Heisenberg Uncertainty Principle
$$測不準原理關係式:\;\;\;\Delta x\Delta p\geqslant \frac{\hbar}{2}$$
$$in\; ring\; theory,\; commutator[a,b]\equiv ab-ba$$ $$if \;[a,b]= ab-ba =0\; commute$$ $$if \;[a,b] = ab-ba \neq 0\; non-commute$$In March 1926,Heisenberg showed that the commutation relation implies an uncertainty, or in Bohr's language a
實驗觀點:
粒子的位置與動量不可同時被確定，位置的不確定性越小，則動量的不確定性越大，反之亦然。* Heisenberg's microscope:
If the photon has a short wavelength, the position can be measured accurately.
If the photon has a long wavelength, the collision does not disturb the electron's
momentum very much, but the scattering will reveal its position only vaguely.
$$\Delta p_x= p\sin(\theta)\sim p\theta \:\:\:光子動量在x分量上的不準度$$ $$\Delta x = \frac{\lambda}{2\theta} \;\;位置在透鏡的解析度$$ $$\Delta x \Delta p \sim \frac{\lambda}{2\theta} p \theta = \frac{\lambda}{2p}=\frac{h}{2}\;\;位置在透鏡的解析度$$
* 觀察者效應(observer effect):
In physics, the observer effect is the theory that simply observing a situation
or phenomenon necessarily changes that phenomenon
* From wave-particle duality to Heisenberg Uncertainty Principle:
Single Slit diffraction
$$\sin \theta = \lambda /w: \;\; 中央極大亮區的張角$$ $$\Delta p_y\sim p\sin(\theta)\sim p\lambda /w: \:\:\:光子動量在y分量上的不準度$$ $$\Delta y \sim w:\;\;\;\; 粒子通過狹縫時的不準度等於狹縫寬度w$$ $$\Delta y \Delta p_y \sim p\lambda $$ $$\lambda=\frac{h}{p}:\;\;\;\;\; de\; Broglie\; wave $$ $$\Delta y \Delta p_y \sim h $$
波動力學觀點:
在量子論中, 粒子的位置可以用波函數.
假設粒子波函數的空間部分是單色平面波
$$\psi (x)\propto e^{ik_{0}x}=e^{ip_{0}x/\hbar }；$$ $$P[a\leq x\leq b]=\int _{a}^{b}|\psi (x)|^{2}\,\mathrm {d} x:\;\;\;在位置a與b之間找到粒子的機率$$ 但平面波為空間均勻分佈,我們如要考慮粒子的位置, 應考慮以波包的方式表示(多個正弦波疊加)
$$\psi (x)\propto \sum _{n}A_{n}e^{ip_{n}x/\hbar }:\;\;\;\;通式: 不同正弦波疊加形成的波函數$$ $$\psi (x)={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }\phi (p)\cdot e^{ipx/\hbar }\,\mathrm {d} p；:\;\;\;\;通式: 連續傅立葉轉換$$
廣義的數學解釋:
任意一組正則共軛物理量(Canonical Conjugate quantities)不可同時被確定，其中一個物理量的不確定性越小，*Fourier transform
Canonical Conjugate quantities
Momentum Position
other examples: t & E; A and i, V and q
*考慮極限情形:高斯波包 (Gaussian wave packet)
$$\psi (x)=\left({\frac {A}{\pi }}\right)^{1/4}e^{-{Ax^{2}/2}}:\;\;\; Gaussian\; wave\; packet$$ $$\sigma _{x}^{2}=\langle x^{2}\rangle =\left({\frac {A}{\pi }}\right)^{1/2}\int_{-\infty }^{\infty} x^{2}e^{-Ax^{2}} \mathrm {d} x= \frac {1}{2A}:\;\;\;$$ $$ {\begin{aligned}\phi (k)&={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }\left({\frac {A}{\pi }}\right)^{1/4}e^{-{A \over 2}x^{2}}e^{-ikx}\,\mathrm {d} x\\&={\frac {1}{\sqrt {2\pi }}}\left({\frac {A}{\pi }}\right)^{1/4}\int _{-\infty }^{\infty }e^{-{A \over 2}(x+ik/A)^{2}-{k^{2}/2A}}\,\mathrm {d} x\\&={\frac {1}{\sqrt {2\pi }}}\left({\frac {A}{\pi }}\right)^{1/4}e^{-{k^{2}/2A}}\int _{-\infty }^{\infty }e^{-{A \over 2}(x+ik/A)^{2}}\,\mathrm {d} x\\& ={\frac {1}{\sqrt {2\pi }}}\left({\frac {A}{\pi }}\right)^{1/4}e^{-{k^{2}/2A}}\int _{-\infty}^{\infty}e^{-{A \over 2}x^{2}}\,\mathrm {d} x\\& =\left({\frac {1}{A\pi }}\right)^{1/4}e^{-k^{2}/2A}\\\end{aligned}} 。 $$ $$\sigma _{k}^{2}=\left({\frac {1}{A\pi }}\right)^{1/2}\int _{-\infty }^{\infty }k^{2}e^{-k^{2}/A}\,\mathrm {d} k={\frac {A}{2}}$$ $$p=\frac{h}{\lambda}=\hbar k:\;\;\;\;\; de\; Broglie\; wave $$ 因此 $$\sigma _{p}^{2}={\frac {A\hbar ^{2}}{2}}$$ $$\sigma _{x}\sigma _{p}={\sqrt {1 \over 2A}}{\sqrt {A\hbar ^{2} \over 2}}={\frac {\hbar }{2}} \sigma _{x}\sigma _{p}={\sqrt {1 \over 2A}}{\sqrt {A\hbar ^{2} \over 2}}={\frac {\hbar }{2}}$$ 得到位置和動量的不確定性關係式
由傅立葉轉換的特性,高斯函數經過轉換後仍為高斯函數(有最小的變化) $$\sigma _{x}\sigma _{p}\geq{\frac {\hbar }{2}}:\;\;\; Heisenberg\; Uncertainty\; Principle$$
