守恒量

Derivation

($In \,Schr \ddot{o}dinger ‘s \, Picture$)

1. $$i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle$$
2. $$\langle F\rangle=\langle\psi(t)| F|\psi(t)\rangle$$
3. $$\begin{align*}\dfrac d{dt}\langle F\rangle=&\dfrac d{dt}\langle\psi(t)| F|\psi(t)\rangle\\=&\frac{\partial}{\partial t}\langle\psi(t)|\cdot F|\psi(t)\rangle+\langle\psi(t)| \cdot \frac{\partial F}{\partial t}\cdot |\psi(t)\rangle+\langle\psi(t)| F\cdot \frac{\partial}{\partial t}|\psi(t)\rangle\\=&\frac{1}{-i\hbar}\langle\psi(t)| HF|\psi(t)\rangle+ \langle\frac{\partial F}{\partial t}\rangle+\frac{1}{i\hbar}\langle\psi(t)| FH |\psi(t)\rangle\\=&\langle\frac{\partial F}{\partial t}\rangle+\frac{1}{i\hbar}\langle\psi(t)| [F,H]|\psi(t)\rangle\\=&\langle\frac{\partial F}{\partial t}\rangle+\frac{1}{i\hbar}\langle[F,H]\rangle\end{align*}$$
   如无特殊声明，一般认为$F$不含时，即$\dfrac{\partial F}{\partial t}=0$，所以
   If $[F,H]\equiv0\Rightarrow \dfrac d{dt}\langle F\rangle=0$
   i.e. $F$ is a conserved variable.

Properties

1. Average is invariant at any state $|\psi\rangle$.
2. Probability is invariant, $\dfrac{d}{dt}|a_k(t)|^2\equiv0$
3. Distinction of conserved quantities between classical and quantum mechanics.
   a) Not having a certain value(state), but a certain probability.
   b) A quantum system may have several conserved variables.
   c) Nothing to do with stationary state.
4. Degeneracy stongly related to conserved quantities.
   a) If $[F,H]\equiv 0,[G,H]\equiv 0, \text{but } [F,G]\equiv 0$, in most cases degeneracy exist.
   b) Lemma: if $[F,H]\equiv C(\neq 0)$, then the system is degenerated with $\infty$ degrees.

Lemma

Suppose

$$|\psi\rangle\to |\psi\rangle=Q|\psi\rangle$$
$$i\hbar\dfrac{\partial}{\partial t}|\psi(t)\rangle=H|\psi(t)\rangle\Rightarrow i\hbar \dfrac{\partial}{\partial t}|\psi‘(t)\rangle=H|\psi‘(t)\rangle$$
$$\langle \psi|\psi\rangle=1,\langle \psi‘|\psi‘\rangle=1$$

Then

1. $\begin{cases}H|\psi‘\rangle=i\hbar \dfrac{\partial}{\partial t}|\psi‘ \rangle=i\hbar \dfrac{\partial}{\partial t}Q|\psi \rangle=Qi\hbar \dfrac{\partial}{\partial t}|\psi\rangle=QH|\psi\rangle\\ H|\psi‘\rangle=HQ|\psi\rangle\end{cases}$
   $\Rightarrow [Q,H]=0$
2. $\langle\psi‘|\psi‘\rangle=\langle\psi‘|Q^{\dagger}Q|\psi\rangle=\langle\psi|\psi\rangle=1\Rightarrow Q^{\dagger}Q=1\Rightarrow Q^{\dagger}=Q^{-1}\text{(Unitary Operaor)}$

Assume

$\hat Q=1+i\varepsilon \hat F(\varepsilon\ll 1)$
1. $[Q,H]=0\Rightarrow [F,H]=0$
2. $Q^{\dagger}Q=1\Rightarrow 1+i\varepsilon(F-F^{\dagger})+ +\varepsilon^2F^{\dagger}F=0\Rightarrow F^{\dagger}=F$

hence $F$ is capable for a observable, particularly, a conserved observable.

动量守恒和角动量守恒

Conservation of Momentum

Invariance in Displacement $\Leftrightarrow$ Conservation of momentum

$\bf\begin{cases} x\to x‘= x +\delta x\\ |\psi\rangle\to |\psi‘\rangle=D|\psi\rangle\\ \bf|\psi‘(x‘)\rangle=|\psi(x)\rangle\end{cases}$

$\Rightarrow D\bf|\psi(x+\delta x)\rangle=|\psi(x)\rangle$

$$\begin{align*}D\bf|\psi(x)\rangle=&D\bf|\psi(x-\delta x)\rangle\\=&\bf|\psi(x)\rangle-\delta x\cdot \frac{\partial}{\partial x}|\psi(x)\rangle+\cdots+\frac{(\delta x)^n}{n!}\cdot \frac{\partial^n}{\partial x^n}|\psi(x)\rangle+\cdots\\=&e^{-i\frac{\delta \bf x\cdot p}{\hbar}}|\psi(\bf x)\rangle\\ \approx&[1-i\frac{\delta \bf x\cdot p}{\hbar}]|\psi(\bf x)\rangle\end{align*}$$

$D=[1-i\dfrac{\bf \delta x\cdot p}{\hbar}]$
Since $[D,H]=0$, hence $[\mathbf{p},H]=0$

Conservation of Angular Momentum

Invariance in Rotation $\Leftrightarrow$ Conservation of angular momentum

$\bf\begin{cases} x\to x‘= x +\delta x,\delta x=\delta\vec{\varphi}\times x\\ |\psi\rangle\to |\psi‘\rangle=R|\psi\rangle\\ \bf|\psi‘(x‘)\rangle=|\psi(x)\rangle\end{cases}$

$\Rightarrow R\bf|\psi(x+\delta x)\rangle=|\psi(x)\rangle$

$$\begin{align*}R\bf|\psi(x)\rangle=&R\bf|\psi(x-\delta x)\rangle\\=&\bf|\psi(x)\rangle-\delta x\cdot \frac{\partial}{\partial x}|\psi(x)\rangle+\cdots+\frac{(\delta x)^n}{n!}\cdot \frac{\partial^n}{\partial x^n}|\psi(x)\rangle+\cdots\\=&e^{-i\frac{\delta \bf x\cdot p}{\hbar}}|\psi(\bf x)\rangle\\=&e^{i\bf \frac{(\delta\vec{\varphi}\times x)\cdot p}{\hbar}}|\psi(\bf x)\rangle\\=&e^{i\bf \frac{\delta\vec{\varphi}\cdot （x\times p）}{\hbar}}|\psi(\bf x)\rangle\\ \approx&[1-i\bf \frac{\delta\vec{\varphi}\cdot (x\times p)}{\hbar}]|\psi(\bf x)\rangle\end{align*}$$

$R=[1-i\bf \dfrac{\delta\vec{\varphi}\cdot (x\times p)}{\hbar}]=[1-i\bf \dfrac{\delta\vec{\varphi}\cdot L}{\hbar}]$
Since $[R,H]=0$, hence $[\mathbf{L},H]=0$

Thanks to Prof. Guo Hong