The state is a ket \(\left|\psi(\vec r,t)\right\rangle\) in the Hilbert space \(\mathcal{H}\) of the system.
Superposition: if \(\left|\psi_1\right\rangle,\ldots,\left|\psi_n\right\rangle\) are states, so is \(\left|\psi\right\rangle=c_1\left|\psi_1\right\rangle+c_2\left|\psi_2\right\rangle+\cdots+c_n\left|\psi_n\right\rangle\).
If \(\displaystyle\int|\psi|^2d\tau=1\) (all space), then \(|\psi|^2=\psi^*\psi\) is the probability density: \[\rho(\vec r,t)=|\psi(\vec r,t)|^2=|\psi(x,y,z,t)|^2.\]
One-dimensional analogy
Probability of finding the particle between \(x=a\) and \(x=b\): \[P(a\le x\le b)=\frac{\int_a^b|\psi(x,t)|^2dx}{\int_{-\infty}^{\infty}|\psi(x,t)|^2dx}.\]
Average of any function \(f(x)\): \[\left\langle f(x)\right\rangle=\frac{\int_{-\infty}^{\infty}f(x)|\psi(x,t)|^2dx}{\int_{-\infty}^{\infty}|\psi(x,t)|^2dx}.\]
Uncertainty (error) in \(f\): \(\Delta f=\sqrt{\left\langle f^2\right\rangle-\left\langle f\right\rangle^2}\); always \(\left\langle f^2\right\rangle\ge\left\langle f\right\rangle^2\).
Example 3.1 Linear probability density
\(\rho(x)=x\) for \(0<x<1\), zero otherwise. (a) Find \(P(0<x<\tfrac12)\). (b) Find \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\) and \(\Delta x\).
Solution
(a) \(P=\dfrac{\int_0^{1/2}x\,dx}{\int_0^1x\,dx}=\dfrac{1/8}{1/2}=\dfrac14\).
(b) \(\left\langle x\right\rangle=\dfrac{\int_0^1x^2dx}{\int_0^1x\,dx}=\dfrac{1/3}{1/2}=\dfrac23\); \(\left\langle x^2\right\rangle=\dfrac{\int_0^1x^3dx}{1/2}=\dfrac{1/4}{1/2}=\dfrac12\).
\(\Delta x=\sqrt{\dfrac12-\dfrac49}=\sqrt{\dfrac1{18}}\).
Example 3.2 Exponential density
\(\rho(x)=x\,e^{-x/\lambda}\) for \(x\ge0\) (\(\lambda>0\)), zero otherwise. (a) Find \(P(x\ge\lambda)\). (b) Find \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\). (c) Find the most probable \(x\).
Solution
Normalization: \(\int_0^\infty xe^{-x/\lambda}dx=\lambda^2\Gamma(2)=\lambda^2\).
(a) \(P(x\ge\lambda)=\dfrac{1}{\lambda^2}\displaystyle\int_\lambda^\infty xe^{-x/\lambda}dx=\dfrac{1}{\lambda^2}\cdot2\lambda^2e^{-1}=\dfrac2e\).
(b) \(\left\langle x\right\rangle=\dfrac{\lambda^3\Gamma(3)}{\lambda^2\Gamma(2)}=2\lambda\); \(\left\langle x^2\right\rangle=\dfrac{\lambda^4\Gamma(4)}{\lambda^2\Gamma(2)}=6\lambda^2\).
(c) \(\dfrac{d\rho}{dx}=e^{-x/\lambda}\Big(1-\dfrac x\lambda\Big)=0\Rightarrow x=\lambda\); \(\rho''(\lambda)<0\) \(\Rightarrow\) maximum, \(\rho(\lambda)=\lambda/e\).
Example 3.3 Broken speedometer needle
A needle is equally likely to stop at any angle \(\theta\) between \(0\) and \(\pi\). Find (a) the distribution \(\rho(\theta)\), (b) \(\left\langle \theta\right\rangle\), (c) \(\Delta\theta\), (d) \(\left\langle \cos\theta\right\rangle\).
Solution
(a) \(\rho=A\) on \((0,\pi)\); \(A\int_0^\pi d\theta=1\Rightarrow\rho(\theta)=\dfrac1\pi\).
(b) \(\left\langle \theta\right\rangle=\dfrac1\pi\displaystyle\int_0^\pi\theta\,d\theta=\dfrac\pi2\).
(c) \(\left\langle \theta^2\right\rangle=\dfrac1\pi\displaystyle\int_0^\pi\theta^2d\theta=\dfrac{\pi^2}{3}\) \(\Rightarrow\Delta\theta=\sqrt{\dfrac{\pi^2}{3}-\dfrac{\pi^2}{4}}=\dfrac{\pi}{\sqrt{12}}\).
(d) \(\left\langle \cos\theta\right\rangle=\dfrac1\pi\displaystyle\int_0^\pi\cos\theta\,d\theta=0\).
Example 3.4 Falling rock
A rock is dropped from a cliff of height \(h\); a million photographs are taken at random times during the fall. Find the probability density of the distance fallen and its (time) average.
Solution
\(x(t)=\tfrac12gt^2\), \(v=gt\), flight time \(T=\sqrt{2h/g}\).
Probability of a photo in \(dt\) is \(dt/T\): \[\frac{dt}{T}=\frac{dx}{gt}\sqrt{\frac{g}{2h}}=\frac{dx}{2\sqrt{hx}}.\]
\(\rho(x)=\dfrac{1}{2\sqrt{hx}}\), \(0\le x\le h\); check \(\int_0^h\rho\,dx=1\).
\(\left\langle x\right\rangle=\displaystyle\int_0^h\frac{x\,dx}{2\sqrt{hx}}=\frac{1}{2\sqrt h}\cdot\frac23h^{3/2}=\frac h3\).
Less than \(h/2\), as expected: the rock spends more time near the top.
Example 3.5 Gaussian distribution
\(\rho(x)=Ae^{-\lambda(x-a)^2}\), with \(A,a,\lambda>0\). (a) Find \(A\). (b) Find \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\), \(\sigma\). (c) Sketch \(\rho(x)\).
Solution
Use \(\displaystyle\int_0^\infty x^ne^{-\alpha x^2}dx=\frac{\Gamma\big(\frac{n+1}{2}\big)}{2\alpha^{(n+1)/2}}\); put \(y=x-a\).
(a) \(A\displaystyle\int_{-\infty}^{\infty}e^{-\lambda y^2}dy=A\sqrt{\frac\pi\lambda}=1\Rightarrow A=\Big(\frac\lambda\pi\Big)^{1/2}\).
(b) \(\left\langle x\right\rangle=A\displaystyle\int(y+a)e^{-\lambda y^2}dy=0+a=a\).
\(\left\langle x^2\right\rangle=A\displaystyle\int(y^2+2ay+a^2)e^{-\lambda y^2}dy=\frac{1}{2\lambda}+a^2\).
\(\sigma^2=\left\langle x^2\right\rangle-\left\langle x\right\rangle^2=\dfrac{1}{2\lambda}\).
(c) Bell curve centred at \(x=a\) (right).
Standard form: \(\rho(x)=\dfrac{1}{\sqrt{2\pi\sigma^2}}\exp\Big[-\dfrac12\Big(\dfrac{x-\mu}{\sigma}\Big)^2\Big]\), with \(\left\langle x\right\rangle=\mu\), \(\left\langle x^2\right\rangle=\sigma^2+\mu^2\).
Comparing: \(\mu=a\), \(\sigma^2=\dfrac1{2\lambda}\).
Example 3.6 Exponential wave function
\(\psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}\), with \(A,\lambda,\omega>0\). (a) Normalize \(\psi\). (b) Find \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\). (c) Find \(\sigma\); sketch \(|\psi|^2\) marking \(\left\langle x\right\rangle\pm\sigma\); find the probability of lying outside this range.
Solution
Use \(\displaystyle\int_0^\infty x^ne^{-\beta x}dx=\frac{n!}{\beta^{n+1}}\).
(a) \(|A|^2\cdot2\displaystyle\int_0^\infty e^{-2\lambda x}dx=\frac{|A|^2}{\lambda}=1\Rightarrow A=\sqrt\lambda\).
(b) \(\left\langle x\right\rangle=\lambda\displaystyle\int_{-\infty}^\infty xe^{-2\lambda|x|}dx=0\) (odd integrand).
\(\left\langle x^2\right\rangle=2\lambda\displaystyle\int_0^\infty x^2e^{-2\lambda x}dx=2\lambda\cdot\frac{2!}{(2\lambda)^3}=\frac{1}{2\lambda^2}\).
(c) \(\sigma=\dfrac{1}{\sqrt2\,\lambda}\). Probability outside \([-\sigma,\sigma]\): \[P=2\lambda\int_\sigma^\infty e^{-2\lambda x}dx=e^{-2\lambda\sigma}=e^{-\sqrt2}\approx0.2431.\]
