量子力学Quantum Mechanics答案解析代写

Solution:
(a) Photoelectric Effect
This refers to the emission of electrons observed when one irradiates a metal under vacuum with ultraviolet light. It was found that the magnitude of the electric current thus produced is proportional to the intensity of the striking radiation provided that the frequency of the light is greater than a minimum value characteristic of the metal, while the speed of the electrons does not depend on the light intensity, but on its frequency. These results could not be explained by classical physics.

Einstein in 1905 explained these results by assuming light, in its interaction with matter, consisted of corpuscles of energy $h \nu$, called photons. When a photon encounters an electron of the metal it is entirely absorbed, and the electon, after receiving the energy $h \nu$, spends an amount of work $\boldsymbol{W}$ equal to its binding energy in the metal, and leaves with a kinetic energy
$$
\frac{1}{2} m v^2=h \nu-W .
$$
This quantitative theory of photoelectricity has been completely verified by experiment, thus establishing the corpuscular nature of light.
(b) Black Body Radiation
A black body is one which absorbs all the radiation falling on it. The spectral distribution of the radiation emitted by a black body can be derived from the general laws of interaction between matter and radiation. The

expressions deduced from the classical theory are known as Wien抯 law and Rayleigh抯 law. The former is in good agreement with experiment in the short wavelength end of the spectrum only, while the latter is in good agreement with the long wavelength results but leads to divergency in total energy.

Planck in 1900 succeeded in removing the difficulties encountered by classical physics in black body radiation by postulating that energy exchanges between matter and radiation do not take place in a continuous manner but by discrete and indivisible quantities, or quanta, of energy. He showed that by assuming that the quantum of energy was proportional to the frequency, $\varepsilon=h \nu$, he was able to obtain an expression for the spectrum which is in complete agreement with experiment:
$$
E_\nu=\frac{8 \pi h \nu^3}{c^3} \frac{\mathrm{L}}{e^{\frac{h \nu}{k T}}-1},
$$
where $\boldsymbol{h}$ is a universal constant, now known as Planck抯 constant.
Planck抯 hypothesis has been confirmed by a whole array of elementary processes and it directly reveals the existence of discontinuities of physical processes on the microscopic scale, namely quantum phenomena.
(c) F捻anck-Hertz Experiment
The experiment of Franck and Hertz consisted of bombarding atoms with monoenergetic electrons and measuring the kinetic energy of the scattered electrons, from which one deduced by subtraction the quantity of energy absorbed in the collisions by the atoms. Suppose $E_0, E_1, E_2, \ldots$ are the sequence of quantized energy levels of the atoms and $T$ is the kinetic energy of the incident electrons. As long as $T$ is below $\mathrm{A}=E_1-E_0$, the atoms cannot absorb the energy and all collisions are elastic. As soon as $T>E_1-E_0$, inelastic collisions occur and some atoms go into their first excited states. Similarly, atoms can be excited into the second excited state as soon as $T>E_2-E_0$, etc. This was exactly what was found experimentally. Thus the F捻anck-Hertz experiment established the quantization of atomic energy levels.
(d) Davisson-Germer Experiment
L. de Broglie, seeking to establish the basis of a unified theory of matter and radiation, postulated that matter, as well as light, exhibited both wave and corpuscular aspects. The first diffraction experiments with matter

waves were performed with electrons by Davisson and Germer (1927). The incident beam was obtained by accelerating electrons through an electrical potential. Knowing the parameters of the crystal lattice it was possible to deduce an experimental value for the electron wavelength and the results were in perfect accord with the de Broglie relation $\lambda=\mathrm{h} / \mathrm{p}$, where $h$ is Planck担 constant and $p$ is the momentum of the electrons. Similar experiments were later performed by others with beams of helium atoms and hydrogen molecules, showing that the wavelike structure was not peculiar to electrons.
(e) Compton Scattering
Compton observed the scattering of X-rays by free (or weakly bound) electrons and found the wavelength of the scattered radiation exceeded that of the incident radiation. The difference $\Delta \lambda$ varied as a function of the angle $\theta$ between the incident and scattered directions:
$$
\Delta \lambda=2 \frac{h}{m c} \sin ^2 \frac{\theta}{2},
$$
where $h$ is Planck抯 constant and $m$ is the rest mass of the electron. Furthermore, $\Delta \lambda$ is independent of the incident wavelength. The Compton effect cannot be explained by any classical wave theory of light and is therefore a confirmation of the photon theory of light.

Solution:
(a) The Schrödinger equation for the system is
$$
i \hbar \partial_t \psi(x, t)=\hat{H} \psi(x, t),
$$
where $\psi(\mathrm{x}, t)$ takes the given value $\psi(\mathrm{x}, 0)$ at $t=0$. As $\hat{H}$ does not depend on $t$ explicitly,
$$
\psi_n(x, t)=\psi_n(x) e^{-i E_n t / \hbar},
$$
where $\psi_n(x)$ is the energy eigenfunction satisfying
$$
\hat{H} \psi_n(x)=E_n \psi_n(x)
$$
Expanding $\psi(x, 0)$ in terms of $\psi_n(\mathrm{x})$ :
$$
\psi(x, 0)=\sum_n a_n \psi_n(x)
$$
where
$$
a_n=\int \psi_n^*(x) \psi(x, 0) d x
$$

Thus
$$
\psi(2, t)-\sum_n a_n \psi_n(x, t)=\sum_n a_n \psi_n(x) e^{-i E_n t / \hbar} .
$$
For a harmonic oscillator,
$$
\psi_n(x)=N_n e^{-\alpha^2 x^2 / 2} H_n(\alpha x)
$$
so that
$$
\begin{aligned}
\mathbf{a},= & \int N_n e^{-\alpha^2 x^2 / 2} H_n(\alpha x) \cdot A e^{-\alpha^2 x^2 / 2} \
& \times\left[\cos \beta H_0(\alpha x)+\frac{\sin \beta}{2 \sqrt{2}} H_2(\alpha x) \mid \boldsymbol{d} \boldsymbol{x} .\right.
\end{aligned}
$$
As the functions $\exp \left(-\frac{1}{2} x^2\right) H_n$ (dare orthogonal, all $\mathrm{a},=0$ except
$$
\begin{aligned}
& a_0=A N_0 \sqrt{\frac{\pi}{\alpha^2}} \cos \beta, \
& a_2=A N_2 \sqrt{\frac{\pi}{\alpha^2}} 2 \sqrt{2} \sin \beta .
\end{aligned}
$$
Hence
$$
\begin{aligned}
& \psi(x, t)=A \sqrt{\frac{\pi}{\alpha^2}\left[N_0 \cos \beta \psi_0(x) e^{-i E_0 t / \hbar}\right.} \
& \left.+2 \sqrt{2} N_2 \sin \beta \psi_2(x) e^{-i E_2 t / \hbar}\right] . \
& =\boldsymbol{A}\left(\frac{\pi}{\alpha^2}\right)^{\frac{1}{4}}\left[\cos \beta \psi_0(x) e^{-i \frac{E_0 t}{h}}+\sin \beta \psi_2(x) e^{-i \frac{E_2 t}{\hbar}}\right] \text {, } \
&
\end{aligned}
$$
as $N_n$ are given by $\int\left[\psi_n(x)\right]^2 \boldsymbol{d x}=1$ to be $N_0=\left(\frac{\alpha^2}{\pi}\right)^{\frac{1}{4}}, N_2=\frac{1}{2 \sqrt{2}}\left(\frac{\alpha^2}{\pi}\right)^{\frac{1}{4}}$.
(b) The observable energy values for this state are $E_0=\hbar \omega / 2$ and $\mathbf{E 2}=5 \hbar \omega / 2$, and the relative probability of getting these values is
$$
P_0 / P_2=\cos ^2 \beta / \sin ^2 \beta=\cot ^2 \beta \text {. }
$$
(c) As $\psi(x, 0)$ is a linear combination of only $\psi_0(x)$ and $\psi_2(x)$ which have even parity,
$$
\psi(-x, 0)=\psi(x, 0)
$$

Hence for $t=0$,
$$
\langle x\rangle=\int \psi(x, 0) x \psi(x, 0) \boldsymbol{d x}=\boldsymbol{0} .
$$
It follows that the average value of $x$ does not change with time.