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Mean position of a parti le submitted to a

27 pages
Mean position of a parti le submitted to a potential barrier D. Mer ier, V. Régnier ? Abstra t A one-dimensional Klein-Gordon problem, whi h is a physi al model for a quantum parti le submitted to a potential barrier, is studied numeri ally : using a variational formulation and a Newmark numeri al method, we ompute the mean position and standard deviation of the parti le as well as their time evolution. Key words Klein-Gordon equation, Newmark method, mean and standard devia- tion, kernel smoothing, linear regression. AMS 35A15, 65M06, 65M12, 81Q05, 81Q10. 1 Introdu tion It has been well-known for a few years now that in quantum me hani s a parti le an limb up a step even if it has not enough energy a priori and it will be ree ted then with a delay. In lassi al me hani s it would just try and go ba k to its position. Everything happens here as if the parti le ould go through a wall ( f. Fig. 5.1 in [17?) ! This phenomenon is alled tunnel ee t and has been a subje t of interest for physi ians and mathemati ians : the delay has been measured by the physi ists A.

  • iennes

  • tion

  • initial boundary

  • tly supported initial

  • exa tly

  • gaussian wave

  • tral theory

  • mer ieruniv-valen

  • delay whi

  • numeri al


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35A15,
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in
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(AP)

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(t)+Au(t) = 0 H(AP) 2dt du u(0) = Φ, (0) = Ψ
dt
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D( A) g : Ω → k∈{1;2}k k
d = 1, ∀j∈{1···n}j
c a Aj j
√Q+ + +2 n 2 1 0C ( , L (Ω ))∩C ( ,D( A))∩C ( ,D(A))kk=1
√ Q
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g =g = 01 2
f : Ω → k∈{1;2}k k
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+∗(t,x) ×Ω1
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for

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to
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rst
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4√
cos( At)Φ
(AP) Ψ√ √ √
−1cos( At)Φ+( A) sin( At)Ψ
M(t) V(t) t
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 Z 1 2 2 R V(t) = (x−M(t)) |u(t,x)| dx
2|u(t,x)| dx
(P) t∈I = [0;T]

2 d (u(t),v) +a(u(t),v) = 0, ∀ v∈VH t
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sc
is
2.1
one
ariational
of
ulation
them,
Newmark
the
d
more
n

treatmen
is
of
the
problem,
other
h
one.
dev
So
ed
only
the

wing,
based
an
on
of
Gaussian
w
functions
ab


b
olution
e
(cf.

8
hosen
[10],
and
197-204).
that
v
is
form
our
of

time
hoice.
at
Our
for
main
osition
result
its
is
ariance
that
the
the
:
time
o-
series
mean
of

the
need
mean
w
p
the
osition
viour
of
b
the
transien
particle
study
has
T
a
is
linear
:
trend
sine
as
an
w
solution
ell
Otherwise
as
v
that
of
of
oretic
the
the
v
h
ariance.
of


from
in
their
the
sin
is
It
sucien
tly
5(·;·) H =H
2 1L ((−L;L)) V =H ((−L;L)) a (u,v)∈V×V
Z L
2a(u,v) = c ∂ u(x)·∂ v(x)+a(x)·u(x)·v(x) dxx x
−L
S ⊂S ⊂S ···S ⊂V0 1 2 m
u ∈S (P )m m m

2 d (u (t),v) +a(u (t),v) = 0, ∀ v∈Sm H m H m t
u (0) =f (f ∈S )m m m m
u (0) =g (g ∈S )m,t m m m
f g f gm m
x u um
t Δt = T/n t = nΔtmax n
n = 0;1;2···n n mmax max
n nU u (t ) V =u (t )m n m,t n
 1 n+1 n n 2 n+1 n (U −U ;v) −(Δt)·(V ;v) +(Δt) a βU + −β U ;v = 0 H H 2
n+1 n n+1 n(V −V ;v) +(Δt)·a(γU +(1−γ)U ;v) = 0 H 0 U =f m 0V =gm
β γ
m m{Ψ } Smj j=0
G = (Ψ ;Ψ ′) a A = (a(Ψ ;Ψ ′))m j j ′ m j j ′j,j j,j
n n 0 0 n nC D C D U Vm m
m mf g {Ψ }m m j j=0

1 n+1 n n 2 n+1 n G (C −C −(Δt)D )+(Δt) A βC + −β C = 0 m m 2
n+1 n n+1 nG (D −D )+(Δt)A (γC +(1−γ)C ) = 0 m m 0 0 C =C m 0 0D =Dm
P
m n nu (t ,x) = C ψ (x) Cm n jj=0 j j
u x = x t = nT/n ψj n max j
2ψ (x ) =δ (j,k)∈{0;1;2;···;m} δ = 1 j =kj k jk jk
then
the
sc
.
to
ximation
heme
and
is
v
:
when
of
is
ordinates
of

in
the
mated
taining
w

y
ectors
and
v
with
the
holds
and
and
and
ximations
,
the
,
tain
y
is
b
the
Denote
in
.
's
is
and
form
The
bilinear
If
the
n
of
of
matrix
Since
the
a
and
:
is

basis
e
this
and
of
space
matrix

Gram
),
The
appro
.
alues
space
an
the
,
in
.
basis
then
a
time
ose
ariable
ho
dened

,
e
with
w
the
w
hoices
No
dep

to
wing
sc
follo

the
basis
(cf.
spaces
stable
to
e
it
b
nd
to
solution
heme
Problem
sc
Since
the
are
for
hosen
further
b
put
appro
e
of
b

will
the

,
h

whic
ts
on
ariable
parameters

are
the
and
xi-
the
v
scalar
of
pro
at

appro
is
of
the
at

time
one
No
otherwise.
a
in
discretization
the
the
Hilb
v
ert
the
space
is
,
are
and
b
the
done
bilinear
:
form
and
is
and
dened,
in
for
for
,

b
of
y
is
No
end
w
Newmark.
the
due
idea
heme
,
umerical
ted
The
if
of
the
precision
sequel
is
to
an

w
a
on
,
0
where
6Δt γ β
3/2f ∈D(A ) g∈D(A) β = 1/4 γ = 1/2
3−j j/2u∈C ([0;T);D(A )) j = 0;1;2;3
Πm
Sm
0 0 1 1|u (t )−u(t )|≤C{|u −Π u |+|u −Π u |m n n m mm m Rt (3)n+|(I−Π )u(t )|+ |(I−Π )u (s)|+Δt|u (s)| ds}m n m tt t0
A D(A)
u
2 3u ∈ C ([0,T];V)∩C ([0,T];H)
Δt β = 1/4 γ = 1/2
(k+1)/2 k/2f ∈D(A ) g∈D(A )
k+1−j j/2 1/2u C ([0;T);D(A ) j = 0;1;2;···k+1 V =D(A )
0 3/2H =D(A ) f ∈D(A ) g∈D(A)
c
a (t,x) a (t,x)1 2
a = 01
f
m σ m
1/σ
f g g f
m 2O(|x| exp(−0.5x ))
f g
f
′g =−cf u (0)t
that
and
ose
)
function
b
Supp
F
1
ev
osition
functions
Prop
to
d.
ourier
metho
functions
Newmark
for
the
is
,
mak
then
the
the
space
solution
e
to
ther
b
unique
elongs
Hardy's
to
est
y
ourier
stabilit
No
brings
b
and
m
,
zero.
for
elo
alues
initial
v
tly
adapted
the
of
oth
hoice
No

Gaussian
translate
e
,
and
or
is
der
d
The
a
estimates
they
error
only
2.
that,
for
)
Denoting
of
by
Newmark
the
stable
pr

oje
order

the
on

the

ve
for


sp
that
ac
is
e
e
,

.
frequency
Since
and
the
transform
err
to
or

is
the
The
of
and
with
y
standard
Stabilit
a
op
mean
and
deviation
erator
.
and
y
for
go
is,
for
that
ab
,
of
its
states
,
the
enough
since
regular
of
for
if
that,
the
ys
of
sa
h
[10]
are
of
Lemma
8.6.2
and
and
d
Theorem
onditional
of
giv
part
nite
rst
of
The
if
of.
[18]).
are
b
sucien
function
t
the

anish
3
us

of
results
of
and
initial
ph
y
ysical
y
in

terpretation
the
3.1


required
of
b
the
sucien
initial
lo

in
F
and
or
i.e.
all
function
the
its
n
ourier
umerical
b

need
the
b
v
lo
elo
enough.

w
y
F
of
transform
the
a
signal
function
Pro
mean
is
and

deviation
hosen
is
to
Gaussian
b
with
e
exists
equal
standard
to
a
1
solution
and
That
the
wh
p
Gaussian
oten
are
tials
o


an
us.
tro
note
in
out
the
remark
in
Wiener
dened
en
and
that
are
are
without
b
domain

(i.e.
the
stable
pair

functions
is
and
are
(where

is
t
F
functions.
transform
On
states
the

rst
that

oth
h

w
1.1.6
e
w

.
ho
the
ose
metho
a
and
v

anishing
ly
p
is
oten
en
tial
y
d
linear
and
binations
metho
Hermite
,
for
and
and
.
(see
mak
The

ultiplication
Then
y
is
square
a
then
Gaussian
es
function
function
m
v
ultiplied
at
b
Th
y
the
the
hoice
square
and
function.
and
This
on

the
hoice
v
is

justied
(
b
the
y
.
The
)
initial
es
the
signal
2.2
7x = 0
u(t,0) =u u (t,0) =v0 t 0
Z
x+ct1 1
[u (x+ct)+u (x−ct)]+ v (y)dy0 0 0
2 2c x−ct
− +u (t,x) u (t,x)
 Z
x−ct1 1 + u (t,x) = u (x−ct)− v (y)dy 0 0 2 2c 0
 Z x+ct 1 1 − u (t,x) = u (x+ct)− v (y)dy0 0
2 2c 0
′ − +v =−cu u u (0) = 0 u u (t,x) =u (x−ct)0 0 0 00
f
−3 1
f f

1 12 2√f(x) = x − (x+3)
210 2π
u (0,x)t

1 12 2√g(x) = x(x +3x−2) − (x+3)
210 2π
L 60
x = 0 f Ω2
−3.5
∞f g C
f g

1 1 2 2 √f(x) = x − (x+3)
210 2π
1 1 2 2 √g(x) = x(x +3x−2) − (x+3)
210 2π
3/2f ∈D(A ) g∈D(A)
is
it
Then
functions.
is
t
1.
ysically
to
un
equal
p
is
I
of
dened
R

I
left
er
hes
and
in
v
"falling"
o
k
tegral
h
in
a
the
a
.
oten
exp
particle
is
(around
y
y
probabilit
and
a
Note
ts
rapidly
represen
The
that
om
so
from
added
v
is
anishes

with
t
a

sup
A
with
.
of
deviation
at
standard
in
and
the
mean
on
with
h
function
o
Gaussian
its
exp

the
maxim
y
meets
b
tial
function
to
square
b
the
itself
of
osition

and
pro
d
the
R
is
by
The
the
length

of
adapted
the
.

,
h
exp

y
is
t-going

term
hosen
osition
to
is
b

e
e
equal
the
to
the
the
Let
or
tial
whic
the
h
it
a
t
v
the
oids
is
the
the


at
initially
the
parts
exterior
w
no
where
des
probabilit
to
of
tak
reac
e
its
place
um)
to
it
o
a
fast.
oten
They
step.
w
that
ould
and
in
are
terfere
oth
with

what
separating
happ
Prop
ens
2
at
functions
the
and
discon
dene
tin
fr
uit
I
y
to
F
R

:
osite
signal
,
eeps
whic
y
h
elo
is
initial
our
an
p
Ph
oin
and
t
v
here.
that
F

urther
and
since
So
the
b
v
one
alue
righ
of
and
opp
left-going
on
of
to
erp
going
the
is
It
negligible,
and
the
initial
probabilit
equation
y
v
for
w
the
exp
particle
solution
to
that
b
us
e
.
on
step
the
satisfy
righ
p
t
to

bumps
h
til
is
and
negligible.
righ
So
densit
to
y
i.e.
8(k+1)/2u∈D(A )
u (K ),(K ),...,(K )0 1 k
Q2 j+1u (K )⇐⇒u∈ H (Ω )j ii=1
(
j/2A u (T ), j0
(j−1)/2A u (T ), j0
Q
2 2u∈D(A)⇐⇒u∈ H (Ω ) u (T ) (T )i 0 1i=1
x = 0
∞g C
Q3/2 2 3u ∈ D(A )⇐⇒ u ∈ H (Ω ) u (T ) (T ) Aui 0 1i=1
(T )0
a2′′ + ′′ −u (0 )−u (0 ) = u(0)
2c
∞f f C
f(0) = 0 f
u(t,·)
t (P) M(t)
σ(t) u(t,·) t
 Z
1 2 RM(t) = x|u(t,x)| dx 2 |u(t,x)| dx
 Z  1 2 2 σ(t) = R (x−M(t)) |u(t,x)| dx
2|u(t,x)| dx
M σ t
a = 0 a > 0 g =g = 01 2 1 2
f : Ω → k∈{1;2}k k
quan
tum

resp
where
the

visaged
the

p
.
osition
of
of
less
a
the
particle
of.
is
and
nev
The
er
here
kno
an
wn
step
with
(
precision.
satises
Only
of
its
equiv
probabilit
e
y
y
densit

y
Then
is
if
kno
are
wn
situations.
and

giv

en
of
b
w
y
p
the
satises
expression
in
dieren
ts
the
e
in
as
e
Assume
(P),

Pr
,
at
deriv
the
R
time
tin
data
uit
,

solution
satises
of
o
our
ariations
problem
satises
,
to
In
n
b
dieren
.

W
n
e
In


the
e
mean
the
satised
oten
is
the

b
transmission
With
and
tial
the
is
standard
and
deviation
),
this
particular
,
(i)
since
t
and

of
in
the
ourier
signal
ws
function)
Theorem

alen
I
a
is
with
(
1.5.2
uous
are
tin
at
at
ativ
the
rst
time
I

of
.
uit
Let

us
y

tin
that
of
they
Those
are
and
dened
and
as
dd
is
is
function
v
the
of
of
and
e
with
ativ
ect
deriv
en

studied
the
umerically
Since
in
i.e.
t
0
3.2.1
at
y
e
a
ativ
umerical
deriv
h


the
analytic
of
h
jump
b
I
en
R
in
the
absence
es
p
giv
tial
it
since
one,
situation
third
ould
the
e
for

As
a
more.
oten

step
satised
ev
are
if
I
and
R
satises

,
o
and
w
the
t

rst
(ii)
The
:
.
statemen
satises
,
and
solution
and
b
satises
rewritten
,
terms
w
F
No
transforms
function.
follo
a
(cf.
is
:
h
2
whic
that
t
function
I
the
R
es
the
R
y
giv
b
of
satised
Theorem

y
Pro
initial
.
of
3.2
oblem
In
describ
terpretation
d
of
9(f ,f )∈D(A) f (−∞;0)1 2 1
f ≡ 02
Ω (u ,u )1 1 2√Q2 + 2 2 1 + 0 +C ( , L (Ω ))∩C ( ,D( A))∩C ( ,D(A))kk=1
+∗(t,x) ×Ω1
h √ i1 −1 2 2u (t,x) = F cos( c ω t)Ff (ω)1 1ω → xπ

2 2c ω −a2
 q√
a22 2 c ω −a ω≥ 2 2 c q q√
a a2 2 2 2i a −c ω − ≤ω≤2 2 2c c q√ a2 2 2 − c ω −a ω≤−2 2c
+∗(t,x) ×Ω u (t,x) =2 2
√ !Z ∞ √ 1 2 21 i c ω −a x 2ωc2c2 2− cos( c ω t)e √ Ff (ω)dω12 22π −∞ ωc+ c ω −a2
2a =a =m1 2
q1 −1 2 2 2u (t,x) = F cos( m +c ξ t)Ff(ξ)1 ξ → x
π
f
Z Z
1 1i(ξx−ω(ξ)t) i(ξx+ω(ξ)t)u (t,x) = e Ff(ξ)dξ + e Ff(ξ)dξ1
2π 2π

2 2 2ω(ξ) = m +c ξ (ξx−ω(ξ)t)
′x/t =ω (ξ)
2c ξ′ω (ξ) = √
2 2 2m +c ξ
ω
[ξ ;ξ ] Ff [ξ ;ξ ]1 2 1 2
′ ′t u(.,t) [ω (ξ )t;ω (ξ )t]1 2
t
Pr
oblem
I
d
the
exp
R
of
(P)
Then
(such
sum
that
time.
b
problem
elongs
in
to
of
I
at
R
tially
I
tial
R
the
I
W
R
an
)

is
to
given,
if
for
a
in
w
I
of
R

,
.
by
actly
:
gro
wher
is
e
rate
I
problem.
R
in
the
F.

function.
omplex

squar
to
e
frequency
r
r
o
to
ot
supp
has
tials
b
of
e

en
,
dene
xed
d
the

half
,
lies
that
Rewriting
=
initial
if
supp
where
p
if
due
if
linearly
R

ev
to
olution

I
the
of
Klein-Gordon
in
Strauss
the
w
mean
in
Likewise,
[22
and
is
for

.
If
The
initial
expression
is
of
hosen
the
lie
v
the
ariance
band
is
estriction
not
the
simple.
:
If
i.e.
the
the
p
ort
is
leads
the
is
phase
subset
of
onen
solution

Since
o

t
and
then
amplitude
a
this
time
terv
,
b
solution
gro
the
linearly
as
of
the
the
essen
v
in
unique
.
the

and
and
standard
with
of
in
particle
orte
the
to

of
the
.
rst
the
in
ter
tegral
the
and
of
it
in
is
al
stationary
oth
when
w
oten
with
tial
,
is
mean
:
alue
is
the
solution
deviation
the
the
),
submitted
intr
a
.
t
Moreo
oten
v
are
er
to

w
t
with
(
This

h
is
useful
,
study

time
o
y

of
that
solution

the
e
transmission
ar
Marshall,
So
and
analysing
ainger
the
ere
Mehmeti
terested
w
that
([3]
(cf.
us

omp
Ali
time
as
(1)
ell
Th
and
the
10

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