WHY it happens: A particle is described by a wavefunction ψ(x). Schrödinger's equation forces ψ and its slope ψ′ to be continuous everywhere. Inside a barrier where E<V0, ψ cannot oscillate — but it also cannot vanish abruptly. The only continuous, normalizable behaviour left is an exponential decay. Decay ≠ zero, so some amplitude reaches the other side.
Why? This is the energy eigenvalue equation; solving it gives the spatial shape of the matter wave.
Region I (x<0, V=0):ψ′′=−ℏ22mEψ≡−k2ψ,k=ℏ2mE
Solution: ψI=Aeikx+Be−ikx (incoming + reflected waves).
Region II (0<x<L, V=V0):ψ′′=+ℏ22m(V0−E)ψ≡+κ2ψ,κ=ℏ2m(V0−E)Why a +sign? Because V0>E, the bracket is positive, so the equation gives real exponentials, not oscillations:
ψII=Ce−κx+De+κx
Region III (x>L, V=0):ψIII=FeikxWhy only e+ikx? Nothing comes back from +∞; only a transmitted wave moves rightward.
Imagine throwing a tennis ball at a wall. Normally it bounces back. But picture the ball as a fuzzy cloud instead of a hard dot. When the cloud hits the wall, most of it bounces, but a tiny faint bit of the fog seeps through the wall and reappears faintly on the other side. If the wall is thin, a little fog gets across — and once in a while the ball is found there! Thin walls and light balls = more leaking. That seepage is tunneling.
Recall Active-recall checklist
Why is ψ exponential (not oscillatory) inside the barrier?
What is κ and what controls it?
Where does the factor of 2 in e−2κL come from?
Does the emerging particle have more energy? Less? Same?
A particle passing through a barrier with V0>E, impossible classically, allowed because ψ is continuous and nonzero (exponentially decaying) inside the forbidden region.
Why is ψ exponential inside the barrier?
Because V0>E makes Schrödinger's equation read ψ′′=+κ2ψ, whose real solutions are e±κx, not oscillations.
Define the decay constant κ.
κ=2m(V0−E)/ℏ, the rate at which ψ decays inside the barrier.
Approximate transmission coefficient?
T≈Ge−2κL with G=16E(V0−E)/V02.
Exact transmission coefficient?
T=[1+4E(V0−E)V02sinh2(κL)]−1.
Why the factor of 2 in e−2κL?
Amplitude decays as e−κL; probability ∝∣ψ∣2 squares it to e−2κL.
Does a tunneling particle gain energy?
No — it emerges with the same energy E<V0; energy is conserved.
How does T depend on mass?
Through κ∝m; heavier particles tunnel exponentially less, so macroscopic objects effectively never tunnel.
Dekho, classical physics kehti hai ki agar particle ke paas barrier se kam energy hai (E<V0), to wo kabhi paar nahi kar sakta — wapas bounce ho jayega. Lekin quantum world mein particle ek wavefunctionψ hai, aur Schrödinger equation force karti hai ki ψ aur uska slope har jagah continuous ho. Barrier ke andar, kyunki V0>E, ψ oscillate nahi karta — wo exponentially decay karta hai (e−κx). Decay ka matlab zero nahi hota! Thoda sa amplitude doosri side tak pohanch jata hai. Yahi tunneling hai.
Transmission coefficient T batata hai ki kitni probability se particle paar hota hai. Mota/important result hai T≈Ge−2κL, jahan κ=2m(V0−E)/ℏ. Yaad rakho: T exponential hai — yani barrier ki width L thodi si badhao to T bahut tezi se girta hai. Example mein dekha, L double karne par T ~170 guna kam ho gaya. Yeh sensitivity STM (Scanning Tunneling Microscope) ki super-resolution ka raaz hai.
Mass ka effect bhi samajh lo: κ∝m, to bhaari particle (jaise proton ya cricket ball) ka T practically zero ho jata hai — isliye macroscopic cheezein kabhi tunnel nahi karti. Halki cheez jaise electron asaani se tunnel karti hai. Aur ek important baat: tunneling mein energy conserve hoti hai — particle jis energy se aata hai, usi energy se nikalta hai, koi energy "udhaar" nahi leta. Bas wavefunction ka magic hai.
