Today: more about “When is
small ‘small enough’ for QM”?
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Last class: Energy levels for electron in short wire
Step 1 (solving Schr. eqn): Figure out V(x)
What is V(x)
of this wire?
Look at energy level spacing compared to thermal energy,
kBT= ~25 meV at room temp. kB= Boltzmann’s constant
L
0
E
E
Use time independ. Schrod. eq.
?
Want this to be
~25 meV. What L
do we need?
h 2 ∂ 2ψ ( x)
+ V ( x)ψ ( x) = Eψ ( x)
−
2m ∂x 2
Step 1: Figure out V(x
Then: figure out how to solve, and
what solutions mean physically.
x
Depth of the potential = Energy needed to remove
the electron from the metal = work function Φ.
(Φ ~ 4.7 eV for copper)
Remember: for atoms L very
small (~aB); spacing very
large (several eV)
Q1
4.7 eV
0 eV
mathematically
V(x) = 4.7 eV for x<0 and x>L
V(x) = 0 eV for 0>x<L
0
L x
Reasoning to simplify how to solve.
Electron energy not much more than ~kT=0.025 eV.
Where is electron likely to be?
What is the chance it will be outside of the well?
A. zero chance
B. very small chance (<50% chance)
C. 50:50 chance
D. very likely (>50% chance)
E. 100% chance it will be outside of the well
Q2
4.7 eV
mathematically
V(x) = 4.7 eV for x<0 and x>L
V(x) = 0 eV for 0<x<L
0 eV
0
L x
What does that say about boundary condition on ψ(x) ?
A. ψ(x) is about the same everywhere
B. ψ(x)~0 everywhere, except for 0<x<L
C. ψ(x)=0 everywhere, except for 0<x<L
D. ψ(x)~0 for 0<x<L, and ψ(x)=0 everywhere else.
E. Can’t tell anything yet. Need to find ψ(x) first.
Q3
Simplification: What if the potential would be infinity
outside the wire? (Very, very large work function)
∞
Approximation: the infinite square well
mathematically
V(x) = ∞ for x<0 and x>L
V(x) = 0 eV for 0>x<L
0
L
What can we say about ψ(x) in this situation?
A. ψ(x) is about the same everywhere
B. ψ(x) ~0 everywhere
C. ψ(x) ~0 everywhere, except for 0<x<L
D. ψ(x) = 0 everywhere, except for 0<x<L
E. Can’t tell anything yet. Need to find ψ(x) first.
V(x)
0 eV
So, a clever physicist just has
to solve (for 0<x<L)
h 2 ∂ 2ψ ( x)
= Eψ ( x)
−
2m ∂x 2
V(x)
Energy
with boundary conditions,
ψ(0)=ψ(L) =0
0
0
L
Good Approximation:
“Infinite square well” or “rigid box”
Electrons never get out of wire
ψ(x<0 or x>L) =0.
x (OK when Energy << work function)
0
0
L
NOTE:
Book uses “rigid box” for “infinite square well”
The “infinite square well”
x<0, V(x) = ∞
x> L, V(x) = ∞
0<x<L, V(x) = 0
Exact Potential Energy curve (V)
“Finite square well” or “non-rigid box”
small chance electrons get out of wire
ψ(x<0 or x>L)~0, but not exactly 0!
V(x)
∞
0 eV
∞
−
0
L
h 2 ∂ 2ψ ( x)
= Eψ ( x)
2m ∂x 2
functional form of solution: ψ ( x ) =
A cos(kx ) + B sin(kx )
Apply boundary conditions:
x=0 ?
x=L ψ (0) = A
ψ ( L) = B sin(kL) = 0
A=0
why no ‘0’?
⇒? kL=nπ (n=1,2,3,4 J)
k=nπ/L
2π
2L
k=
λ=
λ
n
x
What is the momentum p?
2
1
p = hk = h (nπ / L)
Wow!! Electron has non-zero momentum even in ground state!
p = hk = h (nπ / L)
λ=
2L
n
−
What is E?
a. can be any value (not quantized).
π h
2 2
nπ h
b.
c.
2
nmL
2mL2
n 2π 2h
nπ 2 h
+ PE e.
d.
mL
mL
2
2 2
Does this L dependence make sense?
What value of L when E2 - E1 = kT?
( when motion of e’s depends on wire size)
You should check,
I estimate L ~6.4nm (~50 atoms)
Why bother? Remember Moore’s law?
2009: 32nm, 2011: 22nm, 2013: 16nm,
2015: 11nm, 2017: 8nm, 2019: 6nm
h 2 ∂ 2ψ ( x)
= Eψ ( x)
2m ∂x 2
E quantized by B. C.’s
p 2 n 2π 2 h 2
En =
=
2m
2mL2
Finishing the story: Solving for everything there is to
know about an electron in a small metallic object (flat V(x) with
very high walls).
Ψ( x, t ) = ψ ( x )φ (t ) = B sin(
nπx −iEnt / h
)e
L
This fulfills the Schr. eqn. and the boundary conditions. But
we still need toJ normalize the wavefunction!
Probability of finding electron between -∞ and ∞ must be 1.
∫
∞
−∞
L
L
0
0
| Ψ ( x, t ) |2 dx = ∫ Ψ *Ψdx = ∫ B 2 sin 2 (nπx / L)dx = 1
B=
2
L
(Do in Homework)
Ψ( x, t ) = ψ ( x )φ (t ) =
nπx −iEnt / h
2
sin(
)e
n=1,2,3,J.
L
L
Q4
Quantized: kn= nπ/L
Quantized En :
Results:
Ψ ( x, t ) = ψ ( x)φ (t ) =
nπx −iEt / h
2
sin(
)e
L
L
Real(Ψ)
En = n 2
π 2h 2
2mL2
= n 2 E1
n=1
0
ψ
nπx −iEnt / h
2
sin(
)e
L
L
Quantized: En = n
What you expect classically:
Electron can have any
energy
Electron is localized
If it has KE, it bounces
back and forth between the
walls
nπx −iEnt / h
2
sin(
)e
L
L
Quantized: k=nπ/L
Quantized:
π 2h 2
2mL2
= n 2 E1
L
2
π 2h2
2mL2
How does probability of finding electron close to L/2 in n =3 state
compare to probability at L/2 for when in n=2 state?
a. much more likely for n=3.
b. equal prob. for both n = 2 and 3.
c. much more likely for n=2
Answers to clicker questions
Quantized: kn = nπ/L
n=2
Ψ ( x, t ) =
En = n 2
What is potential energy of
electron in the lowest energy state
(n=1)?
a. 2E1
b. E1
c. ½ E1
d. 0
e. ∞
Ψ ( x, t ) =
Q5
= n E1
2
What you get quantum
mechanically:
Electron can only have specific
energies. (quantized)
Electron is delocalized
J spread out between 0 and L
Electron does not “bounce”
from one wall to the other.
Q1: B. 0.025 eV<< 4.7eV. So very small chance (e-4.7/.025) an electron could
have enough energy get out.
Q2: B
Q3: D
Q4: D
Q5: A