Chapter 4: One-Dimensional Potentials The Potential Step a) When a neutron with an external kinetic energy K enters a nucleus, it experiences a potential b) While a charged particle moves along the axis of two cylindrical electrodes held at different voltages, its potential energy changes very rapidly when passing from one to the other. Potential energy function can be approximated by a step potential. For the step potential, the x-axis breaks up into two regions 0 x V 0 x 0 ) x ( V 0 Since the potential is time-independent, problem is to solve the time-independent Schrodinger equation. There are two cases, 0 V E and 0 V E For 0 V E : Schrodinger equations for both regions are 0 x for (x) Eu (x) u V dx (x) u d 2m - 0 x for (x) Eu dx (x) u d 2m - II II 0 2 II 2 2 I 2 I 2 2
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Chapter 4: One-Dimensional Potentials The Potential Step
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Chapter 4: One-Dimensional Potentials
The Potential Step
a) When a neutron with an external kinetic energy K enters a nucleus, it experiences
a potential
b) While a charged particle moves along the axis of two cylindrical electrodes held at
different voltages, its potential energy changes very rapidly when passing from
one to the other. Potential energy function can be approximated by a step potential.
For the step potential, the x-axis breaks up into two regions
0x V
0x 0)x(V
0
Since the potential is time-independent, problem is to solve the time-independent
Schrodinger equation.
There are two cases, 0VE and 0VE
For 0VE :
Schrodinger equations for both regions are
0xfor (x)Eu(x)uVdx
(x)ud
2m-
0xfor (x)Eudx
(x)ud
2m-
IIII02
II
22
I2
I
22
The physically acceptable solutions are
2ikx -ikxI
I I2 2
2iqx iqx0II
II II2 2
d u (x) 2mEu (x) 0 u (x) e e
dx
2m(E-V )d u (x)u (x) 0 u (x) e e
dx
R
T D
where 2
22
1
2mEkk
and
2
022
2
)V-2m(Eqk
At the point x=0, the continuity property of the wave functions requires
I x 0 I x 0
I II0 0
u (x)| u (x)| 1 R T
du (x) du (x)| | (1 R) T
dx dxx x ik iq
E
V11
E
V1-1
qk
q-k R
qk
2k T
0
0
• Reflection probability 2|R| is not zero for 0VE . Only tends to zero in high energy
limit, 0VE (correspondence principle again). Notice that reflection and transmission
probabilities satisfy the relation 2 2| | | | 1R T . This is a simple example of a scattering
problem.
For 0VE :
Schrodinger equations for both regions are
0xfor (x)Eu(x)uVdx
(x)ud
2m-
0xfor (x)Eudx
(x)ud
2m-
IIII02
II
22
I2
I
22
The physically acceptable solutions are
2ikx -ikxI
I I2 2
2q'x0II
II II2 2
d u (x) 2mEu (x) 0 u (x) e e
dx
2m(V )d u (x)u (x) 0 u (x) e
dx
R
EC
-q'xeT
where 2
22
1
2mEkk
and
2
022
2
)E2m(V'qk
At the point x=0, the continuity property of the wave functions requires
I x 0 I x 0
I II0 0
u (x)| u (x)| 1 R T
du (x) du (x)| | (1 R) 'T
dx dxx x ik q
q'ik
iq'-k Rand
q'ik
2k T
x-q'eT illustrates an important difference between classical and quantum physics. While
none of the particles can be found classically in the region x>0, quantum mechanically
there is a nonzero probability density that requires the wave function penetrates this
classically forbidden region
x2q'-
22
2x2q'-22
II e'qk
k4e|T||)x(u|)x(P
The penetration distance xΔ at which the probability density is
E)2m(V22q'
1Δxe|T|
e
1e|T|
)xP(xe
1Δx)xP(x
0
x2q'-2Δx)(x2q'-2
00
00
In classical limit, the term E)2m(V0 is so large compared to 2 , thus xΔ is
immeasurably small.
The probability density 2II |)x(u| for a group wave function incident on the step with
0VE . The group moves up to the step, penetrates slightly into the classically
forbidden region, and then is completely reflected from the step.
• Some tunneling of particles into classically forbidden region even for energies below
step height.
• Tunneling depth depends on energy difference EV0 , E)2m(V22q'
1Δx
0
• But no transmitted particle flux, 100% reflection, like classical case.
The Potential Well
The square well potential is often used in quantum mechanics to represent a situation in
which a particle moves in a restricted region of space under the influence of forces.
III regiona x 0
II regiona xa- V
I region-a x 0
)x(V 0
a) The motion of a neutron in a nucleus can be approximated by assuming that the
particle is in a square well potential with a depth about 50 MeV.
b) A square well potential results from superimposing the potential acting on a
conducting electron in a metal.
The scattering solutions ( E > 0 ):
When E>0 the particle is unconfined and this case corresponds to the scattering problem.
There are unbound states with a continuum range of energies.
2ikx -ikxI
I I2 2
2iqx -iqx0II
II II2 2
2ikx -ikxIII
III III2 2
d u (x) 2mEu (x) 0 u (x) e e
dx
2m(E V )d u (x)u (x) 0 u (x) e e
dx
d u (x) 2mEu (x) 0 u (x) e e
dx
R
A B
T G
At the point ax , the continuity property of the wave functions requires
-ika ika -iqa iqa
I x -a I x -a
-ika ika -iqa iqaI II
u (x)| u (x)| e e e e
du (x) du (x)| | (e e ) ( e e )
dx dxx a x a
R A B
ik R iq A B
At the point ax ,
iqa -iqa ika
II x a III x a
iqa -iqa ikaII III
u (x)| u (x)| e e e
du (x) du (x)| | ( e e ) e
dx dxx a x a
A B T
iq A B ikT
As a result,
2-2ika
22 220
0
2 2-2ika
2 2
2 1e
2 cos 2 ( )sin 21 sin (2 )
4 ( )
( )sin 2e
2 cos 2 ( )sin 2
kqT T
Vkq qa i q k qaqa
E E V
q k qaR i
kq qa i q k qa
Note:
a) If 0Rkq2)kq(VE 220 , no reflection
b) As 0T0k0E 2
c) 1,2,3,...n wherenπqa20R0qa2sin
12
0 En-VE This is called the transmission resonance (Ramsaver-Townsend)
Bound states in a potential well ( E<0 ):
The case E<0 corresponds to a particle which is confined in a bound state. When well is
deep enough, there are bound states with discrete energies.
ax ; eCeC(x)u0(x)u2mE
dx
(x)ud
axa- ;Bsinqx Acosqx(x)u0(x)u|)E|2m(V
dx
(x)ud
-ax ; eCeC(x)u0(x)u|E|2m
dx
(x)ud
αx-'2
αx2IIIIII22
III2
IIII2
0
2
II2
αx-'1
αx1II22
I2
The physically acceptable solutions:
ax ; Te(x)u
axa- ;Bsinqx Acosqx(x)u
-ax ; eC(x)u
αxIII
II
αx1I
At the point ax , the continuity property of the wave functions requires
- a
I x -a I x -a 1
- aI II1
u (x)| u (x)| e cos sin
du (x) du (x)| | e sin cos
dx dxx a x a
C A qa B qa
C qA qa qB qa
At the point ax ,
- a
II x a III x a 2
- aII III2
u (x)| u (x)| cos sin e
du (x) du (x)| | sin cos e
dx dxx a x a
A qa B qa C
qA qa qB qa C
As a result,
qcotgqa α Acosqx (x)u solutions even : 0 Band 0A 1)