The Time-Independent Schrödinger Equation
For a nonrelativistic particle in a time-independent potential, the stationary Schrödinger equation is an energy eigenvalue equation. Its solutions must satisfy the differential equation and the physical boundary conditions.
What a learner can do afterwards
- Identifies kinetic, potential and total energy terms in the equation
- States the conditions a physical solution has to satisfy at a boundary
- Explains what makes a state stationary when the particle is still described by a wave
1 · Read
The wavefunction predicts probabilities, as in the packet topic. For a nonrelativistic particle of constant mass m in one dimension, with time-independent potential energy U(x), the stationary equation is -(ħ²/2m)ψ'' + Uψ = Eψ. Here ψ'' means d²ψ/dx². This quantum equation does not follow from classical energy conservation alone. Solve for ψ and allowed energies E.
The kinetic-energy operator acting on ψ gives -(ħ²/2m)ψ''. Uψ is the potential term and Eψ the total-energy term. These are functions, not three energy numbers. In constant U, ψ = A sin(kx) has ψ'' = -k²ψ. Its kinetic eigenvalue is T = ħ²k²/(2m), so tripling k multiplies T by nine. Absolute curvature alone does not give a local kinetic energy.
An infinite well from 0 to L has U = 0 inside and ψ(0) = ψ(L) = 0. Its normalised solutions are ψ_n = √(2/L) sin(nπx/L), with n = 1, 2, 3, ... . Each satisfies the equation and the wall conditions. The index n counts half-waves across the well. E_n = n²π²ħ²/(2mL²), so n = 3 has three half-waves and nine times the ground-state energy. The bars show energies in units of E_1, not wave shapes.
Check a candidate by substituting ψ'' into the equation: one constant E must work throughout the interval. Also check the wall conditions and confirm that the squared norm is finite and nonzero. The n = 0 sine has norm zero, so the well starts at n = 1. For constant mass at a finite, nonsingular potential boundary, ψ and ψ' are continuous. At an infinite well wall, ψ is zero; ψ' need not match the zero function outside.
A stationary energy eigenstate has Ψ(x,t) = ψ(x) exp(-iEt/ħ). Its time factor has magnitude one, so |Ψ(x,t)|² = |ψ(x)|². The overall phase can change while the density stays fixed. Stationary does not mean fixed at one position. Every infinite-well eigenstate has positive kinetic energy. A static potential also permits superpositions of different energies whose position densities may change with time.
Solve the energy eigenvalue equation and check its boundary conditions and norm. A stationary state's density stays fixed even when its overall phase changes.
Lesson, questions and flashcards written for this dataset using standard stationary-equation and infinite-well methods.
Readings and flashcards written for this dataset.
More to read
Solving for a function and an energy
The stationary equation asks for ψ(x) and an energy E that satisfy -(ħ²/2m)ψ'' + Uψ = Eψ. Here the model is one nonrelativistic particle of constant mass in one dimension, with a potential that does not depend on time.
Classical kinetic plus potential energy motivates the structure, but classical conservation alone does not derive the quantum equation. The derivative term is an operator acting on a function. It cannot generally be read as a kinetic-energy number at a position.
For a sine A sin(kx) in constant U, differentiation gives ψ'' = -k²ψ. The kinetic operator therefore returns ψ multiplied by ħ²k²/(2m). This is the kinetic eigenvalue for that sine solution.
To obtain the stationary time-dependent state, multiply ψ by exp(-iEt/ħ). This factor can change the complex phase but has magnitude one. The position probability density stays unchanged.
Which waves fit the infinite well?
Place impenetrable walls at x = 0 and x = L. The infinite-well conditions are ψ(0) = ψ(L) = 0. With U = 0 inside, the equation becomes -(ħ²/2m)ψ'' = Eψ.
The sine family sin(nπx/L) meets both wall conditions for positive integer n. There are n half-waves across the well. Differentiating twice returns the sine multiplied by -(nπ/L)². Substitution then gives E_n = n²π²ħ²/(2mL²).
The wall conditions select the wave numbers. Normalisation sets the amplitude's magnitude to √(2/L). These are separate jobs. A function that vanishes at the walls is not automatically an energy eigenfunction.
The value n = 0 produces the zero function. Its squared norm is zero, so no constant multiplier can turn it into a normalised state. The lowest allowed value is n = 1, with positive energy.
Three checks for a bound-state candidate
First substitute the proposed function and its derivatives into the stationary equation. One constant E must make the equation hold throughout the region. Agreement at a single point is not enough.
Next apply the boundary conditions for the stated potential. At a finite, nonsingular interface, constant-mass solutions have continuous ψ and ψ'. An infinite-well wall requires ψ = 0, but the derivative need not match the zero function outside. Singular potentials require matching rules specific to that potential.
Finally calculate the squared norm. For a bound state, its integral must be finite and nonzero before normalisation. A function can remain finite everywhere and still have an infinite norm on an infinite interval.
These checks are independent requirements. Normalisation and correct endpoints do not prove that a candidate solves the equation. Also, a time-independent potential permits states made from different energy eigenfunctions; their position densities need not remain fixed.
2 · Watch
En anglaisTake it off screen
Where it sits
8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.