8.2. Wave-particle duality
The de Broglie energy and momentum are given by:
\[\begin{split}\begin{align*}
E &= h f = \hbar \omega, \\
p &= h/\lambda = \hbar k.
\end{align*}\end{split}\]
8.3. Schrödinger equation
The general Schrödinger equation in three dimensions is given by:
(8.1)\[
i \hbar \frac{\partial \Psi(\bm{x}, t)}{\partial t} = - \frac{\hbar^2}{2m} \nabla^2 \Psi(\bm{x}, t) + V(\bm{x}) \Psi(\bm{x}, t).
\]
In one dimension, the time-independent version of the Schrödinger equation is:
(8.2)\[
\hat{H} \psi(x) = -\frac{\hbar^2}{2m} \frac{\mathrm{d}^2 \psi(x)}{\mathrm{d}x^2} + V(x) \psi(x) = E \psi(x).
\]
8.4. Operators
The position and momentum operators are given by
\[\begin{split}\begin{align*}
\hat{x} \Psi(x,t) &= x \Psi(x,t), \\
\hat{p} \Psi(x,t) &= - i \hbar \frac{\partial \Psi(x,t)}{\partial x}.
\end{align*}\end{split}\]
The expectation value of any operator \(\hat{Q}\) can be calculated through
(8.3)\[
\Braket{\hat{Q}} = \int_{-\infty}^\infty \Psi^*(x,t) \hat{Q} \Psi(x,t) \mathrm{d}x.
\]
The commutator \(\left[ \hat{A}, \hat{B} \right]\) of two operators is defined as
\[
\left[ \hat{A}, \hat{B} \right] = \hat{A}\hat{B} - \hat{B}\hat{A}.
\]
For any two Hermitian operators \(\hat{A}\) and \(\hat{B}\), we have the general uncertainty principle
\[
\sigma_A \sigma_B \geq \left| \frac{1}{2i} \Braket{\left[ \hat{A}, \hat{B} \right]} \right|.
\]
The time evolution of the expectation value of an operator is given by
(8.4)\[
\frac{\mathrm{d}}{\mathrm{d}t} \Braket{\hat{Q}} = \frac{i}{\hbar} \Braket{\left[ \hat{H}, \hat{Q} \right]} + \Braket{ \frac{\partial \hat{Q}}{\partial t}},
\]
where \(\hat{H}\) is the Hamiltonian of the system.
8.5. Spin
The operators for the \(x\), \(y\) and \(z\) component of the spin satisfy the commutation relations
(8.5)\[\begin{split}\begin{align*}
\left[ \hat{S}_x, \hat{S}_y \right] &= i \hbar \hat{S}_z, \\
\left[ \hat{S}_y, \hat{S}_z \right] &= i \hbar \hat{S}_x, \\
\left[ \hat{S}_z, \hat{S}_x \right] &= i \hbar \hat{S}_y.
\end{align*}\end{split}\]
For a spin-1/2 particle in the basis of the eigenvectors \(\chi_{+} = \left(\begin{array}{c}1\\ 0 \end{array}\right)\) and \(\chi_{-} = \left(\begin{array}{c}0\\ 1 \end{array}\right)\) of the \(\hat{S}_z\) operator, the spin operators are given by the matrix expressions:
(8.6)\[\begin{split}\begin{align*}
\hat{S}_x &= \frac{\hbar}{2} \hat{\bm{\sigma}}_x = \frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \\
\hat{S}_y &= \frac{\hbar}{2} \hat{\bm{\sigma}}_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \\
\hat{S}_z &= \frac{\hbar}{2} \hat{\bm{\sigma}}_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
\end{align*}\end{split}\]
8.6. Mathematical identities
8.6.1. Goniometric functions
\[\begin{split}\begin{align*}
e^{i\phi} &= \cos(\phi) + i \sin(\phi) \\
\sin(\phi) &= \frac{1}{2i} \left( e^{i \phi} - e^{-i \phi} \right) \\
\cos(\phi) &= \frac12 \left( e^{i \phi} + e^{-i \phi} \right) \\
1 &= \sin^2(\phi) + \cos^2(\phi) \\
\sin(2 \phi) &= 2 \sin(\phi) \cos(\phi)\\
\cos(2\phi) &= \cos^2(\phi)-\sin^2(\phi) \\
\sin(a \pm b) &= \sin(a) \cos(b) \pm \sin(b) \cos(a) \\
\cos(a \pm b) &= \cos(a) \cos(b) \pm \sin(a) \sin(b)
\end{align*}\end{split}\]
8.6.2. Exponential integrals
(8.7)\[
\int_0^\infty x^n e^{-x/a} \mathrm{d}x = n!\, a^{n+1} \qquad \text{for any integer } n.
\]
8.6.3. Some goniometric integrals
The following hold for any integers \(m\) and \(n\).
\[\begin{split}\begin{align*}
\int_0^L \sin\left(\frac{m \pi x}{L} \right) \sin\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= \frac{L}{2} \delta_{mn}.\\
\int_0^L \cos\left(\frac{m \pi x}{L} \right) \cos\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= \frac{L}{2} \delta_{mn}. \\
\int_0^L \sin\left(\frac{n \pi x}{L} \right) \cos\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= 0. \\
\int_0^L \sin\left(\frac{m \pi x}{L} \right) \cos\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= \frac{1-(-1)^{m+n}}{m^2-n^2} \frac{L m}{\pi}. \\
\int_0^L x \sin^2\left(\frac{n \pi x}{L} \right) \mathrm{d}x = \int_0^L x \cos^2\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= \frac{L^2}{4}. \\
\int_0^L x \sin\left(\frac{\pi x}{L} \right) \sin\left(\frac{2 \pi x}{L} \right) \mathrm{d}x &= - \frac{8 L^2}{9 \pi^2}. \\
\int_0^L x \cos\left(\frac{\pi x}{L} \right) \cos\left(\frac{2 \pi x}{L} \right) \mathrm{d}x &= - \frac{10 L^2}{9 \pi^2}. \\
\int_0^L x \sin\left(\frac{n \pi x}{L} \right) \cos\left(\frac{n \pi x}{L} \right) \mathrm{d}x &= -\frac{L^2}{4 n \pi}.
\end{align*}\end{split}\]
For any \(a \neq 0\), we have the indefinite integrals
\[\begin{split}\begin{align*}
\int x \sin(ax) \,\mathrm{d}x &= \frac{1}{a^2} \sin(ax) - \frac{x}{a} \cos(ax), \\
\int x \cos(ax) \,\mathrm{d}x &= \frac{1}{a^2} \cos(ax) + \frac{x}{a} \sin(ax).
\end{align*}\end{split}\]
8.6.4. Some Gaussian integrals
Gaussian integrals are integrals over the Gaussian distribution \(\exp(-x^2)\) (aka the normal distribution). In general, these cannot be evaluated in closed form, and we define the error function as the integral:
(8.8)\[
\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-y^2} \mathrm{d}y.
\]
There are however a number of Gaussian integrals that can be evaluated explicitly: those over all of \(\mathbb{R}\). Of course, as the Gaussian distribution is symmetric, we can also find the integrals from plus or minus infinity to the point of symmetry.
(8.9)\[\begin{split}\begin{align*}
\int_{-\infty}^\infty e^{-a x^2} \mathrm{d}x &= \sqrt{\frac{\pi}{a}}. \\
\int_{-\infty}^\infty e^{-a x^2 + b x + c} \mathrm{d}x &= \sqrt{\frac{\pi}{a}} \exp\left(\frac{b^2}{4a}+c\right). \\
\int_{-\infty}^\infty x^n e^{-a x^2} \mathrm{d}x &= 0 \qquad \text{for any odd value of } n. \
\end{align*}\end{split}\]
(8.10)\[\begin{align*}
\int_{-\infty}^\infty x^n e^{-a x^2} \mathrm{d}x &= \frac{1 \cdot 3 \cdot 5 \cdots (n-1) \sqrt{\pi}}{2^{n/2} a^{(n+1)/2}} \qquad \text{for any even value of } n.
\end{align*}\]
Note that we can get (8.10) by repeated differentiation of (8.9) with respect to \(a\).
8.6.5. Vector derivatives
The expressions below are in Cartesian (\(x, y, z\)), cylindrical (\(\rho, \phi, z\)) and spherical (\(r, \theta, \phi\)) coordinates.
Gradient:
(8.11)\[\begin{split}\begin{align*}
\bm{\nabla} f(\bm{r}) &= \bm{\nabla} f(x, y, z) = \begin{pmatrix} \partial_x f \\ \partial_y f \\ \partial_z f \end{pmatrix} = \frac{\partial f}{\partial x} \bm{\hat{x}} + \frac{\partial f}{\partial y} \bm{\hat{y}} + \frac{\partial f}{\partial z} \bm{\hat{z}}, \\
&= \bm{\nabla} f(\rho, \phi, z) = \frac{\partial f}{\partial \rho} \bm{\hat{\rho}} + \frac{1}{\rho} \frac{\partial f}{\partial \phi} \bm{\hat{\phi}} + \frac{\partial f}{\partial z} \bm{\hat{z}}, \\
&= \bm{\nabla} f(r, \theta, \phi) = \frac{\partial f}{\partial r} \bm{\hat{r}} + \frac{1}{r} \frac{\partial f}{\partial \theta} \bm{\hat{\theta}} + \frac{1}{r \sin(\theta)} \frac{\partial f}{\partial \phi} \bm{\hat{\phi}}.
\end{align*}\end{split}\]
Divergence:
(8.12)\[\begin{split}\begin{align*}
\bm{\nabla} \cdot \bm{v} &= \left( \partial_x, \partial_y, \partial_z \right) \cdot \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}, \\
&= \frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \rho v_\rho \right) + \frac{1}{\rho} \frac{\partial v_\phi}{\partial \phi} + \frac{\partial v_z}{\partial z}, \\
&= \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 v_r \right) + \frac{1}{r \sin(\theta)} \frac{\partial }{\partial \theta} \left(\sin(\theta) v_\theta\right) + \frac{1}{r \sin(\theta)} \frac{\partial v_\phi}{\partial \phi}.
\end{align*}\end{split}\]
Curl:
(8.13)\[\begin{split}\begin{align*}
\bm{\nabla} \times \bm{A} &= \left( \partial_x, \partial_y, \partial_z \right) \times \begin{pmatrix} A_x \\ A_y \\ A_z \end{pmatrix} = \begin{pmatrix} \partial_y A_z - \partial_z A_y \\ \partial_z A_x - \partial_x A_z \\ \partial_x A_y - \partial_y A_x \end{pmatrix}, \\
&= \left( \frac{1}{\rho} \frac{\partial v_z}{\partial \phi} - \frac{\partial v_\phi}{\partial z} \right) \bm{\hat{\rho}} + \left( \frac{\partial v_\rho}{\partial z} - \frac{\partial v_z}{\partial \rho} \right) \bm{\hat{\phi}} + \frac{1}{\rho} \left( \frac{\partial }{\partial \rho} \left( \rho v_\phi \right) - \frac{\partial v_\rho}{\partial \phi} \right) \bm{\hat{z}}, \\
&= \frac{1}{r \sin(\theta)} \left[ \frac{\partial }{\partial \theta} \left(v_\phi \sin(\theta) \right) - \frac{\partial \phi}{\partial v_\theta} \right] \bm{\hat{r}} + \frac{1}{r} \left[ \frac{1}{\sin(\theta)} \frac{\partial v_r}{\partial \phi} - \frac{\partial }{\partial r}\left(r v_\phi\right) \right] \bm{\hat{\theta}} + \frac{1}{r} \left[ \frac{\partial }{\partial r}\left(r v_\phi\right) - \frac{\partial v_r}{\partial \theta} \right] \bm{\hat{\phi}}
\end{align*}\end{split}\]
Laplacian:
(8.14)\[\begin{split}\begin{align*}
\nabla^2 f &= \bm{\nabla} \cdot \left(\bm{\nabla} f \right) = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}, \\
&= \frac{1}{\rho} \frac{\partial }{\partial \rho} \left( \rho \frac{\partial f}{\partial \rho} \right) + \frac{1}{\rho^2} \frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2}, \\
&= \frac{1}{r^2} \frac{\partial }{\partial r} \left( r^2 \frac{\partial f}{\partial r} \right) + \frac{1}{r^2 \sin(\theta)} \frac{\partial }{\partial \theta} \left(\sin(\theta) \frac{\partial f}{\partial \theta}\right) + \frac{1}{r^2 \sin^2(\theta)} \frac{\partial^2 f}{\partial \phi^2}.
\end{align*}\end{split}\]
