,
1971.
36