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# Notes on the Schrödinger Equation

## Derivation of the Time-Dependent Schrödinger Equation

## The Time-Independent Schrödinger Equation

The two key ingredients are the *Planck equation* for the energy of a photon,
and the *de Broglie wavelength* of a matter wave.

The Planck equation for the energy, \(E\) of a photon with frequency \(f\) is:

\[ E = h f \]

The de Broglie wavelength, \(\lambda\), of the matter wave asoociated with a particle with momentum \(p\) is:

\[ \lambda = \frac{h}{p} \]

It's convenient to recast this in terms of the wave number \(k = 2\pi / \lambda\) and the normalized frequency \(\omega = 2\pi f\). (Conceptually, \(k\) represents the number of wave cycles in an interval of length \(2\pi\), and \(\omega\) represents the numbers of cycles in a time interval of length \(2\pi\).) In these terms, the Planck equations and the de Broglie wavelength formulas become respectively \[ E = \hbar \omega \qquad\text{and}\qquad p = \frac{h}{\lambda} = \hbar\frac{2\pi}{\lambda} = \hbar k \] where \(\hbar = h/2\pi\) is the adjusted Planck constant, approximately \(1.05 \times 10^{-34} J\cdot s\).

The Schrödinger Equation is motivated by looking for a wave-type equation which has as a solution the simplest possible candidate for a matter wave, i.e., a plane wave of the form \[ \Psi(x,t) = e^{i(kx - \omega t)} \]

Then compute \[ \frac{\partial\Psi}{\partial t} = -i\omega \Psi = \frac{-iE}{\hbar}\Psi \] and so \[ i\hbar\frac{\partial\Psi}{\partial t} = E\Psi \]

Next we calculate \[ \frac{\partial^2\Psi}{\partial x^2} = -k^2 \Psi \] and then, using the classical (non-relativistic) relationship \[ E = \frac{p^2}{2m} \] together with the de Broglie formula, we obtain \[ \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{\hbar^2 k^2}{2m} \Psi = \frac{p^2}{2m} \Psi = E\Psi \]

Putting together the two formulas for \(E\Psi\) we obtain the free Schrödinger Equation: \[ i\hbar\frac{\partial\Psi}{\partial t} = E\Psi = \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} \]

Finally, since we are interpreting \[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \] as the energy operator for a free particle, it makes sense that the energy operator for a particle in potential \(V(x, t)\) is \[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \] and so we reassemble the equations for a particle in a potential as \[ i\hbar\frac{\partial\Psi}{\partial t} = \left[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \right] \Psi \]

Suppose that \(V=V(x)\) does not depend on \(t\). Then the Schrödinger Equation
becomes a separable PDE. Write \(\Psi(x,t) = \phi(t)\psi(x)\) and substitute:
\[
i\hbar\frac{d\phi(t)}{dt}\psi(x) =
- \phi(t)\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}
+ V(x)\phi(t)\psi(x)
\]
Rearraging this gives
\[
i\hbar\frac{1}{\phi(t)}\frac{d\phi(t)}{dt} =
\left(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}
+ V(x)\psi(x)\right) \frac{1}{\psi(x)}
\]
The lhs doesn't depend on \(x\) and the rhs doesn't depend on \(t\) and so
both must equal a common constant value, \(E\). From this
\[
\frac{d\phi(t)}{dt} = \frac{-iE}{\hbar}\phi(t)
\]
and so
\[
\phi(t) = e^{-iEt/\hbar}
\]
(the multiplicative constant of integration is absorbed into \(\psi\) for
convenience). This leaves the *time-independent Schrödinger equation*:
\[
-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)
\]