notes/Schrodinger explained.md

Schrödinger Equation Derivation

Why the Kinetic Term is -\hbar^2/(2m)\,\nabla^2

Goal

Start from the classical energy equation, E=total energy, p=momentum, m=mass, V=potential energy. So the Hamiltonian (not the Larangian)...

E = \frac{p^2}{2m} + V

Note this is E=\frac{1}{2} m v^2. P is momentum so p^2 = {m^2v^2}, so dividing by 2m converts the kinetic energy a momentum equation

and show how this becomes the time-dependent Schrödinger equation

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

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1. Slowly with classical mechanics....

For a non-relativistic particle, total energy is

E = \frac{p^2}{2m} + V

where:

• E = total energy
• p = momentum
• m = mass
• V = potential energy

The term

\frac{p^2}{2m}

is the usual classical kinetic energy.

Since

p = mv

we have

\frac{p^2}{2m} = \frac{m^2 v^2}{2m} = \frac{1}{2}mv^2

So the divide by $2m$ simply comes from the classical kinetic energy formula.

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2. Introduce a wave

Quantum mechanics uses a wavefunction \psi(x,t).

Take a simple plane wave in 1D:

\psi(x,t) = e^{i(kx-\omega t)} 

This is a plane wave solution. In quantum mechanics, position is not definite — the probability of finding the particle is given by |\psi|^2. This is similar to the position in classical physics.

This is useful because derivatives acting on exponentials bring down constants.

Quantum Phase Corkscrew

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3. Differentiate with respect to position

\frac{\partial \psi}{\partial t} = -i\omega\psi

Note this differentiates by time. It's looking at how fast the particle waveform oscillates which is directly related to the energy via the plank constant. As for a photon this means the frequency is \propto E.

Second derivative w.r.t. distance:

\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi

So the second derivative returns the same wave multiplied by -k^2 because i^2 is -1. The first derivative gives momentum, and the second derivative gives momentum squared, which corresponds to kinetic energy.

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4. Relate k to momentum

From de Broglie:

p = \hbar k

so

k = \frac{p}{\hbar}

Substitute this into the second derivative result:

\frac{\partial^2 \psi}{\partial x^2}
=
-\left(\frac{p}{\hbar}\right)^2 \psi
=
-\frac{p^2}{\hbar^2}\psi

Rearrange:

p^2 \psi = -\hbar^2 \frac{\partial^2 \psi}{\partial x^2}

Key idea:

Momentum squared acting on a wave becomes minus \hbar^2 times the second derivative.

So in operator form,

\hat{p}^2 = -\hbar^2 \frac{\partial^2}{\partial x^2}

and in 3D,

\hat{p}^2 = -\hbar^2 \nabla^2

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5. Put that into the energy equation

Start from

E = \frac{p^2}{2m} + V

Multiply through by \psi:

E\psi = \frac{p^2}{2m}\psi + V\psi

Replace p^2\psi:

E\psi

\frac{1}{2m}\left(-\hbar^2 \frac{\partial^2 \psi}{\partial x^2}\right) + V\psi

So

E\psi =
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V\psi

This is where the factor

-\frac{\hbar^2}{2m}

comes from.

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6. Replace energy by a time derivative

For a wave

\psi(x,t) = e^{i(kx-\omega t)}

Differentiate with respect to time:

\frac{\partial \psi}{\partial t} = -i\omega \psi

Multiply by i\hbar:

i\hbar \frac{\partial \psi}{\partial t}
=
i\hbar (-i\omega)\psi
=
\hbar\omega \psi

But Planck's relation says

E = \hbar \omega

So

E\psi = i\hbar \frac{\partial \psi}{\partial t}

Thus

\hat{E} = i\hbar \frac{\partial}{\partial t}

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7. Substitute into the equation

We had

E\psi =
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
+ V\psi

Replace E\psi:

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}
+ V\psi

This is the 1D time‑dependent Schrödinger equation.

In 3D:

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

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8. Physical meaning of the second derivative

The second derivative measures curvature of the wavefunction.

• Gentle curvature → long wavelength → small k → small momentum
• Rapid oscillation → short wavelength → large k → large momentum

Since kinetic energy depends on p^2, it naturally connects to a second derivative.

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9. Operator summary

Classical equation:

E = \frac{p^2}{2m} + V

Quantum substitutions:

E \rightarrow i\hbar \frac{\partial}{\partial t}

p \rightarrow -i\hbar \nabla

Therefore

p^2 \rightarrow (-i\hbar \nabla)^2 = -\hbar^2 \nabla^2

and

\frac{p^2}{2m}
\rightarrow
-\frac{\hbar^2}{2m}\nabla^2

So we obtain

i\hbar \frac{\partial \psi}{\partial t}
=
-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi

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10. Intuition in one sentence

The Schrödinger equation is simply

\text{Energy} = \text{kinetic} + \text{potential}

rewritten so energy and momentum become differential operators acting on a wavefunction.

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