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Quantum Physics Lecture 8
Applications of Steady state Schroedinger Equation
Box of more than one dimension
Harmonic oscillator
Particle meeting a potential step
Waves/particles in a box of >1 dimension
b
Consider 2-D box
Potential U = 0 between x = 0 and x = a
And between y = 0 and y = b
U infinite elsewhere
( )
y
U=0
0
x
() ( )
Wavefunction Ψ expressed as Ψ x.y = f x g y
- Separation of variables
Steady state Schrödinger Equation inside box, where U = 0
( )
( )
( )
() ( )
2
2
 2 ⎛ ∂ Ψ x.y ∂ Ψ x.y ⎞
−
+
⎜
⎟ = EΨ x.y
2
2
2m ⎝ ∂x
∂y
⎠
Put in
Ψ x.y = f x g y
( )
a
( Recall HΨ = EΨ )
- noting partial differentials!
Waves/particles in a 2-D box (cont.)
()
( )
∂2 f x
∂2 g y ⎞
2 ⎛
−
+ f x
⎜g y
⎟=Ef x g y
2m ⎝
∂x 2
∂y 2 ⎠
( )
()
() ( )
Thus
Only one term is x dependent, and it equals a constant
()
∂2 f x
∂x
f x
2
()
= −C
So
( ) = −Cf
∂2 f x
∂x
2
( x)
()
( )⎞
()
( )
⎛ ∂2 f x
∂2 g y
⎜
 2 ⎜ ∂x 2
∂y 2
−
+
2m ⎜ f x
g y
⎜
⎝
⎟
⎟=E
⎟
⎟
⎠
Which we have seen before…
2 2
nπ x
n
Solutions, using boundary conditions, are f x = Asin
with C = π2
a
a
()
mπ y
Similarly, for y dependence g y = Bsin
b
( )
Hence the energy levels in the box are
En,m
 2 ⎛ n2π 2 m2π 2 ⎞
=
+ 2 ⎟
⎜
2
2m ⎝ a
b ⎠
With TWO quantum numbers n,m needed to specify the state
Waves/particles in a 2-D box (cont.)
Ψ is specified by the quantum numbers n & m
There are as many states as there are possible
n,m combinations
(N.B. n & m are positive)
Two distinct wave functions are DEGENERATE
if they have the same energy.
e.g. the states 1,3 and 3,1 are degenerate if a = b
If a/b is irrational there are no degeneracies
Readily extended to 3-D….
Useful, especially when filling box with more
than one particle. C.f. black body cavity
Examples:
2,3 and 3,2
wavefunctions
Harmonic Oscillator
Examples: mass on spring, diatomic molecule…
2 Hooke’s Law: “restoring force” F = -kx
d x
−kx = m 2
dt
d 2x k
+ x=0
2
m
dt
x = Acos ω t
Where ω =
k
m
Potential energy U = (1/2)kx2 (potential well, parabolic)
Expect (1) quantised energy, (2) Emin ≠ 0
(3) particle outside the classical limits A
Apply SSSE to Harmonic Oscillator
d ψ 2m
2
1
+
E
−
kx
ψ
=
0
2
2
2

dx
2
d ψ ⎛ 2mE km 2 ⎞
or
+ ⎜ 2 − 2 x ⎟ψ = 0
2
⎠
dx ⎝ 

2
(
)
1
⎛ km ⎞ 2
write y = ⎜
⎟ x
⎝  ⎠
⎛ km ⎞ 2
and dy = ⎜
⎟ dx
⎝  ⎠
2mE
2
2
dψ
2E
km 2E
2E
2E
2

+
α
−
y
ψ
=
0
α
=
=
=
(
)
=
2
km
dy
 mk hω

Book typo!
2
ν
Results for Harmonic Oscillator
Solution requires:
α = 2n + 1 for n = 0,1,2, 3...
(
d 2ψ
2
+
α
−
y
ψ =0
(
)
2
dy
)
⎛
νω
h
ων
1⎞ 1
h
+ 1)+=1⎜)n=+ n⎟+ω hν
(2n
EnEn== 2αα== ( 2n
2
2
2
2
2⎠
⎝
EoE ==112ω
hν a.k.a. “zero point energy
0
2
Recall Planck’s assumption & blackbody formula (Lecture 6):
Oscillator energies assumed En = nω
Later, found additional detail & statistics, Cv etc. but…
Right ideas on quantisation: “fortunate guesswork!”
Wavefunctions of Harmonic Oscillator
Comparing with infinite square well case, ψ has approximately same shape except:
(1) width of well is changing
(2) ψ extends beyond classical limit
(3) amplitude increases at well edge
Look at (2) and (3) via the probability
density…..
Probability Density of Harmonic Oscillator
(1) Finite probability of particle
outside the classical limits
(2) the quantum picture only
approaches the classical picture at
large n values
(classical - probability maximum
at extremities of oscillation - slower)
- Correspondence Principle Footnote: compare square well [En ∝ n2] and harmonic oscillator [En ∝ (n + 1/2)]
and Bohr H atom [En ∝ -1/n2] for energies. Differing shapes of the potential wells. Particle meeting step potential
2 types:
“step-up”
➙
U=Uo
U=Uo
U=0
“step-down”
➙
U=0
Particle direction “left-to-right”; potential flat apart from step region;
Particle energy E and step size Uo; 3 distinct situations:
“up” E < Uo “up” E > Uo
“down” E > Uo Find elements of “propagating” and “decaying” wavefunctions;
Solutions may not be standing waves (as previously)
cannot draw (∵ complex ψ) but can always draw |ψ |2 (real).
Apply SSSE to step-up potential (E < Uo)
Classically, particle is reflected!
d 2ψ 2m
+
2 E −U ψ = 0
2

dx
(
)
For region I, solution is complex exponentials:
ψ I = Aexp ( ik1 x ) + Bexp (−ik1 x )
2mE
where k1 =

For region II, solutions are real exponentials:
ψ II = C exp ( k2 x ) + D exp (−k2 x )
Also, require C = 0, to keep ψ finite at large +ve x:
ψ II = D exp( −k2 x )
where k2 =
(
2m U o − E

)
Apply boundary matching
ψII = ψI at x = 0 Dexp(0) = Aexp(0) + Bexp(0)
D = A + B
dψII/dx = dψI/dx at x = 0
-k2Dexp(0) = ik1Aexp(0) - ik1Bexp(0)
(ik2/k1)D = A - B
Convenient to write A and B in terms of D:
ik2 ⎞
ik2 ⎞
⎛
⎛
D
D
ψ I = 2 ⎝ 1 + k ⎠ exp(ik1 x ) + 2 ⎝ 1 − k ⎠ exp (−ik1 x )
1
1
meaning?
incident particle(s)
ψ II = D exp( −k2 x )
reflected particle(s)
particle(s) probability decays into wall!
Particle reflection
Reflection Coefficient R = B*B/A*A
*
⎛1 − i k2 ⎞ ⎛1 − i k2 ⎞ ⎛ 1 + i k2 ⎞ ⎛ 1− i k2 ⎞
⎝
k1 ⎠ ⎝
k1 ⎠ ⎝
k1 ⎠ ⎝
k1 ⎠
R=
=
=1
*
⎛1 + i k2 ⎞ ⎛1 + i k2 ⎞ ⎛ 1 − i k2 ⎞ ⎛ 1+ i k2 ⎞
k1 ⎠ ⎝
k1 ⎠
⎝
k1 ⎠ ⎝
k1 ⎠ ⎝
As expected, in agreement with classical picture!
Exercise: use [exp(iθ) = cosθ + isinθ] to show
k2 ⎞
⎛
ψ I = D cos(k1 x ) − D⎝ k ⎠ sin(k1 x )
1
can be generalised as sin(k1x + φ) i.e. standing wave! (combination of equal incident and reflected waves)
Plotting ψ and
ψ
2
|ψ |
|ψ |2 Because in this case ψI is a pure standing wave, it can still be
plotted; note how ψI matches onto ψII (red) As Uo increases, k2 increases (less penetration of wall) and ψI moves closer to simple sin(k1x). Plot of |ψ|2 always possible
Signature of perfect reflection (R = 1): minimum value of |ψ |2 = 0
Apply SSSE to step-up potential (E > Uo)
Classically: kinetic energy decrease and particle not reflected!
d 2ψ 2m
+
2 E −U ψ = 0
2

dx
(
)
For region I, solution is complex exponentials as before:
ψ I = Aexp ( ik1 x ) + Bexp (−ik1 x )
where
k1 = 2mE

For region II, solution is also complex exponentials:
ψ II = C exp ( ik2 x ) + D exp (−ik2 x )
where k2 =
(
2m E −U o
)

Require D = 0, no negative-going wave for x > 0, as all particles are incident in +ve x
direction! Not so for x < 0 , there is some reflection!
ψ II = Cexp(ik 2 x )
Apply boundary matching
ψII = ψI at x = 0 Cexp(0) = Aexp(0) + Bexp(0)
C = A + B
dψII/dx = dψI/dx at x = 0
ik2Cexp(0) = ik1Aexp(0) - ik1Bexp(0)
(k2/k1)C = A - B
Convenient to write B and C in terms of A:
k1 − k 2
ψ I = A exp(ik1 x ) + A
exp(−ik1 x )
k1 + k 2
meaning?
incident particle(s) reflected particle(s)
2k1
(ik 2x ) particle(s) with reduced energy!
ψ II = A exp
k1 + k2
Particle reflection
Recall Reflection Coefficient R = B*B/A*A
2
⎛ 1 − k2 ⎞
2
⎛ k1 − k2 ⎞
k1
⎜
⎟
R=
= ⎜⎜
⎟
k
⎟
⎝ k1 + k2 ⎠
2
⎝1 + k ⎠
1
Recall the equations for k1 and k2
k2
k1 =
2m( E − U0 )
⎛
1 − 1 − U0 E ⎞
R = ⎜⎜
⎟
⎟
U
0
⎝1 + 1 − E ⎠
for E > Uo
U
= 1− 0 E
2mE
2
and R = 1 for E < Uo Particle meeting step-down potential
compare:
“step-up”
with
“step-down”
Reverse situations, simply exchange k1 and k2:
ψ I = A exp(ik 2 x ) + Bexp( −ik2 x )
ψ II = Cexp(ik1 x)
Proceed to determine matching etc.
Significant point is that some reflection occurs here too;
Origin of reflectivity is “change in potential U”
For quantum ‘lemmings’, some reflect from cliff edge….
General conclusions
Step-up, where E < Uo
- total reflection
- but can exist in wall!
Step-up, where E > Uo
- decreased kinetic energy
- and partial reflection!
Step-down, where E > Uo
- increased kinetic energy
- also partial reflection!
“solve, match, R and plot”