1. 3D Schrödinger Equation
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Extending quantum mechanics to three spatial dimensions.
Time-Independent Schrödinger Equation
$$-\frac{\hbar^2}{2m}\nabla^2\psi(\vec{r}) + V(\vec{r})\psi(\vec{r}) = E\psi(\vec{r})$$
where the Laplacian is $\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$
Cartesian Coordinates
In Cartesian coordinates ($x, y, z$):
$$-\frac{\hbar^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) + V(x,y,z)\psi = E\psi$$
Spherical Coordinates
For spherically symmetric problems, use $(r, \theta, \phi)$:
$$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}$$
Coordinate transformations:
$$x = r\sin\theta\cos\phi, \quad y = r\sin\theta\sin\phi, \quad z = r\cos\theta$$
Separation of Variables
For potentials with separable symmetry, try:
$$\psi(x,y,z) = X(x)Y(y)Z(z)$$
This works when:
$$V(x,y,z) = V_x(x) + V_y(y) + V_z(z)$$
Total energy: $E = E_x + E_y + E_z$
3D Infinite Box
Particle confined to $0 \leq x,y,z \leq L$:
$$\psi_{n_x,n_y,n_z}(x,y,z) = \left(\frac{2}{L}\right)^{3/2}\sin\left(\frac{n_x\pi x}{L}\right)\sin\left(\frac{n_y\pi y}{L}\right)\sin\left(\frac{n_z\pi z}{L}\right)$$
Energy levels:
$$E_{n_x,n_y,n_z} = \frac{\hbar^2\pi^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2)$$
Degeneracy: Different $(n_x, n_y, n_z)$ can give same energy
Momentum Operators
$$\hat{\vec{p}} = -i\hbar\nabla = -i\hbar\left(\frac{\partial}{\partial x}\hat{x} + \frac{\partial}{\partial y}\hat{y} + \frac{\partial}{\partial z}\hat{z}\right)$$
Commutation relations:
$$[\hat{x}_i, \hat{p}_j] = i\hbar\delta_{ij}$$
Free Particle in 3D
Plane wave solution:
$$\psi_{\vec{k}}(\vec{r}) = Ae^{i\vec{k}\cdot\vec{r}}$$
Energy-momentum relation:
$$E = \frac{\hbar^2k^2}{2m} = \frac{p^2}{2m}$$
where $\vec{p} = \hbar\vec{k}$
Probability Current Density
$$\vec{j}(\vec{r},t) = \frac{\hbar}{2mi}\left(\psi^*\nabla\psi - \psi\nabla\psi^*\right) = \frac{1}{m}\text{Re}(\psi^*\hat{\vec{p}}\psi)$$
Continuity equation:
$$\frac{\partial\rho}{\partial t} + \nabla\cdot\vec{j} = 0$$