The Schrödinger equation is the fundamental equation of quantum mechanics, describing how the quantum state of a physical system changes over time. Unlike classical mechanics where Newton's second law ($F = ma$) determines particle trajectories, quantum mechanics uses the Schrödinger equation to determine the evolution of the wavefunction ($\Psi$), which contains all measurable information about the particle.
The wavefunction itself has no direct physical meaning, but its absolute square $|\Psi|^2$ represents the probability density of finding the particle at a specific position and time (Born's interpretation).
We start with a free particle moving in the +x direction with momentum $p$ and energy $E$. The de Broglie hypothesis and Planck's relation give:
$$ p = \hbar k \quad \text{and} \quad E = \hbar \omega $$The plane wave representing such a particle is:
$$ \Psi(x,t) = A e^{i(kx - \omega t)} $$Differentiating $\Psi$ with respect to space (x):
$$ \frac{\partial \Psi}{\partial x} = ik \Psi \implies \frac{\partial^2 \Psi}{\partial x^2} = -k^2 \Psi = -\frac{p^2}{\hbar^2}\Psi $$This gives us the momentum operator: $\hat{p} = -i\hbar \frac{\partial}{\partial x}$. Therefore, the kinetic energy is:
$$ \frac{\hat{p}^2}{2m}\Psi = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} = \frac{p^2}{2m}\Psi $$Now, differentiating $\Psi$ with respect to time (t):
$$ \frac{\partial \Psi}{\partial t} = -i\omega \Psi = -i\frac{E}{\hbar}\Psi \implies i\hbar\frac{\partial \Psi}{\partial t} = E\Psi $$In classical mechanics, total energy $E = \frac{p^2}{2m} + V(x,t)$. Replacing these classical observables with their corresponding quantum mechanical operators, we get the TDSE:
$$ i\hbar \frac{\partial \Psi(x,t)}{\partial t} = \left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \right] \Psi(x,t) $$If the potential energy $V$ does not depend on time, we can use the method of separation of variables: $\Psi(x,t) = \psi(x) \phi(t)$.
$$ i\hbar \frac{1}{\phi}\frac{d\phi}{dt} = \frac{1}{\psi}\left[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi \right] = E $$The left side gives the time evolution: $\phi(t) = e^{-iEt/\hbar}$. The right side gives the TISE:
$$ -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) $$Or in Hamiltonian form: $\hat{H}\psi = E\psi$, which is an eigenvalue equation.
Apply the TISE: $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi$
First derivative: $\psi' = A(-\frac{x}{a^2})e^{-x^2 / 2a^2} = -\frac{x}{a^2}\psi$
Second derivative: $\psi'' = -\frac{1}{a^2}\psi - \frac{x}{a^2}\psi' = -\frac{1}{a^2}\psi + \frac{x^2}{a^4}\psi = \psi\left(\frac{x^2}{a^4} - \frac{1}{a^2}\right)$
Substitute back into TISE:
$$ -\frac{\hbar^2}{2m} \left(\frac{x^2}{a^4} - \frac{1}{a^2}\right)\psi + V(x)\psi = \frac{\hbar^2}{4ma^2}\psi $$Cancel $\psi$ and solve for $V(x)$:
$$ V(x) = \frac{\hbar^2}{4ma^2} + \frac{\hbar^2 x^2}{2ma^4} - \frac{\hbar^2}{2ma^2} = \frac{\hbar^2 x^2}{2ma^4} - \frac{\hbar^2}{4ma^2} $$Thus, $V(x) = \frac{\hbar^2}{2ma^4}x^2 - \frac{\hbar^2}{4ma^2}$, which resembles a harmonic oscillator potential.
Momentum operator $\hat{p} = -i\hbar \frac{d}{dx}$. Expectation value $\langle p \rangle = \int \psi^* \hat{p} \psi dx$.
For a standing wave formed by a real superposition of sine and cosine, the probability current is zero, and thus the expected momentum is zero. As long as $A$ and $B$ are purely real (or have the same phase), $\langle p \rangle = 0$.
The probability current density is $J = \frac{i\hbar}{2m}(\psi \nabla\psi^* - \psi^*\nabla\psi)$.
For a plane wave $\psi = Ae^{i(kx-\omega t)}$, where $p=\hbar k$, $J = |A|^2 \frac{\hbar k}{m} = |A|^2 v$.
Here $|A| = 1$, so $J = \frac{p}{m} = v$, the classical velocity.