Quantum Mechanic

The Quantum Nature of Light

The Photoelectric Effect(光电效应)

photoelctrons 光电子
photoelectric current 光电流

The quantum of a light wave of frequency \(f\) has the energy, which is also the energy of a single photon

\[E=hf=\hbar\omega \]

where Planck constant \(h = 2\pi \hbar = 6.63 \times 10^{-34} J \cdot s\)

work function: \(hf = K + W\)

photon momentum: \(k = \frac{2\pi}{\lambda}\)

\[p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda} = \hbar k\]

Compton Scattering(康普顿散射)

Compton Shift:

\[\begin{aligned} \frac{h}{\lambda}-\frac{h}{\lambda^{\prime}}+mc=&\gamma mc\\ \frac{h}{\lambda}-\frac{h}{\lambda^{\prime}}\cos\phi=&\gamma mv\cos\theta\\ \frac{h}{\lambda^{\prime}}\sin\phi=&\gamma mv\sin\theta \end{aligned}\]

\[\Delta \lambda = \frac{h}{mc}(1-\cos \phi) = \frac{2h}{mc} \sin^2 \frac{\phi}{2}\]

Compton wavelength:\(h/mc\)

photon angular momentum

Matter Waves

De Broglie Hypothesis

de Broglie wavelength: \(\lambda = \frac{h}{p} = \frac{h}{m_0v}\sqrt{1-\frac{v^2}{c^2}}\)

Probability amplitude \(\psi\) (somplex number)

The probability of an event in an ideal experiment is then given by: \(|\psi|^2 = \psi * \psi\)

The Interference of Electrons

When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately \(\psi = \psi_1 +\psi_2\)

\[P=|\psi|^2=|\psi_1|^2+|\psi_2|^2+2\mathfrak{R} (\psi_1^{\dagger}\psi_2)+\cdots\]

Heisenberg’s Uncertainty Principle

You cannot measure the momentum and coordinate of a particle simultaneously to arbitrary accuracy.

\[\begin{aligned} \Delta x\cdot \Delta p_x \ge \hbar\\ \Delta y\cdot \Delta p_y \ge \hbar\\ \Delta z\cdot \Delta p_z \ge \hbar \end{aligned}\]

Wave Function:

\[ \psi(x)=e^{i(kx-\omega t)}\]

The major significance of the \highlight{wave-particle duality} is that all behavior of light and matter can be explained through the use of a \highlight{complex wave function \(\psi(x,y,z,t)\)}. The probability of finding a particle somewhere at a particular time is proportional to

\[P(x,y,z,t)=\psi^{\dagger}(x,y,z,t)\psi(x,y,z,t)\]

Schroedinger’s Equation

For a free classcical one-dimentional particle

\[E = \frac{p^2}{2m} = \hbar\omega\]

In the presence of potential, e.g. a harmonic potential \(U(x)\) , the classical relation is modified to

\[E = \frac{p^2}{2m} + U(x)\]

Schroedinger’s Equation：

\[i\hbar\frac{\partial\psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi(x,t)}{\partial x^2}+U(x,t)\psi(x,t)\]

Solutions: \(\psi(x,t)=\phi(x)e^{-iEt/\hbar}\)

\[\begin{aligned} &\frac{\partial^2\phi(x)}{\partial x^2}+\frac{2m}{\hbar^2}[E-U(x)]\phi(x)=0\\ &\phi(x)=Ae^{ikx}+B^{-ikx}, \quad k=\sqrt{2mE}/\hbar\\ &\psi(x,t)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)} \end{aligned}\]

Reflection from a Potential Step

• Region 1(x < 0>): \(k=\sqrt{2mE}/\hbar\)

\[\psi_1=Ae^{ikx}+Be^{-ikx}\]

• Rigion 2(x > 0): \(k_b=\sqrt{2m(E-qV_b)}/\hbar\)

\[\psi_2=Ce^{ik_b x}+De^{-ik_b x}\]

Boundary Condition at \(x = 0\):

\[\begin{aligned} A+B=C &\text{ (matching of values)}\\ Ak-Bk=Ck_b &\text{ (matching of slopes)} \end{aligned}\]

Reflection and Transmission Coefficients

\[\begin{aligned} &R=\frac{|\psi_{reflected}|^2}{|\psi_{incident}|} = \frac{|B|^2}{|A|^2}\\ &T = \frac{|\psi_{transimitted}|^2}{|\psi_{indent}|^2} = \frac{|C|^2}{|A|^2}\\ &T = 1 - R \end{aligned}\]

Tunneling through a Potential Barrier

Quantum Wells

An Electron in an Infinite Potential Well

Wave Function

\[\psi_n(x) = A\sin(\frac{n\pi x}{L})\]

Boundary Condition: \(\psi_n(0) = \psi_n(L) = 0\)

Probability of Detection

\[P_n(x) = |\psi_n(x)|^2 = |A|^2 \sin^2(\frac{n\pi }{L}x)\]

Normalization Condition

\[\int_{-\infty}^\infty|\psi_n(x)|^2\mathrm{d}x=\int_0^L|\psi_n(x)|^2\mathrm{d}x=1 \Rightarrow A = \sqrt{2/L}\]

Energies of the Trapped Electron

\[\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{2L}{n}\]

kinetic energy\(K = \frac{p^2}{2m}\)

For free partical:

\[E_n=\frac{h^2}{8mL^2}n^2, \text{for} n = 1,2,3 \cdots\]

\[\hbar\omega=\frac{hc}{\lambda}=\Delta E=E_{high}-E_{low}\]

An Electron in a Finite Potential Well

The Hydrogen Atom

\[\frac{1}{4\pi \epsilon_0}\frac{e^2}{r^2}=m\frac{v^2}{r}\]

Multiplying by \(-r\), we obtain

\[ E_C=-\frac{e^2}{4\pi \epsilon_0 r}=-mv^2=-2E_K\]

Alternatively, the total energy of the electron is

\[E=E_K+E_C=\frac{E_C}{2}=-E_K\]

The Bohr Model

The angular momentum \(\vec{L} = n\hbar\), for \(n = 1,2,3 \cdots\)

\[L = rmv = n\hbar \rightarrow v = \frac{n\hbar}{mv}\]

\(r_n = n^2 a_B,\) where the characteristic length \(a_B = \frac{\hbar^2}{\frac{me^2}{4\pi \epsilon_0}} = 0.529 \text{\AA}\)

Energy of the hydrogen atom:

\[E_n=\frac{1}{2}mv^2-\frac{1}{4\pi \epsilon_0}\frac{e^2}{r}=-\frac{E_R}{n^2},\]

where the characteristic energy \(E_R=\frac{\frac{me^4}{(4\pi \epsilon_0)^2}}{2\hbar^2}=13.6\text{eV}\)

The Hydrogen Spectrum

\[\begin{aligned} \hbar \omega_{nm}=E_R\left(\frac{1}{n^2}-\frac{1}{m^2}\right) \end{aligned}\]

for integers \(m>n\). The wavelengths of the emitted or absorbed light are given by

\[\begin{aligned} &\frac{1}{\lambda}=\frac{E_R}{hc}\left(\frac{1}{n^2}-\frac{1}{m^2}\right)\\ &\frac{hc}{E_R}=912\text{\AA} \end{aligned}\]

Ground-State Energy from Uncertainty Principle

\[E\sim\frac{(\Delta p)^2}{2m}-\frac{e^2}{4\pi\epsilon_0 \Delta_r}=\frac{\hbar^2}{2m(\Delta r)^2}-\frac{e^2}{4\pi\epsilon_0 \Delta_r}\]

To find the minimal energy, we solve, for \(\Delta r\)

\[\begin{aligned} \frac{\mathrm{d}E}{\mathrm{d}(\Delta r)}=0 \end{aligned}\]

After some algebra, we find

\[\begin{aligned} \Delta r=&\frac{\hbar^2}{\frac{me^2}{4\pi\epsilon_0}}\equiv a_B\\ E=&-\frac{\frac{me^2}{4\pi\epsilon_0}}{2\hbar^2}\equiv -E_R \end{aligned}\]

\[ \Delta t \cdot \Delta E \ge \hbar/2s\]