Helpful Math

Group Properties:

• Closure: A, B in G, then AB is in G.
• Associative: (AB)C = A(BC)
• Identity: $\mathbb{1}$A = A $\mathbb{1}=A(\forall A \in G)$
• Inverse: A$\in$G, $A^{-1}$=B; $AA^{-1}=A^{-1}A=\mathbb{1}$

Cylindrical Laplacian Solution

$$\nu\neq 0; R = a\rho^{\nu} + b \rho^{-\nu}$$ (168)

$$\psi = A \cos{\nu\phi} + B \sin{\nu \phi}$$

$$\nu=0; R=a_0 + b_0 \ln{\rho}$$ (169)

$$\psi = A_0 + B_0\phi$$

$$\frac{\rho}{R}\frac{\partial}{\partial \rho}\big(\rho\frac{\partial R}{\partial \rho}\big) = \nu^2$$ (170)

$$\frac{1}{\psi}\frac{\partial^2\psi}{\partial\phi^2}$$ (171)

$$\nabla^2 \Phi = 0$$ (172)

Spherical Laplacian Solution

$$\nabla^2 \Phi = 0$$

$$\Phi = \sum_{l,m}Y^m_l(\Omega)\bigg[A_lr^l + \frac{B_l}{r^{l+1}}\bigg]$$ (173)

$$= \sum_{l,m} P_l(\cos{\theta})e^{im\phi}\bigg[A_lr^l + \frac{B_l}{r^{l+1}}\bigg]$$

Legendre Polynomials

$$P_0(\cos{\theta}) = 1$$

$$P_1(\cos{\theta}) = \cos{\theta}$$

$$P_2(\cos{\theta}) = \frac{1}{2}\big(3 \cos^2{\theta} - 1 \big)$$ Orthogonality relationship:

$$\int_{-1}^1 P_m(x) P_n(x) dx = \frac{2}{2n+1}\delta_{mn}$$ (174)

Gaussian Integrals

$$\boxed{\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}}$$ (175)

Square the full integral over a second variable, pass this to polar coordinates, and perform the (now easy!) integral.

$$\int_{-\infty}^{\infty}e^{-x^2}dx \int_{-\infty}^{\infty}e^{-y^2}dy = \int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy$$

$$= \int_0^{\infty} dr \int_0^{2\pi}rd\theta e^{-r^2}$$

$$u = r^2, du = 2rdr$$

etc.

$$\boxed{\int_{\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}}$$ (176)

Scale $x\rightarrow \sqrt{\alpha}x',dx=\sqrt{\alpha}dx'$. Easy.

$$\boxed{\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2}dx = \frac{\sqrt{\pi}}{2a^{3/2}}}$$ (177)

The trick here is to differentiate under the integral sign (a professed favorite trick of a certain someone).

$$\int_{-\infty}^{\infty}x^2 e^{-\alpha x^2}dx$$

$$= - \int_{-\infty}^{\infty} \frac{d}{d\alpha} e^{-\alpha x^2} dx$$

$$= - \frac{d}{d\alpha}\int_{-\infty}^{\infty} e^{-\alpha x^2} dx$$

$$= - \frac{d}{d\alpha} \sqrt{\frac{\pi}{\alpha}}$$

Advanced Gaussian Integrals
Of the type used in paths, density matrices, etc.

$$\boxed{\int_{-\infty}^{\infty}dx e^{-\frac{a}{2}x^2 + Jx} = \sqrt{\frac{2\pi}{\alpha}}e^{J^2/2\alpha} }$$ (178)

$$\int_{-\infty}^{\infty}dx e^{- \frac{\alpha}{2} + Jx}$$ complete the square in the exponent.

$$-\frac{\alpha}{2}x^2 + Jx = -\frac{\alpha}{2}\bigg(x^2 - \frac{2Jx}{\alpha}\bigg)$$

$$= -\frac{\alpha}{2}\bigg(x - \frac{J}{\alpha}\bigg)^2 + \frac{J^2}{2\alpha}$$

$$\rightarrow \int_{-\infty}^{\infty}e^{-\frac{\alpha}{2}(x-\frac{J}{\alpha})^2} dx e^{J^2/2\alpha}$$

$$u = x-\frac{J}{\alpha}, du=dx$$

Now this is a Gaussian integral of the 'easy' type from the previous section.

$$\boxed{ \int_{-\infty}^{\infty}dx e^{- \frac{\alpha}{2} + iJx}= \sqrt{\frac{2\pi}{\alpha}}e^{-J^2/2\alpha}}$$ (179)

Just sub $J\rightarrow iJ$ in the previous solution.

$$\boxed{ \int_{-\infty}^{\infty}dx e^{i\frac{\alpha}{2} + iJx}= \sqrt{\frac{2\pi i}{\alpha}}e^{-J^2/2\alpha}}$$ (180)

Sub $\alpha \rightarrow -i\alpha$ in the previous solution.