Physics for Information Science. Module 2, matter waves to quantum tunnelling.

Page 1 of 15

Module 2 at a glance

Quantum mechanics is the branch of mechanics that gives the mathematical description of the motion and interaction of very small particles such as electrons, protons and atoms. Classical (Newtonian) mechanics works perfectly for cricket balls and planets, but it fails at the atomic scale. It cannot explain why atoms are stable, why their spectra have sharp lines, or why electrons are diffracted by crystals. Quantum mechanics takes its place there. It is based on three ideas that have no place in everyday physics:

1. Quantisation of energy. A confined particle cannot have any energy it likes. Only certain discrete energies are allowed.
2. Wave–particle duality. Every moving particle has a wave associated with it, called the matter wave. The wave is described by a wave function $\Psi$, and $|\Psi|^2$ tells where the particle is likely to be, not where it definitely is.
3. The uncertainty principle. Position and momentum (and also energy and time) cannot both be known exactly at the same time.

From these ideas we build the Schrödinger equation, the equation that the wave function follows. We solve it for two standard problems: a particle trapped in a box (which explains why energy comes in discrete levels and why a confined particle can never be at rest) and a particle meeting a barrier that it does not have enough energy to climb (which explains tunnelling).

What you will be able to do

• Derive the de Broglie relation $\lambda=h/p$ and calculate the wavelength of any particle, and of an electron accelerated through a voltage.
• Explain the physical meaning of the wave function (Born's interpretation), write the conditions for a well-behaved wave function, and normalise a wave function.
• Write the wave function of a free particle and show that $\omega=E/\hbar$ and $k=p/\hbar$.
• Explain wave packets, phase and group velocity, and how a wave packet leads to the uncertainty principle.
• State the uncertainty principle for the three conjugate pairs and use it to show that electrons cannot exist inside the nucleus and that spectral lines have a natural width.
• Derive the time-dependent Schrödinger equation by two methods, and derive the time-independent equation from it.
• Solve the Schrödinger equation for a particle in a one-dimensional box: find the energy eigenvalues and the normalised eigenfunctions, and interpret them.
• Explain quantum mechanical tunnelling, write the wave function in each region, define the transmission coefficient and list its applications.

How to use these notes

• Each page follows idea → explanation or derivation → worked example → common mistakes.
• Green boxes show key ideas and definitions. Amber boxes marked Watch out show mistakes that make you lose marks. Boxes marked Note on the source show where the lecture notes were corrected.
• The derivations of the Schrödinger equation and the particle in a box are the most frequently asked essay questions. Read them, then close the notes and write them yourself, step by step, until you can do it without looking.
• The last page has practice problems with full answers, the past KTU exam questions and a formula sheet.

Where each topic sits

Before you start. You should know wavelength $\lambda$, frequency $\nu$, period $T=1/\nu$, angular frequency $\omega=2\pi\nu=2\pi/T$ and wave number $k=2\pi/\lambda$; the photon relations $E=h\nu$ and $p=h/\lambda$; complex numbers, with $e^{i\theta}=\cos\theta+i\sin\theta$, the conjugate $z^*$ and $|z|^2=z^*z$; partial derivatives and simple integration, including $\sin^2\theta=\tfrac12(1-\cos2\theta)$.

Page 2 of 15

Matter waves

The idea

By the early twentieth century it was clear that light has a dual nature. Interference, diffraction and polarisation show that light is a wave. The photoelectric effect and the Compton effect show that light also behaves as a stream of particles, called photons, each carrying energy $E=h\nu$ and momentum $p=h/\lambda$.

In 1924 the French physicist Louis de Broglie argued by analogy and symmetry: if waves (light) can behave like particles, then particles should also be able to behave like waves. He proposed that every moving object has a wave associated with it under suitable conditions. These waves are called matter waves or de Broglie waves.

Matter waves. The waves associated with a moving particle are called matter waves or de Broglie waves. Their wavelength is called the de Broglie wavelength.

Deriving the de Broglie wavelength (route 1: from the photon's energy)

For a photon the energy can be written in two ways, one from its wave nature and one from its particle nature:

Since both describe the same photon, they must be equal. This shows its wave–particle dual nature:

Cancel one factor of $c$:

Here $mc$ (mass times velocity) is the momentum $p$ of the photon. So

Deriving it (route 2: from the photon's momentum)

The momentum of a photon is known directly from experiment and theory to be $p=h/\lambda$, which can be written as $\lambda=h/p$. De Broglie's step was to assume that this equation is completely general. It applies not only to photons but to electrons, protons, atoms and every other particle.

The de Broglie wavelength of a particle

A particle of mass $m$ moving with speed $v$ has momentum $p=mv$. Its kinetic energy is $KE=\tfrac12mv^2$, which can be written as

so the de Broglie wavelength can be written in three equivalent ways:

de Broglie wavelength.

$\lambda=\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2m\,KE}}$

An electron accelerated through a voltage

An electron of charge $e$ and mass $m$, starting from rest and accelerated through a potential difference of $V$ volts, gains kinetic energy $KE=eV$. So

Then the wavelength of the electron is

This is the same as putting $KE=eV$ in $\lambda=h/\sqrt{2m\,KE}$. Now put $h=6.626\times10^{-34}\ \text{J s}$, $m=9.11\times10^{-31}\ \text{kg}$ and $e=1.6\times10^{-19}\ \text{C}$:

Since $1\ \text{Å}=10^{-10}\ \text{m}$, the same result in ångström is

(because $12.27^2\approx150$). Both forms appear in textbooks; they are the same formula.

Watch out. These shortcut formulas are only for electrons, because the mass and charge of the electron are already inside the number $1.227\times10^{-9}$. For a proton, neutron, alpha particle or anything else, go back to $\lambda=h/\sqrt{2m\,KE}$ with the correct mass.

Some properties of matter waves

• The lighter the particle, the longer its wavelength for a given speed ($\lambda\propto1/m$). This is why wave effects are noticeable for electrons but not for everyday objects.
• The faster the particle, the shorter its wavelength ($\lambda\propto1/v$).
• A particle at rest ($v=0$) would have an infinite wavelength. Matter waves are associated only with moving particles.
• Matter waves are not electromagnetic waves. They are produced by any moving particle, charged or neutral (neutrons are also diffracted).
• The wave is not a physical disturbance of a medium. As the wave function pages explain, it is a probability wave.

Experimental evidence

De Broglie had no experimental evidence when he proposed the idea. But he could show that it explained the stationary orbits and the quantised energy levels that Bohr had assumed for the hydrogen atom in 1913. An electron orbit is allowed only if a whole number of de Broglie wavelengths fits around it, $2\pi r=n\lambda$, which is exactly Bohr's condition on angular momentum.

The direct proof came in 1927. Davisson and Germer in the USA, and independently G. P. Thomson in Britain, showed that electrons are diffracted by crystals exactly as X-rays are, with the wavelength given by $\lambda=h/p$. Today, electron diffraction and the electron microscope depend on this.

Key idea. Wave and particle are two aspects of the same object. Which aspect we see depends on the experiment. Diffraction shows the wave, and a collision with a detector shows the particle.

Page 3 of 15

de Broglie wavelength: worked examples

Constants: $h=6.626\times10^{-34}\ \text{J s}$, $m_e=9.11\times10^{-31}\ \text{kg}$, $m_p=1.673\times10^{-27}\ \text{kg}$, $m_n=1.675\times10^{-27}\ \text{kg}$, $e=1.6\times10^{-19}\ \text{C}$.

Example 1. Why we never notice matter waves. Find the de Broglie wavelength of (a) a ball of mass $500\ \text{g}$ moving at $50\ \text{km/h}$, and (b) an electron accelerated through $100\ \text{V}$.

(a) First convert the speed: $v=50\ \text{km/h}=\dfrac{50\times1000}{3600}=13.9\ \text{m/s}$. Then

This is roughly $10^{-34}\ \text{m}$, which is extremely small compared with the size of the ball (and far smaller than even a nucleus). No experiment can ever detect a wave aspect in the behaviour of the ball, so for everyday objects classical mechanics is perfectly good.

(b) Using the electron formula,

This is about the size of an atom, and it is much larger than the classical radius of the electron ($\approx10^{-15}\ \text{m}$). So the wave nature of the electron cannot be neglected. It is also comparable to the spacing between atoms in a crystal, which is why crystals diffract electrons.

Note on the source. One lecture set gave the wavelength of the electron in part (b) as $0.123\times10^{-10}\ \text{m}$. The correct value is $1.23\times10^{-10}\ \text{m}$, ten times larger, as the formula $\lambda=1.227\times10^{-9}/\sqrt V$ shows for $V=100$.

Example 2. The Davisson–Germer experiment. In the original experiment the electrons were accelerated through $54\ \text{V}$. Find their wavelength.

This matched the wavelength measured from the diffraction pattern of a nickel crystal, and this is how the experiment confirmed de Broglie's hypothesis.

Example 3. A slow (thermal) neutron. Find the de Broglie wavelength of a neutron with kinetic energy $0.025\ \text{eV}$.

The electron formula cannot be used. Change the energy to joules, $KE=0.025\times1.6\times10^{-19}=4.0\times10^{-21}\ \text{J}$, and use the general form:

Again about $1.8\ \text{Å}$. Slow neutrons are diffracted by crystals and are used to study crystal structures.

Example 4. A proton versus an electron. A proton and an electron are each accelerated through $100\ \text{V}$. Compare their wavelengths.

Both gain the same kinetic energy $eV$, so $\lambda=h/\sqrt{2meV}\propto1/\sqrt m$:

The proton's wavelength is about 43 times shorter, because it is 1836 times heavier.

Example 5. Reverse problem. Through what voltage must an electron be accelerated to have a wavelength of $1\ \text{Å}$?

From $\lambda=\sqrt{150/V}\ \text{Å}$: $1=\sqrt{150/V}$, so $V=\mathbf{150\ \text{V}}$.

Watch out. Always work in SI units: mass in kg (500 g is $0.5$ kg), speed in m/s (divide km/h by 3.6) and energy in joules (multiply eV by $1.6\times10^{-19}$). Most wrong answers here come from a unit that was not converted.

Page 4 of 15

The wave function and probability

The idea

In every wave, something varies periodically with space and time:

The quantity whose variations make up matter waves is called the wave function, written $\Psi$ (the Greek capital letter psi, pronounced "sigh"). In general it is a function of all three space coordinates and of time:

The wave function is a mathematical function that represents the state of the particle. The mathematical theory of quantum mechanics tells how to use $\Psi(x,y,z,t)$ to find the position, velocity, momentum, energy, angular momentum and so on of the particle. Everything that can be known about the particle is contained in $\Psi$.

Physical significance (Born's interpretation)

1. $\Psi$ itself cannot be measured. The wave function is in general a complex quantity (it contains $i=\sqrt{-1}$). A complex number cannot be the result of a measurement, so $\Psi$ has no direct physical meaning by itself.
2. $|\Psi|^2$ is real and measurable. The square of the absolute value of $\Psi$,

where $\Psi^*$ is the complex conjugate of $\Psi$, is always real and positive.

3. $|\Psi|^2$ is a probability density. $|\Psi(x,y,z,t)|^2$ represents the probability of finding the particle at the point $(x,y,z)$ at time $t$, per unit volume. It is called the probability density or the probability distribution function. Where $|\Psi|^2$ is large, the particle is likely to be found. Where it is small, the particle is unlikely to be found.
4. Probability in a region. The probability of finding the particle in a small volume $dV$ around $(x,y,z)$ at time $t$ is $|\Psi|^2\,dV$. For a finite region it is the integral

$P$ always lies between $0$ (never there) and $1$ (certainly there). For example, if $P=0.4$ for a region, there is a 40 % chance of finding the particle in that region.

5. Probability amplitude. $\Psi$ itself is not a probability, but its square is. So $\Psi$ is also called the probability amplitude.
6. Statistical link between wave and particle. The wave function connects the particle nature and the wave nature statistically. It does not tell exactly where the particle is. It tells where the particle is likely to be. If we repeat the same experiment many times, the particle is found most often where $|\Psi|^2$ is largest.

Key idea. Quantum mechanics does not predict the exact location of a particle. It predicts the probability of finding it in each region of space. The matter wave is a probability wave.

A one-dimensional picture

For a particle moving only along $x$, the wave function is $\Psi(x,t)$ and $|\Psi(x,t)|^2\,dx$ is the probability of finding the particle between $x$ and $x+dx$. The probability of finding it between $x=a$ and $x=b$ is

Watch out. The probability density is $|\Psi|^2=\Psi^*\Psi$, not $\Psi^2$. For a complex $\Psi$, $\Psi^2$ is itself complex and cannot be a probability. Only when $\Psi$ is real (as for the particle in a box) are the two the same.

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Normalisation and well-behaved wave functions

Normalisation

A particle that exists in a particular region must be found somewhere in that region. So the probability of finding it anywhere in the region must be exactly 1 (100 %):

A wave function that satisfies this is said to be normalised (normalised to unity).

Now suppose we have a wave function $\Psi$ that solves the Schrödinger equation but is not normalised. If we multiply a solution by a constant $A$ (not depending on $x$, $y$, $z$), we get another solution, $\Psi_N=A\Psi$. We choose $A$ so that the new wave function is normalised. Adding the probability at all points in the volume $V$,

Normalisation. The process of multiplying a wave function by a suitable constant so that the probability of finding the particle within the region becomes exactly 1 is called normalisation. The constant is called the normalisation constant.

Why a wave function must be normalisable

$|\Psi|^2$ is supposed to be a probability density. A probability can only be interpreted if the total adds up to 1, so the integral $\int|\Psi|^2\,dV$ must be finite. A wave function whose integral is infinite cannot be normalised and cannot describe a real particle.

Conditions for a well-behaved wave function

The physically acceptable ("well-behaved") wave functions satisfy these conditions:

1. $\Psi$ is finite everywhere. An infinite value would mean an infinite probability.
2. $\Psi$ is single-valued. At any point and time there must be only one value of the probability.
3. $\Psi$ is continuous. The probability cannot jump suddenly from place to place.
4. The first derivatives $\dfrac{\partial\Psi}{\partial x}$, $\dfrac{\partial\Psi}{\partial y}$, $\dfrac{\partial\Psi}{\partial z}$ are continuous and single-valued, wherever the potential is finite. These derivatives decide the momentum of the particle, which must change smoothly.
5. $\Psi$ is normalisable, so $\Psi\to0$ as $x,y,z\to\pm\infty$.

Worked example

Example 1. A particle confined between $x=0$ and $x=L$ has the wave function $\psi=A\,x(L-x)$ inside, and $\psi=0$ outside. Find the normalisation constant $A$.

Normalisation requires

Expand $x^2(L-x)^2=L^2x^2-2Lx^3+x^4$ and integrate term by term:

So $|A|^2\dfrac{L^5}{30}=1$, giving

This function is finite, single-valued, continuous and zero at both walls, so it is well behaved.

Watch out. When normalising, square the absolute value of the constant ($|A|^2$), integrate over the whole region where the particle can be, and remember to take the square root at the end.

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The wave function of a free particle

Setting it up

Consider a particle moving freely along the $x$ direction with energy $E$ and momentum $p$. Its matter wave is a travelling wave, which we write in complex form as

where $a$ is the amplitude, $\omega$ is the angular frequency and $k$ is the wave number:

Expressing $\omega$ and $k$ through energy and momentum

Angular frequency. Use Planck's relation $E=h\nu$, so $\nu=E/h$:

where $\hbar=\dfrac{h}{2\pi}=1.055\times10^{-34}\ \text{J s}$ (read "h-bar").

Wave number. Use the de Broglie relation $\lambda=h/p$:

Put (2) and (3) in (1):

This is the wave function of a free particle of energy $E$ and momentum $p$. It is the starting point for the Schrödinger equation.

A second way to reach the same form

The same result can be reached through frequency and wavelength (this is the method used in the second lecture set). Write the wave as

where $v$ is the wave speed. With $\omega=2\pi\nu$ and $v=\nu\lambda$:

Now use $E=h\nu$ (so $\nu=E/h$) and $\lambda=h/p$ (so $1/\lambda=p/h$):

Since $\dfrac{2\pi}{h}=\dfrac1\hbar$,

exactly as before.

Its probability density is constant

This is a constant. It does not depend on position or time. The particle is equally likely to be found at any point along the $x$ axis. This makes sense, because this wave has one exact wavelength and so describes a particle with an exactly known momentum but a completely unknown position. The next page shows how a particle that is located somewhere is described.

Watch out. The sign convention matters only for consistency. These notes always use $e^{-\frac{i}{\hbar}(Et-px)}$, which represents a wave moving in the $+x$ direction. Use the same form in the whole derivation.

Page 7 of 15

Wave packets

A single wave cannot represent a localised particle

The wave function of a moving body shows the probability of finding it at a particular place at a particular time. Consider the free-particle wave function $\Psi=Ae^{-i(\omega t-kx)}$. Its real part, $A\cos(\omega t-kx)$, is a sine-shaped wave of constant amplitude that goes on for ever in both directions.

As the previous page showed, its probability density is

which is the same everywhere. So this wave says the particle is equally likely to be found at any point along the $x$ axis. That describes a particle with an exact wavelength, and hence an exact momentum, but with no position at all.

A wave packet

A real particle is more likely to be found in some places than in others at a given time. Its wave function must therefore be large in a small region of space and nearly zero everywhere else. Graphically this is a short burst of waves, a wave packet (also called a wave group).

How a wave packet is formed. A wave packet is formed by the superposition (adding together) of a large number of waves of slightly different wavelengths and frequencies. In one small region the waves are in step and reinforce each other, which gives a large amplitude. Away from that region they go out of step and cancel, so the amplitude dies away.

• The larger the number of component waves (the wider the range of wavelengths), the smaller the length of the wave packet.
• The smaller the number of component waves, the longer the packet.

When the particle moves, the wave packet moves as a whole.

Phase velocity and group velocity

Two different velocities describe a wave packet:

• The phase velocity $v_p$ is the velocity of the individual component waves (the crests inside the packet).
• The group velocity $v_g$ is the velocity with which the wave packet as a whole (its envelope) moves.

The group velocity is equal to the velocity of the particle, so it is also called the particle velocity. It is the group, not the individual waves, that carries the particle along.

From wave packets to uncertainty

This picture leads directly to the uncertainty principle:

• If the wave function is a single sine wave of wavelength $\lambda$, the momentum $p=h/\lambda$ is known exactly. But $|\Psi|^2$ has the same value all over space, so the position at a given time cannot be determined. The particle is "spread out" in space: exact momentum means a very large uncertainty in position.
• If the wave function is a narrow packet of width $\Delta x$, then $|\Psi|^2$ is zero everywhere except in that region, so the position is known to within $\Delta x$. But the packet is formed by the interference of a large number of waves of slightly different wavelengths. Since $\lambda$ can take any value in that range, the momentum $p=h/\lambda$ can also take a range of values. So there is an uncertainty in momentum $\Delta p$.
• For a narrower packet the number of component waves must be larger, so $\Delta p$ is larger.

Key idea. You cannot make $\Delta x$ and $\Delta p$ both small at the same time. This trade-off is not a fault of our instruments. It is part of the wave nature of matter. It is Heisenberg's uncertainty principle (next page).

Watch out. The individual waves inside a packet (phase velocity) and the packet itself (group velocity) move at different speeds. When an exam asks for the speed of the particle, it is the group velocity.

Page 8 of 15

Heisenberg's uncertainty principle

Statement

When we treat a moving particle as a wave group, there are basic limits to the accuracy with which we can measure particle properties such as position and momentum.

Heisenberg's uncertainty principle. It is impossible to know both the position and the momentum of a particle exactly at the same time. If $\Delta x$ is the uncertainty in the position and $\Delta p$ the uncertainty in the corresponding component of momentum ($p_x$), then

$\Delta x\,\Delta p\ \ge\ \frac{\hbar}{2}$

where $\hbar=\dfrac{h}{2\pi}=1.055\times10^{-34}\ \text{J s}$, so that $\dfrac\hbar2=\dfrac{h}{4\pi}$.

In simple words: the product of the uncertainty in position and the uncertainty in momentum can never be smaller than $\hbar/2$. The more precisely one is known, the less precisely the other can be known.

Other conjugate pairs

A pair of variables connected by an uncertainty relation is called a conjugate pair. Besides position and momentum, the important pairs are energy and time, and angular momentum and angular position:

Here $\Delta E$ and $\Delta t$ are the uncertainties in finding the energy and time, and $\Delta L$ and $\Delta\theta$ are the uncertainties in the angular momentum and the angle.

There is no uncertainty relation between $\Delta x$ and a component of momentum other than the $x$ component. For example, $x$ and $p_y$ can both be known exactly at the same time.

Note on the source. One lecture set wrote these relations with $\hbar$ on the right ($\Delta x\,\Delta p\ge\hbar$, $\Delta E\,\Delta t\ge\hbar$, $\Delta J\,\Delta\theta\ge\hbar$) and the other with $\hbar/2$. The exact lower limit is $\hbar/2$, used in all these notes. Some textbooks write $\hbar$ or even $h$ because they make an estimate of the order of magnitude. Then the answers change by a factor of 2 or $4\pi$. Write which form you use.

What the principle does and does not say

• It is not about poor instruments or experimental error. Even with perfect instruments, $\Delta x\,\Delta p$ cannot be smaller than $\hbar/2$. It is a property of nature, coming from the wave nature of matter.
• It applies to simultaneous knowledge. You can measure position very precisely, but then the momentum becomes very uncertain, and the other way round.
• Because $\hbar$ is tiny, the limits matter only on the atomic scale. For everyday objects the uncertainties are far too small to notice.

Why we never notice it in everyday life

Example 1. A bullet of mass $25\ \text{g}$ moves with a speed of $400\ \text{m/s}$. The speed is measured to an accuracy of $0.02\,\%$. How precisely can the position of the bullet be found?

The uncertainty in speed is $0.02\,\%$ of $400\ \text{m/s}$:

The uncertainty in momentum is

So

This uncertainty is far smaller than even the size of an atom (about $10^{-10}\ \text{m}$), so for a bullet it is completely negligible.

Note on the source. The lecture notes gave $2.626\times10^{-32}\ \text{m}$. When we calculate $h/(4\pi\,\Delta p)$ with $h=6.625\times10^{-34}$, we get $2.64\times10^{-32}\ \text{m}$.

Example 2. An electron in an atom. An electron is confined within an atom of size about $10^{-10}\ \text{m}$. Find the minimum uncertainty in its speed.

An uncertainty of more than half a million metres per second. For an electron in an atom, the uncertainty principle is very large, and the idea of a definite orbit with a definite speed breaks down.

Watch out. Change a percentage accuracy into an actual $\Delta v$ before multiplying by the mass, and use the mass in kilograms. Also remember that $\Delta x$ comes out as a minimum ($\ge$), not an exact value.

Page 9 of 15

Applications of the uncertainty principle

Planck's constant is so small that the limitations given by the uncertainty principle are important only at the subatomic level. On that scale, the principle is a powerful tool for understanding many phenomena.

1. Electrons cannot exist inside the nucleus

In the early days, people thought that the electrons coming out of a nucleus as $\beta$ rays in radioactivity actually existed inside the nucleus before they were emitted. The uncertainty principle shows that this is impossible.

Step 1: the uncertainty in position. The diameter of a nucleus is typically about $10^{-14}\ \text{m}$. If an electron existed inside it, the maximum uncertainty in its position would be the size of the nucleus:

Step 2: the minimum uncertainty in momentum.

Step 3: the minimum momentum. If this is the uncertainty in the momentum, the momentum itself must be at least as large:

Step 4: the minimum energy. At such a large momentum the electron moves at nearly the speed of light, so its rest-mass energy can be neglected and its kinetic energy is about $pc$:

Step 5: compare with experiment. An electron inside the nucleus would need a kinetic energy of at least about $10\ \text{MeV}$. But the electrons emitted in $\beta$ decay have energies of only a fraction of this, a few MeV at most. Hence electrons do not exist inside the nucleus. The electrons emitted in $\beta$ decay are created at the moment of decay, when a neutron changes into a proton.

Note on the source. One lecture set used $\Delta x\,\Delta p\ge\hbar$ with $\Delta x=10^{-14}\ \text{m}$ taken as the nuclear radius, and got $\Delta p=1.1\times10^{-20}\ \text{kg m/s}$ and about $20\ \text{MeV}$. The other used $\hbar/2$ and got $10\ \text{MeV}$. In both cases the conclusion is the same, because both values are far above the energies seen in $\beta$ decay.

2. Natural width of spectral lines (natural line broadening)

When excited atoms return to the ground state, they emit radiation. The emitted lines do not have a perfectly sharp frequency. Each line has a small spread of frequencies, called its natural width. The uncertainty principle explains this.

Step 1: the uncertainty in time. The average time between the excitation of an atom and the moment it radiates is called the lifetime of the excited state. It is typically about $10^{-8}\ \text{s}$. This is the uncertainty in the time of the de-excitation:

Step 2: the uncertainty in energy. From $\Delta E\,\Delta t\approx\hbar/2$,

Step 3: the uncertainty in frequency. Since $E=h\nu$, $\Delta E=h\,\Delta\nu$:

So the emitted frequency is not sharp: the spectral line has a finite width of a few megahertz. The shorter the lifetime of the excited state, the broader the line.

Note on the source. The lecture set that used $\Delta E\,\Delta t\ge\hbar$ got $\Delta\nu=1/(2\pi\Delta t)\approx16\ \text{MHz}$ for the same lifetime. It differs from $8\ \text{MHz}$ only by the factor of 2 in the convention.

Key idea. Both applications use the same idea: a quantity confined to a small range ($\Delta x$ or $\Delta t$) forces a large spread in its conjugate ($\Delta p$ or $\Delta E$).

Watch out. Use $\Delta\nu=\Delta E/h$ (with $h$), not $\Delta E/\hbar$. If you combine it with $\Delta E=\hbar/(2\Delta t)$, you get the neat result $\Delta\nu=1/(4\pi\Delta t)$, which is worth remembering.

Page 10 of 15

The time-dependent Schrödinger equation

The idea

How can we find $\Psi$ for a particle? The answer is the Schrödinger equation. It is to quantum mechanics what Newton's second law is to classical mechanics: given the forces acting (described by the potential energy $V$), it tells how the wave function behaves in space and time.

The Schrödinger equation cannot be derived from older physics. We get it by starting from the wave function of a free particle and using Planck's relation $E=h\nu$ and de Broglie's relation $p=h/\lambda$. The equation we reach is then taken as the general law, valid for any potential. Two equivalent methods are given below. Learn one thoroughly. The second one is useful because it also gives the energy and momentum operators.

Derivation, route 1 (differentiating twice)

Consider a particle moving in the $x$ direction with momentum $p$ and energy $E$. Its wave function is (see the free-particle page)

Step 1: differentiate (1) with respect to $t$. Only the exponent depends on $t$, and its derivative is $-\dfrac{iE}{\hbar}$:

Multiply both sides by $i\hbar$. Since $i\hbar\cdot\left(-\dfrac{iE}{\hbar}\right)=-i^2E=E$:

Step 2: differentiate (1) twice with respect to $x$. Each differentiation brings down a factor $\dfrac{ip}{\hbar}$:

Multiply both sides by $-\hbar^2$:

Step 3: total energy. The total energy of the particle is the sum of its kinetic energy and potential energy $V$. With $KE=\tfrac12mv^2=\dfrac{(mv)^2}{2m}=\dfrac{p^2}{2m}$:

Multiply by $\Psi$:

Step 4: put (2) and (3) in (4).

This is the time-dependent Schrödinger equation in one dimension. It is the most general form of the Schrödinger equation. $V$ and $\Psi$ may depend on both position and time, which is why it is called time-dependent.

Derivation, route 2 (energy and momentum operators)

Start from the same wave function, $\Psi=Ae^{-\frac{i}{\hbar}(Et-px)}$.

Differentiating partially with respect to $t$:

(using $-\dfrac1i=i$). Differentiating partially with respect to $x$:

Equations (6) and (7) say that, when they act on $\Psi$, energy and momentum can be replaced by operators:

Energy and momentum operators.

$E\ \to\ i\hbar\,\frac{\partial}{\partial t}\qquad\qquad p\ \to\ -i\hbar\,\frac{\partial}{\partial x}$

These are used to find the energy and momentum of a particle when its wave function is known.

Now write the total energy, $E=\dfrac{p^2}{2m}+V$, multiply by $\Psi$, and replace $E$ and $p$ by their operators ($p^2$ means applying the momentum operator twice):

Since $(-i\hbar)^2=i^2\hbar^2=-\hbar^2$,

which is equation (5) again.

Three dimensions

If $\Psi$ depends on $x$, $y$ and $z$, the second derivative is replaced by the sum of all three:

where $\nabla^2=\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial y^2}+\dfrac{\partial^2}{\partial z^2}$ is the Laplacian operator.

What $V$ does

The potential energy function $V$ represents the effect of the surroundings on the particle. Any restriction on the motion of the particle (walls, barriers, the attraction of a nucleus) appears in $V$. Once $V$ is known, the Schrödinger equation can be solved for the wave function $\Psi$ of the particle, and from $\Psi$ we get $|\Psi|^2$ and all the other physical quantities.

Watch out. Be careful with the signs: there is a minus sign in front of $\dfrac{\hbar^2}{2m}\dfrac{\partial^2\Psi}{\partial x^2}$, and $i\hbar$ (not $-i\hbar$) on the left. They come from $\left(\tfrac{ip}{\hbar}\right)^2=-\tfrac{p^2}{\hbar^2}$ and from multiplying $-\tfrac{iE}{\hbar}\Psi$ by $i\hbar$. In an exam, write each step. Examiners give marks for the steps, not only for the final equation.

Page 11 of 15

The time-independent Schrödinger equation

The idea

In many situations, the potential energy of the particle does not depend on time. The forces acting on the particle, and hence $V$, change only with the position of the particle, $V=V(x)$. The particle is then in a steady state. When this is true, the Schrödinger equation can be simplified by separating the wave function into a part that depends only on position and a part that depends only on time. The result is the steady-state or time-independent form.

Derivation

Step 1: separate the wave function. Start from the free-particle wave function and split the exponential into two factors:

Let $\psi=A\,e^{\frac{i}{\hbar}px}$, a function of position only (it is sometimes written $\Psi_0$ or $\psi_0$). Then

So the wave function $\Psi$ has a time-independent part $\psi$ and a time-dependent part $e^{-iEt/\hbar}$.

Step 2: differentiate (1) with respect to $t$. $\psi$ does not depend on $t$, so only the exponential is differentiated:

Step 3: differentiate (1) twice with respect to $x$. The exponential does not depend on $x$, so only $\psi$ is differentiated:

Step 4: put in the time-dependent equation. The time-dependent Schrödinger equation is

Putting (1), (2) and (3) in (4):

On the left, $i\hbar\cdot\left(-\dfrac{iE}{\hbar}\right)=E$, so

Step 5: cancel the time factor. Divide throughout by $e^{-iEt/\hbar}$, which is never zero:

Step 6: rearrange. Take everything to one side:

and multiply through by $\dfrac{2m}{\hbar^2}$:

This is the time-independent Schrödinger equation in one dimension. Since $\psi$ depends only on $x$, there is no need for partial derivatives. Ordinary derivatives $\dfrac{d^2\psi}{dx^2}$ are used.

Three dimensions

Stationary states

For a steady state, the probability density is

because $|e^{-iEt/\hbar}|=1$. So the probability density does not change with time. This is why these solutions are called stationary states. All the problems in this module (the box and the barrier) use the time-independent equation, because their potentials do not change with time.

Key idea. The time-independent equation can also be written as $-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}+V\psi=E\psi$: some operator acting on $\psi$ gives a number $E$ times $\psi$. Only certain values of $E$ (the energy eigenvalues) allow acceptable solutions $\psi$ (the eigenfunctions). The particle in a box shows this clearly.

Watch out. An equivalent and equally correct way to write the result is $\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}+(E-V)\psi=0$ (multiply through by $\hbar^2/2m$). Do not mix the two forms in the middle of a derivation.

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Particle in a one-dimensional box: solving the equation

The set-up

Consider a particle of mass $m$ moving inside a box along the $x$ direction, bouncing back and forth between two walls a distance $L$ apart. The walls at $x=0$ and $x=L$ are infinitely high (potential barriers that cannot be crossed), so the particle can never get past them, however much energy it has. This is the one-dimensional infinite square well potential, also called a particle in a one-dimensional box. The potential energy varies as:

The position of the particle at any moment lies in $0<x<L$. Since the particle cannot exist outside the box, the probability of finding it there is zero, so $\Psi^*\Psi=0$ and hence $\psi=0$ for $x\le0$ and $x\ge L$. Our task is to find $\psi$ inside the box. Since the potential does not depend on time, we use the time-independent Schrödinger equation.

The equation inside the box

The time-independent Schrödinger equation is

Inside the box $V=0$, so

Put

so the equation becomes

This is the equation of simple harmonic motion. Its general solution can be written in two equivalent ways.

Method A: sine and cosine form

The general solution of (2) is

with two constants $A$ and $B$ fixed by the boundary conditions. The wave function must be continuous, and it is zero outside the box, so it must vanish at both walls:

Boundary condition at $x=0$:

So $B=0$ and

Boundary condition at $x=L$:

We cannot have $A=0$, because then $\psi=0$ everywhere and there would be no particle. So

Method B: exponential form

The general solution of (2) can also be written as

At $x=0$: $\psi(0)=A+B=0$, so $B=-A$ and

using $e^{ikx}-e^{-ikx}=2i\sin kx$.

At $x=L$: $2iA\sin kL=0$. Since $2iA\ne0$, $\sin kL=0$, so again $kL=n\pi$ and $k=n\pi/L$. The wave function is