Quantum Mechanics — Lecture notes for PHYS223 XII Quantum mechanics in three dimensions XIV Central potentials

XIII Three-dimensional examples and applications

XIII.1 Example: free particle in three dimensions

The Hamiltonian of a free particle is given by

\hat{H}=\frac{\hat{p}_{x}^{2}+\hat{p}_{y}^{2}+\hat{p}_{z}^{2}}{2m}=-\frac{% \hbar^{2}}{2m}\Delta,

(200)

and the associated Schrödinger equation is

E\psi({\bf r})=-\frac{\hbar^{2}}{2m}\Delta\psi({\bf r}).

(201)

The solutions are the momentum eigenstates $\psi_{\bf p}({\bf r})$ given in Eq. (190):

E\psi_{\bf p}({\bf r})=\frac{p^{2}}{2m}\psi_{\bf p}({\bf r})

(202)

where $p^{2}=p_{x}^{2}+p_{y}^{2}+p_{z}^{2}$ .

XIII.2 Separation of variables in cartesian coordinates

In this course we will only consider problems which can be reduced to one-dimensional sub-problems. In the simplest case, this reduction can be carried out in cartesian coordinates: if $V({\bf r})=V_{1}(x)+V_{2}(y)+V_{3}(z)$ then we can find solutions of the form $\psi({\bf r})=X(x)Y(y)Z(z)$ .

Proof: Insert $\psi({\bf r})=X(x)Y(y)Z(z)$ into the Schrödinger equation:

	$\displaystyle EX(x)Y(y)Z(z)$
	$\displaystyle=\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}X(% x)+V_{1}(x)X(x)\right)Y(y)Z(z)$
	$\displaystyle+\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial y^{2}}Y(% y)+V_{2}(y)Y(y)\right)X(x)Z(z)$
	$\displaystyle+\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial z^{2}}Z(% z)+V_{3}(z)Z(z)\right)X(x)Y(y).$		(203)

Devide both sides by $X(x)Y(y)Z(z)$ :

$\displaystyle E$	$\displaystyle=$	$\displaystyle\frac{1}{X(x)}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{% \partial x^{2}}X(x)+V_{1}(x)X(x)\right)$	(204)
		$\displaystyle+\frac{1}{Y(y)}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{% \partial y^{2}}Y(y)+V_{2}(y)Y(y)\right)$
		$\displaystyle+\frac{1}{Z(z)}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{% \partial z^{2}}Z(z)+V_{3}(z)Z(z)\right).$

Since the left hand side of this equation is constant and each of the terms on the right hand side only depends on one of the three variables, each of those terms has to be constant. This gives three one-dimensional Schrödinger equations

$\displaystyle E_{1}X(x)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}X(x)+V_{1}(x)X(x),$	(205)
$\displaystyle E_{2}Y(y)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dy^{2}}Y(y)+V_{2}(y)Y(y),$	(206)
$\displaystyle E_{3}Z(z)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dz^{2}}Z(z)+V_{3}(z)Z(z).$	(207)

The total energy is $E=E_{1}+E_{2}+E_{3}$ .

XIII.3 Example: particle in the box

A three-dimensional box is defined by the potential $V({\bf r})=0$ for $0<x<L_{1}$ and $0<y<L_{2}$ and $0<z<L_{3}$ while $V({\bf r})=\infty$ elsewhere.

This potential can be written in the form $V({\bf r})=V_{1}(x)+V_{2}(y)+V_{3}(z)$ where

$\displaystyle V_{1}(x)$	$\displaystyle=$	$\displaystyle 0\mbox{ for }0<x<L_{1},\quad V_{1}(x)=\infty\mbox{ elsewhere},$
$\displaystyle V_{2}(y)$	$\displaystyle=$	$\displaystyle 0\mbox{ for }0<y<L_{2},\quad V_{2}(y)=\infty\mbox{ elsewhere},$
$\displaystyle V_{3}(z)$	$\displaystyle=$	$\displaystyle 0\mbox{ for }0<z<L_{3},\quad V_{3}(z)=\infty\mbox{ elsewhere}.$

The three equations (205,206,207) are of the form of one-dimensional particles in the box and are solved by

$\displaystyle X(x)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{2}{L_{1}}}\sin\frac{n_{1}\pi x}{L_{1}},\quad E_{1}=% \frac{n_{1}^{2}\pi^{2}\hbar^{2}}{2mL_{1}^{2}}$	(208)
$\displaystyle Y(y)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{2}{L_{2}}}\sin\frac{n_{2}\pi y}{L_{2}},\quad E_{2}=% \frac{n_{2}^{2}\pi^{2}\hbar^{2}}{2mL_{2}^{2}}$	(209)
$\displaystyle Z(z)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{2}{L_{3}}}\sin\frac{n_{3}\pi z}{L_{3}},\quad E_{3}=% \frac{n_{3}^{2}\pi^{2}\hbar^{2}}{2mL_{3}^{2}}$	(210)

The total energy is

E=E_{1}+E_{2}+E_{3}=\frac{\pi^{2}\hbar^{2}}{2m}\left(\frac{n_{1}^{2}}{L_{1}^{2% }}+\frac{n_{2}^{2}}{L_{2}^{2}}+\frac{n_{3}^{2}}{L_{3}^{2}}\right).

(211)

The numbers $n_{1}=1,2,3,\ldots$ , $n_{2}=1,2,3,\ldots$ , $n_{3}=1,2,3,\ldots$ which enumerate the energies are also called quantum numbers.

XIII.4 Degeneracy

For a symmetric box with $L_{1}=L_{2}=L_{3}=L$ the energies are

E=\frac{\pi^{2}\hbar^{2}}{2mL^{2}}(n_{1}^{2}+n_{2}^{2}+n_{3}^{2}).

(212)

It is then possible to have degeneracy: Different sets of quantum numbers $(n_{1},n_{2},n_{3})$ (hence, different eigenfunctions) can have the same energy. The degeneracy factor is generally denoted as $g$ . For example: the six combinations $(n_{1},n_{2},n_{3})=(1,2,3)$ , $(2,3,1)$ , $(3,1,2)$ , $(1,3,2)$ , $(2,1,3)$ , $(3,2,1)$ all give the same energy $E=14\frac{\pi^{2}\hbar^{2}}{2mL^{2}}$ . This is then called a six-fold degenerate energy level ( $g=6$ ).

XIII.5 Density of states

For many physical problems (such as thermodynamics, or transitions in scattering or decay processes, to be encountered later in this course), we need to know the number of states with energies $E_{n}\approx E$ , without resolving the details of the energy quantisation. This number can be estimated by first considering the number of states with energy $E_{n}<E$ ,

N(E)=\sum_{n}\Theta(E-E_{n})

(213)

[where $\Theta(x)$ is the unit step function, with $\Theta(x)=0$ for $x<0$ and $\Theta(x)=1$ for $x>1$ ], and then smoothing this out over energy to obtain a continuous function $\bar{N}(E)$ to finally obtain the density of states

\rho(E)=d\bar{N}/dE.

(214)

This quantity is large in regions where levels are closely spaced [the level spacing is $\Delta E=1/\rho(E)$ ].

For the particle in the three-dimensional box, we can carry out this program by replacing the sum over the quantum numbers by an integral over a continuous three-dimensional vector ${\bf n}=(n_{1},n_{2},n_{3})$ (with all components positive), and interpreting $E=\pi^{2}\hbar^{2}|{\bf n}|^{2}/2mL^{2}$ as a continuous function of this vector. In the space of ${\bf n}$ , the allowed states populate a volume

N(E)=\frac{1}{8}\frac{4\pi}{3}|{\bf n}|^{3}=\frac{(2m)^{3/2}L^{3}}{6\pi^{2}% \hbar^{3}}E^{3/2},

(215)

and the density of states becomes

\rho(E)=\frac{(2m)^{3/2}L^{3}}{4\pi^{2}\hbar^{3}}E^{1/2}.

(216)

Since this expression is proportional to the volume of the box, it is advantageous to introduce the local density of states

\nu(E)=\rho(E)/L^{3}=\frac{(2m)^{3/2}}{4\pi^{2}\hbar^{3}}E^{1/2}.

(217)

Analogously, in two dimensions one obtains

\nu(E)=\frac{m}{2\pi\hbar^{2}}

(218)

(i.e., a constant), while in one dimension

\nu(E)=\frac{1}{2\pi\hbar}\sqrt{\frac{2m}{E}}.

(219)

To a very good approximation, these expressions also apply to charge carriers in metals or semiconductors, if one only replaces the mass $m$ by a suitable effective mass $m^{*}$ which accounts for the forces from the ionic background in the material. E.g., in GaAs, $m^{*}\approx 0.067m_{e}$ for electron-like carriers, where $m_{e}$ the electron mass. Such materials allow to realize particle boxes of various dimensions (quantum wells, nanowires, and quantum dots) by suitable position-dependent doping and gating. Furthermore, since electrons carry an additional degree of freedom called spin (which we discuss in section XVI), the density of state has to be multiplied by a factor of two.

For massless particles like photons, the density of states can be constructed analogously by using the dispersion relation $E=\hbar\omega=\hbar ck$ . In three dimensions, accounting for the two polarisation directions of the photon we then obtain the local density of states

\nu(E)=\frac{E^{2}}{\pi^{2}\hbar^{3}c^{3}}

(220)

(as we will discuss later, each of these states can carry multiple photons). A related example is graphene, a two-dimensional sheet of carbon atoms, where electrons have dispersion relation $E=\hbar v_{F}|{\bf k}|$ with constant Fermi velocity $v_{F}\sim 10^{6}m/s$ . Accounting for all internal degrees of freedom (spin and pseudospin), the local density of states in this material is $\nu(E)=2|E|/(\pi\hbar^{2}v_{F}^{2})$ .

XIII.6 Example: harmonic oscillator

XIII.6.1 Revision of the one-dimensional harmonic oscillator

In one dimension, the Schrödinger equation

E\psi(x)=-\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+\frac{1}{2}m\omega^{% 2}x^{2}\psi(x),

(221)

of the harmonic oscillator is solved by

E_{n}=\hbar\omega(n+1/2),

(222)

\psi_{n}(x)=\sqrt{\frac{\alpha}{\pi^{1/2}2^{n}n!}}H_{n}(\alpha x)e^{-\alpha^{2% }x^{2}/2}

(223)

where $\alpha=\sqrt{\frac{m\omega}{\hbar}}$ and the first Hermite polynomials are $H_{0}(s)=1$ , $H_{1}(s)=2s$ , $H_{2}(s)=4s^{2}-2$ .

XIII.6.2 The three-dimensional harmonic oscillator

The three-dimensional oscillator has potential energy

$\displaystyle V({\bf r})$	$\displaystyle=$	$\displaystyle V_{1}(x)+V_{2}(y)+V_{3}(z),$	(224)
$\displaystyle V_{1}(x)$	$\displaystyle=$	$\displaystyle\frac{1}{2}m\omega_{1}^{2}x^{2},$	(225)
$\displaystyle V_{2}(y)$	$\displaystyle=$	$\displaystyle\frac{1}{2}m\omega_{2}^{2}y^{2},$	(226)
$\displaystyle V_{3}(z)$	$\displaystyle=$	$\displaystyle\frac{1}{2}m\omega_{3}^{2}z^{2}.$	(227)

Hence the three equations (205,206,207) are of the form of three one-dimensional harmonic oscillators

$\displaystyle E_{1}X(x)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}X(x)+\frac{1}{2}m\omega_% {1}^{2}x^{2}X(x),$	(228)
$\displaystyle E_{2}Y(y)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dy^{2}}Y(y)+\frac{1}{2}m\omega_% {2}^{2}y^{2}Y(y),$	(229)
$\displaystyle E_{3}Z(z)$	$\displaystyle=$	$\displaystyle-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dz^{2}}Z(z)+\frac{1}{2}m\omega_% {3}^{2}z^{2}Z(z).$	(230)

and are solved by

$\displaystyle X(x)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{\alpha_{1}}{\pi^{1/2}2^{n_{1}}n_{1}!}}H_{n_{1}}(% \alpha_{1}x)e^{-\alpha_{1}^{2}x^{2}/2},$	(231)
$\displaystyle Y(y)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{\alpha_{2}}{\pi^{1/2}2^{n_{2}}n_{2}!}}H_{n_{2}}(% \alpha_{2}y)e^{-\alpha_{2}^{2}y^{2}/2},$	(232)
$\displaystyle Z(z)$	$\displaystyle=$	$\displaystyle\sqrt{\frac{\alpha_{3}}{\pi^{1/2}2^{n_{3}}n_{3}!}}H_{n_{3}}(% \alpha_{3}z)e^{-\alpha_{3}^{2}z^{2}/2}.$	(233)

where $\alpha_{i}=\sqrt{\frac{m\omega_{i}}{\hbar}}$ . The individual energies are $E_{i}=\hbar\omega_{i}(n_{i}+1/2)$ and the total energy is

E=\hbar\omega_{1}(n_{1}+\frac{1}{2})+\hbar\omega_{2}(n_{2}+\frac{1}{2})+\hbar% \omega_{3}(n_{3}+\frac{1}{2}),\quad n_{i}=0,1,2,\ldots

(234)

XIII.6.3 Degeneracy and density of states of the isotropic oscillator

When all three frequencies $\omega_{1}=\omega_{2}=\omega_{3}=\omega$ are identical then the potential $V({\bf r})=\frac{1}{2}m\omega^{2}|{\bf r}|^{2}$ depends only on the radial distance from the origin, hence it is spherical symmetric. This is called the isotropic harmonic oscillator (isotropic means independent of the direction). The energy levels are now given by $E=\hbar\omega(n_{1}+n_{2}+n_{3}+3/2)$ . Hence, different states with the same sum of quantum numbers $n_{1}+n_{2}+n_{3}$ have the same energy. Therefore, each level with energy $E_{n}=\hbar\omega(n+3/2)$ has degeneracy $g_{n}=(n+1)(n+2)/2$ . It follows that the (smoothed) number of states with energy less than $E$ is given by $\bar{N}(E)=E^{3}/6(\hbar\omega)^{3}$ , and the density of states is $\rho(E)=E^{2}/2(\hbar\omega)^{3}$ . (In two dimensions, $\rho(E)=E/\hbar^{2}\omega^{2}$ , and in one dimension $\rho(E)=1/\hbar\omega$ , where the latter is a constant in accordance with the constant level spacing $\Delta E=\hbar\omega$ .)