不确定原理

bilibili

1.Uncertainty Principle

①Heisenberg Uncertainty Principle(HUP)

$$\Delta x\Delta p\geq\frac{\hbar}{2}$$

②Generalized Uncertainty Principle(GUP)

$$\Delta A\Delta B\geq\frac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|$$

Derivation:

$$\Delta\hat{A}^2 = \langle\hat{A}^2\rangle - \langle\hat{A}\rangle^2$$

Expectation value:

$$\langle\hat{A}\rangle = \bra{\psi}\hat{A}\ket{\psi}$$

2.Proof of GUP

偏差算符：

$$\hat{\sigma}_A = \hat{A} - \langle\hat{A}\rangle$$ $$\hat{\sigma}_B = \hat{B} - \langle\hat{B}\rangle$$

$\hat{\sigma}_A$,$\hat{\sigma}_B$性质
Hermitian:

$$\hat{\sigma}_A^\dagger = \left(\hat{A} - \langle\hat{A}\rangle\right)^\dagger = \hat{A}^\dagger - \langle\hat{A}\rangle^\dagger = \hat{A} - \langle\hat{A}\rangle = \hat{\sigma}_A$$

=>

$$\hat{\sigma}_A^\dagger\hat{\sigma}_A = \hat{\sigma}_A^2$$

Commutator:

$$\left[\hat{\sigma}_A,\hat{\sigma}_B\right] = \left[\hat{A},\hat{B}\right]$$ $$\hat{\sigma}_A\hat{\sigma}_B = \frac{1}{2}\left[\hat{\sigma}_A,\hat{\sigma}_B\right]+\frac{1}{2}\left\{\hat{\sigma}_A,\hat{\sigma}_B\right\}$$

Expectation value of $\left[\hat{\sigma}_A,\hat{\sigma}_B\right]$,$\left\{\hat{\sigma}_A,\hat{\sigma}_B\right\}$:

$$\left[\hat{\sigma}_A,\hat{\sigma}_B\right]^\dagger = \left(\hat{\sigma}_A\hat{\sigma}_B - \hat{\sigma}_B\hat{\sigma}_A\right)^\dagger = \hat{\sigma}_B^\dagger\hat{\sigma}_A^\dagger - \hat{\sigma}_A^\dagger\hat{\sigma}_B^\dagger = -\left[\hat{\sigma}_A,\hat{\sigma}_B\right]$$

(skew-Hermitian)

if $\hat{A}$ is skew-Hermitian,then $\langle\hat{A}\rangle=c\in\mathbb{I}$

任意算符可分解为两个算符，其中一个是厄米算符，另一个是反厄米算符

$$\left\{\hat{A},\hat{B}\right\}^\dagger = \left(\hat{A}\hat{B} + \hat{B}\hat{A}\right)^\dagger = \hat{B}^\dagger\hat{A}^\dagger + \hat{A}^\dagger\hat{B}^\dagger = \left\{\hat{A},\hat{B}\right\}$$

(Hermitian)
=>

$$\langle\hat{\sigma}_A\hat{\sigma}_B\rangle = \langle\frac{1}{2}\left[\hat{\sigma}_A,\hat{\sigma}_B\right]+\frac{1}{2}\left\{\hat{\sigma}_A,\hat{\sigma}_B\right\}\rangle = \frac{1}{2}\langle\left[\hat{\sigma}_A,\hat{\sigma}_B\right]\rangle+\frac{1}{2}\langle\left\{\hat{\sigma}_A,\hat{\sigma}_B\right\}\rangle$$

=>

$$\begin{aligned} \left|\langle\hat{\sigma}_A\hat{\sigma}_B\rangle\right| &=\frac{1}{2}\left|\langle\left[\hat{\sigma}_A,\hat{\sigma}_B\right]\rangle\right|+\frac{1}{2}\left|\langle\left\{\hat{\sigma}_A,\hat{\sigma}_B\right\}\rangle\right| \\ &\geqslant\frac{1}{2}\left|\langle\left[\hat{\sigma}_A,\hat{\sigma}_B\right]\rangle\right|=\frac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right| \end{aligned}$$

Now proof:

$$\langle\hat{\sigma}_A^2\rangle\langle\hat{\sigma}_B^2\rangle\geqslant\frac{1}{4}\left|\langle\left[\hat{\sigma}_A,\hat{\sigma}_B\right]\rangle\right|^2$$

Cauthy-Schwarz inequality:
①

$$uv \geqslant \vec{u}\cdot\vec{v}$$

②

$$\boxed{\braket{u|u}\braket{v|v} \geqslant \left|\braket{u|v}\right|^2}$$

equality iff $\vec{u} = \lambda\vec{v}$

$$\begin{align*} \langle\hat{\sigma}_A^2\rangle\langle\hat{\sigma}_B^2\rangle&=\bra{\psi}\hat{\sigma}_A^2\ket{\psi}\bra{\psi}\hat{\sigma}_B^2\ket{\psi} \\ &= \bra{\psi}\hat{\sigma}_A^\dagger\hat{\sigma}_A\ket{\psi}\bra{\psi}\hat{\sigma}_B^\dagger\hat{\sigma}_B\ket{\psi} \\ &= \braket{\psi_A|\psi_A}\braket{\psi_B|\psi_B} \\ &\geqslant \left|\braket{\psi_A|\psi_B}\right|^2 \\ &= \left|\bra{\psi}\hat{\sigma}_A\hat{\sigma}_B\ket{\psi}\right|^2 \\ \end{align*}$$

=>

$$\left\{\begin{gather*} \Delta\hat{A}&=\langle\hat{\sigma}_A^2\rangle^{\frac{1}{2}} \\ \Delta\hat{B}&=\langle\hat{\sigma}_B^2\rangle^{\frac{1}{2}} \\ \end{gather*} \right.$$

So we have:

$$\Delta\hat{A}\Delta\hat{B}\geqslant\frac{1}{2}\left|\langle\left[\hat{\sigma}_A,\hat{\sigma}_B\right]\rangle\right|$$

then:

$$\boxed{\Delta\hat{A}\Delta\hat{B}\geqslant\frac{1}{2}\left|\langle\left[\hat{A},\hat{B}\right]\rangle\right|}$$

存在一个态$\ket{\psi}$使得等号成立，则$\ket{\psi}$称为最小不确定态(minimum uncertainty state)

Minimum uncertainty states / Minimum uncertainty wave packets

$\hat{x},hat{p}$=> $\Delta\hat{x}\Delta\hat{p}\geqslant\frac{\hbar}{2}$

Define:

$$\hat{\sigma}_x = \hat{x}-\langle\hat{x}\rangle,\quad\hat{\sigma}_p = \hat{p}-\langle\hat{p}\rangle$$ $$\ket{\alpha} = (\hat{\sigma}_x+i\hat{\sigma}_p)\ket{\psi}$$

$\ket{\psi}$为最小不确定态
$\braket{\alpha|\alpha} \geqslant 0$ equality iff $\ket{\alpha} = \vec{0}$

$$\braket{\alpha|\alpha} = \bra{\psi}(\hat{\sigma}_x-i\hat{\sigma}_p)(\hat{\sigma}_x+i\hat{\sigma}_p)\ket{\psi} \geqslant 0$$ $$\begin{align*} l.h.s &= \bra{\psi}(\hat{\sigma}_x-i\lambda\hat{\sigma}_p)(\hat{\sigma}_x+i\lambda\hat{\sigma}_p)\ket{\psi} \\ &= \bra{\psi}(\hat{\sigma}_x^2+\hat{\sigma}_p^2-i\lambda\left[\hat{\sigma}_x,\hat{\sigma}_p\right])\ket{\psi} \\ &= \langle\hat{\sigma}_x^2\rangle+\langle\hat{\sigma}_p^2\rangle\cdot\lambda^2+i\lambda(i\hbar) \\ &= \langle\hat{\sigma}_p^2\rangle\cdot\lambda^2-\hbar\lambda +\langle\hat{\sigma}_x^2\rangle\geqslant 0 \end{align*}$$

二次多项式，所以

$$\Delta = (-\hbar)^2-4\langle\hat{\sigma}_x^2\rangle\langle\hat{\sigma}_p^2\rangle \leqslant 0$$

所以

$$\langle\hat{\sigma}_x^2\rangle\langle\hat{\sigma}_p^2\rangle \geqslant \frac{\hbar^2}{4}$$

即

$$\Delta\hat{x}\Delta\hat{p}\geqslant\frac{\hbar}{2}$$

Heisenberg uncertainty principle
when $\Delta=0$ => $ \Delta\hat{x}\Delta\hat{p}=\frac{\hbar}{2}$

$$\lambda = \frac{2\Delta\hat{x}^2}{\hbar}$$

=>

$$\ket{\alpha}\equiv 0 = (\hat{\sigma}_x+i\lambda\hat{\sigma}_p)\ket{\psi}$$ $$\bra{x}(\hat{\sigma}_x+i\lambda\hat{\sigma}_p)\ket{\psi} = 0$$

=>

$$\bra{x}\left[\hat{x}-\langle\hat{x}\rangle+i\lambda\left(\hat{p}-\langle\hat{p}\rangle\right)\right]\ket{\psi} = 0$$ $$\begin{align*} \hat{x}\ket{x} &=x\ket{x} \\ \hat{p} &= -i\hbar\frac{\partial}{\partial x} \end{align*}$$

=>

$$\left(x-\langle\hat{x}\rangle\right)\psi(x)+\lambda\left(\hbar\frac{d}{d x}-i\langle\hat{p}\rangle\right)\psi(x) = 0$$

=>

$$\psi(x) = \frac{1}{\sqrt{\lambda}}\exp\left(-\frac{x-\langle\hat{x}\rangle}{2\lambda\hbar}\right)$$

=>

$$\boxed{|\psi(x)|^2 \sim \exp\left[-\left(\frac{x-\langle\hat{x}\rangle}{\Delta \hat{x}}\right)^2\right]}$$

(Gauss function)

Two Remarks:
①只要算符不对易，就有不确定性
②可以找到最小不确定态

4.Energy-time uncertainty principle

$$\Delta E\Delta T \geqslant\frac{\hbar}{2}$$

Time is NOT an observable

Ehrenfest’s theorem:

$$\frac{d}{dt}\langle\hat{A}\rangle = \frac{1}{i\hbar}\langle\left[\hat{A},\hat{H}\right]\rangle+\left\langle\frac{\partial\hat{A}}{\partial t}\right\rangle$$

if $\left[\hat{A},\hat{H}\right]\neq0$ and $\frac{\partial\hat{A}}{\partial t}=0$
=>

$$\Delta\hat{A}\Delta\hat{H} \geqslant\frac{1}{2}\left|\langle\left[\hat{A},\hat{H}\right]\rangle\right|$$

=>

$$\frac{\Delta \hat{A}}{\left|\frac{d}{dt}\langle\hat{A}\rangle\right|}\Delta \hat{H} \geqslant\frac{\hbar}{2}$$ $$\Delta T = \frac{\Delta \hat{A}}{\left|\frac{d}{dt}\langle\hat{A}\rangle\right|}$$