Wavefunction in Forbidden Region Equation, Derivation & Rearrangements

Q: What are common mistakes with the Wavefunction in Forbidden Region formula?

Using an oscillatory solution where the energy is below the barrier. Forgetting to match both the wavefunction and its derivative at the boundaries. Underestimating how quickly the tunneling signal drops with barrier width. Using an oscillatory form where the energy is below the barrier.

Q: What is a real-world example of the Wavefunction in Forbidden Region formula?

In writing the left-side tail of a finite square barrier or well, Wavefunction in Forbidden Region is used to calculate \psi_1(x) from the measured values. The result matters because it helps size components, compare operating conditions, or check a design margin.

Q: What are some study tips for the Wavefunction in Forbidden Region formula?

Inside the classically allowed region you get sine and cosine solutions. In the forbidden region the physically acceptable branch is exponential decay. Barrier width matters exponentially, not linearly. Pick the exponential branch that stays finite as x goes toward negative infinity.

Core idea

Overview

In a classically forbidden region, the physically acceptable branch must decay away from the barrier rather than oscillate.

When to use: Use this when the wavefunction must be matched across a finite barrier or finite well.

Why it matters: The tunneling picture explains why wavefunctions oscillate in allowed regions and decay exponentially in forbidden regions.

Walkthrough

Derivation

Derivation of Wavefunction in a Forbidden Region

The solution is derived from the time-independent Schrödinger equation for a region where the potential energy V(x) is greater than the total energy E.

The particle is in a region where the potential energy V > E.
The Schrödinger equation is the time-independent one-dimensional form: -ħ²/(2m) * d²ψ/dx² + Vψ = Eψ.
The wavefunction must remain finite as x → -∞.
The particle mass m and potential V are constant in this region.

1

Rearrange the Schrödinger Equation

Isolate the second derivative term. Since V > E, let k'² = 2m(V - E)/ħ², where k' is real.

\frac{d ^{2} ψ}{d x ^{2}} = \frac{2 m ( V - E )}{ℏ ^{2}} ψ

Note: This turns the equation into d²ψ/dx² = k'²ψ.

2

General Solution

The general solution to the second-order linear differential equation is a sum of two exponential functions.

ψ (x) = A e^{k^{'} x} + B e^{- k^{'} x}

3

Apply Physical Boundary Conditions

For the region to the left of the barrier (x < 0), the term $e^{- k^{'} x}$ would grow exponentially as x becomes more negative, which is physically non-normalizable. Therefore, we set B = 0.

ψ (x) = C_{1} e^{k^{'} x}

Note: The constant $C_{1}$ is determined by boundary matching at the barrier interface.

Result

ψ (x) = C_{1} e^{k^{'} x}

Source: Engineering LibreTexts, finite square well and tunneling-barrier notes

Free formulas

Rearrangements

Solve for $C_{1} = ψ_{1} (x) e^{- k^{'} x}$

Coefficient $C_{1}$

C_{1} = ψ_{1} (x) e^{- k^{'} x}

Isolate the amplitude coefficient $C_{1}$ by dividing both sides by the exponential term.

Difficulty: 2/5

Solve for $k^{'} = \frac{1}{x} ln (\frac{ψ _{1} ( x )}{C _{1}})$

Wavefunction Decay Constant k'

k^{'} = \frac{1}{x} ln (\frac{ψ _{1} ( x )}{C _{1}})

Isolate the decay constant k' using logarithms.

Difficulty: 3/5

Solve for $x = \frac{1}{k ^{'}} ln (\frac{ψ _{1} ( x )}{C _{1}})$

Position x

x = \frac{1}{k ^{'}} ln (\frac{ψ _{1} ( x )}{C _{1}})

Solve for the position x by isolating the exponential term and applying logarithms.

Difficulty: 3/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Why it behaves this way

Intuition

Imagine a particle hitting a wall it doesn't have the energy to climb. Instead of vanishing instantly at the boundary, the wavefunction 'seeps' into the wall. Because the exponent is positive (k'x) and we are looking at the region to the left of the barrier (where x is negative), the function gets smaller and smaller as you move deeper into the wall (more negative x), representing a fading probability of finding the particle far inside the forbidden zone.

ψ_{1} (x)

Wavefunction in the first region

The 'state' or probability amplitude of the particle in the area before or within the left side of a potential barrier.

C_{1}

Normalization constant / Amplitude

A scaling factor determined by boundary conditions that ensures the total probability across all space adds up to 1.

e^{k^{'} x}

Exponential decay term

Describes how the particle's presence 'leaks' into the forbidden zone; it shows that the likelihood of finding the particle drops off rapidly rather than oscillating like a wave.

k'

Decay constant

A measure of how 'impenetrable' the barrier is; higher energy barriers or heavier particles lead to a larger k', causing the wavefunction to die out much faster.

Signs and relationships

Positive exponent (k'x): In the region to the left of the barrier (where x < 0), a positive exponent ensures that as we move further into the forbidden region (as x becomes more negative), the value approaches zero. A negative exponent here would cause the wavefunction to explode to infinity, which is physically impossible.

Free study cues

Insight

Canonical usage

This equation describes the exponential decay of the wavefunction in a region where the particle's kinetic energy is less than the potential energy barrier, with the decay rate determined by the barrier height and.

Common confusion

Students may incorrectly assume the wavefunction is directly a probability or that the exponential term is dimensionless without considering the units of k' and x.

Dimension note

While the wavefunction itself is often treated as dimensionless in many contexts (as its square gives probability density), the parameter k' has units of inverse length, and x has units of length.

Unit systems

$ψ_{1} (x)$ dimensionless (probability amplitude) · The wavefunction itself is not directly observable, but its square, |\psi|^2, represents probability density.

$C_{1}$ dimensionless (normalization constant) · This constant is determined by boundary conditions and normalization.

k'm^-1 · The parameter k' is defined as \sqrt{2m(V_0 - E)}/\hbar, where m is mass, V_0 is barrier height, E is particle energy, and \hbar is the reduced Planck constant.

$x$ m · Represents position along the axis perpendicular to the barrier.

One free problem

Practice Problem

In a classically forbidden region where E < V, why does the wavefunction exhibit exponential behavior rather than oscillatory behavior?

Solve for: $ψ_{1} (x)$

Hint: Consider the sign of the term (E - V) in the Schrodinger equation.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

In writing the left-side tail of a finite square barrier or well, Wavefunction in Forbidden Region is used to calculate \psi_1(x) from the measured values. The result matters because it helps size components, compare operating conditions, or check a design margin.

Study smarter

Tips

Inside the classically allowed region you get sine and cosine solutions.
In the forbidden region the physically acceptable branch is exponential decay.
Barrier width matters exponentially, not linearly.
Pick the exponential branch that stays finite as x goes toward negative infinity.

Avoid these traps

Common Mistakes

Using an oscillatory solution where the energy is below the barrier.
Forgetting to match both the wavefunction and its derivative at the boundaries.
Underestimating how quickly the tunneling signal drops with barrier width.
Using an oscillatory form where the energy is below the barrier.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

The solution is derived from the time-independent Schrödinger equation for a region where the potential energy V(x) is greater than the total energy E.

Use this when the wavefunction must be matched across a finite barrier or finite well.

The tunneling picture explains why wavefunctions oscillate in allowed regions and decay exponentially in forbidden regions.

Using an oscillatory solution where the energy is below the barrier. Forgetting to match both the wavefunction and its derivative at the boundaries. Underestimating how quickly the tunneling signal drops with barrier width. Using an oscillatory form where the energy is below the barrier.

In writing the left-side tail of a finite square barrier or well, Wavefunction in Forbidden Region is used to calculate \psi_1(x) from the measured values. The result matters because it helps size components, compare operating conditions, or check a design margin.

Inside the classically allowed region you get sine and cosine solutions. In the forbidden region the physically acceptable branch is exponential decay. Barrier width matters exponentially, not linearly. Pick the exponential branch that stays finite as x goes toward negative infinity.

References

Sources

Engineering LibreTexts, finite square well and tunneling-barrier notes, accessed 2026-04-09
Peverati, The Live Textbook of Physical Chemistry 2, quantum weirdness/tunneling section, accessed 2026-04-09
Engineering LibreTexts, field enhanced emission and tunnelling effects, accessed 2026-04-09
Griffiths, David J. (2018). Introduction to Quantum Mechanics (3rd ed.). Cambridge University Press.
NIST CODATA Value: Reduced Planck constant (ħ)
Liboff, Richard L. (2003). Introductory Quantum Mechanics (4th ed.). Addison-Wesley.
Engineering LibreTexts, finite square well and tunneling-barrier notes

Wavefunction in Forbidden Region

Overview

Derivation

Rearrange the Schrödinger Equation

General Solution

Apply Physical Boundary Conditions

Rearrangements

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Finite Square Well Potential

Barrier Decay Constant

General Oscillatory Wave Function

Work function with trapezoidal approximation at junction

Frequently Asked Questions

Sources