# # Fundamental Principles of Quantum Mechanics

Bohr(1913) is a half classical atomic model. This is not quantum mechanics.

1. de Broglie (1925) De broglie's hypothesis.
2. Uncertainty principle.

## # Louis de Broglie's hypothesis 1924

The motion of a particle is governed by the wave propagation properties of a 'pilot' wave called matter wave. The wavelength and the frequency of the pilot wave associated with a particle of momentum and energy are

$\lambda = \frac{h}{p}, \nu = \frac{E}{h}$

He was rewarded the Nobel Prize in 1929 because of his Phd thesis.

TIP

From Einstein's thory of relativity, $E = mc^2 = mcc = pc = p\lambda \nu$ and from his hypothesis, $E = h\nu$. One can deduce the the wavelength of the pilot wave.

Consider a bullet of mass m = 0.1 kg moving with a velocity v=10^3 m/sec. Acoording to de Broglie, the wavelength of the pilot wave will be,

$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{J-sec}}{0.1 \text{kg} \times 10^3 \text{m/sec}} = 6.63 \times 10^{-36} \text{m} = 6.63 \times 10^{-26} \text{Angstrom}$

In order to observe interference effects, the size of the diffracting slit $d$ must be comparable to the wavelength of the wave, i.e., de Broglie wavelength $\lambda \geq$ Angstrom. To 'see' the de Broglie's wave. As $\lambda = h/p$, to increase the wave length, we need to decrease the momentum, and the most practical way of achieving this condition is to choose a particle of small mass such as the electron.

## # Davisson-Germer Experiment

1927

Bragg condition

From geometric considerations, $l=d\sin \theta$, and the path difference is $2l = 2d\sin \theta$. By analogy to the other interference experiments, if this path differency is equal to an intergral number of wavelengths, the two beams will add constructively; that is, the radiation reflected by two adjacent layers of atoms will add constructively if

$2d \sin \theta = n\lambda, n = 1, 2, 3, ...$

This equation is known as the Bragg condition.

What is electronvolt

In physics, an electronvolt(symbol eV) is the measure of an amount of kinetic energy gained by a single electron accelerating from rest though an electric potential difference of on volt in vacuum.[1] Or the work required to move an electron through a potential difference of on volt.[2]

What is difference between X-ray and electron beam with the same wavelength?

Electron ray is a matter wave, while X-ray is not. ??

## # J. J. Tomthon's son

The wavelength range of visible lights is from 380 to 700 naonometers ($10^{-9}$ m). The wavelength of electron can be controled to angstrams ($10^{-10}$ m). The resolution of a microscope is propotional to the the $1/\lambda$.

## # Nature of the matter wave

Electron version of double-slit experiment In classicl physics, wave is propagation of energy. Howerver, the matter wave is to describe the existence probability of the particle. 哥本哈根学派，Bohr gathered lots the people. The matter of wave is a mathematical term rather than the energy term in physics. The wave magnitude indicates probobility of the discovery of this particle at a position.

## # The uncertainty principle

An experiment cannot simultaneously determine a component of the momentum of a particle, for exmaple $p_x$ and exact value of the corresponding coordinate $x$. The best one can do is

$\Delta p_x \Delta x \geq \hbar$

where $\Delta$ represents the uncertainty of measurement.

How does this inequanlity come from?

It is said the uncertainty principle is based on experiment. Yet, up till now, I don't understand.

Werner Heisenberg (1901-1976)

Einstein, Bohr, Heisenberg

## # The nature of wave function

Let the particle wave be represented by

$\psi (r, t) = A \sin (kx - \omega t)$

1. The amplitude $A$ is the same at all points in space.
2. It has a well-defined wavelength, $\lambda = 2\pi/k$
3. It has a well-defined frequency, $\nu = \omega/2\pi$
4. It travels toward increasing values of $x$ with a velocity

$v = \lambda \nu = \frac{\omega}{k}$

Because the amplitude of the wave is the same for all values of $x$, the particle can be found with equal probability at any point in space, that is, the particle is completely unlocalized, $\Delta x = \infty$. Actually, we should not be surprised by this result. A well-defined $\lambda$ implies a well-defined momentum for the particle($\lambda = h/p$), that is, $\Delta p_x = 0$. This inturn, according to the uncertainty principle, should lead to an infinite uncertainity in the coordinate of the particle. This particle is free particle with well-defined monentum and energy, the price paid is a complete lack of localization.

To describe a particle that is partially localized, we need a wave with an amplitude that is different from zero only over a small region of space where there is a chance of finding the particle. We need a wave that looks like the wave packet. The wavepacket can be written as

$\psi(x, t)=\int_{0}^{\infty} \int_{0}^{\infty} A(k, \omega) \sin (k x-\omega t) d k d \omega$

The coefficient $A$ basically determines how much of a particular wavelength and frequency we mix with the others. The greater the range of $k$'s (and therefore $\lambda$'s) that we mix, the narrower is the width of the resulting wave packet. This is why we cannot determine a component of the momentum of and coordinate simultaneously, no matter you measuring it or not.

Energy transfor of a wave

The wave has given energy to the jparticle because the wave carries energy with it. Let us consider the sinusoidal traveling wave represented by

$y = A \sin (kx - \omega t)$

The rate of energy production, consumption, or transmission was defined as power $P$. We can obtain $P$ by calculating the energy crossing a given point in a string in 1 sec. This will be equal to the wave energy of the string particles in one wavelength multiplied by the number of wavelengths passing this point in 1 sec, that is, by the frequency $\nu$.

$P = (\text{energy per wavelength}) \times \nu$

At any given time the energy of a particular particle may be all kinetic or all potential or a micture. When a point passing through the equilibrium point, the energy of this particle is all kinetic. To obtain the kinetic energy of the particles in the string we need an expression for the transverse velocity $v_y$.

$v_y = \frac{\partial y}{\partial t} = -\omega A \cos(kx - \omega t)$

We now calculate the energy of the particle passing through the equilibrium point of mass $\Delta m$.

$E = E_k + E_p = E_k = \frac{1}{2} \Delta m v_\text{ymax}^2 = \frac{1}{2} \Delta m (-\omega A )^2 = \frac{1}{2} \Delta m \omega^2 A^2$

Because the energy of all the particle is the same, iif the amplitude of the wave is small compared with the wavelength (a situation often satisfied), we can obtain the energy in one wavelength by replacing $\Delta m$ with the mass contained in one wavelength, $\mu \lambda$, where $\mu$ is the mass per unit length of the string. Therfore, energy per wavelength

$= \frac{1}{2} \mu \lambda \omega^2 A^2$

We obtain

$P = \frac{1}{2} \mu\lambda\nu\omega^2 A^2 = 2 \pi^2 \mu v \nu^2 A^2$

where we have made use of the fact that $\lambda \nu = v$ and $\omega = 2\pi\nu$. The power transported by a wave is proportional to the square of the amplitude and the velocity of propagation of the wave. When considering waves that propagate in three dimensions, such as sound wave or light waves, it is convenient to talk about the energy flowing through a given area of the medium traversed by the wave. The term intensity, with symbol $I$, is used for this purpose, which is defined as the power transmitted per unit area perpendicular to the directon of propagation of the wave. Clearly, intensity and power are related by a simple geometric factor. Thus, the intensity of the wave is also proportional to the square of the amplitude. In the SI system, intensity has units of $\text{W/}\text{s}^2$.

$I \propto A^2$

The physical significance of the wavefunction is the following:

At some instatn $t$, a measurement is made to locate the particle associated with the wavefunction $\psi$. The probability $P(r, t) dV$ that the particle will be found within a small volume $dV$ centered around a point with position vector $r$ is equal to $|\psi|^2 dV$, that is

$P(r,t) dV = |\psi|^2 dV$

The probability of finding the particle somewhere in space must be unity. Therefore, the wavefunction $\psi$ must be normalized; that is, it must satisfy the condition

$\int_{-\infty}^{\infty} |\psi|^2 dV = 1$

where the integration is carried over all space.

WARNING

$|\psi|$ 代表的是模，而非增幅。要弄清楚。非週期的函數是沒有增幅的，只有週期的信號才有增幅。