# 线性代数入门 Introduction to Linear Algebra MATH103

0

Let $[., .]$ denote an indefinite inner product defined on $\mathrm{R}^{n}$ by a real symmetric invertible matrix $H$ of size $n$, i.e.,,
$$[x, y]=(H x, y), \quad \text { for all } x, y \in \mathrm{R}^{n} .$$
Recall that (… ) stands for the standard inner product in $\mathrm{R}^{n}$ :
$$(x, y)=\sum_{j=1}^{n} x^{(j)} y^{(j)}$$
where $x=\left\langle x^{(1)}, \ldots, x^{(n)}\right\rangle, y=\left\langle y^{(1)}, \ldots, y^{(n)}\right\rangle$ are vectors in $\mathrm{R}^{n}$. The adjoint $A^{[]}$ of a real $n \times n$ matrix $A$ is defined just as in (4.1.2) and, as in (4.1.3) it is easily seen that $A^{[]}=H^{-1} A^{} H$. Since $A^{}$ is now just the transpose of $A, A^{[]}$ is obviously real. The following facts and definitions are all formally identical with predecessors in earlier chapters: A real matrix $A$ is $H$-selfadjoint if $A^{}=A$, i.e., if $H A=A^{} H$. A real matrix $A$ is $H$-unitary if $A^{} A=I$, i.e., if $A^{} H A=H$. A real matrix $A$ is $H$-normal if $A^{} A=A A^{}$, i.e., if $\left(H^{-1} A^{} H\right) A=A\left(H^{-1} A^{*} H\right)$.

## MATH103COURSE NOTES ：

only if there is a $(B, G) \in S_{r}$ which can be transformed to $\left(J, P_{e, J}\right)$ by both $T_{+}$ and $T_{-}$, say, with $\operatorname{det} T_{+}>0$ and $\operatorname{det} T_{-}<0$. Thus,
$$B=T_{+}^{-1} J T_{+}=T_{-}^{-1} J T_{-}, \quad G=T_{+}^{} P_{\varepsilon, J} T_{+}=T_{-}^{} P_{\varepsilon, J} T_{-},$$
and it follows immediately that $U S_{+}=U S_{-}$if and only if there is a real $T$ with negative determinant such that
$$J=T^{-1} J T, \quad P_{\varepsilon, J}=T^{*} P_{\varepsilon, J} T .$$

# 计算物理学概论 Introduction to Computational Physics PHYS105

0

$$f^{\prime}\left(x_{0} \pm h\right)=\frac{\mp f\left(x_{0} \pm h\right) \pm f\left(x_{0}\right)}{h}+O(h)$$
and if we stop the Taylor expansion at that point our function becomes,
$$f(x)=f_{0}+\frac{f_{h}-f_{0}}{h} x+O\left(x^{2}\right) .$$
for $x=x_{0}$ to $x=x_{0}+h$ and
$$f(x)=f_{0}+\frac{f_{0}-f_{-h}}{h} x+O\left(x^{2}\right),$$
for $x=x_{0}-h$ to $x=x_{0}$. The error goes like $O\left(x^{2}\right)$. If we then evaluate the integral we obtain
$$\int_{-h}^{+h} f(x) d x=\frac{h}{2}\left(f_{h}+2 f_{0}+f_{-h}\right)+O\left(h^{3}\right),$$

## PHYS105 COURSE NOTES ：

$$f(x)=f_{0}+\frac{f_{h}-f_{-h}}{2 h} x+\frac{f_{h}-2 f_{0}+f_{-h}}{h^{2}} x^{2}+O\left(x^{3}\right)$$
Inserting this formula in the integral we obtain
$$\int_{-h}^{+h} f(x) d x=\frac{h}{3}\left(f_{h}+4 f_{0}+f_{-h}\right)+O\left(h^{5}\right)$$

# 量子物理学的基础 Foundations of Quantum Physics PHYS104

0

$$\sum_{\lambda} F_{\lambda ‘ \lambda}^{j} C_{\lambda j}=\epsilon_{j} \sum_{\lambda} S_{\lambda^{\prime} \lambda} C_{\lambda j},$$
where
\begin{aligned} F_{\lambda^{\prime} \lambda}^{j}=&\left\langle\chi_{\lambda^{\prime}}|h| \chi_{\lambda}\right\rangle+\sum_{\delta \kappa}\left[\gamma_{\delta_{k}}\left\langle\chi_{\lambda^{\prime}} \chi_{\delta}|g| \chi_{\lambda} \chi_{\chi^{\prime}}\right\rangle\right.\ &\left.-\gamma_{\delta \kappa}^{\text {exch }}\left\langle\chi_{\lambda^{\prime} \cdot} \chi_{\delta}|g| \chi_{\kappa} \chi_{\lambda}\right\rangle\right] . \end{aligned}
Here, $h$ is the one-electron part of the full Hamiltonian, $g$ is an electron-electron repulsion potential energy, and
$$\begin{gathered} \gamma_{\delta x}=\sum_{i}^{\prime} C_{\delta i} C_{\kappa i}, \ \gamma_{\delta \kappa}^{\text {exch }}=\sum_{i}^{n \prime} C_{\delta i} C_{\kappa i}, \end{gathered}$$

## PHYS104COURSE NOTES ：

Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$

# 微积分 Calculus MATH101/MATH102

0

For $h \neq 0$ and $x+h$ in the domain of $f$,
$$f(x+h)-f(x)=\frac{f(x+h)-f(x)}{h} \cdot h$$
With $f$ differentiable at $x$,
$$\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$$ Since $\lim {h \rightarrow 0} h=0$, we have
$$\lim {h \rightarrow 0}[f(x+h)-f(x)]=\left[\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right] \cdot\left[\lim _{h \rightarrow 0} h\right]=f^{\prime}(x) \cdot 0=0 .$$

## MATH101/MATH102COURSE NOTES ：

Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$

# 广义相对论 General Relativity PHYS5390M01

0

$$g\left(\lambda^{3} \gamma, \lambda \pi_{A^{\prime}}\right)=\lambda^{-1} g\left(\gamma, \pi_{A^{\prime}}\right) \forall \lambda \in C-{0} .$$
This enables one to define the spinor field
$$p^{A} \equiv p^{A A^{\prime} B^{\prime}} \pi_{A^{\prime}} \pi_{B^{\prime}} \quad,$$
and the patching function
$$f^{A} \equiv p^{A} g\left(p_{B} \omega^{B}, \pi_{B^{\prime}}\right),$$
and the function
$$F\left(x^{a}, \pi_{A^{\prime}}\right) \equiv g\left(i p_{A} x^{A C^{\prime}} \pi_{C^{\prime}}, \pi_{A^{\prime}}\right) .$$

## PHYS5390M01COURSE NOTES ：

and such that
$$\psi_{A^{\prime} B^{\prime} C^{\prime}}=\mathcal{D}{A A^{\prime}} \gamma{B^{\prime} C^{\prime}}^{A}$$
The second potential is a spinor field symmetric in its unprimed indices
$$\rho_{C^{\prime}}^{A B}=\rho_{C^{\prime}}^{(A B)}$$
subject to the equation
$$\mathcal{D}^{C C^{\prime}} \rho_{C^{\prime}}^{A B}=0$$
and it yields the $\gamma_{B^{\prime} C^{\prime}}^{A}$ ‘ potential by means of
$$\gamma_{B^{\prime} C^{\prime}}^{A}=\mathcal{D}{B B^{\prime}} \rho{C^{\prime}}^{A B}$$
If we introduce the spinor fields $v_{C^{\prime}}$ and $x^{B}$ obeying the equations
$$\begin{gathered} \mathcal{D}^{A C^{\prime}} \nu_{C^{\prime}}=0 \ \mathcal{D}{A C^{\prime}} \chi^{A}=2 i \nu{C^{\prime}} \end{gathered}$$

# 量子现象 Physics 4- Quantum Phenomena PHYS231101

0

To calculate $R$ and $T$ we need to evaluate the current $J$ for $z<0$ and for $z>0$. Using the wavefunction from we have
$$J(z<0)=-\frac{|C|^{2} g \hbar k_{1}}{m^{}}\left(1-|r|^{2}\right)=J_{i}-J_{\mathrm{r}}$$ Similarly, using the wavefunction from eq. $(1.39)$, $$J(z>0)=-\frac{|C|^{2} q \Phi k_{2}}{m^{}}|t|^{2}=J_{1}$$
Hence from
\begin{aligned} &R=J_{1} / J_{1}=|r|^{2} \ &T=J_{3} / J_{1}=|r|^{2} k_{2} / k_{1} \end{aligned}

## PHYS231101COURSE NOTES ：

Continuity Equation: Consider a single electron whose probability density is given
$$n(\mathbf{r}, t)=\Psi *(\mathbf{r}, t) \Psi(\mathbf{r}, t)$$
and whose probability current density is given
$$\mathbf{J}(\mathbf{r}, t)=-\frac{i q h}{2 m *}[(\nabla \Psi r) * \Psi-\Psi *(\nabla \Psi)]$$

# 高能量天体物理学 High Energy Astrophysics PHYS201501

0

$$H=\frac{p^{2}}{2 m_{e}}-V(r)$$
Making the usual substitution: $\boldsymbol{p}=-i \hbar \nabla$ we get the relevant form of (114):
$$\left(-\frac{\hbar^{2}}{2 m_{e}} \nabla^{2}-V(r)\right) \psi(\boldsymbol{r})=E \psi(\boldsymbol{r})$$
It is convenient to use atomic units where the natural unit of length is the Bohr radius: $a_{0} \equiv \hbar^{2} / m e^{2}=0.52910^{-8} \mathrm{~cm}$, and the natural unit of energy is twice the Rydberg constant: $e^{2} / a_{0} \equiv 2 R y=27.2 \mathrm{eV}=4.3610^{-11}$ erg. In these units, $e=\hbar=m=1$.
Equation (119) then takes the form:
$$\left(\frac{1}{2} \nabla^{2}+E+V(r)\right) \psi(\boldsymbol{r})=0$$

## PHYS201501COURSE NOTES ：

$$\frac{1}{2} m v^{2}=\frac{Z e^{2}}{2 r}$$
For the ground-state:
$$r \simeq \frac{a_{0}}{Z},$$
and thus:
$$v \simeq\left(\frac{Z^{2} e^{2}}{m a_{0}}\right)^{1 / 2}=(Z \alpha) c$$

# 纳米技术入门 Introduction to Nanotechnology PHYS131101

0

This is often written in the abbreviated form,
$$i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t)=\left[-\frac{\hbar^{2}}{2 m} \nabla^{2}+V(\mathbf{r})\right] \psi(\mathbf{r}, t),$$
and the momentum operator $\hat{p}$ itself as
$$\hat{\mathbf{p}}=\frac{\hbar}{i}\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)=\frac{\hbar}{i} \boldsymbol{\nabla} .$$

## PHYS131101COURSE NOTES ：

$$n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\mathbf{E}(\mathbf{r}, t)|^{2} .$$
Naturally, the total number of photons $N(t)$ at time $t$ is found by integrating over the entire volume $\mathcal{V}$,
$$N(t)=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t) .$$
Returning to the quantum wavefunction $\psi(\mathbf{r}, t)$ of a single particle we postulate in analogy withthat the intensity $|\psi(\mathbf{r}, t)|^{2}$ of the wavefunction is related to the particle density $n$ and write
$$n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\psi(\mathbf{r}, t)|^{2} .$$
But since we have only one particle the analogy reduces to
$$1=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t)$$

# 数学 Maths 2 PHYS130001/PHYS238001

0

Let $b_{\alpha \beta \mid \sigma}:=\partial_{\sigma} b_{\alpha \beta}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}-\Gamma_{\beta \sigma}^{\tau} b_{\alpha \tau}$ denote the first-order covariant derivatives of the curvature tensor, defined here by means of its covariant components. Show that these covariant derivatives satisfy the Codazzi-Mainardi identities
$$b_{\alpha \beta \mid \sigma}=b_{\alpha \sigma \mid \beta}$$
which are themselves equivalent to the relations (Thm. 2.8-1)
$$\partial_{\sigma} b_{\alpha \beta}-\partial_{\beta} b_{\alpha \sigma}+\Gamma_{\alpha \beta}^{\tau} b_{\tau \sigma}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}=0$$
Hint: The proof is analogous to that given in for establishing the relations $\left.b_{\beta}^{\tau}\right|{\alpha}=\left.b{\alpha}^{\tau}\right|_{\beta}$.

## PPHYS130001/PHYS238001COURSE NOTES ：

$u_{i}^{\varepsilon}\left(x^{\varepsilon}\right)=u_{i}(\varepsilon)(x)$ for all $x^{\varepsilon}=\pi^{\varepsilon} x \in \bar{\Omega}^{\varepsilon}$,
where $\pi^{c}\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}, x_{2}, \varepsilon x_{3}\right)$. We then assume that there exist constants $\lambda>0, \mu>0$ and functions $f^{i}$ independent of $\varepsilon$ such that
$$\begin{gathered} \lambda^{\varepsilon}=\lambda \text { and } \mu^{\varepsilon}=\mu, \ f^{i, \varepsilon}\left(x^{\varepsilon}\right)=\varepsilon^{p} f^{i}(x) \text { for all } x^{\varepsilon}=\pi^{e} x \in \Omega^{\varepsilon}, \end{gathered}$$

# 震动与热物理 Vibrations & Thermal Phys (JH) PHYS128001

0

It can also be written
$$\frac{D \rho}{D t}+\rho \nabla \cdot \mathbf{u}=0$$
where a widely-used abbreviation for the differential operator is
$$\frac{D}{D t} \equiv \frac{\partial}{\partial t}+\mathbf{u} . \nabla$$
An important special case is that of an incompressible fluid in which there is no change of density following the motion of the fluid. If a small volume of fluid is labeled, then the fluid bearing that label always has the same density wherever it is in the future, although its neighbors may have different densities. The idea can be expressed in mathematical form by considering a short interval of time $\delta t$; the idea of incompressibility means that
$$\rho(\mathbf{r}+\mathbf{u} \delta t, t+\delta t)=\rho(\mathbf{r}, t)+O\left(\delta t^{2}\right)$$

## PHYS128001COURSE NOTES ：

$$\frac{\partial S}{\partial t}+\mathbf{u} \cdot \nabla S=\nabla . \kappa \nabla S+q$$
The special case when $\rho, c_{p}$, and $k$ are all constants reduces the equation to
$$\frac{\partial T}{\partial t}+\mathbf{u} . \nabla T=\kappa \nabla^{2} T+\frac{q}{\rho c_{p}}$$
If, in addition, the entire medium is moving with a constant velocity $U$ in the direction of the positive $x$ axis,
$$\frac{\partial T}{\partial t}+U \frac{\partial T}{\partial x}=\kappa \nabla^{2} T+\frac{q}{\rho c_{p}}$$