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Tunneling: wave function in different regions

Short answer

Tunneling occurs when a particle passes through a potential barrier higher than its energy. In regions I and III (V=0), wavefunctions are oscillatory (e^(±jkx)); inside region II (barrier, V>E) they decay exponentially as e^(-αx). Transmission probability T ∝ e^(-2αa).

What is tunneling in quantum mechanics? With necessary mathematical expression, explain the nature of wave function in different regions in case of tunneling.

Quantum tunneling is the passing of a particle through a potential barrier whose height V0V_0 is greater than the particle's energy EE. In classical mechanics this is impossible, but in quantum mechanics the wave function does not become zero inside the barrier. It decays, so a small part of it comes out on the other side and there is a finite chance of finding the particle there.

Barrier and regions

 V(x)
  ^
  |            +----------+   V = V0
  |            |          |
  |  - - - - - |- - - - - |- - - - -  E < V0
  |  Region I  | Region II| Region III
  |  V = 0     | V = V0   | V = 0
--+------------+----------+-----------> x
               0          a

The time-independent Schrödinger equation is

d2ψdx2+2mℏ2(E−V)ψ=0\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E - V)\psi = 0

Region I (x<0x < 0, V=0V = 0)

ψ1=Aejkx+Be−jkx,k=2mEℏ\psi_1 = A e^{jkx} + B e^{-jkx}, \qquad k = \frac{\sqrt{2mE}}{\hbar}

AejkxA e^{jkx} is the incident wave and Be−jkxB e^{-jkx} the reflected wave. The wave function is oscillatory.

Region II (0≤x≤a0 \le x \le a, V=V0>EV = V_0 > E)

d2ψ2dx2−α2ψ2=0,α=2m(V0−E)ℏ\frac{d^2\psi_2}{dx^2} - \alpha^2\psi_2 = 0, \qquad \alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar} ψ2=Ceαx+De−αx\psi_2 = C e^{\alpha x} + D e^{-\alpha x}

The solution is exponential, not oscillatory. For a wide barrier the decaying term dominates, so ψ\psi falls off roughly as e−αxe^{-\alpha x} inside the barrier.

Region III (x>ax > a, V=0V = 0)

ψ3=Fejkx\psi_3 = F e^{jkx}

Only a transmitted wave travels to the right (there is nothing to reflect it back). It is oscillatory with the same kk (same energy) but a much smaller amplitude.

Boundary conditions

ψ\psi and dψ/dxd\psi/dx must be continuous at x=0x = 0 and x=ax = a. These four conditions fix B,C,D,FB, C, D, F in terms of AA.

Transmission probability

T=∣F∣2∣A∣2≈T0 e−2αa,T0=16EV0(1−EV0)T = \frac{|F|^2}{|A|^2} \approx T_0\, e^{-2\alpha a}, \qquad T_0 = 16\frac{E}{V_0}\left(1 - \frac{E}{V_0}\right)

(valid when αa≫1\alpha a \gg 1). TT falls very quickly as the barrier width aa, the mass mm or (V0−E)(V_0 - E) increases.

 psi(x)
     Region I       Region II     Region III
  /\    /\    /\ |\            |
 /  \  /  \  /  \| \           |
/    \/    \/    |  \__        |/\/\/\/\/\/\
                 |     ``--..__|  (small amp.)
                 0             a

Examples

Alpha decay of nuclei, the tunnel diode, the scanning tunneling microscope (STM), and conduction through thin oxide layers between metal contacts all rely on tunneling.

✓ Answer checked by Claude Opus 5.5 (high effort)

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