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202203170840 Half Picture of Quantum Mechanics 01A

(This half picture is based on the first half of manuscript of Chapter One, 2015-2016 PHYS3351 Quantum Mechanics Lecture Notes.)

Section 01

Given a particle of mass m and a force F(x,t), we classify

\begin{cases}\textrm{conservative }F =\displaystyle{-\frac{\partial V(x)}{\partial x}}\enspace\big( V(x)\textrm{: potential energy}\big)\\ \textrm{nonconservative} \\ \end{cases}

In classical mechanics, description of dynamics is given by

i. trajectory (i.e., position as a function of time):

x(t)

ii. velocity:

\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}}

iii. momentum:

p=m\displaystyle{\frac{\mathrm{d}x}{\mathrm{d}t}}

iv. kinetic energy:

T=\displaystyle{\frac{1}{2}m\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2}

v. solving from Newton’s second law \textrm{Net }F=ma:

\displaystyle{F=ma\qquad\Longrightarrow\qquad m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=F=-\frac{\partial V}{\partial x}}

In quantum mechanics, description of dynamics is given by wavefunction \Psi (x,t) which is a single-valued, finite, and continuous complex function.

Statistical interpretation.

|\Psi (x,t)|^2\,\mathrm{d}x: the probability of particle being between x and x+\mathrm{d}x at time t

In 3D case, |\Psi (x,y,z,t)|^2\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z: the probability of particle being in the volume between x and x+\mathrm{d}x, y and y+\mathrm{d}y, and z and z+\mathrm{d}z.

\Psi (x) is also known as the probability amplitude.

Imagine the particle as a “cloud” which is moving and changing its shape, instead of a point object following a trajectory curve in space. The particle has a larger probability to be at the denser point of the “cloud”. Even if you know everything about the particle dynamics, that is, the wavefunction \Psi (\mathbf{r},t), you cannot predict with certainty the outcome of an experiment to measure its position. Instead, you can say the probability you will detect the particle in the volume between x and x+\mathrm{d}x, y and y+\mathrm{d}y, and z and z+\mathrm{d}z is

|\Psi (x,y,z,t)|^2\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z.

Normalization.

\displaystyle{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\, |\Psi (x,y,z,t)|^2=1}

Q. Which one of the following is \textrm{\scriptsize{NOT}} a possible wavefunction in the region x\geqslant 0 for \beta >0:

(I) Axe^{\beta x}; (II) Axe^{-\beta x}; (III) Ae^{-\beta x}.

A. (I)

Important characteristics of a wavefunction

A discrete example

There are a number of positions x_1, x_2, … ,x_{10} on a 1D line, at which the particle can appear with probability \rho_1, \rho_2, … , \rho_{10}.

Normalization: \rho_1 + \rho_2+\cdots +\rho_{10}=1

If we throw dice for N\gg 1 times, each time we get a particle at one of the positions:

x(j)\in\{ x_1, x_2, \dots ,x_{10}\} for j=1,2,\dots ,N

the times we get the particle at position x_i is given by N\rho_i,

i.e., the number of x(j)\big|_{j=1,\dots ,N}=x_i is N\rho_i.

Average position:

\begin{aligned} \langle x\rangle & = \sum_{j=1}^{N}x(j)\frac{1}{N}\\ & = \sum_{i=1}^{10}\frac{x_iN\rho_i}{N} \\ & = \sum_{i=1}^{10}x_i\rho_i \\ \end{aligned}

Numerical measure of the amount of spread:

\begin{aligned} \Delta x(j) & \equiv x(j)-\langle x\rangle \\ \sum_{j}\Delta x(j) & = 0 \\ \langle (\Delta x)^2\rangle & \equiv \sum_{j}\frac{1}{N}(\Delta x(j))^2 \\ & = \sum_{j}\frac{1}{N}(x^2(j)+\langle x\rangle^2-2x(j)\langle x\rangle )\\ & = \sum_{j}\frac{1}{N}x^2(j) - \langle x\rangle^2 \\ & = \langle x^2\rangle -\langle x\rangle^2 \textrm{ (always +ve)} \end{aligned}

Owing to the statistical description of quantum mechanics, we can no longer speak of the precise position of a particle, nevertheless, we still care about two quantities: (I) the expectation value of position; and (II) the uncertainty in position.

Expectation value is the weighted average of all possible values that a physical observable can take.

Expectation value of position:

\langle \mathbf{r}\rangle \equiv \displaystyle{\iiint\mathrm{d}V\,\mathbf{r}|\Psi (\mathbf{r},t)|^2}

Deviation.

\begin{aligned} \Delta\mathbf{r} & = \mathbf{r}-\langle \mathbf{r}\rangle \\ \langle \Delta\mathbf{r}\rangle & \equiv \langle\mathbf{r}\rangle - \langle\langle\mathbf{r}\rangle\rangle =0 \\ \langle (\Delta\mathbf{r})^2\rangle & = \langle\mathbf{r}^2-2\mathbf{r}\langle \mathbf{r}\rangle + \langle \mathbf{r}\rangle^2\rangle \\ & = \langle \mathbf{r}^2\rangle - \langle\mathbf{r}\rangle^2 \\ \end{aligned}

For the x-, y-, z-components of position, observe that

\begin{aligned} \langle (\Delta x)^2\rangle & = \langle x^2\rangle - \langle x\rangle^2 \\ \langle (\Delta y)^2\rangle & = \langle y^2\rangle - \langle y\rangle^2 \\ \langle (\Delta z)^2\rangle & = \langle z^2\rangle - \langle z\rangle^2 \\ \langle (\Delta\mathbf{r})^2\rangle & = \langle (\Delta x)^2\rangle + \langle (\Delta y)^2\rangle + \langle (\Delta z)^2\rangle \\ \end{aligned}

Standard deviation in position (aka uncertainty)

\sigma is defined by \sigma^2=\langle (\Delta\mathbf{r})^2\rangle;

The expectation value of position \langle \mathbf{r}\rangle gives the central position of the particle “cloud”, and \sigma tells how much the “cloud” spreads out in space. Roughly speaking, you will find the particle in the space interval of

(\langle x\rangle\pm\sigma_x, \langle y\rangle\pm\sigma_y, \langle z\rangle\pm\sigma_z)

Example (Gaussian wavepacket)

Consider a wavefunction

\Psi (x,0)=A\exp \bigg( \displaystyle{-\frac{x^2}{4\sigma^2}+iKx} \bigg).

By the normalization requirement,

\begin{aligned} \int\mathrm{d}x\, |\Psi (x,0)|^2 & = 1 \\ |A|^2\int\mathrm{d}x\,\exp\bigg( -\frac{x^2}{2\sigma^2}\bigg) & = 1 \\ A & = (2\pi\sigma^2)^{1/4} \end{aligned}

\begin{aligned} \langle x\rangle & = \int_{-\infty}^{\infty}\mathrm{d}x\, x(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) = 0 \\ \langle x^2\rangle & = \int_{-\infty}^{\infty}\mathrm{d}x\, x^2(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) \\ & = \int_{-\infty}^{\infty}\frac{1}{2}\,\mathrm{d}x^2\, x(2\pi\sigma^2)^{-1/2}\exp \bigg( -\frac{x^2}{2\sigma^2}\bigg) \\ & = \int_{-\infty}^{\infty}(-\sigma^2)\,\mathrm{d}\bigg(\exp\bigg( -\frac{x^2}{2\sigma^2} \bigg)\bigg)x(2\pi\sigma^2)^{-1/2} \\ & = \int_{-\infty}^{\infty}\sigma^2(2\pi\sigma^2)^{-1/2}\exp\bigg( -\frac{x^2}{2\sigma^2} \bigg)\,\mathrm{d}x \\ & = \sigma^2 \\ \Rightarrow \langle (\Delta x)^2\rangle & = \sigma^2 \\ \end{aligned}

Section 02

The dynamics of the wavefunction \Psi (\mathbf{r},t) is described by Schrödinger equation:

\displaystyle{i\hbar\frac{\partial \Psi}{\partial t}=-\frac{\hbar^2}{2m}\nabla^2\Psi + V(\mathbf{r},t)\Psi = \hat{H}\Psi}.

Hamiltonian operator:

\hat{H}=\hat{T}+\hat{V}

Kinetic energy operator:

\displaystyle{\hat{T}=-\frac{\hbar^2}{2m}\nabla^2}

Potential energy operator:

\hat{V}=V(\mathbf{r},t)

Planck constant:

\hbar =1.0546\times 10^{-34}\,\mathrm{J\, s}

Given the wavefunction \Psi (\mathbf{r},t_0) at time t_0, the Schrödinger equation determines \Psi (\mathbf{r},t) for all future time.

What is quantum mechanics? The simplest answer is wavefunction (and its statistical interpretation) plus Schrödinger equation.

In integral form, Schrödinger equation reads:

\Psi (x,t)=\Psi (x,t_0)\underbrace{\exp \bigg(-i\frac{\hat{H}}{\hbar}(t-t_0) \bigg)}_{\textrm{evolution operator}}

Probability density and flux

According to the probability interpretation, wavefunction has to be normalized. In a 1D example:

\displaystyle{\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x=1}

Motivation.

Does \Psi (x,t) remain normalized in the evolution according to Schrödinger equation? I.e.,

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x\stackrel{?}{=}0}

The answer to this question is affirmative.

Proof.

To begin with,

\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x & = \int_{-\infty}^{\infty}\bigg[ \frac{\partial}{\partial t}|\Psi (x,t)|^2\bigg]\,\mathrm{d}x \end{aligned}

evaluating the square-bracketed term in the integrand,

\displaystyle{\frac{\partial}{\partial t}|\Psi (x,t)|^2=\Psi^*\frac{\partial \Psi}{\partial t}+\frac{\partial\Psi^*}{\partial t}\Psi}

by identities of Schrödinger’s

\begin{aligned} \frac{\partial \Psi}{\partial t} & = \frac{i\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi\\ \frac{\partial \Psi^*}{\partial t} & = -\frac{i\hbar}{2m}\frac{\partial^2\Psi^*}{\partial x^2} +\frac{i}{\hbar}V\Psi^*\\ \end{aligned}

it follows that

\begin{aligned} \frac{\partial}{\partial t}|\Psi |^2 & = \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi \bigg) \\ & = \frac{\partial}{\partial x}\bigg[ \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^*}{\partial x}\Psi \bigg) \bigg]\\ \end{aligned}

and

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x = \frac{i\hbar}{2m}\bigg( \Psi^*\frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\bigg)\bigg|_{-\infty}^{\infty}}

since \Psi (x,t) is normalized, i.e.,

\Psi (x=\pm\infty ,t)=0,

so

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^{\infty}|\Psi (x,t)|^2\,\mathrm{d}x=0}.

\therefore The Schrödinger equation guarantees the wavefunctions remain normalized.

QED

Defining the probability density by:

\rho (x,t)\equiv |\Psi (x,t)|^2

and the probability current by:

\displaystyle{j=-\frac{i\hbar}{2m}\bigg(\Psi^*\frac{\partial \Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\bigg)}

we write the continuity equation:

\displaystyle{\frac{\partial}{\partial t}\rho =-\frac{\partial}{\partial x}j}.

Note that

\underbrace{\frac{\partial}{\partial t}\int_{a}^{b}\rho\,\mathrm{d}x}_{\textrm{(A)}}= \underbrace{-j(b)}_{\textrm{(B)}} +\underbrace{j(a)}_{\textrm{(C)}}

Physical meanings of (A), (B), and (C):

\textrm{(A)}: change in the probability of being between a and b;
\textrm{(B)}/\textrm{(C)}: positive (/negative) value means inward (/outward) flux into the region between a and b.

In 3D case:

\begin{aligned} \mathbf{j}(\mathbf{r}) & =-\frac{i\hbar}{2m}[\Psi^*\nabla\Psi - (\nabla\Psi^*)\Psi ] \\ \frac{\partial}{\partial t}|\Psi |^2 & = -\nabla\cdot\mathbf{j} \\ \end{aligned}

Probability flux through a surface area is \mathbf{j}\cdot\mathrm{d}\mathbf{s}.

(to be continued)