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Wave function and time dependent Schrödinger equation
Short answer
The wave function Ψ(x,t) describes a particle's quantum state; |Ψ|² is the probability density and must be normalized. Starting from a plane wave Ψ = A e^{i(kx−ωt)} and using E=ħω, p=ħk, substituting into E = p²/2m + V yields the time-dependent Schrödinger equation: iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ.
What is the significance of the wave function? Using the wave function derive the expression for the time dependent Schrödinger wave equation.
Significance of the wave function
The wave function Ψ(x, t) describes the quantum state of a particle.
- Ψ itself is not measurable. It can be complex.
- |Ψ|² = Ψ*Ψ is the probability density: |Ψ|² dx is the probability of finding the particle between x and x + dx at time t.
- Normalisation: ∫|Ψ|² dx = 1 (the particle is somewhere).
- Ψ must be single-valued, finite and continuous, and its derivative must be continuous (where V is finite).
- It contains all information about the system, such as the expectation value ⟨x⟩ = ∫Ψ* x Ψ dx.
Time-dependent Schrödinger equation
Take a free particle travelling along x as a plane wave (de Broglie):
Ψ(x, t) = A e^{i(kx − ωt)}, with p = ħk and E = ħω.
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Differentiate with respect to t: ∂Ψ/∂t = −iωΨ. Multiply by iħ: iħ ∂Ψ/∂t = ħωΨ = EΨ.
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Differentiate twice with respect to x: ∂²Ψ/∂x² = −k²Ψ. Multiply by −ħ²: −ħ² ∂²Ψ/∂x² = ħ²k²Ψ = p²Ψ.
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For a particle with potential energy V(x), total energy E = p²/2m + V. Multiply by Ψ:
EΨ = (p²/2m) Ψ + VΨ
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Substitute from steps 1 and 2:
iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x) Ψ
In three dimensions: iħ ∂Ψ/∂t = −(ħ²/2m)∇²Ψ + VΨ, that is, iħ∂Ψ/∂t = ĤΨ, with Ĥ the Hamiltonian (energy) operator.
The equation is first order in time, so Ψ at one instant fixes its later evolution. It is a postulate, justified by its agreement with experiments.
✓ Answer checked by Claude Opus 5.5 (high effort)
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