Quantum Tunneling: Rectangular Barrier
Quantum Tunneling: Rectangular Barrier
potential barrier
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From Wikipedia, the free encyclopedia
Although classically a particle behaving as a point mass would be reflected if its energy
is less than
V0
Calculation[edit]
                             Scattering at a finite potential barrier of height
V0
         . The amplitudes and direction of left and right moving waves are indicated. In red, those
               waves used for the derivation of the reflection and transmission amplitude.
E>V0
ψ(x)
reads
H^ψ(x)=[−ℏ22md2dx2+V(x)]ψ(x)=Eψ(x)
where
H^
  is the Hamiltonian,
ℏ
is the mass,
V(x)=V0[Θ(x)−Θ(x−a)]
V0>0
and width
Θ(x)=0,x<0;Θ(x)=1,x>0
x=0
         and
x=a
position without changing the results. The first term in the Hamiltonian,
−ℏ22md2dx2ψ
x<0,0<x<a,x>a
E>V0
     {ψL(x)=Areik0x+Ale−ik0xx<0ψC(x)=Breik1x+Ble−ik1x0<x<aψR(x)=Crei
     k0x+Cle−ik0xx>a
     {k0=2mE/ℏ2x<0orx>ak1=2m(E−V0)/ℏ20<x<a.
The index
r/l
on the coefficients
and
  denotes the direction of the velocity vector. Note that, if the energy of the particle is
below the barrier height,
k1
   becomes imaginary and the wave function is exponentially decaying within the
barrier. Nevertheless, we keep the notation
r/l
even though the waves are not propagating anymore in this case. Here we assumed
E≠V0
. The case
E=V0
is treated below.
The coefficients
A,B,C
x=0
and
x=a
ψL(0)=ψC(0)dψLdx|x=0=dψCdx|x=0ψC(a)=ψR(a)dψCdx|x=a=dψRdx|x=a.
Inserting the wave functions, the boundary conditions give the following restrictions on
the coefficients
Ar+Al=Br+Bl
ik0(Ar−Al)=ik1(Br−Bl)
   Breiak1+Ble−iak1=Creiak0+Cle−iak0
     ik1(Breiak1−Ble−iak1)=ik0(Creiak0−Cle−iak0).
V0
E<V0
To study the quantum case, consider the following situation: a particle incident on the
barrier from the left side (
Ar
). It may be reflected (
Al
) or transmitted (
Cr
     ).
To find the amplitudes for reflection and transmission for incidence from the left, we put
in the above equations
Ar=1
(incoming particle),
Al=r
(reflection),
Cl=0
Cr=t
Bl,Br
and
         t=4k0k1e−ia(k0−k1)(k0+k1)2−e2iak1(k0−k1)2
   r=(k02−k12)sin(ak1)2ik0k1cos(ak1)+(k02+k12)sin(ak1).
Due to the mirror symmetry of the model, the amplitudes for incidence from the right are
the same as those from the left. Note that these expressions hold for any energy
E>0
E≠V0
. If
E=V0
, then
k1=0
E < V0[edit]
                    Transmission probability through a finite potential barrier for
2mV0a/ℏ
The surprising result is that for energies less than the barrier height,
E<V0
T=|t|2=11+V02sinh2(k1a)4E(V0−E)
k1=2m(V0−E)/ℏ2
k0
1/k1
      . If the barrier is much wider than this decay length, the left and right part are
virtually independent and tunneling as a consequence is suppressed.
E > V0[edit]
In this case
T=|t|2=11+V02sin2(k1a)4E(E−V0),
where
k1=2m(E−V0)/ℏ2
Equally surprising is that for energies larger than the barrier height,
E>V0
, the particle may be reflected from the barrier with a non-zero probability
     R=|r|2=1−T.
The transmission and reflection probabilities are in fact oscillating with
k1a
T=1
R=0
E≫V0
k1a=nπ
, where
n=1,2,…
E/V0=1.2
               and 1.8 in the above figure). Note that the probabilities and amplitudes as
written are for any energy (above/below) the barrier height.
E = V0[edit]
E=V0
                  [1]
             is
T=11+ma2V0/2ℏ2.
This expression can be obtained by calculating the transmission coefficient from the
constants stated above as for the other cases or by taking the limit of
as
approaches
V0
x=EV0
f(x)
     f(x)=sinh(v01−x)1−x
In the last equation
v0
is defined as follows:
v0=2mV0a2ℏ2
E<V0
T(x)=11+f(x)24x
f(x)
T(x)
as
limx→1T(x)=limx→111+f(x)24x=11+v024
v0
   in the evaluated value for the limit, the above expression for T is successfully
reproduced.
   . Often the surfaces of such materials are covered with oxide layers or are not ideal
for other reasons. This thin, non-conducting layer may then be modeled by a barrier
potential as above. Electrons may then tunnel from one material to the other giving rise
to a current.
The operation of a scanning tunneling microscope (STM) relies on this tunneling effect.
In that case, the barrier is due to the gap between the tip of the STM and the underlying
object. Since the tunnel current depends exponentially on the barrier width, this device
is extremely sensitive to height variations on the examined sample.
Ψ(x,y,z)=ψ(x)ϕ(y,z)
For another, related model of a barrier, see Delta potential barrier (QM), which can be
regarded as a special case of the finite potential barrier. All results from this article
immediately apply to the delta potential barrier by taking the limits
V0→∞,a→0
while keeping
V0a=λ
constant.
See also[edit]
       ● Morse/Long-range potential
       ● Step potential
       ● Finite potential well
References[edit]
            ●   ^ McQuarrie DA, Simon JD (1997). Physical Chemistry - A molecular Approach
                (1st ed.). University Science Books. ISBN 978-0935702996.
       ● Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.).
         Prentice Hall. ISBN 0-13-111892-7.
       ● Cohen-Tannoudji, Claude; Diu, Bernard; Laloë, Franck; et al. (1996).
         Quantum mechanics. transl. from the French by Susan Reid Hemley. Wiley-
         Interscience: Wiley. pp. 231–233. ISBN 978-0-471-56952-7.
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Categories:
   Quantum models
   Scattering theory
   Schrödinger equation
   Quantum mechanical potentials
   This page was last edited on 22 April 2024, at 15:36 (UTC).
   Text is available under the Creative Commons Attribution-ShareAlike License 4.0; additional terms may apply.
       By using this site, you agre
Rectangular
potential barrier
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From Wikipedia, the free encyclopedia
Although classically a particle behaving as a point mass would be reflected if its energy
is less than
V0
Calculation[edit]
                             Scattering at a finite potential barrier of height
V0
         . The amplitudes and direction of left and right moving waves are indicated. In red, those
               waves used for the derivation of the reflection and transmission amplitude.
E>V0
ψ(x)
reads
H^ψ(x)=[−ℏ22md2dx2+V(x)]ψ(x)=Eψ(x)
where
H^
  is the Hamiltonian,
ℏ
is the mass,
V(x)=V0[Θ(x)−Θ(x−a)]
V0>0
and width
Θ(x)=0,x<0;Θ(x)=1,x>0
x=0
         and
x=a
position without changing the results. The first term in the Hamiltonian,
−ℏ22md2dx2ψ
x<0,0<x<a,x>a
E>V0
     {ψL(x)=Areik0x+Ale−ik0xx<0ψC(x)=Breik1x+Ble−ik1x0<x<aψR(x)=Crei
     k0x+Cle−ik0xx>a
     {k0=2mE/ℏ2x<0orx>ak1=2m(E−V0)/ℏ20<x<a.
The index
r/l
on the coefficients
and
  denotes the direction of the velocity vector. Note that, if the energy of the particle is
below the barrier height,
k1
   becomes imaginary and the wave function is exponentially decaying within the
barrier. Nevertheless, we keep the notation
r/l
even though the waves are not propagating anymore in this case. Here we assumed
E≠V0
. The case
E=V0
is treated below.
The coefficients
A,B,C
x=0
and
x=a
ψL(0)=ψC(0)dψLdx|x=0=dψCdx|x=0ψC(a)=ψR(a)dψCdx|x=a=dψRdx|x=a.
Inserting the wave functions, the boundary conditions give the following restrictions on
the coefficients
Ar+Al=Br+Bl
ik0(Ar−Al)=ik1(Br−Bl)
   Breiak1+Ble−iak1=Creiak0+Cle−iak0
     ik1(Breiak1−Ble−iak1)=ik0(Creiak0−Cle−iak0).
V0
E<V0
To study the quantum case, consider the following situation: a particle incident on the
barrier from the left side (
Ar
). It may be reflected (
Al
) or transmitted (
Cr
     ).
To find the amplitudes for reflection and transmission for incidence from the left, we put
in the above equations
Ar=1
(incoming particle),
Al=r
(reflection),
Cl=0
Cr=t
Bl,Br
and
         t=4k0k1e−ia(k0−k1)(k0+k1)2−e2iak1(k0−k1)2
   r=(k02−k12)sin(ak1)2ik0k1cos(ak1)+(k02+k12)sin(ak1).
Due to the mirror symmetry of the model, the amplitudes for incidence from the right are
the same as those from the left. Note that these expressions hold for any energy
E>0
E≠V0
Categories:
   Quantum models
   Scattering theory
   Schrödinger equation
   Quantum mechanical potentials
   This page was last edited on 22 April 2024, at 15:36 (UTC).
   Text is available under the Creative Commons Attribution-ShareAlike License 4.0; additional terms may apply.
       By using this site, you agre
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