Quantum Mechanics Revision Notes, Lecture notes of Physics

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Quantum Mechanics

Revision Notes

C.R.D. Guetta

April 7, 2008

• 1 Failure of Classical Physics
  • 1.1 Waves are particles
  • 1.2 Particles are waves
• 2 Basics
  • 2.1 Basic postulates
  • 2.2 Wavepackets
    • 2.2.1 Introduction
    • 2.2.2 Momentum Representation of a Wavepacket
    • 2.2.3 The Dispersion Relation for a Wavepacket
    • 2.2.4 Time Evolution of a general Wavepacket
    • 2.2.5 Time evolution of a wavepacket representing a particle
  • 2.3 The Heisenberg Uncertainty Principle
  • 2.4 Probability current
  • 2.5 Beams of particles
    • 2.5.1 Position representation
    • 2.5.2 Momentum representation
  • 2.6 Practical stuff
• 3 The Schr¨odinger Equation
  • 3.1 Derivation
  • 3.2 Stationary states
  • 3.3 Boundary Conditions
  • 3.4 Solutions for Constant V
    • 3.4.1 Plane wave solution
    • 3.4.2 Non-plane wave solutions
• 4 The Wave Approach to QM
  • 4.1 Unbound states – scattering
    • 4.1.1 Applications of tunnelling
  • 4.2 Bound states
    • 4.2.1 The Infinite Well
    • 4.2.2 The δ-function potential
    • 4.2.3 Finite square well
    • 4.2.4 The 1D Harmonic Oscillator
    • 4.2.5 Three Dimensions
• 5 The Basic Postulates & Operators
  • 5.1 The Basic Postulates of QM 4 CONTENTS
  • 5.2 Operators
    • 5.2.1 Dirac notation
    • 5.2.2 Properties of Observables
  • 5.3 Expectation values
  • 5.4 The Generalised Uncertainty Principle
    • 5.4.1 Compatible Observables
    • 5.4.2 Incompatible Observables
    • 5.4.3 Conjugate Observables
    • 5.4.4 Minimum Uncertainty States
  • 5.5 Examples of operators
  • 5.6 Ladder Operators
    • 5.6.1 The Harmonic Oscillator
• 6 Time Dependence
  • 6.1 Measurements
  • 6.2 Ehrenfest’s Theorem
    • 6.2.1 Classical limit
  • 6.3 Conserved Quantities
    • 6.3.1 Parity
  • 6.4 The Time-Energy Uncertainty Relation
• 7 3 Dimensions
  • 7.1 Angular Momentum
  • 7.2 Commutation relations
• 8 Angular Momentum
  • 8.1 Orbital Angular Momentum Eigenvalues
  • 8.2 Eigenstates
  • 8.3 Bits and bobs
  • 8.4 The Stern-Gerlach Experiment
• 9 Spin
  • 9.1 Experimental Evidence
  • 9.2 Eigenvalues and Eigenvectors
  • 9.3 Uncertainties
  • 9.4 Spin in any direction
  • 9.5 Combining Orbital & Spin Angular Momentum
    • 9.5.1 Introduction
    • 9.5.2 The values of J
    • 9.5.3 Combined wavefunctions
  • 9.6 Conservation of total angular momentum
• 10 Central Potentials
  • 10.1 Conservation of angular momentum
  • 10.2 Quantum numbers
  • 10.3 Separation of Variables
  • 10.4 Normalisation & Probabilities
  • 10.5 Practical tips
• CONTENTS
• 11 Two-Particle Systems
  • 11.1 Position Probabilities
  • 11.2 Observables
  • 11.3 The Hamiltonian
  • 11.4 Conservation of Total Momentum
  • 11.5 Centre of Mass Motion
  • 11.6 Separation of CoM Motion and Relative Motion
  • 11.7 Combining spins
• 12 Examples of real systems
  • 12.1 The rigid rotor
  • 12.2 The Harmonic Oscillator
  • 12.3 The Hydrogenic Atom
• 13 Identical Particles
  • 13.1 Exchange Symmetry
  • 13.2 The Spin-Statistics Theorem
  • 13.3 Non-interacting particles
  • 13.4 N particles
    • 13.4.1 Distinguishable
    • 13.4.2 Indistinguishable
  • 13.5 N =
    • 13.5.1 Diatomic molecules
  • 13.6 Correlation and Exchange Forces
  • 13.7 Interacting Particles
    • 13.7.1 Exchange energy
    • 13.7.2 Energy Spectra
  • 13.8 Practical tips
• 14 Matrix Methods
  • 14.1 Introduction
  • 14.2 Orbital angular momentum
  • 14.3 Spin angular momentum
• 15 Heat capacities, etc.
  • 15.1 Vibrational Specific Heat of Diatomic molecules
  • 15.2 Rotational Specific Heat of Diatomic molecules
  • 15.3 Identical Particles and Rotational Heat Capacities

Chapter 1

Failure of Classical Physics

1.1 Waves are particles

There is a plethora of evidence that waves can be- have as particles, with the following properties:

E = hν p = h λ

Black body radiation

Classically:

• For a cavity, the number of EM modes per unit volume with frequency between ν and ν + dν is given by

N (ν) dν =

8 πν^2 c^3

dν

Derive as follows:

• Let k be the wavevector of our wave. Boundary conditions dictate that it must be quantised as |k | = π|p|/L, where p is a vector with integer compoenents in 3D space, and L is the dimension of the (square) cavity.
• In p-space, the volume density of states is, by definition, 1 (because p must have integer components). Therefore, the number of states in the range p to p + dp is given by

dN (p) = N (p) dp =

4 πp^2 dp

(Given that we are only interested in p- vectors with positive components.

• Feeding this into the formula |k | = π|p|/L, taking into account both polar- isation of EM waves, and dividing by L^3 to find the density, we obtain the relation above.
• Classically, the average energy of each har- monic mode (2 degrees of freedom) is kB T – so the energy density (ie: energy per unit vol- ume) is given by the Rayleigh-Jeans Law :

ρ(ν, T ) =

8 πν^2 c^3

kB T

This, however, predicts that ρ increases with- out bound at high ν – the ultraviolet catastro- phe.

• Planck hypothesised that the energy of each mode is not just kT , but is quantised in units of hν – in other words, the average energy per mode is given by (note the major assumption that there is no zero-point energy – we say that this is the ‘energy of the vacuum’)

∑n=∞ n=0 nhν

Boltzmann ︷ ︸︸ ︷ e−nhν/kB^ T ∑n=∞ n=0 e

−nhv/kB T ︸ ︷︷ ︸ Partition func

hν ehν/kB^ T^ − 1

This expression can easily be obtained by not- ing that the denominator is an infinite geomet- ric series (

∑n=∞ n=0 an

r (^) = a/(1−r)) and that the top is the negative differential of the bottom, with respect to β = kB T.

• We then get

ρ(ν, T ) =

8 πν^2 c^3

hν ehν/kB^ T^ − 1

7

8 CHAPTER 1. FAILURE OF CLASSICAL PHYSICS

This fits data, and reduces to Rayleigh-Jeans as ν → 0.

Conclusion:

The energy of each mode of electromagnetic radiation is quantised in units of hν

The Photoelectric Effect

When light falls on a metal, electrons emerge with different energies, which can be measured by mak- ing the metal the positive-end of an electrode, and varying V , the voltage across the electrode. The stopping voltage, Vstop is the voltage above which no electrons reach the cathode, and is proportional to the maximum kinetic energy of electrons that leave the metal (Emax).

The phenomenon is characterised by the following properties

• Emax = hν − W , where W is the work function of the metal (the minimum energy required to release an electron from the metal). Classi- cally, one would expect the energy in the wave to vary with intensity, not frequency.
• The number of electrons emitted (the satura- tion current) is proportional to the intensity of light. Higher intensity means more packets.
• Electron emission is instantaneous – classi- cally, one would expect a time delay to absorb enough energy to emit an electron.

These can be explained by the following conclusion:

Light exists as quantised packets, each of energy hν

The Compton Effect

In Compton Scattering, X − Rays are fired onto a metal sample and scattered by electrons. It is found that the wavelength of the scattered X-Rays, λ′^ varies with φ, the angle between the new beam and the undeflected beam.

The correct relation between λ′^ and φ can be de- rived directly from conservation of momentum and energy as long as we assume that

The X-rays are made out of photons with well-defined momentum given by

p =

h λ

Interference of light at low intensity

In a Young’s Double Slit experiment at a very low intensity of light, we detect the light as discrete photons. However, the resulting pattern is what is expected of a double-slit pattern – each quantum particle interacts with both slits, and its trajectory is indeterminate – trying to identify which slit the electron passes through destroys the pattern.

1.2 Particles are waves

Evidence exists that suggests particles with mo- mentum p have wave-like properties, with wave- length given by

λ = h p

and angular momentum quantized in units of ℏ

Atomic structure & the de Broglie Hypoth- esis

Incandescent gasses have characteristic line spec- tra. The positive charge in the atom is known to be concentration in the nucleus (α-scattering ex- periments), and the electrons orbit the nucleus. To explain why the electrons do not radiate and spiral in, Bohr suggested that

The electron’s angular momentum L is quantized in units of ℏ

Starting from this assumption, he was able to re- cover the correct form for the line spectrum of hy- drogen.

De Broglie hypothesised that

10 CHAPTER 1. FAILURE OF CLASSICAL PHYSICS

Chapter 2

Basics

2.1 Basic postulates

Some of the basic postulates of Quantum Mechan- ics can be expressed as follows:

1. A particle is represented by a ‘wavefunction’, ψ(x, t), which contains all possible information about the particle.
2. The probability density is given by

ρ(x, t) = |ψ(x, t)|^2

The probability that the position of the parti- cle is between x and x + dx is given by

P (x, t) dx = ρ(x, t) dx

The wavefunction needs to be normalised such that the probability of finding the particle in all space is 1. As such ∫ (^) ∞

−∞

ρ(x, t) dx = 1

3. A free particle of mass m, momentum p and energy E is represented by a plane wave of the form

Ψ(x, t) = Aei(kx−ωt)^ = Aei(^

p ℏ x−^ E ℏ t)^ (2.1)

The form of each term in the wavefunction arises from experimental evidence:

• The De Broglie Hypothesis, supported by the Compton and Davidson-Germer ex- periments, gives us k = p/ℏ
• The photoelectric effect gives rise to ω = E/ℏ

2.2 Wavepackets

2.2.1 Introduction

The problem with this representation of a parti- cle is that it P (x, t) is completely uniform over all space – there is no information about the particle’s position. If we want to know something about the particle’s position, ψ must to some degree be localised, such that ψ(x, t) → 0 as x → ±∞. This is achieved by the superposition of planes waves of different wavenumbers and frequencies

ψ(x, t) =

2 π

−∞

g(k)ei(kx−ωt)^ dk (2.2)

For t = 0, this becomes a Fourier Integral.

2.2.2 Momentum Representation of

a Wavepacket

Let us define

Φ(p, t) =

= g(k, t) = g

( (^) p ℏ

, t

We then have, in accordance with Equation 2.2:

Ψ(x, t) =

2 πℏ

−∞

Φ(p, t)e(^

ix ℏ p) dp

and

Φ(x, t) =

2 πℏ

−∞

Ψ(x, t)e(−^

ip ℏ x) dx

2.6. PRACTICAL STUFF 13

We then obtain the following expression:

j (r , t) =

2 mi

[Ψ∗(r , t)∇Ψ(r , t) − Ψ(r , t)∇Ψ∗(r , t)]

For stationary states:

j (r ) =

2 mi

[ψ∗(r )∇ψ(r ) − ψ(r )∇ψ∗(r )]

Re-arranging:

j (r ) =

2 mi

[ψ∗(r )∇ψ(r ) − (ψ∗(r )∇ψ(r ))∗]

j = <

ψ∗^

im

∇ψ

ψ∗ p̂ m

ψ

We note two important points:

• For a stationary state, the probability current j is time-independent.
• From the equation above, we have ∇ · j = ·ρ. However, we saw above that for a stationary state, ·ρ = 0, and so ∇·j = 0. In 1-Dimension, this means that

dj dx

Which means that the flux j is independent of position as well. In more than 1 dimension, however, this is no longer the case.

2.5 Beams of particles

The plane wave

Ψ(x, t) = Aei(kx−ωt)

is readily normalised if it only extends over a finite volume. Otherwise, it represents a particle of well- defined momentum but completely unknown posi- tion.

By choosing A properly, Ψ can be made to repre- sent a beam of particles with momentum ℏk.

2.5.1 Position representation

In the position representation of a beam ∫

unit length

|Ψ|^2 dx = |A|^2

So |A|^2 is the number of particles per unit length (or volume in 3D) – the particle number density. The particle flux (Equation 2.5) is then given by

j(x) = |A|^2

p m (In other words, the product of the number density and the velocity – as expected).

2.5.2 Momentum representation

Feeding the expression for Ψ into the momentum representation, we find that the time-independent momentum-representation of the beam is

φ(p) ∝ δ(p − p 0 )

The momentum is known exactly for the beam.

2.6 Practical stuff

• When trying to find the minimum uncertainty in a position measurement as a result of a wavepacket spreading, find an expression for the uncertainty in terms of the original uncer- tainty ∆x 0 and differentiate.
• Alternatively, for the harmonic oscillator, write the expectation value of the energy as

〈E〉 = k

x^2

〈p〉^2 2 m And note that

x^2

= (∆x)^2 , because 〈x〉 = 0, since the potential is symmetric.

p^2

= (∆p)^2 , because 〈p〉 = 0, since the particle isn’t drifting away. Then, use the uncertainty relation to obtain ∆p in terms of ∆x, feed it into the expression for 〈E〉 above, and hence obtain an expression for ∆x.

14 CHAPTER 2. BASICS

16 CHAPTER 3. THE SCHR ODINGER EQUATION¨

3.4 Solutions for Constant V

3.4.1 Plane wave solution

When dealing with unbound particles, we seek a plane wave solution. Feeding the trial solution into the Schr¨odinger Equation gives

ψ(x, t) = Aei(kx−ωt)

Where

k = ±

2 m[E − V (x)] ℏ^2

• For E > V (ie: the particle’s has real kinetic energy), ψ oscillates.
• For E < V , the kinetic energy is imaginary, and ψ decays exponentially. In that case, we write κ = −ik.

A general approach to solving the Schr¨odinger Equation for any potential V is simply to solve it in each region of constant V , and to match boundary conditions.

3.4.2 Non-plane wave solutions

For bound states, we need non-plane wave solu- tions, and we therefore use the most general solu- tion of the equation (involving both the positive and negative exponential). The boundary condi- tions on the wave can then often we used to elimi- nate one of the exponentials.

Chapter 4

The Wave Approach to QM

4.1 Unbound states – scatter-

ing

We first consider a beam of particles, represented by a plane wave, travelling through some potential and being reflected/transmitted from an obstruc- tion. Possible cases include a potential step, a po- tential barrier or a potential well.

In each case the strategy is the same:

• Find a plane-wave solution in each case. Note that reflected waves will have negative wavenumbers
• Match boundary conditions, bearing in mind that the wave at any point is given by the su- perposition of all the waves present there.

A few notes:

• Two kinds of reflection coefficients exist:
  • Amplitude reflection coefficients, ˜r and ˜t, are simply given by dividing the reflected amplitude over the incident amplitude.
  • Flux reflection coefficients, R and T are found by diving the reflected flux (Equa- tion 2.5) over the incident flux. If the wavenumber is the same in the incident and reflected regions, T = ˜t^2 , and R = ˜r^2.

In all these problems, it is simplest to assume that the incident wave has an amplitude of 1, and it is useful to remember, in that case, that T + R = 1.

• Negative reflection coefficients indicate a phase change of π. - In cases of, for example, a potential barrier, one could consider waves infinitely bouncing back and forth within the barrier – but it turns out to be just as general to consider one forward-going wave and one backward going wave. - For a potential barrier, it is found that T = 1 when k 2 a = nπ, in other words, when a = nλ/2 – at that point, the wave reflected from one side of the barrier and from the other side interfere destructively, resulting in R = 0. - For a potential barrier, V > E, tunnelling oc- curs. If κ 2 a is very large (very high or thick barrier), we are in the weak tunnelling limit, where we ignore all e−κ^2 a^ terms, and we have

T ≈

16 k 12 κ^22 e−^2 κ^2 a (k 12 + κ^22 )^2

Note that

• This is a strong function of a, m and V 0 − E.
• A single evanescent wave carries no par- ticle current, but a superposition of two opposite evanescent waves does. This is how tunnelling occurs.
• Note, in all cases, that the energy of the beam is not specified – a continuous range of energies is possible.

4.2. BOUND STATES 19

4.2.1 The Infinite Well

The energy of the states is

E =

ℏ^2 n^2 π^2 2 ma^2

Where a is the width of the well, and n is an integer, n ≥ 1.

4.2.2 The δ-function potential

When faced with a δ-function potential:

• Integrate the Schr¨odinger Equation over the range x 0 +  to x 0 − .
• Any contribution on the RHS not involving the δ-function will vanish as  → 0.
• The result will be an expression for the discon- tinuity in ψ′^ at x 0 , in terms of ψ(x 0 ).

Thereafter, the tactic is to find trial solutions around the δ-function and then to match the boundaries using continuity of ψ and the discon- tinuity in ψ′^ found above.

4.2.3 Finite square well

For the finite square well, the ‘trial solution — boundary conditions’ approach can be used, and this leads to the following two equations:

Y = X tan X

Y = −X cot X

Where X = ka/2 and Y = κa/2, and k and κ are the wavenumbers in the classically allowed and forbidden regions respectively.

From the form of κ and k, we can deduce that

X^2 + Y 2 =

mV 0 a^2 2 ℏ^2

Finding the intersection of this circle with the func- tions above gives the bound states, the energy of which can be recovered by using either the expres- sion for k or κ.

4.2.4 The 1D Harmonic Oscillator

The energies of the oscillator are

E =

n +

ℏω

and their wavefunctions are

ψn = AnHn(q)e−q

(^2) / 2

Where q = x

mω/ℏ and the H are the Hermite Polynomials. The constant ω is defined as

ω =

α m

Where the potential is given by

V (x) =

αx^2

4.2.5 Three Dimensions

The Harmonic Oscillator

In that case, each potential is harmonic, and the overall wavefuntion is a product of three 1D wave- functions. The energy is characterised by three quantum numbers, and is given by

E = (nx + ny + nz + 3/2)ℏω

The infinite well

If the infinite well has sides a, b and c, the total energy is given by

E =

π^2 ℏ^2 2 m

[

n^2 x a^2

n^2 y b^2

n^2 z c^2

]

Again, the wavefunction is a product of the 1D ones.

20 CHAPTER 4. THE WAVE APPROACH TO QM