Download Lecture 15: Simple problems in 1D and Probability

Lecture 15: Simple problems in 1D
and Probability Current I
Phy851 Fall 2009
Continuity Theorem
From previous Lecture:
Theorem:
• the wavefunction and its first derivative
must be everywhere continuous.
– Exception: where there is a δ(x-x0) or
δ’(x-x0) in the potential.
• δ(x-x0) potential  discontinuity in ψ’(x) at
x=x0
• δ’(x-x0) potential  discontinuity in ψ(x) at
x=x0
Solution to the Step Potential
Scattering Problem
•
Assuming an incoming flux from the left only,
we make the ansatz:
ψ I ( x) = ei k1x + r e − i k1x
•
ψ II ( x) = t ei k2 x
As there is no δ or δ’ potential, we need to
impose two boundary conditions at x=0:
ψ I (0) = ψ II (0)
1+ r = t
(1)
ψ I′ (0) = ψ II′ (0)
ik1 (1 − r ) = ik 2t
(2)
Insert (1) into (2)
Collect r terms
together
Solve for r
Plug solutions into
(1) and solve for t
ik1 (1 − r ) = ik 2 (1 + r )
− i (k1 + k 2 )r = −i (k1 − k 2 )
k1 − k 2
r=
k1 + k 2
k1 − k 2
2k1
t = 1+
t=
k1 + k 2
k1 + k 2
Case I: Tunneling into the Barrier
• Consider the case where E < V0:
ψ I ( x) = ei k1x + r e − i k1x
ψII ( x) = t ei k2 x
V0
E
A classical particle would reflect from x=0
k1 =
2mE
:= k
h
Q: why did we
choose “+i”,
instead of “–i”?
A: If we had chosen
“-i”, solution would
‘blow up’ as x→∞.
− 2m(V0 − E )
2m(V0 − E )
k2 =
=i
:= iγ
h
h
ψ II ( x) = te −γ x
k −k
r= 1 2 →
k1 + k 2
k − iγ
r=
k + iγ
2k1
t=
→
k1 + k 2
2k
t=
k + iγ
Note that:
2
r =1
4k 2
t = 2
k +γ 2
That would describe a
particle at x =∞, but
not the particle we
are interested in
2
Q: What is physical meaning of |r|2 and |t|2 ?
Case II: Quantum Reflection
• Consider the case where E > V0:
E
ψ I ( x) = ei k1x + r e − i k1x
ψII ( x ) = t ei k2 x
V0
A classical particle would have 100% transmission
2mE
:= k
h
k1 =
2m(E − V0 )
2 MV0
2
2
k2 =
= k−
:
=
k
−
k
0
h
h2
k − k 2 − k02
k1 − k 2
r=
→r =
k1 + k 2
k + k 2 − k02
2k1
2k
t=
→t=
k1 + k 2
k + k 2 − k02
2
r =
Note:
2k 2 − k02 − 2k k 2 − k02
2
2
0
2
2k − k + 2k k − k
2
2
r + t ≠1
2
0
2
t =
Note that k2 < k1,
as it should
Quantum
particle has
non-zero
probability to
reflect!
4k 2
2k 2 − k02 + 2k k 2 − k02
So |r|2 and |t|2 are not reflection
and transmission probabilities !
Limiting case: Infinite Barrier
• Consider an infinitely high potential barrier:
∞
ψ I ( x) = ei k x + r e −i k x
ψ II ( x) = te − γ x
E
x
x=0
2m(V0 − E )
2m(∞ − E )
γ=
=
=∞
h
h
(
γ + ik )
r=−
= −1
(γ − ik )
2ik
t=−
=0
γ − ik
ψ I ( x) = sin( k x)
ψ II ( x) = 0
π phase-shift upon
reflection.
no tunneling
(up to a constant)
Wave function goes to
zero at infinite barrier
Bound states: Infinite Square Well Solution
• Extend this result to the infinite square well
I
II
III
ψ I ( x) = 0
E
ψ III ( x) = 0
0
From continuity
of ψ:
Most general
free-space
solution:
ψ II (0) = 0
L
x
ψ II ( L) = 0
ψ II (x) = a e i kx + b e−i kx = asin(kx) + bcos(kx)
Apply boundary
conditions:
€
ψ II (0) = 0 → b = 0
nπ
ψ II ( L) = 0 → k =
; n = 1,2,3,...
L
• Momentum and Energy are quantized by the
boundary conditions at 0 and L:
nπ
Bound States are
normalizable:
L
h 2 k n2 h 2π 2 2 ϕ ( x) = 2 sin  nπ x 


En =
=
n n
2
L  L 
2m 2mL
kn =
Probability Current
• From studying quantum reflection at a step
potential, we saw that |r|2+|t|2=1 is not always
true
–
–
–
–
Let R:= reflection probability
Let T:= transmission probability
Clearly we must have R +T = 1
So R = |r|2 and T = |t|2 must not always be correct
• The problem is that in this case the velocity in
region I is not the same as in II
• This suggests that we need to think in terms of
a probability current
• Derivation of probability current:
– Start from the probability density:
ρ ( x, t ) = ψ ( x, t )
2
– Consider a tiny region of length 2ε located a
position x:
P( x, t ) = ρ ( x, t )2ε
– Imagine currents are flowing through points x-ε
and x+ε: (A positive current flows from left to right)
Probability in
region can
increase or
decrease
dP( x, t )
= j( x − ε ) − j( x + ε )
dt
Current at -ε
Current at +ε
Continuity Equation
dP( x, t )
j( x − ε ) − j( x + ε ) =
dt
dρ ( x, t )2ε
j( x − ε ) − j( x + ε ) =
dt
j ( x − ε ) − j ( x + ε ) dρ ( x, t )
=
2ε
dt
d
d
−
j ( x, t ) = ρ ( x, t )
dx
dt
•
This is the standard continuity equation, valid for
any kind of fluid
•
For energy eigenstates (stationary states), we
need:
d
ρ ( x, t ) = 0 ⇒ ρ ( x, t ) = ρ ( x,0)
dt
d
j(x,t) = 0 ⇒ j(x,t) = j(x,0)
dt
•
This gives:
•
€ have spatially uniform current in steady
Must
state (of course j0 can be zero)
€
d
j(x,t) = 0 ⇒
dx
j(x,0) = j 0