2006

From Mathsoc wiki

Course Name: 2006 Quantum Physics
Lecturer: Prof. Iggy McGovern
Course Page: http://www.tcd.ie/Physics/Local/Students/SF/2006QuantumPhysics/Index.php
Textbook: Concepts of Modern Physics by Arthur Beiser (S-LEN 539 K72*5;2)

This is (or was) a painful course, mostly because I was expecting quantum physics taught by a physics department to be interesting... Anyway rather than read the rest of these notes which will teach you no real quantum mechanics, why don't you consider the following (which I came across at the start of Tony Zee's nutshell QFT book):

Hopefully you have heard of the two-slit experiment: if we fire an electron (say) at a screen with two slits in it, we end up with a diffraction pattern on a detecting screen beyond the first screen (as long as we do not actually measure which slit the electron went through). This is consistent with considering the electron to have a certain probability to go through either slit, and is the classic example of wave-particle duality. In fact we describe the electron by a wave-function, which gives us probability amplitudes for the electron to go through one slit or the other: the two probability amplitudes interfere to give the diffraction pattern we observe.

Now, if we add a third slit we can suppose that now there is a probability for the electron to go through one of the three slits, with the different wave functions for each probability interfering. The question is: what happens if we remove the screen entirely?

Richard Feynman was great, by the way.

[edit]

Black Body Radiation

A blackbody is an ideal body which absorbs all radiation incident on it. When in thermal equilibrium with its surroundings, such a body also radiates energy at an equal rate to the absorption. A blackbody can be modelled as a thermal cavity - this is a hollow body with a tiny hole into it. When electromagnetic waves are incident on the body they enter through the hole and set up standing waves inside the cavity. These can be thought of as being emitted by oscillators in the walls of the cavity.

We can experimentally measure the energy density per unit volume $u ( \nu )$ where $\nu$ is the frequency of the emitted radiation. We have that $u ( \nu ) d \nu$ is the energy per unit volume in the frequency range $\nu$ to $\nu + d \nu$ .

[edit]

Rayleigh-Jeans Model

The Rayleigh-Jeans model assumed that a blackbody was a cavity at temperature $T$ , filled with standing waves being emitted by oscillators in the walls, and assumed the classical equipartition of energy.

Now, the wavelength of a standing wave is given by

$\lambda = \frac{2L}{n}$

where $L$ is the distance from one wall of the cavity to the other, and $n = 1,2,3 \dots$ Hence in the x,y and z-directions we have:

$j_x = \frac{2L}{\lambda} = 1,2,3 \dots$

$j_y = \frac{2L}{\lambda} = 1,2,3 \dots$

$j_z = \frac{2L}{\lambda} = 1,2,3 \dots$

and in three-dimensions then

$j_x^2 + j_y^2 + j_z^2 = \left(\frac{2L}{\lambda}\right)^2$

(note that now the $j_i$ can take values starting at 0). We can now imagine a so-called j-space, consisting of $j_x$ , $j_y$ and $j_z$ axes, with each point in the j-space corresponding to a permissible set of $j_x$ , $j_y$ and $j_z$ values, and hence to a standing wave.

We let the points in j-space be so close together that they are continuous, and wish to find the number of waves with wavelengths between $\lambda$ and $\lambda + d \lambda$ . This equals the number of points in j-space between two spheres of radii $j$ and $j + dj$ , and so is given by:

$g(j) dj = 4 \pi j^2 dj$

However, we only want the portion of the sphere corresponding to positive $j_x$ , $j_y$ and $j_z$ values, hence we divide by eight, and for each wave included there are two perpendicular directions of polarisation, so we multiply by two. This gives

$g(j) dj = \pi j^2 dj$

Now,

$j = \frac{2L}{\lambda} = \frac{2L \nu}{c} \Rightarrow dj = \frac{2L}{c} d \nu$

$g (\nu) d \nu = \frac{8 \pi L^3}{c^3} \nu^2 d \nu$

and the density of standing waves in the cavity is hence

$G(\nu) d \nu = \frac{8 \pi \nu^2}{c^3} d \nu$

Now, the energy density = (average energy)x(standing waves density). From statistical physics, the average amount of energy per degree is freedom is $\frac{kT}{2}$ , and for an oscillator, there are two degrees of freedom. This gives us the Rayleigh-Jeans formula for the energy density of blackbody radiation:

$u (\nu) d \nu = \frac{8 \pi k T}{c^3} \nu^2 d \nu$

which however fails, except at low frequencies. The formula gives an energy density tending to infinity for high frequencies (this was known as the ultraviolet catastrophe).

[edit]

Wien Formula

This formula was fashioned so as to fit the experimental data:

$u (\nu) d \nu = \alpha \nu^3 \mbox{exp} \left( - \beta \frac{\nu}{T} \right) d \nu$

It however failed at low frequencies.

[edit]

Planck's Formula

The correct expression was derived by Max Planck in 1900, and is:

$u (\nu) = \frac{8 \pi h}{c^3} \frac{\nu^3}{\mbox{exp} \left(\frac{h \nu}{kT}\right) - 1 } d \nu$

where $h$ is now known as Planck's constant.

Planck's model implies that energy was quantised, as $E_n = n h \nu$ as for the oscillators in the cavity walls, and also gives back Rayleigh-Jeans and Wien formulae in the low and high frequency limit.

[edit]

The Photoelectric Effect

The photoelectric effect. Incoming EM radiation on the left ejects electrons, depicted as flying off to the right, from a substance.

The photoelectric effect is the emission of electrons from matter upon the absorption of electromagnetic radiation. We can consider the electrons in the material as being in an energy well, at an energy depth $\phi$ known as the work function. Einstein's model of light assumed that the energy of light was quantised in packets (known as photons) of energy

$E = h \nu$

which implied that the incoming photon transfers all its energy to the electron, hence

$h \nu = \mbox{kinetic energy} + \mbox{depth in well}$

The standard model of the photoelectric effect involves light incident on an emitter in a electrical circuit, causing a flow of electrons which is opposed by a variable voltage $V$ . At $V = V_0$ , no current flows, and this occurs at the cut-off frequency $\nu_0$ . Consider the highest-lying electrons:

$h \nu = KE_{max} + \phi$

$\Rightarrow e V_0 = h \nu - \phi$

$\Rightarrow V_0 = \frac{h}{e} \nu - \frac{\phi}{e}$

$\Rightarrow V_0 = \frac{h}{e} ( \nu - \nu_0)$

[edit]

The Compton Effect

[edit]

Momentum of Photons

Relativistic energy is given by $E = \frac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$ , which implies that as photons travel with velocity $c$ they must have zero rest mass (otherwise they would have infinite energy). They do however have momentum:

$E = m_0^2 c^4 + p^2 c^2 \Rightarrow E = pc = h \nu$

so we have that

$p = \frac{h \nu}{c} = \frac{h}{\lambda}$

[edit]

Compton Effect

The Compton effect.

Consider the collision of a photon with a stationary electron.

As a result of the collision, the photon's frequency decreases from $\nu$ to $\nu'$ , and the electron gains kinetic energy equal to $h \nu - h \nu'$ . Now, from conservation of momentum,

$\frac{h \nu}{c} = \frac{h \nu'}{c} \cos \phi + p \cos \theta$ (horizontally)

$\Rightarrow pc \cos \theta = h \nu - h \nu' \cos \phi$

and

$0 = \frac{h \nu'}{c} \sin \phi - p \sin \theta$ (vertically)

$\Rightarrow pc \sin \theta = h \nu' \sin \phi$

giving

$p^2 c^2 = h^2 \nu^2 - 2 h\nu h \nu' \cos \phi + h^2 \nu'^2$

Now, for a particle the energy is given by

$E = \sqrt{m_0^2 c^4 + p^2 c^2} = KE + m_0 c^2$

$\Rightarrow p^2 c^2 = (KE)^2 + 2 m_0 c^2 KE$

meaning that

$(h \nu - h \nu')^2 + 2 m_0 c^2 (h \nu - h \nu') = h^2 \nu^2 - 2 h\nu h \nu' \cos \phi + h^2 \nu'^2$

$\Rightarrow m_0 c^2 (h \nu - h \nu') = h \nu h \nu' (1 - \cos \phi)$

or in terms of wavelength,

$\lambda' - \lambda = \frac{h}{m_0 c} (1 - \cos \phi)$

This increase in wavelength is known as the Compton effect. The quantity $\frac{h}{m_0c}$ is sometimes called the Compton wavelength.

[edit]

Wave Properties of Particles

[edit]

De Broglie Wavelength

All particles with momentum have an associated de Broglie wavelength:

$\lambda = \frac{h}{p} = \frac{h}{mv}$

An experiment by Davisson and Germer found that electrons were diffracted by a crystal, just like x-rays.

[edit]

Wave Description and Group Velocity

Consider the standard wave formula

$y = A \cos (\omega t - k x)$

The phase velocity is given by

$v_p = \nu \lambda = \frac{\omega}{2 \pi} \frac{2 \pi}{k} = \frac{\omega}{k}$

Now consider a particle, then

$v_p = \nu \lambda = h \nu \frac{\lambda}{h} = \frac{h \nu}{\frac{h}{\lambda}} = \frac{E}{p} = \frac{m c^2}{m v}$

so $v_p = \frac{c^2}{v}$ and it follows that the phase velocity does not equal the velocity of the particle.

A better description of a particle as a wave is given by beats. Consider two waves of nearly equal frequencies,

$y_1 = A \cos ( \omega t - k x)$

$y_2 = A \cos \Big[ (\omega + \Delta \omega) t - (k + \Delta k) x \Big]$

combining to give $y = y_1 + y_2$ or, using a trigonometric identity,

$y = 2A \cos \left[ \frac{1}{2} \Big( \left[2 \omega + \Delta \omega \right]t - \left[2 k + \Delta k \right]x \Big) \right] \cos \left[\frac{1}{2} (\Delta \omega t - \Delta k x)\right]$

and as $2 \omega + \Delta \omega \approx 2 \omega$ ,

$y = 2A \cos \left[\omega t - kx \right] \cos \left[\frac{\Delta \omega}{2}t - \frac{\Delta k}{2} x\right]$

describing a wave with amplitude modulated at beat angular frequency $\frac{\Delta \omega}{2}$ and wave number $\frac{\Delta k}{2}$ .

We now define the group velocity

$v_g = \frac{\frac{\Delta \omega}{2}}{\frac{\Delta k}{2}} = \frac{\Delta \omega}{\Delta k} = \frac{d \omega}{d k}$

Now,

$\omega = 2 \pi \nu = 2 \pi \frac{mc^2}{h} = \frac{2 \pi m_0 c^2}{h} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

$k = \frac{2 \pi}{\lambda} = 2 \pi \frac{mv}{h} = \frac{2 \pi m_0}{h} \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}$

and

$v_g = \frac{\frac{d \omega}{d v}}{\frac{dk}{dv}}$

and upon differentiating we find that

$v_g = v$

hence the group velocity of the wave group equals the velocity of the particle.

Note that for a wave group to represent a particle the phase velocity must also vary with wavelength.

[edit]

Wave Function and Expectation Value

[edit]

Wave function

We can describe a particle using a wave function $\psi$ . The probability of finding a particle in a particular state is equal to $| \psi |^2$ .

The wave function is single-valued, continuous and has a single-valued and continuous derivative. It is also normalisable such that $\int |\psi|^2 dV =1$ i.e. the integrated probability density over all space is unity. In one dimension we have that $\int_{- \infty}^{\infty} |\psi|^2 dx = 1$

[edit]

Expectation Value

The most probable value of a variable is known as the expectation value, and for a general variable $G(x)$ is given by

$\langle G(x) \rangle = \frac{\int_{- \infty}^{\infty} G(x) |\psi|^2 dx}{\int_{- \infty}^{\infty} |\psi|^2 dx}$

$\int_{- \infty}^{\infty} G(x) |\psi|^2 dx$

if the wave function is normalised.

[edit]

The Uncertainty Principle

[edit]

Uncertainty Principle

Consider a wave group representing a single particle. We could assume that the particle is "actually located" in the centre of the wave group. To be more precise, we could make the wave group narrower, but this leads to the wavelength being poorly defined, and hence the momentum $p = \frac{h}{\lambda}$ is poorly defined. Conversely, increasing the length of the wavegroup gives us a better idea of the momentum, but a worse idea of the position.

The Heisenberg Uncertainty Principle states that it is impossible to know exact position and exact momentum of an object at the same time. Expressed mathematically,

$\Delta x \propto \frac{1}{\Delta p}$

If we plot graphs of the wave function $\psi$ against $x$ and also of the Fourier transform $g(k)$ of the wave-function we find that the product $\Delta x \Delta k$

is minimised for the Gaussian curve, for which

$\Delta x \Delta k = \frac{1}{2}$

hence in general

$\Delta x \Delta k \geq \frac{1}{2}$ .

Using $k = \frac{2 \pi}{\lambda}$ and $\lambda = \frac{h}{p} \Rightarrow k = \frac{2 \pi p}{h}$ , we get

$\Delta x \Delta p \geq \frac{h}{4 \pi}$

$\Delta x \Delta p \geq \frac{\hbar}{2}$

where $\hbar = \frac{h}{2 \pi}$

[edit]

Thought Experiments: Optical Microscope and Slits

Optical Microscope: Consider using an optical microscope to find a particle's position, by scattering a photon into the lens. The photon therefore arrives anywhere within the lens angle $2 \alpha$ . The photon momentum $p = \frac{h}{\lambda}$ causes the electron to recoil.

Along the horizontal, the momentum changes in the range $- p \sin \alpha$ to $p \sin \alpha$ , hence

$\Delta p = 2 p \sin \alpha = 2 \frac{h}{\lambda} \sin \alpha$

The uncertainty in position is associated with the diffraction limit of the microscope. The minimum separation of points is

$\Delta x = \frac{\lambda}{\sin \alpha}$

hence

$\Delta x \Delta p = 2 h \ge \frac{\hbar}{2}$ .

Single Slit: Here the uncertainty in position is the slit width $s$ . The first diffraction minimum occurs at $s \sin \theta = \lambda$ so

$\Delta x = \frac{\lambda}{\sin \theta} = \frac{h}{p \sin \theta}$

Now an electron or photon arriving within the central maximum must be deflected through the angle range $0$ to $\theta$ , giving an uncertainty in momentum across the slit of

$\Delta p = p \sin \theta$

so that

$\Delta x \Delta p = h \ge \frac{\hbar}{2}$

We can also extend this analysis to the Young's slit experiment, where the uncertainty in position is the slit separation. Note that in this case, if we close one of the slits (or even install detectors at the slits to see which slit the particle passes through) then we lose the diffraction pattern.

[edit]

Bohr Model of The Hydrogen Atom

Consider an electron in a circular orbit around a nucleus. From Coulomb's law,

$F_e = \frac{e^2}{4 \pi \epsilon_0 r^2} = m \frac{v^2}{r}$

$\Rightarrow v = \frac{e}{\sqrt{4 \pi \epsilon_0 m r}}$

and the electron's energy is given by

$E = \frac{1}{2} m v^2 + q V$

where $V$ is the potential of the proton. This gives:

$E = \frac{1}{2} m \frac{e^2}{4 \pi \epsilon_0 m r} - \frac{e^2}{4 \pi \epsilon_0 r}$

$\Rightarrow E = - \frac{e^2}{8 \pi \epsilon_0 r}$

Classically, any radius and hence energy is possible. However, we find that the energy required to remove an electron from the hydrogen atom by photoelectric is always the same. Also, classical electromagnetic theory predicts that an accelerating charge will radiate energy, and hence lose kinetic energy and spiral inwards in the case of the orbiting electron.

According to the quantum model of the hydrogen atom, an electron can circle a nucleus without losing energy if its orbital circumference contains an integral multiple of de Broglie wavelengths, ie

$2 \pi r_n = n \lambda$

From $\lambda = \frac{h}{mv}$ we get

$\lambda = \frac{h}{e} \sqrt{\frac{4 \pi \epsilon_0 r_n}{m}} \Rightarrow \frac{nh}{e} \sqrt{\frac{4 \pi \epsilon_0 r_n}{m}} = 2 \pi r_n$

so the allowed values of radii are

$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$

The innermost orbit is called the Bohr radius of the hydrogen atom, and is denoted $a_0 = 5.292 \times 10^{-11}m$ .

We then have that the electron energy is

$E_n = - \frac{e^2}{8 \pi \epsilon_0 r_n} = - \frac{me^4}{8 \epsilon_0^2 h^2} \frac{1}{n^2} = \frac{E_1}{n^2}$

where we have let $E_1 = - \frac{me^4}{8 \epsilon_0^2 h^2} = - 13.6 eV$ .

If we now consider the emission of light due to an electron transition from a higher energy level to a lower we find that

$E_{photon} = h \nu = E_{n_i} - E_{n_f} = - \frac{me^4}{8 \epsilon_0^2 h^2} \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$

$\frac{1}{\lambda} = \frac{me^4}{8 \epsilon_0^2 h^3 c}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

which gives the expression for the Balmer, Paschen etc. series.

Note that to be completely precise we must take into account the fact that the electron actually orbits the centre of mass of the electron-proton system. Hence we must replace $m$ with the reduced mass, $\frac{mM}{m + M}$ , where $M$ is the mass of the proton.

[edit]

Correspondence Principle

The correspondence principle states that that quantum physics approaches classical physics in the limit of large quantum number. We can demonstrate this using the Bohr model of the hydrogen atom.

According to classical electromagnetic theory, a charge moving in a circular orbit will radiate electromagnetic radiation of frequency equal to the frequency of revolution.

$f = \frac{v}{2 \pi r}$

Now, $v = \frac{e}{\sqrt{4 \pi \epsilon_0 m r}}$ and $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$ giving

$f = \frac{m e^4}{8 \epsilon_0^2 h^3} \frac{2}{n^3}$ .

We compare this with the emitted photon frequency predicted by Bohr's model.

$\nu = \frac{me^4}{8 \epsilon_0^2 h^3}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$

We let $n_i = n$ and $n_f = n - p$ , then

$\frac{1}{n_f^2} - \frac{1}{n_i^2} = \frac{1}{(n-p)^2} - \frac{1}{n^2} = \frac{2np - p^2}{n^2(n-p)^2}$

If $n>>p$ then $2np - p^2 \approx 2np$ and $(n-p)^2 \approx n^2$ , so

$\nu = \frac{m e^4}{8 \epsilon_0^2 h^3} \frac{2 p}{n^3}$

which for $p=1$ gives back the classical frequency.

[edit]

Schrödinger Equation

[edit]

General Form

Consider the wave function

$\psi = A \exp \Big[- i(\omega t - kx)\Big]$

and recall that $\omega = 2 \pi \nu = 2 \pi \frac{E}{h} = \frac{E}{\hbar}$ and $k = \frac{2 \pi}{\lambda} = \frac{2 \pi p}{h} = \frac{p}{\hbar}$ . Hence,

$\psi = A \exp \Big[-i(\frac{E}{\hbar} t - \frac{p}{\hbar}x)\Big]$

$\Rightarrow \frac{\partial \psi}{\partial x} = \frac{i p}{\hbar} \psi \Rightarrow \frac{\partial^2 \psi}{\partial x^2} = - \frac{p^2}{\hbar^2} \psi$

so $p^2 \psi = - \hbar^2 \frac{\partial^2 \psi}{\partial x^2}$

and

$\Rightarrow \frac{\partial \psi}{\partial t} = - \frac{i E}{\hbar} \psi \Rightarrow E \psi = i \hbar \frac{\partial \psi}{\partial t}$

and as

$E = KE + PE = \frac{p^2}{2m} + U(x,t)$

we have

$E \psi = \frac{p^2}{2m} \psi + U \psi$

$\Rightarrow i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + U \psi$

the one dimensional Schrödinger Equation.

The three dimensional form is

$i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m} \vec{\nabla} \psi + U \psi$

Bear in mind that this was "derived" by assuming all wave functions are of the form above, and so is basically nonsense (what it's doing is starting with the solution to the Schrödinger equation for a free particle and then working backwards). The Schrödinger equation is actually derived by demanding that time evolution of the system be a unitary transformation on the Hilbert space of possible states...

[edit]

Steady State Schrödinger Equation

Now let $U = U(x)$ only, and write

$\psi = \psi' \exp\left(-i \frac{E}{\hbar} t \right)$

where $\psi' = A \exp\left(i \frac{p}{\hbar}x\right)$

Subbing this into the Schrödinger equation gives

$E \psi' \exp\left(-i \frac{E}{\hbar} t \right) = - \frac{\hbar^2}{2m} \exp\left(-i \frac{E}{\hbar} t \right) \frac{\partial^2 \psi'}{\partial x^2} + U \psi' \exp\left(-i \frac{E}{\hbar} t \right)$

Dropping the dashes and rearranging gives the steady state Schrödinger equation

$\frac{d^2 \psi}{d x^2} + \frac{2m}{\hbar^2} (E - U) \psi = 0$

[edit]

Particle in an Infinite Square Well

The infinite square well potential is described by

$U = \begin{cases}\infty & x=0, x=L \\ 0 & 0<x<L\end{cases}$

$\Rightarrow \psi = 0, x \leq 0, x \geq L$

The Schrödinger equation becomes

$\frac{d^2 \psi}{d x^2} - \frac{2m}{\hbar^2} E \psi = 0$

and the general solution is

$\psi = A' \exp \left( i \frac{p}{\hbar} x \right) + B' \exp \left( - i \frac{p}{\hbar} x \right)$

and consider the two cases $A' = B' = \frac{B}{2}$ and $A' = -B' = - \frac{A}{2i}$ , giving general solution

$\psi = A \sin \left( \frac{p}{\hbar} x \right) + B \cos \left( \frac{p}{\hbar} x \right)$

Applying the boundary condition $\psi = 0$ at $x=0$ gives $B=0$ , and the condition $\psi = 0$ at $x=L$ gives

$\sin \left( \frac{p}{\hbar} L \right) = 0 \Rightarrow \frac{p}{\hbar}L = n \pi \,\,\,\, n = 1,2 \dots$

then

$\psi_n = A \sin \left(\frac{n \pi x}{L}\right)$

which agrees with our previous standing waves description of the particle in a box.

As $E = \frac{p^2}{2m}$ we find that

$E_n = \frac{n^2 h^2}{8 m L^2}$

We now normalise $\psi$ .

$\int_{-\infty}^{\infty} | \psi_n |^2 dx = \int_0^L | \psi_n |^2 dx = A^2 \int_0^L \sin^2 \left(\frac{n \pi x}{L}\right) dx$

and the integration leads to

$\frac{A^2 L}{2} = 1 \Rightarrow A = \sqrt{\frac{2}{L}}$

We can also work out the expectation value

$\langle x \rangle = \int_{-\infty}^{\infty} x | \psi_n |^2 dx = \frac{2}{L} \int_0^L x \sin^2 \left(\frac{n \pi x}{L}\right)dx = \frac{1}{L} \int_0^L \Bigg(x - x \cos \left( \frac{2 n \pi x}{L} \right) \Bigg) dx$

$\Rightarrow \langle x \rangle = \frac{1}{L} \Bigg[ \frac{x^2}{2} - x \sin \left( \frac{2 n \pi x}{L} \right) \frac{L}{2 \pi n} - \cos \left( \frac{2 n \pi x}{L} \right) \frac{L^2}{(2n \pi)^2}\Bigg]_0^L$

and this gives

$\langle x \rangle = \frac{L}{2}$

showing the expectation value of $x$ is the centre of the box, irrespective of quantum number.

[edit]

Finite Square Well

The finite square well potential is described by

$U = \begin{cases} U_0 & x \leq0 , x\geq L \\ 0 & 0 < x < L \end{cases}$

We consider only the cases where $E \leq U_0$ . In the region $x \leq 0$ , the Schrödinger equation is

$\frac{d^2 \psi}{d x^2} + \frac{2m}{\hbar^2} (E - U_0) \psi = 0$

$\frac{d^2 \psi}{d x^2} - a^2 \psi = 0$

where $a = \frac{\sqrt{2m(U_0 - E)}}{\hbar}$ . This gives solution

$\psi_1 = A \exp(ax) + B \exp (-ax)$

and for $\psi_1$ to be finite as $x \rightarrow - \infty$ we require $B=0$ , thus

$\psi_1 = A \exp(ax)$

which represents exponential decay of the wavefunction. In the region where $x\geq L$ we similarly have that

$\psi_2 = A' \exp(-ax)$

In the region $0 < x < L$ the potential energy is zero, and the analysis is similar to that for the infinite square potential. We have general solution in this region

$\psi = C \sin \left(\frac{p}{\hbar} x \right) + D \cos \left(\frac{p}{\hbar} x \right)$

and as $\psi \neq 0$ at $x = 0$ we have that $D$ is non-zero; however we would expect it to be small.

A more thorough mathematical analysis is needed to ensure that $\psi$ and its first derivative agree at $x=0$ and $x=L$ . It is found that the wavelengths are slightly longer in this case than in that of the infinite square potential.

[edit]

Simple Harmonic Oscillator

The equation of motion of the simple harmonic oscillator is

$\frac{d^2 x}{dt^2} + \frac{k}{m} x = 0$

and a solution is

$x = A \cos (2 \pi \nu t)$ where $\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

The potential energy is $U = \frac{1}{2} k x^2$ hence the Schrödinger equation is

$\frac{d^2 \psi}{d x^2} + \frac{2m}{\hbar^2} \left( E - \frac{1}{2} k x^2 \right) \psi = 0$

$\Rightarrow \frac{d^2 \psi}{d x^2} + \left(\frac{2mE}{\hbar^2} - \frac{km}{\hbar^2} x^2 \right) \psi = 0$

We let $y =\left(\frac{\sqrt{km}}{\hbar}\right)^{\frac{1}{2}} x \Rightarrow dy^2 = \frac{\sqrt{km}}{\hbar} dx^2$ , and we get

$\frac{d^2 \psi}{dy^2} \frac{\sqrt{km}}{\hbar} + \left(\frac{2mE}{\hbar^2} - \frac{\sqrt{km}}{\hbar}y^2\right)\psi = 0$

$\frac{d^2 \psi}{dy^2} + (\alpha - y^2) \psi = 0$

where $\alpha = \frac{\frac{2mE}{\hbar^2}}{\frac{\sqrt{km}}{\hbar}} = \frac{2E}{\hbar} \sqrt{\frac{m}{k}} = \frac{2E}{h \nu}$ using $\nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

The more detailed solution requires that $\alpha = 2n + 1$ , $n=0,1,2 \dots$ , hence

$E_n = \frac{h \nu}{2} \alpha = (n + \frac{1}{2}) h \nu$

where

$E_0 = \frac{1}{2} h \nu$

is called the zero point energy. Note that Planck's derivation of the blackbody radiation formula used $E_n = n h \nu$ , which we now see is incorrect, but still gave him the correct answer.

[edit]

Step Potential

We have three distinct cases:

Step-up Potential $U_0$ , $E < U_0$

In region I (left of the step-up potential region), we have $U=0$ , hence

$\frac{d^2 \psi}{dx^2} + \frac{2mE}{\hbar^2} \psi = 0$

giving solution

$\psi_I = A \exp(i k_1 x) + B \exp (-ik_1x)$

where $k_1 = \frac{\sqrt{2mE}}{\hbar}$ .

In region II, $U = U_0$ , hence

$\frac{d^2 \psi}{dx^2} + \frac{2m}{\hbar^2}(E-U_0) \psi = 0$

giving solution

$\psi_{II} = C \exp(k_2 x) + D \exp (- k_2 x)$

where $k_2 = \frac{\sqrt{2m(U_0 - E)}}{\hbar}$ and we require $C=0$ so that $\psi_{II}$ is finite as $x \rightarrow \infty$

We now apply our boundary conditions:

$\psi_{I} = \psi_{II}$ at $x=0 \Rightarrow A + B = D$

$\psi_{I}^{\prime} = \psi_{II}^{\prime}$ at $x=0 \Rightarrow i k_1 A - i k_1B = -k_2D$

$\Rightarrow A = \frac{1}{2} \left(1 + \frac{i k_2}{k_1} \right) D$

$\Rightarrow B = \frac{1}{2} \left(1 - \frac{i k_2}{k_1} \right) D$

hence

$\psi_I = \frac{D}{2} \left(1 + \frac{i k_2}{k_1} \right) \exp(i k_1 x) + \frac{D}{2} \left(1 - \frac{i k_2}{k_1} \right) D \exp(- ik_1 x)$

where the first term describes the incident particles and the second term the reflected particles, and

$\psi_{II} = D \exp(-k_2x)$

showing that the particles' probability decays into the wall.

We also have the reflection coeffecient

$R = \frac{\overline{B} B}{\overline{A}A} = 1$

in agreement with the classical picture.

Step-up Potential $U_0$ , $E > U_0$

In region I we find that

$\psi_I = A \exp(i k_1 x) + B \exp (-ik_1x)$

where $k_1 = \frac{\sqrt{2mE}}{h}$ , and in region II

$\psi_{II} = C \exp(i k_2 x) + D\exp (-ik_2x)$

where $k_2 = \frac{\sqrt{2m(E- U_0)}}{\hbar}$ and we require $D=0$ as there is no physical meaning to the negative definition of travel. We again apply our boundary conditions,

$\psi_{I} = \psi_{II}$ at $x=0 \Rightarrow A + B = C$

$\psi_{I}^{\prime} = \psi_{II}^{\prime}$ at $x=0 \Rightarrow i k_1 A - i k_1B = ik_2C$

$\Rightarrow C = \frac{2k_1}{k_1 + k_2} A$

$\Rightarrow B = \frac{k_1 - k_2}{k_1 + k_2}A$

$\psi_I = A \exp(i k_1 x) + \frac{k_1 - k_2}{k_1 + k_2}A \exp(- ik_1 x)$

where the first term again describes the incident particles and the second term the reflected particles, and

$\psi_{II} = \frac{2k_1}{k_1 + k_2}A \exp(i k_2 x)$

The reflection coefficient now is

$R = \frac{\overline{B} B}{\overline{A}A} = \left(\frac{k_1-k_2}{k_1+k_2}\right)^2 = \left(\frac{1 -\frac{k_2}{k_1}}{1+ \frac{k_2}{k_1}}\right)^2$

and $\frac{k_2}{k_1} = \frac{\sqrt{2m(E- U_0)}}{\sqrt{2mE}} = \sqrt{1 - \frac{U_0}{E}}$ , hence

$R = \left(\frac{1 -\sqrt{1 - \frac{U_0}{E}}}{1+ \sqrt{1 - \frac{U_0}{E}}}\right)^2$

Step-down Potential $U_0$ , $E > U_0$

This situation is the reverse of the step-up situation, so we can just exchange $k_1$ and $k_2$ , hence

$\psi_I = A \exp(i k_2 x) + B \exp(- ik_2 x)$

$\psi_{II} = C \exp(i k_1 x)$

[edit]

Potential Barrier

Consider a potential barrier of height $U_0$ and width $L$ . Our solutions are

$\psi_I = A \exp(i k_1 x) + B \exp (-ik_1x)$

$\psi_{II} = C \exp(k_2 x) + D \exp (k_2x)$

$\psi_{III} = F \exp(i k_1 x) + G \exp (-ik_1x)$

and we require $G=0$ so that $\psi_{III}$ is finite as $x \rightarrow \infty$

We have transmission coefficient

$T = \frac{\overline{F} F}{\overline{A}A}$

and if $E << U_0$ we get

$T \approx \frac{16}{4 + \left(\frac{k_2}{k_1}\right)^2 } \exp(-2k_2 L) \approx \exp(-2k_2 L)$

Applications: scanning tunnelling microscopes, nuclear decay.

If $E>U_0$ we get reflections from both edges of the barrier, and for certain wavelengths $\lambda = \frac{2L}{n}$ we get destructive interference (see Ramsauer-Townsend effect).

[edit]

Quantum Theory of the Hydrogen Atom

Retrieved from "http://www.maths.tcd.ie/~mathsoc/wiki/2006"

Categories: SF physics courses | Physics courses

2006

From Mathsoc wiki

Contents

Black Body Radiation

Rayleigh-Jeans Model

Wien Formula

Planck's Formula

The Photoelectric Effect

The Compton Effect

Momentum of Photons

Compton Effect

Wave Properties of Particles

De Broglie Wavelength

Wave Description and Group Velocity

Wave Function and Expectation Value

Wave function

Expectation Value

The Uncertainty Principle

Uncertainty Principle

Thought Experiments: Optical Microscope and Slits

Bohr Model of The Hydrogen Atom

Correspondence Principle

Schrödinger Equation

General Form

Steady State Schrödinger Equation

Particle in an Infinite Square Well

Finite Square Well

Simple Harmonic Oscillator

Step Potential

Potential Barrier

Quantum Theory of the Hydrogen Atom

current page

other pages

personal

Search