Quantum Mechanics (8hrs) · Engineering Physics

Potential barrier

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Potential barrier

1. What a Potential Barrier Is

Consider 1D potential step/barrier:

V(x)=0forx<0(regionI),V=V0>0for0<x<a(regionII,barrierwidtha),V=0forx>a(regionIII)V(x)=0 for x<0 (region I), V=V_0>0 for 0<x<a (region II, barrier width a), V=0 for x>a (region III)

Particle energy EE incident from left. Classically: if E>V0E>V_0, transmit over; if E<V0E<V_0, reflect, cannot enter barrier. Quantum: wave penetrates, even E<V0E<V_0 has non-zero transmission (tunneling, next topic); E>V0E>V_0 has partial reflection even above barrier due to wave mismatch.

2. TISE in Three Regions

Let

k1=√(2mE)/ħk_1 =√(2mE)/ħ

in I/III (free),

k2=√(2m(E−V0))/ħk_2 =√(2m(E-V_0))/ħ

if E>V0E>V_0 (propagating inside), or

κ=√(2m(V0−E))/ħκ =√(2m(V_0-E))/ħ

imaginary k2=iκk_2=iκ if E<V0E<V_0 (evanescent, decaying).

General solutions:

  • I:
φI=Aeik1x+Be−ik1xφ_I = A e^{ik_1 x} + B e^{-ik_1 x}

(incident + reflected)

  • II:
φII=Ceik2x+De−ik2xφ_{II}=C e^{ik_2 x}+D e^{-ik_2 x}

if E>V0E>V_0, or

Ceκx+De−κxC e^{κx}+D e^{-κx}

decaying/growing if E<V0E<V_0 (physically keep decaying).

  • III:
φIII=Feik1xφ_{III}=F e^{ik_1 x}

(transmitted only, no incoming from +∞)

Apply continuity φφ and dφ/dxdφ/dx at x=0x=0 and x=ax=a (where VV jumps finite) ⇒ solve for B/AB/A (reflection amplitude rr) and F/AF/A (tt) ⇒ reflection

R=∣r∣2R=|r|²

, transmission

T=∣t∣2T=|t|²

with R+T=1R+T=1.

3. Results

  • E>V0E>V_0:
T=1/[1+(V02sin2(k2a))/(4E(E−V0))]T =1/[1+ (V_0² sin²(k_2 a))/(4E(E-V_0))]

— oscillates with aa; T=1T=1 at resonance when k2a=nπk_2 a=nπ (barrier transparent like Fabry-Perot). Even with enough energy, some reflection — unlike classical.

  • E<V0E<V_0: replace k2→iκk_2→iκ ⇒
T=1/[1+(V02sinh2(κa))/(4E(V0−E))]T =1/[1+ (V_0² sinh²(κ a))/(4E(V_0-E))]

⇒

T≈(16E(V0−E)/V02)e−2κaT≈ (16E(V_0-E)/V_0²) e^{-2κa}

for thick barrier κa≫1κa≫1 — exponential decay, not zero (next topic tunneling).

4. Worked Checks

(a) High barrier V0=2E, a=1 nm, electron E=1 eV: κ=√(2m(V0-E))/ħ=√(2·9.1e-31·1.6e-19)/1.05e-34≈5.1e9 m⁻¹ ⇒ κa≈5.1 ⇒ T≈ e^{-10.2}·factor≈3e-5 small but not zero ✓

(b) At resonance k2 a=π ⇒ T=1 even though V0≠0 ⇒ resonant tunneling.

(c) Classical would predict T=0 for E<V0, but quantum gives ~e^{-2κa}.

5. Exam Points

  • Sketch V(x)V(x) step/barrier — 3 marks, frequent.
  • Write TISE solutions in three regions with k1,k2,κk_1,k_2,κ — 5 marks, very frequent.
  • State continuity of φ,dφ/dxφ, dφ/dx at boundaries — 3 marks.
  • Give TT formula for E>V0E>V_0 and mention resonance — 5 marks, frequent.
  • Mention T≠0T≠0 for E<V0E<V_0 preview — 3 marks.
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