# 简明量子力学

## 光子

Time: end of the 19th century. Maxwell’s equations have established Faraday’s hunch that light is an electromagnetic wave. However, by early 20th century, experimental evidence mounted pointing towards the fact that light is carried by ‘particles’ that pack a definite momentum and energy. Here is the crux of the problem: consider the double-slit experiment. Monochromatic light of wavelength $\lambda$ passing through two slits separated by a distance $d \sim \lambda$ forms a diffraction pattern on a photographic plate. If one tunes down the intensity of light in a double-slit experiment, one does not get a ‘dimmer’ interference pattern, but discrete strikes on the photographic plate and illumination at specific points. That means light is composed of ‘particles’ whose energy and momentum are concentrated in one point which leads to discrete hits. But their wavelength extends over space, which leads to diffraction patterns.

Planck postulated that light is composed of discrete lumps of momemtum $\boldsymbol{p} = \hbar \boldsymbol{k}$ and energy $E = \hbar \omega$. Here $\boldsymbol{k} = (2\pi / \lambda) \hat{\boldsymbol{n}}$, $\hat{\boldsymbol{n}}$ the direction of propagation, $\hbar$ is Planck’s constant, and $\omega = c \left | \boldsymbol{k} \right |$ with $c$ the speed of light. Planck’s hypothesis explained spectral features of the blackbody radiation. It was used by Einstein to explain the photoelectric effect. Einstein was developing the theory of relativity around the same time. In this theory, the momentum of a particle of mass $m$ and velocity $v$ is $p = mv / \sqrt{1 − (v / c)^2}$, where $c$ is the speed of light. Thus if a particle has $m = 0$, the only way it can pack a momentum is if its velocity is $v = c$. Nature takes advantage of this possibility and gives us such particles. They are now called photons. Thus photons have no mass, but have momentum. Thus light, which was thought a wave acquired a certain degree of particle attributes. So what about particles with mass - do they have wave nature too? Nature is too beautiful to ignore this symmetry!

## 波粒二象性

de Broglie hypothesized in his PhD dissertation that classical ‘particles’ with mass also have wavelengths associated with their motion. The wavelength is $\lambda = 2\pi \hbar / \left | \boldsymbol{p} \right |$, which is identical to $\boldsymbol{p} = \hbar \boldsymbol{k}$. How could it be proven? The wavelength of light was such that diffraction gratings (or slits) were available at that time. But electron wavelengths were much shorter, since they had substantial momentum due to their mass. Elsassaer proposed using a crystal where the periodic arrangement of atoms will offer a diffraction grating for electrons. Davisson and Germer at Bell labs shot electrons in a vacuum chamber on the surface of crystalline Nickel. They observed diffraction patterns of electrons. The experiment proved de Broglie’s hypothesis was true. All particles had now acquired a wavelength.

The experiment challenged the understanding of the motion or ‘mechanics’ of particles, which was based on Newton’s classical mechanics. In classical mechanics, the question is the following: a particle of mass $m$ has location $x$ and momentum $p$ now. If a force $F$ acts on it, what are $(x’, p’)$ later? Newton’s law $F = m \mathrm{d}^{2}x / \mathrm{d}t^{2}$ gives the answer. The answer is deterministic, the particle’s future fate is completely determined from its present. This is no longer correct if the particle has wave-like nature. The wave-particle duality is the central fabric of quantum mechanics. It leads to the idea of a wavefunction.

## 波函数

If a particle has a wavelength, what is its location $x$? A wave is an extended quantity. If a measurement of the particle’s location is performed, it may materialize at location $x_{0}$. But repeated measurements of the same state will yield $\left \langle x \right \rangle = x_{0} + \Delta x$. Separate measurements of the momentum of the particle prepared in the same state will yield $\left \langle p \right \rangle = p_{0} + \Delta p$. The ‘uncertainty’ relation $\Delta x \Delta p \ge \hbar / 2$ is a strictly mathematical consequence (*more description needed*) of representing a particle by a wave.

Because the ‘numbers’ $(x, p)$ of a particle cannot be determined with infinite accuracy simultaneously, one has to let go of this picture. How must one then capture the mechanics of a particle? Any mathematical structure used to represent the particle’s state must contain information about its location $x$ and its momentum $p$, since they are forever intertwined by the wave-particle duality. One is then forced to use a function, not a number. The function is denoted by $\psi$, and is called the wavefunction.

A first attempt at constructing such a function is $\psi(x) = A cos(px / \hbar)$. This guess is borrowed from the classical representation of waves in electromagnetism, and in fluid dynamics. The wavefunction can represent a particle of a definite momentum $p$. Max Born provided the statistical interpretation of the wavefunction by demanding that ${\left | \psi \right |}^2$ be the probability density, and $\int {\left | \psi \right |}^2 \mathrm{d}x = 1$. In this interpretation, ${\left | \psi \right |}^2 \mathrm{d}x$ is the probability that a measurement of the particle’s location will find the particle in the location $(x, x + \mathrm{d}x)$. It is clear that ${\left | \psi \right |}^2 = {\left | A \right |}^2 cos^{2}(px / \hbar)$ assigns specific probabilities of the location of the particle, going to zero at certain points. Since the momentum $p$ is definite, the location of the particle must be equally probable at all points in space. Thus we reject the attempted wavefunction as inconsistent with the uncertainty principle.

The simplest wavefunction that is consistent with the wave-particle duality picture is $\psi_{p}(x) = A e^{ipx / \hbar}$. The complex exponential respects the wave-nature of the particle by providing a periodic variation in $x$, yet it never goes to zero. The probability (density) is ${\left | \psi_p(x) \right |}^2 = {\left | A \right |}^2$, equal at all $x$. Thus, complex numbers are inevitable in the construction of the wavefunction representing a particle.

## 算符

Every physical observable in quantum mechanics is represented by an operator. When the operator ‘acts’ on the wavefunction of the particle, it extracts the value of the observable. For example, the momentum operator is $\hat{p} = -i \hbar \partial / \partial x$, and for states of definite momentum $\hat{p} \psi_{p}(x) = (\hbar k) \psi_{p}(x)$. We note that $(x \hat{p} - \hat{p} x) f(x) = i \hbar f(x)$ for any function $f(x)$. The presence of the function in this equation is superfluous, and thus one gets the identity
$$x \hat{p} - \hat{p} x = [x, \hat{p}] = i \hbar$$
The square brackets define a commutation relation. The space and momentum operators do not commute. In classical mechanics, $[x, p] = 0$. Quantum mechanics elevates the ‘status’ of $x$ and $p$ to those of mathematical operators, preventing them from commuting. This is referred to as the ‘first quantization’ from classical to quantum mechanics. In this scheme, the dynamical variables $(x, p)$ that were scalars in classical mechanics are promoted to operators, and the wavefunction $\psi$ is a scalar. If the number of particles is not conserved, then one needs to go one step further, and elevate the status of the wavefunction $\psi \to \hat{\psi}$ too, which is called second quantization.

$$x \hat{p} - \hat{p} x = [x, \hat{p}] = i \hbar$$

## 定动量的态和定位置的态

The wavefunction $\psi_{p}(x) = A e^{ipx / \hbar}$ is a state of definite momentum since it is an eigenstate of the momentum operator $\hat{p} \psi_{p}(x) = p \psi_{p}(x)$. One may demand the location of the particle to be limited to a finite length $L$. This may be achieved by putting an electron on a ring of circumference $L$, which yields upon normalization $A = 1 / \sqrt{L}$. In that case, the wavefunction must satisfy the relation $\psi_{p}(x + L) = \psi_{p}(x)$ to be single-valued. This leads to $e^{ikL} = 1 = e^{i 2\pi \times n}$, and $k_{n} = n \times (2\pi / L)$. Here $n = 0, \pm 1, \pm 2, \ldots$. The linear momentum of the electron is then quantized, allowing only discrete values. Since $L = 2\pi R$ where $R$ is the radius of the ring, $k_{n} L = 2\pi n \to pR = n\hbar$, showing angular momentum is quantized to $0, \pm \hbar, \pm 2\hbar, \ldots$. This indeed is the quantum of quantum mechanics! One may then index the wavefunctions of definite linear momentum by writing $\psi_{n}(x)$. Expressing states of definite momentum in terms of states of definite location similarly yields
$$\psi_{n}(x) = \frac{1}{\sqrt{L}} e^{ik_{n} x}$$
The set of wave functions $[\ldots, \psi_{-2}(x), \psi_{-1}(x), \psi_{0}(x), \psi_{1}(x), \psi_{2}(x), \ldots] = [\psi_{n}(x)]$ are special. We note that $\int_{0}^{L} \psi_{m}^{*}(x) \psi_{n}(x) = \delta_{nm}$, i.e., the functions are orthogonal. Any general wavefunction representing the particle $\psi(x)$ can be expressed as a linear combination of this set. This is the principle of superposition, and a basic mathematical result from Fourier theory. Thus the quantum mechanical state of a particle may be represented as $\psi(x) = \sum_{n} A_{n} \psi_{n}(x)$. Clearly, $A_{n} = \int \mathrm{d}x \psi_{n}^{*} \psi(x)$. Every wavefunction constructed in this fashion represents a permitted state of the particle, as long as $\sum_{n} {\left | A_{n} \right |}^{2} = 1$.

$$\psi_{n}(x) = \frac{1}{\sqrt{L}} e^{ik_{n} x}$$

It is useful here to draw an analogy to the decomposition of a vector into specific coordinates. The ‘hybrid’ state function $\psi(x)$ is pictured as a vector $\left | \psi \right \rangle$ in an abstract space. The definite momentum wavefunctions $\psi_{n}(x)$ are pictured as the ‘coordinate’ vectors $\left | n \right \rangle$ in that space of vectors. This set of vectors is called the basis. Since there are an infinite set of integers $n = 0, \pm 1, \pm 2, \ldots$, the vector space is infinite dimensional. It is called the Hilbert space. One may then consider the coefficients $A_{n}$ as the length of the projections of the state on the basis states. The abstract picture allows great economy of expression by writing $\left | \psi \right \rangle = \sum_{n} A_{n} \left | n \right \rangle$. The orthogonality of the basis states is $\left \langle m | n \right \rangle = \delta_{mn}$, and thus $A_{n} = \left \langle n | \psi \right \rangle$. Then it is evident that $\left | \psi \right \rangle = \sum_{n} \left \langle n | \psi \right \rangle \left | n \right \rangle = \sum_{n} \left | n \right \rangle \left \langle n | \psi \right \rangle$, and $\sum_{n} \left | n \right \rangle \left \langle n \right | = 1$.

A vector may be decomposed in various basis coordinates. For example, a vector in 3-d real space may be decomposed into cartesian, spherical, or cylindrical coordinate systems. Similarly, the choice of basis states of definite momentum is not unique. The wavefunctions for states of definite location are those functions that satisfy $x \psi_{x_{0}}(x) = x_{0} \psi_{x_{0}}(x)$, which lets us identify $\psi_{x_{0}}(x) = \delta(x - x_{0})$. Here $\delta(\ldots)$ is the Dirac-delta function, sharply peaked at $x = x_{0}$. It is instructive to expand the states of definite location in the basis of the states of definite momentum. From the uncertainty relation, we expect a state of definite location to contain many momenta. The expansion yields $A_{n} = \int_{-\infty}^{\infty} \mathrm{d}k / (2\pi / L) \times (e^{ik_{n} x} / \sqrt{L}) \delta(x - x_{0}) = e^{ik_{n} x_{0}} / \sqrt{L}$ (**check this!!**), whereby ${\left | A_{n} \right |}^{2} = 1 / L$. Thus, the state of definite location $x_{0}$ is constructed of an infinite number of states of definite momentum $n = 0, \pm 1, \pm 2, \ldots$, each with equal probability $1 / L$.

## 定能量的态：薛定谔方程

States of definite energy $\psi_{E}(x)$ are special. Unlike the states of definite momentum or definite location, we cannot write down their general wavefunction without more information. That is because the energy of a particle depends on its potential and kinetic components. In classical mechanics, the total energy is $p^{2} / 2m + V(x)$, i.e., split between kinetic and potential energy components. Once $x$ & $p$ are known for a classical particle, the energy is completely defined, meaning one does not need to ask another question. However, since $x$ and $p$ cannot be simultaneously defined for a quantum-mechanical particle with arbitrary accuracy, the energy must be obtained through operations performed on the wavefunction.

Schrodinger provided the recipe, and the equation is thus identified with his name. The Schrodinger equation is
$$\left [ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right ] \psi_{E}(x) = E \psi_{E}(x)$$
The solution of this eigenvalue equation for a $V(x)$ identifies the special wavefunctions $\psi_{E}(x)$. These wavefunctions represent states of definite energy. How did we ascertain the accuracy of the Schrodinger equation? The answer is through experiments. A major unresolved problem at the time was explaining the discrete spectral lines emitted from excited hydrogen atoms. Neils Bohr had a heuristic model to explain the spectral lines that lacked mathematical rigor. The triumph of Schrodinger equation was in explaining the precise spectral lines. An electron orbiting a proton in a hydrogen atom sees a potential $V(r) = -q^{2} / 4\pi \epsilon_{0} r$. Schrodinger solved this equation (with help from a mathematician), and obtained energy eigenvalues $E_{n} = -13.6 / n^{2}\ \rm{eV}$. Thus Bohr’s semiqualitative model was given a rigid mathematical basis by Schrodinger’s equation. The equation also laid down the recipe for solving similar problems in most other situations we encounter. Just as the case for states of definite energy or definite location, one may expand any state of a quantum particle in terms of the states of definite energy $\psi(x) = \sum_{E} A_{E} \psi_{E}(x)$, or equivalently $\left | \psi \right \rangle = \sum_{E} A_{E} \left | E \right \rangle$.

$$\left [ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right ] \psi_{E}(x) = E \psi_{E}(x)$$

So why do states of definite energy occupy a special position in applied quantum mechanics? That becomes clear if we consider the time-dependent Schrodinger equation.

## 含时薛定谔方程

Newton’s law $F = dp / dt$ provides the prescription for determining the future $(x’, p’)$ of a particle given its present $(x, p)$. Schrodinger provided the quantum-mechanical equivalent, through the time-dependent equation
$$i\hbar \frac{\partial \Psi(x, t)}{\partial t} = \underbrace{\left [ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right ]}_{\hat{H}} \Psi(x, t)$$
To track the time-evolution of quantum states, one must solve this equation and obtain the composite space-time wavefunction $\Psi(x, t)$. Then physical observables can be obtained by operating upon the wavefunction by the suitable operators. Let’s look at a particular set of solution wavefunctions which allow the separation of the time and space variables, of the form $\Psi(x, t) = \chi(t)\psi(x)$. Inserting it back into the time-dependent Schrodinger equation and rearranging, we obtain
$$i\hbar \frac{\dot{\chi(t)}}{\chi(t)} = \frac{\hat{H} \psi(x)}{\psi(x)} = E$$
Note that since the left side does not depend on space, and the right side does not depend on time, both the fractions must be a constant. The constant is called $E$, and clearly has dimensions of energy in Joules. The right half of the equation lets us identify that $\hat{H} \psi_{E}(x) = E\psi_{E}(x)$ are states of definite energy. Then the left side dictates that the time dependence of these states is described by $\chi(t) = \chi(0) e^{-iEt / \hbar}$. Thus the particular set of solutions
$$\Psi_{E}(x, t) = \psi_{E}(x) e^{-iEt / \hbar}$$
now define the time evolution of the states. Here $\psi_{E}(x)$ are states of definite energy, as obtained by solving the time-independent Schrodinger equation.

$$i\hbar \frac{\partial \Psi(x, t)}{\partial t} = \underbrace{\left [ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x) \right ]}_{\hat{H}} \Psi(x, t)$$

$$i\hbar \frac{\dot{\chi(t)}}{\chi(t)} = \frac{\hat{H} \psi(x)}{\psi(x)} = E$$

$$\Psi_{E}(x, t) = \psi_{E}(x) e^{-iEt / \hbar}$$