KTU S1

Formulation of the Schrodinger Equations

By the end you should be able to: Formulate the time dependent Schrodinger equation from the de Broglie relation and the classical energy equation, and reduce it to the time independent form by separation of variables.

Newton's second law tells you how a classical particle moves. The Schrodinger equation is the quantum replacement: it tells you how the wave function evolves.

It cannot be derived from classical mechanics — it is a new postulate, and Schrodinger arrived at it partly by inspired guesswork. What follows is a formulation: a plausible construction showing where each term comes from. Its justification is that its predictions are correct.

The starting ingredients

De Broglie. A particle of momentum pp has an associated wavelength

λ=hp⟺p=ℏk,k=2πλ\lambda = \frac{h}{p} \quad\Longleftrightarrow\quad p = \hbar k, \qquad k = \frac{2\pi}{\lambda}

Planck-Einstein. Its energy relates to angular frequency by

E=hν=ℏωE = h\nu = \hbar\omega

A free-particle wave. The natural complex travelling wave is

ψ(x,t)=A ei(kx−ωt)\psi(x,t) = A\,e^{i(kx - \omega t)}

Building the equation

Differentiate this trial wave function and see what each derivative produces.

Once with respect to time:

∂ψ∂t=−iω ψ⇒iℏ∂ψ∂t=iℏ(−iω)ψ=ℏω ψ=Eψ\frac{\partial \psi}{\partial t} = -i\omega\,\psi \quad\Rightarrow\quad i\hbar\frac{\partial \psi}{\partial t} = i\hbar(-i\omega)\psi = \hbar\omega\,\psi = E\psi

So the operator iℏ ∂/∂ti\hbar\,\partial/\partial t extracts the energy.

Twice with respect to position:

∂ψ∂x=ikψ,∂2ψ∂x2=(ik)2ψ=−k2ψ\frac{\partial \psi}{\partial x} = ik\psi, \qquad \frac{\partial^{2}\psi}{\partial x^{2}} = (ik)^{2}\psi = -k^{2}\psi −ℏ22m∂2ψ∂x2=ℏ2k22mψ=p22mψ=KE⋅ψ-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} = \frac{\hbar^{2}k^{2}}{2m}\psi = \frac{p^{2}}{2m}\psi = KE \cdot \psi

So −ℏ22m∂2∂x2-\dfrac{\hbar^{2}}{2m}\dfrac{\partial^{2}}{\partial x^{2}} extracts the kinetic energy.

Now impose energy conservation. Classically,

E=KE+PE=p22m+VE = KE + PE = \frac{p^{2}}{2m} + V

Replace each term by the operator that produces it, acting on ψ\psi:

 iℏ∂ψ∂t=−ℏ22m∂2ψ∂x2+Vψ \boxed{\,i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} + V\psi\,}

This is the time dependent Schrodinger equation. Note the structure: it is nothing more than "total energy equals kinetic plus potential", with each quantity written as the operation that measures it.

Two features are worth naming. The ii makes the equation complex, which is why ψ\psi must be complex and why the interpretation uses ∣ψ∣2|\psi|^2. And it is first order in time, so knowing ψ\psi at one instant determines it at all later instants — the theory is deterministic in ψ\psi, even though the measurements it predicts are probabilistic.

Reducing to the time independent form

When VV depends on position only, not on time — which covers almost every problem in this course — the variables separate. Try

ψ(x,t)=ψ(x) ϕ(t)\psi(x,t) = \psi(x)\,\phi(t)

Substituting and dividing through by ψ(x)ϕ(t)\psi(x)\phi(t):

iℏ1ϕdϕdt⏟function of t only=−ℏ22m1ψd2ψdx2+V(x)⏟function of x only\underbrace{i\hbar\frac{1}{\phi}\frac{d\phi}{dt}}_{\text{function of } t \text{ only}} = \underbrace{-\frac{\hbar^{2}}{2m}\frac{1}{\psi}\frac{d^{2}\psi}{dx^{2}} + V(x)}_{\text{function of } x \text{ only}}

A function of tt alone equals a function of xx alone. Vary xx with tt fixed and the left side cannot change, so neither can the right; the same argument runs the other way. Both sides must equal the same constant, and by the first calculation above that constant is the energy EE.

The space part gives

 −ℏ22md2ψdx2+Vψ=Eψ \boxed{\,-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}} + V\psi = E\psi\,}

the time independent Schrodinger equation, usually rearranged as

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

The time part gives ϕ(t)=e−iEt/ℏ\phi(t) = e^{-iEt/\hbar}, so

ψ(x,t)=ψ(x) e−iEt/ℏ\psi(x,t) = \psi(x)\,e^{-iEt/\hbar}

Why these are called stationary states

The time factor is a pure phase of modulus one, so

∣ψ(x,t)∣2=∣ψ(x)∣2∣e−iEt/ℏ∣2=∣ψ(x)∣2|\psi(x,t)|^{2} = |\psi(x)|^{2}\left|e^{-iEt/\hbar}\right|^{2} = |\psi(x)|^{2}

The probability density does not change with time. States of definite energy are called stationary states for exactly this reason: the wave function oscillates in phase, but every measurable prediction is constant.

This is why the time independent equation does the work. Solve it for ψ(x)\psi(x) and EE, and the physics is settled.