In 1966,
Edward Nelson, a Princeton professor of mathematics,
demonstrated a derivation of the Schrödinger equation from Brownian
motion. His 7pages long paper, published in Physical Review,
vol. 150, no. 4, pp. 1079–1085, presented an introduction to
stochastic mechanics, Brownian motion and, eventually, the actual
derivation of the Schrödinger equation, based on a number of
scattered assumptions. A much simpler derivation, the one we're going
to present here, was delivered a year later by Luis de la
PeñaAuerbach, a professor of physics at Universidad Nacional de
México in a brief, 2pages long letter published in Physics Letters,
vol. 24A, no. 11, pp. 603–604.
The starting point in de la Peña's derivation is the continuity
equation, Equation 30,
∂
∂
t
P
(
x
,
t
)

(
x
0
,
t
0
)
=

∂
∂
x
J
(
x
,
t
)

(
x
0
,
t
0
)
,
∂
∂
t
P
(
x
,
t
)

(
x
0
,
t
0
)
=

∂
∂
x
J
(
x
,
t
)

(
x
0
,
t
0
)
,
(79)where, Equation 29
J
(
x
,
t
)

(
x
0
,
t
0
)
=
μ
(
x
,
t
)
P
(
x
,
t
)

(
x
0
,
t
0
)

1
2
∂
∂
x
D
(
x
,
t
)
P
(
x
,
t
)

(
x
0
,
t
0
)
.
J
(
x
,
t
)

(
x
0
,
t
0
)
=
μ
(
x
,
t
)
P
(
x
,
t
)

(
x
0
,
t
0
)

1
2
∂
∂
x
D
(
x
,
t
)
P
(
x
,
t
)

(
x
0
,
t
0
)
.
(80)The probability density function P(x,t)(x0,t0)P(x,t)(x0,t0)
is real and positive or zero everywhere, therefore we can represent it
by
P
(
x
,
t
)

(
x
0
,
t
0
)
=
e
2
R
(
x
,
t
)
.
P
(
x
,
t
)

(
x
0
,
t
0
)
=
e
2
R
(
x
,
t
)
.
(81)Why 2R2R instead of just RR will become clear later. It'll make our equations
down the road prettier. Of course, RR has to be such that
∫

∞
∞
e
2
R
(
x
,
t
)
d
x
=
1
∫

∞
∞
e
2
R
(
x
,
t
)
d
x
=
1
(82)for all tt.
We can rewrite Equation 80
in terms of RR too
J
(
x
,
t
)

(
x
0
,
t
0
)
=
μ
(
x
,
t
)
e
2
R
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)
e
2
R
(
x
,
t
)
=
μ
(
x
,
t
)
e
2
R
(
x
,
t
)

1
2
e
2
R
(
x
,
t
)
∂
∂
x
D
(
x
,
t
)
+
D
(
x
,
t
)
2
e
2
R
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
=
e
2
R
(
x
,
t
)
μ
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)

D
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
,
J
(
x
,
t
)

(
x
0
,
t
0
)
=
μ
(
x
,
t
)
e
2
R
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)
e
2
R
(
x
,
t
)
=
μ
(
x
,
t
)
e
2
R
(
x
,
t
)

1
2
e
2
R
(
x
,
t
)
∂
∂
x
D
(
x
,
t
)
+
D
(
x
,
t
)
2
e
2
R
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
=
e
2
R
(
x
,
t
)
μ
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)

D
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
,
(83)wherefrom we
derive
v
(
x
,
t
)
=
μ
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)

D
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
,
v
(
x
,
t
)
=
μ
(
x
,
t
)

1
2
∂
∂
x
D
(
x
,
t
)

D
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
,
(84)the expression for current velocity, given that
normally, where in our case
ρ
=
P
(
x
,
t
)

(
x
0
,
t
0
)
=
e
2
R
(
x
,
t
)
.
ρ
=
P
(
x
,
t
)

(
x
0
,
t
0
)
=
e
2
R
(
x
,
t
)
.
(86)We return to Equation 79, substituting Equation 81
and Equation 84, which yields
∂
∂
t
P
(
x
,
t
)

(
x
0
,
t
0
)
=
∂
∂
t
e
2
R
(
x
,
t
)
=
2
e
2
R
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
∂
∂
t
P
(
x
,
t
)

(
x
0
,
t
0
)
=
∂
∂
t
e
2
R
(
x
,
t
)
=
2
e
2
R
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
(87)on the left side and

∂
∂
x
e
2
R
(
x
,
t
)
v
(
x
,
t
)
=

2
e
2
R
(
x
,
t
)
v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)

e
2
R
(
x
,
t
)
∂
∂
x
v
(
x
,
t
)

∂
∂
x
e
2
R
(
x
,
t
)
v
(
x
,
t
)
=

2
e
2
R
(
x
,
t
)
v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)

e
2
R
(
x
,
t
)
∂
∂
x
v
(
x
,
t
)
(88)on the right which, upon division by 2e2R(x,t)2e2R(x,t),
yields
∂
∂
t
R
(
x
,
t
)
=

v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)

1
2
∂
∂
x
v
(
x
,
t
)
.
∂
∂
t
R
(
x
,
t
)
=

v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)

1
2
∂
∂
x
v
(
x
,
t
)
.
(89)So far we have merely rewritten the FokkerPlanck equation using
R(x,t)R(x,t) and v(x,t)v(x,t). Equation 89 is fully equivalent
to the forward FokkerPlanck. But now we introduce an assumption that
goes beyond FokkerPlanck (and beyond Markov). We assume
that
v
=
α
∂
∂
x
S
(
x
,
t
)
.
v
=
α
∂
∂
x
S
(
x
,
t
)
.
(90)If we were to carry out this reasoning in 3D, the assumption would be
v
(
r
,
t
)
=
α
∇
S
(
r
,
t
)
.
v
(
r
,
t
)
=
α
∇
S
(
r
,
t
)
.
(91)This would imply that the velocity field is curl
free
∇
×
v
(
x
,
t
)
=
0
.
∇
×
v
(
x
,
t
)
=
0
.
(92)This is also true in the opposite direction, that is, any continuous,
differentiable and curlfree velocity field can be represented by a
gradient.
A vector field that is a gradient of a function is called
conservative. Conservative vector fields have various
interesting properties including path independence. The line
integral of such a field from one point to another depends on end
points only, and not on the path taken.
Having made this assumption, we combine R(x,t)R(x,t) and S(x,t)S(x,t)
into a single complexvalued
function
Ψ
(
x
,
t
)
=
e
R
(
x
,
t
)
+
i
S
(
x
,
t
)
,
Ψ
(
x
,
t
)
=
e
R
(
x
,
t
)
+
i
S
(
x
,
t
)
,
(93)where RR and SS are both real. It is easy to see that
Ψ
¯
(
x
,
t
)
Ψ
(
x
,
t
)
=
e
R
(
x
,
t
)

i
S
(
x
,
t
)
e
R
(
x
,
t
)
+
i
S
(
x
,
t
)
=
e
2
R
(
x
,
t
)
=
P
(
x
,
t
)

(
x
0
,
t
0
)
,
Ψ
¯
(
x
,
t
)
Ψ
(
x
,
t
)
=
e
R
(
x
,
t
)

i
S
(
x
,
t
)
e
R
(
x
,
t
)
+
i
S
(
x
,
t
)
=
e
2
R
(
x
,
t
)
=
P
(
x
,
t
)

(
x
0
,
t
0
)
,
(94)so Ψ(x,t)Ψ(x,t) is the probability amplitude. The above also explains why
we chose to have the 2 in e2R(x,t)e2R(x,t).
To derive a differential equation for ΨΨ we differentiate it first
∂
∂
t
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
+
i
∂
∂
t
S
(
x
,
t
)
,
∂
∂
x
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
,
∂
2
∂
x
2
Ψ
(
x
,
t
)
=
∂
∂
x
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
2
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
2
+
2
i
∂
∂
x
R
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)

∂
∂
x
S
(
x
,
t
)
2
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
,
∂
∂
t
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
+
i
∂
∂
t
S
(
x
,
t
)
,
∂
∂
x
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
,
∂
2
∂
x
2
Ψ
(
x
,
t
)
=
∂
∂
x
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
i
∂
∂
x
S
(
x
,
t
)
2
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
2
+
2
i
∂
∂
x
R
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)

∂
∂
x
S
(
x
,
t
)
2
+
Ψ
(
x
,
t
)
∂
2
∂
x
2
R
(
x
,
t
)
+
i
∂
2
∂
x
2
S
(
x
,
t
)
,
(95)wherefrom
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
=
∂
∂
t
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
∂
∂
t
S
(
x
,
t
)
.
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
=
∂
∂
t
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
∂
∂
t
S
(
x
,
t
)
.
(96)This tells us that by multiplying both sides of Equation 89
by Ψ(x,t)Ψ(x,t) we'll obtain an equation for ΨΨ itself:
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
=
∂
∂
t
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
∂
∂
t
S
(
x
,
t
)
=

Ψ
(
x
,
t
)
v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
∂
x
v
(
x
,
t
)
=

α
Ψ
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
2
∂
x
2
S
(
x
,
t
)
.
Ψ
(
x
,
t
)
∂
∂
t
R
(
x
,
t
)
=
∂
∂
t
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
∂
∂
t
S
(
x
,
t
)
=

Ψ
(
x
,
t
)
v
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
∂
x
v
(
x
,
t
)
=

α
Ψ
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
2
∂
x
2
S
(
x
,
t
)
.
(97)Now we make use of the expression for the second derivative of ΨΨ given
by Equation 95 to rewrite the right side of this equation
in terms of ∂2Ψ/∂x2∂2Ψ/∂x2 minus stuff we don't
want,

α
Ψ
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
2
∂
x
2
S
(
x
,
t
)
=

α
2
i
∂
2
∂
x
2
Ψ
(
x
,
t
)

Ψ
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2
.

α
Ψ
(
x
,
t
)
∂
∂
x
S
(
x
,
t
)
∂
∂
x
R
(
x
,
t
)
+
1
2
∂
2
∂
x
2
S
(
x
,
t
)
=

α
2
i
∂
2
∂
x
2
Ψ
(
x
,
t
)

Ψ
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2
.
(98)We combine this result with Equation 97, lift the imaginary
unit from the denominator to the numerator (which changes the sign) and
transfer iΨ∂S/∂tiΨ∂S/∂t to the right side of the equation
to obtain
∂
∂
t
Ψ
(
x
,
t
)
=
i
α
2
∂
2
∂
x
2
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
α
2
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2

∂
S
∂
t
,
∂
∂
t
Ψ
(
x
,
t
)
=
i
α
2
∂
2
∂
x
2
Ψ
(
x
,
t
)

i
Ψ
(
x
,
t
)
α
2
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2

∂
S
∂
t
,
(99)which, upon the multiplication of both sides by ii
becomes
i
∂
∂
t
Ψ
(
x
,
t
)
=

α
2
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
V
(
x
,
t
)
Ψ
(
x
,
t
)
,
i
∂
∂
t
Ψ
(
x
,
t
)
=

α
2
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
V
(
x
,
t
)
Ψ
(
x
,
t
)
,
(100)where
V
(
x
,
t
)
=
α
2
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2

∂
S
∂
t
.
V
(
x
,
t
)
=
α
2
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2

∂
S
∂
t
.
(101)Equation 100 is the celebrated Schrödinger equation
with a funny potential that depends on the probability density itself
through RR and on the velocity of the probability current
through SS.
In summary, we find that the Schrödinger equation is an equation
of the continuous Markov process theory with some additional assumptions,
one of which is given by Equation 90, that makes the probability
current velocity a conservative field. But we have made another nontrivial assumption
on the way,
about which we're going to say more towards the end of this section. The
actual coefficient in the quantum Schrödinger equation is
α
=
ℏ
m
,
α
=
ℏ
m
,
(102)but we should remember that the Schrödinger equation is not limited
to quantum physics. For example, the nonlinear
Schrödinger equation,
i
∂
∂
t
Ψ
(
x
,
t
)
=

1
2
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
κ
Ψ
(
x
,
t
)
2
Ψ
(
x
,
t
)
,
i
∂
∂
t
Ψ
(
x
,
t
)
=

1
2
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
κ
Ψ
(
x
,
t
)
2
Ψ
(
x
,
t
)
,
(103)which we might say is slightly similar to Equation 100 on account of VV
in Equation 101 being
a complicated function of ΨΨ,
occurs in Manakov systems in fiber optics and in hydrodynamics, where it
describes the formation of freak waves. So αα can be anything,
depending on the context. The Schrödinger equation is an equation of
mathematics that shows up in various situations.
We
are now going to show that the definition of the potential given by
Equation 101 is consistent with the true quantum Schrödinger
equation. Let Ψ(x,t)Ψ(x,t) be an arbitrary solution of such an
equation. We will demonstrate that this Ψ(x,t)Ψ(x,t) is also a solution
of Equation 100 with VV defined by Equation 101, assuming
that the FokkerPlanck equation in the form given by Equation 89
is also satisfied. So, for example, the seeming nonlinearity of Equation 100
should not exclude superpositions of various solutions of the “normal” Schrödinger
equation.
We begin by substituting Ψ(x,t)=eR(x,t)+iS(x,t)Ψ(x,t)=eR(x,t)+iS(x,t) into
i
∂
∂
t
Ψ
(
x
,
t
)
=

ℏ
2
m
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
V
(
x
,
t
)
Ψ
(
x
,
t
)
,
i
∂
∂
t
Ψ
(
x
,
t
)
=

ℏ
2
m
∂
2
∂
x
2
Ψ
(
x
,
t
)
+
V
(
x
,
t
)
Ψ
(
x
,
t
)
,
(104)where V(x,t)V(x,t) is an arbitrary function, not related to (or dependent on)
RR and SS explicitly. In particular, we do not assume that
it is given by Equation 101.
First we find that on such a substitution
∂
∂
t
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
t
+
i
∂
S
∂
t
,
∂
∂
x
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
x
+
i
∂
S
∂
x
,
∂
2
∂
x
2
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
x
2
+
2
i
∂
R
∂
x
∂
S
∂
x

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
i
∂
2
S
∂
x
2
.
∂
∂
t
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
t
+
i
∂
S
∂
t
,
∂
∂
x
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
x
+
i
∂
S
∂
x
,
∂
2
∂
x
2
Ψ
(
x
,
t
)
=
Ψ
(
x
,
t
)
∂
R
∂
x
2
+
2
i
∂
R
∂
x
∂
S
∂
x

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
i
∂
2
S
∂
x
2
.
(105)Now we plug this into Equation 104 to find
i
Ψ
∂
R
∂
t
+
i
∂
S
∂
t
=

ℏ
2
m
Ψ
∂
R
∂
x
2
+
2
i
∂
R
∂
x
∂
S
∂
x

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
i
∂
2
S
∂
x
2
+
V
Ψ
.
i
Ψ
∂
R
∂
t
+
i
∂
S
∂
t
=

ℏ
2
m
Ψ
∂
R
∂
x
2
+
2
i
∂
R
∂
x
∂
S
∂
x

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
i
∂
2
S
∂
x
2
+
V
Ψ
.
(106)For this to be true at all points where Ψ(x,t)≠0Ψ(x,t)≠0 we must have that

∂
S
∂
t
=

ℏ
2
m
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
V
,
∂
R
∂
t
=

ℏ
m
∂
R
∂
x
∂
S
∂
x

ℏ
2
m
∂
2
S
∂
x
2
,

∂
S
∂
t
=

ℏ
2
m
∂
R
∂
x
2

∂
S
∂
x
2
+
∂
2
R
∂
x
2
+
V
,
∂
R
∂
t
=

ℏ
m
∂
R
∂
x
∂
S
∂
x

ℏ
2
m
∂
2
S
∂
x
2
,
(107)where the top equation corresponds to the real part of Equation 106
and the bottom equation corresponds to the imaginary part of Equation 106.
Moving all but VV in the top equation to its left side we obtain Equation 101.
In turn, substituting

ℏ
m
∂
S
∂
x
=
v

ℏ
m
∂
S
∂
x
=
v
(108)in the bottom equation yields
∂
R
∂
t
=

v
∂
R
∂
x

1
2
∂
v
∂
x
,
∂
R
∂
t
=

v
∂
R
∂
x

1
2
∂
v
∂
x
,
(109)which is the same as Equation 89. This is why we can call
the function defined by Equation 101 a VV.
We are now going to have an even closer look at the connection between
the FokkerPlanck equation and the quantum Schrödinger equation and
we will discover that in its construction we have unwittingly
introduced two crucial assumptions that alter the physics of
the underlying stochastic system dramatically.
Let
us consider a ring of radius rr with a constant Markovian probability density PP and a constant
probability current JJ. Equation 85 and Equation 86 imply that the probability
current velocity vv must be constant too. Since
v
=
ℏ
m
∂
S
∂
x
v
=
ℏ
m
∂
S
∂
x
(110)we find that
S
=
m
v
ℏ
x
+
f
(
t
)
.
S
=
m
v
ℏ
x
+
f
(
t
)
.
(111)Let us introduce kk such that ℏk=mvℏk=mv then
S
=
k
x
+
f
(
t
)
S
=
k
x
+
f
(
t
)
(112)We are going to search for a solution of the Schrödingerde la Peña equation
such that V(x,t)=0V(x,t)=0, because then a constant PP would satisfy the remainder of the
equation, as we well know from conventional quantum mechanics.
Because PP is constant, RR is constant too, therefore
∂R∂x=∂2R∂x2=0∂R∂x=∂2R∂x2=0. This way
Equation 101 becomes
0
=

ℏ
k
2
2
m

d
f
(
t
)
d
t
,
0
=

ℏ
k
2
2
m

d
f
(
t
)
d
t
,
(113)which implies that
f
(
t
)
=

ℏ
k
2
2
m
t
+
C
,
f
(
t
)
=

ℏ
k
2
2
m
t
+
C
,
(114)where CC is an additive constant and we can make it zero. Let us introduce ωω such
that
ℏ
ω
=
(
ℏ
k
)
2
2
m
,
ℏ
ω
=
(
ℏ
k
)
2
2
m
,
(115)then f(t)=ωtf(t)=ωt and
S
=
k
x

ω
t
.
S
=
k
x

ω
t
.
(116)But now we have a problem: we are on a ring, so for t=0t=0, as we
increase xx from x=0x=0 the function SS would become multivalued,
first at x=2πrx=2πr, and then everywhere else. Luckily, function SS enters
ΨΨ through eiSeiS, that is, through
cos
S
+
i
sin
S
.
cos
S
+
i
sin
S
.
(117)Therefore if we choose kk such that
k
2
π
r
=
2
π
n
k
2
π
r
=
2
π
n
(118)both the cosScosS and the sinSsinS will wrap onto themselves on having circumnavigated
the ring, wherefrom we find that setting
ensures that ΨΨ at least is going to remain a well defined single valued function
on the ring.
Because ℏω=(ℏk)2/(2m)ℏω=(ℏk)2/(2m) this condition yields
ℏ
ω
=
ℏ
2
n
2
2
m
r
2
.
ℏ
ω
=
ℏ
2
n
2
2
m
r
2
.
(120)In turn Equation 119 tells us that the velocity of the probability current
in the ring must be quantized, too:
v
=
ℏ
k
m
=
ℏ
n
m
r
.
v
=
ℏ
k
m
=
ℏ
n
m
r
.
(121)How did we get from the constant probability density and the constant
probability current in the ring on the FokkerPlanck side to the
quantized velocity on the Schrödinger side? Clearly, we must have
added something on the way to get this result. The physics obtained
from the Schrödinger equation is quite different from physics that
corresponds to the FokkerPlanck equation in this case.
Two additional assumptions we've made on the way are responsible for
the change. The first assumption was that
v
∝
∂
S
∂
x
v
∝
∂
S
∂
x
(122)and this, together with Equation 101 led to
S
=
k
x

ω
t
,
S
=
k
x

ω
t
,
(123)where ωω and kk are related to each other by Equation 115.
But this by itself does not give us the quantization. If we were to use SS alone, we would
have to conclude that k=ω=0k=ω=0, the only solution that would work on the ring.
The second assumption, however, that SS should enter ΨΨ as
eiSeiS, has made SS into the phase of a wave. This is a
new physical assumption. This is not just a substitution we can make
willynilly. Together, the two assumptions, both outside the Markov process theory,
introduce new physics described by
e
i
(
k
x

ω
t
)
where
ℏ
ω
=
(
ℏ
k
)
2
2
m
e
i
(
k
x

ω
t
)
where
ℏ
ω
=
(
ℏ
k
)
2
2
m
(124)This is the original de Broglie's 1924 postulate, the subject of
his doctoral thesis. It is the addition of this postulate, a nontrivial step, that makes the
FokkerPlanck equation into the Schrödinger equation.