Quantum World

Tom Charnock

Contents

1 Quantum Mechanical Postulates 21.1 First Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Second Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Third Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Fourth Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.5.1 Compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5.2 Uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.6 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.6.1 Time-Independent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.6.2 Time-Dependent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.6.3 3D Schrodinger Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.7 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.7.1 Commutation Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.7.2 Eigenvalues and Eigenfunctions of Angular Momentum . . . . . . . . . . . . . . . 5

2 Spin 72.0.3 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.0.4 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.0.5 Stern-Gerlach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.0.6 Spin Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.0.7 Pauli Spin Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1 Quantum Spin Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.1 Stationary States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.1.2 Spin Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Oscillating Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Quantum Cryptography 123.1 Cryptographic Keys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.1.1 Sending Keys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4 Quantum Dynamics 144.1 Wavepackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.1.1 Charged Particle in an Electromagnetic Field . . . . . . . . . . . . . . . . . . . . . 144.2 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.3 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.3.1 Uncertainties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.3.2 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

5 Quantum Systems 165.1 Particle in an Infinite Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5.1.1 Outside the Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.1.2 Inside the Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.1.3 Finite Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5.2 Rectangular Potential Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2

5.2.1 Complex Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.2.2 Quantum Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.3 Quantum Harmonic Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.3.1 Raising and Lowering Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3.2 Ground State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.3.3 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

5.4 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.4.2 Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.4.3 Probability Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.4.4 Reduced Mass Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6 Appendix 246.1 Gaussian Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.2 Quantum Harmonic Expectation Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1

Quantum Mechanical Postulates

1.1 First Postulate

The state of a particle is specified by a complex valued wavefunction, |ψ〉, which can also be representedby:

ψ˜ =

αβγ...ω

The wavefunction is restricted such that it must be smooth and continuous as must its first derivative.

1.1.1 Probability

The modulus squared is the probability density function Pi = |〈ψi|ψ〉|2. Which can be written P = ψ˜†iψ˜.

The condition for normalisation is that:

P = |〈ψi|ψj〉|2 = δij

1.2 Second Postulate

Any observable property can be represented by a hermitian operator which acts on the wavefunction ofthe system. The eigenvalues of the operator are the possible results of a measurement of the observableand the eigenfunctions are the states in which the observable can be found.

Oψ = λψ

Where O is the observable and λ is the eigenvalues. O must be hermitian so that O = O†.

1.2.1 Expectation Value

The expectation value of an operator can be calculated by 〈O〉 = 〈ψ|O|ψ〉. This can be written as:

〈O〉 =

∫ψ(r˜)†Oψ(r˜)δ3r˜

Position Operator

The expectation value of the position is 〈r˜〉 = 〈ψ|r|ψ〉 where the operator rψ(r˜) = r˜ψ(r˜)Momentum Operator

The de Broglie form of the momentum described by a wavevector is p˜ = hk˜. This indicates that the

momentum operator is p = −ih∇.

2

Hamiltonian Operator

The Hamiltonian operator is given by:

H =p2

2m+ V (r˜)

Where the stationary states given by Hψ = Eψ are:

− h2

2m∇2ψ + V (r˜)ψ = Eψ

This is the time-independent Schrodinger equation.

1.3 Third Postulate

The time evolution of the wavefunction is governed by the time-dependent Schrodinger equation:

Hψ(r, t) = ih∂ψ(r, t)

∂t

The Hamiltonian is the operator associated with the energy. If the term evolves in an oscillatory formthen this is given by e−

iEth . The equation of motion for the expectation values can be obtained from

〈O〉 = δδt

(ψ˜(t)†Oψ˜(t)

). This can be expanded to 〈O〉 = ψ˜(t)†Oψ˜(t)+ψ˜(t)†Oψ˜(t) and as ψ˜(t) = − i

hHψ˜(t)

the equations of motion are given by Ehrenfest’s Theorem and can be written:

〈O〉 =i

hψ˜(t)

[H, O

]ψ˜(t)

If [H, O] = 0 then the observable quantity is conserved. The general equation of motion is:⟨O⟩

=

⟨i

h

[H, O

]⟩

1.4 Fourth Postulate

Any state can be expressed as a superposition of the complete set of eigenstates of the observable. Thecoefficients in the expansion are the probability amplitudes for the possible outcomes of a measurement.Once the measurement has been made, the state changes discontinuously into the eigenstate of theobservable which has been measured.

ψ˜ = f1u˜(1) + f2u˜(2) + f3u˜(3) + ... =∑i

fiu˜(i)

Where fi is the probability amplitudes of the state and u˜(i) is a set of vectors which can be used to

write the wavefunction as a unique superposition. As u˜†(i)u˜(j) = δij then the value of the probability

amplitudes can be found by multiplying by u˜†(i).u˜†(i)ψ˜ =

∑j

u˜†(i)fju˜(j) =∑j

fiδij = fi

1.5 Uncertainty Principle

1.5.1 Compatibility

Two observables are compatible if the eigenstates are in cannon. The matrices commute. This can betested by using the commutator [A,B] = AB − BA. When this is zero then the matrices are said tocommute.

3

1.5.2 Uncertainty

For two given observables, their uncertainty is given by (∆A)(∆B) ≥ 12 |〈[A,B]〉|. For an arbitrary

wavefunction then:

[r, p]ψ = rpψ − prψ = r˜ (−ih∇ψ)− r˜ (−ih∇ψ) + ihψ = −ihψ

This leads to the uncertainty relating position and momentum:

∆r˜∆p ≥1

2|〈[r, p]〉| = h

2

When ∆r˜ = 0 then ∆p =∞.

1.6 Schrodinger Equation

1.6.1 Time-Independent

The time-independent Schrodinger equation is:

Hψ = Eψ

1.6.2 Time-Dependent

The time-dependent Schrodinger equation is:

Hψ = ih∂ψ

∂t

1.6.3 3D Schrodinger Wave Equation

In three dimensions there is the probability of the particle circulating around a point. From this arisesthe quantum mechanical angular momentum. The quantum mechanical kinetic energy operator is:

T = − h2

2m∇2

With the inclusion of a potential energy V (x, y, z) the 3D time-dependent Schrodinger equation is:

− h2

2m∇2ψ + V ψ = ih

∂ψ

∂t

The variables can be separated ψ(x, y, z, t) = u(x, y, z)T (t) where T is the time function T = e−iEth . This

then gives the three dimensional time independent Schrodinger equation.

− h2

2m∇2u+ V u = Eu

This can be represented as three separate equations and divided through by the wavefunction to get:

− h2

2m

[(1

X

∂2X

∂x2+ Vx

)+

(1

Y

∂2Y

∂y2+ Vy

)+

(1

Z

∂2Z

∂z2+ Vz

)]= E

The sum of these equations is constant E = E1 + E2 + E3, so each equation must be constant.

1.7 Angular Momentum

Particles can undergo free circulation about a point. This means that the particle has angular momentum.The classical angular momentum is L˜ = r˜× p˜. By replacing the observables by the quantum versions:

Li = −i(xj

∂

∂xk− xk

∂

∂xj

)

4

In spherical polar coordinates:

Lx = i

(sinϕ

∂

∂θ+ cot θ cosϕ

∂

∂ϕ

)

Ly = i

(sinϕ

∂

∂θ+ cot θ sinϕ

∂

∂ϕ

)

Lz = −i ∂∂ϕ

The angular momentum squared operator can provide a measure of the squared magnitude of the mo-mentum:

L2 = L2x + L2

y + L2z

Which in spherical polar coordinates is:

L2 = −i(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

)

1.7.1 Commutation Relation [Ly, Lz

]= iLx

[Lx, Lz

]= iLy

[Lx, Ly

]= iLz

The commutator of the angular momentum squared operator and any of the directions of angular mo-mentum is zero: [

L2, Lx

]=[L2, Ly

]=[L2, Lz

]= 0

The precise values of two different angular momentum components cannot be known simultaneously. Thesquare of the magnitude of the angular momentum can be known at the same time as one of the angular

momentum components. This means that L2 and one of the angular momentum components have acommon set of eigenfunctions.

1.7.2 Eigenvalues and Eigenfunctions of Angular Momentum

The eigenvalue equation is given by:

L2Y (θ, ϕ) = α2Y (θ, ϕ)

LzY (θ, ϕ) = βY (θ, ϕ)

Where α2 and β are the eigenfunctions of the squared angular momentum and the angular momentum inthe z-direction respectively. Lz depends only on ϕ so the variables can be separated Y (θ, ϕ) = χ(θ)f(ϕ).The eigenvalue equation becomes −i ∂∂ϕf = βf . The solutions are complex exponential functions f =

Ceiβϕ, which are single valued so that f(ϕ + 2π) = f(ϕ) therefore eiβ2π = 1 and β must be an integer.This shows that the angular momentum is quantised with an integer value of h.

5

m` = −2

m` = −1

m` = 0

m` = 1

m` = 2

θ

hm`

Figure 1.1: Quantisation of Lz

The eigenvalues and eigenfunctions of L2 can be calculated from L2χ(θ)f(ϕ) = α2χ(θ)f(ϕ). Substituting

in the value of the polar form of L2 and using ∂2f(ϕ)∂ϕ2 = −β2f(ϕ) gives:

−(∂2

∂θ2+ cot θ

∂

∂θ− β2

sin2 θ

)χ(θ)f(ϕ) = α2χ(θ)f(ϕ)

As the operator does not depend on ϕ then:

−(∂2

∂θ2+ cot θ

∂

∂θ− β2

sin2 θ− α2

)χ(θ) = 0

This type of differential equation is known to mathematics as the Associated Legendre Equation. Thesolutions exist for integer values ` and β such that α2 = `(`+1) and ` = 0, 1, 2, 3, .... For a particular value

of ` then β must be constrained to 2`+1 values. The eigenfunctions can be written as χ(θ) = P|m`|` (cos θ)

and so the total eigenfunction can be written as a spherical harmonic function:

Y m`

` (θ, ϕ) = χ(θ)f(ϕ) = P|m`|` (cos θ)eiβϕ

The eigenvalues of L2 are `(`+ 1). The magnitudes of the angular momentum are√`(`+ 1)h.

6

2

Spin

Spin is the intrinsic angular momentum, it is a wholly quantum mechanical process.

Figure 2.1: Classical Analogy of Spin

The spin of a particle is associated with its magnetic moment m˜ = γs˜ where γ is the gyro-magnetic ratioand s˜ is the spin quantum number. The magnitude of spin is fixed for a given type of particle. The

general equation for spin is |S|2 = s(s+ 1)h2 where the spin quantum number can take values of integerand half integer numbers.

2.0.3 Fermions

Fermions have half-odd integer spin(s = 1

2 ,32 , ...

). It it not possible for two Fermions to be in the same

quantum state. Electrons, protons, neutrons, etc. are Fermions with s = 12 and |S|2 = 3

4 h2.

Electrons

The magnetic momentum of an electron is m = µBSh where µB = eh

2m is the Bohr magneton.

2.0.4 Bosons

Bosons have integer spin (s = 0, 1, 2, ...) and can have many particles in the same state. Photons, Gluons,π Mesons, etc. are examples of Bosons with s = 1.

2.0.5 Stern-Gerlach

Particles are passed through Stern-Gerlach magnets to split apart the different spins.

7

N

S

Figure 2.2: Stern-Gerlach

There is a gradient in the magnetic field. The energy of the magnetic moment is U = −m˜ · B˜ . Asthe moment is fixed the field must vary, which gives rise to a force F˜ = −∇U . Inside the Stern-Gerlach

magnets, near the axis which the particles travel, the field is well approximated by B˜ = (Bo+αz˜)k and so

the force is F˜ = αmz k. A particle with a magnetic moment m˜ enters the magnet with a velocity vik and

exits with v =(vi + αmz

LU

)k. Classically, a line of randomly distributed particles would be expected,

but as there are only two values of spin in the z direction, then just two points are observed. The generalresult is that the spin vector can only have discrete values S = σh where σ = −s,−s + 1, ..., s − 1, s inall directions.

2.0.6 Spin Filters

A series of Stern-Gerlach magnets can be used to separate particles with different spins. By placing ablocker in the path of one of the spin paths allows only one type of spin particle to be transmitted andtherefore detected.

Figure 2.3: Series of Stern-Gerlach Magnets with a Blocker

The filter performs a measurement of one component of the spin. If a measurement indicates spin ”up”then the particle passes and if the measurement indicates ”down” the particle is blocked. A filter can berepresented by:

Figure 2.4: Spin Filter

If two of the same filters are placed in series then the same intensity is found after the second filter asafter the first. If the second filter is rotated by a half turn around the axis of the beam then half of theparticles are blocked by the first filter and the rest are blocked by the second filter.

8

SySz

Io4

Io2Io

Figure 2.5: Spin Filter Series

Also spin filters can be placed in different orientations. In this case half od the particles get through thefirst filter and half of those can get through the second filter.

SzSySz

Io8

Io4

Io2Io

Figure 2.6: Spin Filter Series

When using three filters with the first and last aligned to cancel all spin particles half of the particleswhich pass through the second still pass through the third filter. This is analogous to the uncertaintyprinciple as, by measuring the second property creates maximum uncertainty in the measurement of thefirst property. When Sy is know with certainty then Sz is maximally uncertain so both spin ”up” andspin ”down” are equally likely.

2.0.7 Pauli Spin Matrices

The spin vector can be represented by Sj = h2σj where j = x, y, z. The spin matrix in the x direction is:

σx =

(0 11 0

)In the y direction the spin matrix is:

σy =

(0 −ii 0

)And in the z direction the spin matrix is:

σz =

(1 00 −1

)

2.1 Quantum Spin Precession

2.1.1 Stationary States

If a system is in a certain eigenstate of energy at t = 0 then it must stay in that state. The Hamiltonianfor a particle in a magnetic field is:

H = −γB˜ ·m˜If the magnetic field is aligned along the z axis then H = −γBSz = −γB h

2

(1 00 −1

)and so the

eigenvalues are ± h2γB. It is expected that the linear separate equations would give two values which getfurther apart in energy as the magnetic field increases. This was observed by Ornstein et al. and lead toPauli’s invention of spin.

2.1.2 Spin Precession

The Hamiltonian for a half-odd integer spin particle in a static magnetic field is H = −γB˜Si. In the zdirection the Hamiltonian is:

H = −γB h2

(1 00 −1

)

9

The eigenstate for E+ = − h2γB is u˜(+) =

(10

)and for E− = h

2γB is u˜(−) =

(01

)This gives the

Larmor frequency at which the spin precesses, ωL = γB. When the initial state of the system is that ofSx:

ψ˜(t = 0) =1√2

(11

)Writing this in terms of the constituent probabilities and adding the independent oscillatory time evolutiongives:

ψ˜(t) =1√2e

iωLt

2

(10

)+

1√2e−

iωLt

2

(01

)The probability amplitudes can be calculated for different spin states by inserting the eigenstates intothe initial state of the system. Measuring Sx gives:

ψ˜(t) =1√2

(e

iωLt

2

e−iωLt

2

)= cos

(ωLt

2

)1√2

(11

)+ i sin

(ωLt

2

)1√2

(1−1

)So the probability is:

P(Sx =

h

2

)=

∣∣∣∣cos

(ωLt

2

)∣∣∣∣2 = cos2

(ωLt

2

)=

1

2(1 + cos(ωLt))

P(Sx = − h

2

)=

∣∣∣∣sin(ωLt2

)∣∣∣∣2 = sin2

(ωLt

2

)=

1

2(1− cos(ωLt))

This shows that the system evolves such that the state changes every half of the Larmor frequency. Theexpectation of the spin precession can be calculated for Sx, Sy and Sz.

〈Sx〉 = h2 cos(ωLt)

〈Sy〉 = h2 sin(ωLt)

〈Sz〉 = 0

This is the same as the classical angular momentum vector for precession. The equations of motion fora particle in a magnetic field in the z direction is:⟨

Sx

⟩= ωL 〈Sy〉⟨

Sy

⟩= −ωL 〈Sx〉⟨

Sz

⟩= 0

⟨S˜⟩

= −γB˜ × ⟨S˜⟩This is the same as the classical equation of motion for the angular momentum of a rotating body subjectto torque. As the magnitude is h

2 in the x− y plane then the spin vector has a definite value.

2.2 Oscillating Magnetic Field

An oscillating magnetic field added to a static magnetic field alters the Hamiltonian to:

H = −γB˜ 0 · S˜ +B˜ 2 cosωt

Where B1 B0. If ω0 = γB0 and ω1 = γB1 the equation of motion is:

ihψ˜ = − hω0

2σi −

hω1

2cos(ωt)σjψ˜

10

This states that when ω0 = 0 then there is no precession relative to the reference frame which is rotating

with angular velocity ω0. If ψ˜ =

(e

iω0t2 a

e−iω1t

2 b

)then a matrix can be designed where ψ˜ = U

(ab

)and

U =

(ζ 00 ζ∗

)where ζ = e

iω0t2 . The equation of motion can now be written:

ihδ

δt

(U

(ab

))= HU

(ab

)This can be expanded and multiplied by U† to get U†HU = − hω0

2 σi − hω1

2 cos(ωt)U†σjU . Expandingthis and using ∆ = ω − ω0:

ia = −ω1

4

(ei∆t + e−2ωt

)b

ib = −ω1

4

(ei2ωt + e−i∆t

)a

From this it can be seen that the equations have a slowly varying term and a rapidly varying term. Inthe case where ∆ ω the equations can be analytically solved.

ia = −ω1

4ei∆tb

ib = −ω1

4e−i∆ta

By differentiating the first equation and substituting into the second equations gives:

a− i∆a+ω1

4a = 0

If a = CeiΩt then:Ω2 −∆Ω− ω1

4= 0

This leads to the Rabi Frequency ΩR =

√∆2 +

ω21

4 . The general solution for a is now a = Cei(∆+ΩR)t

2 +

Dei(∆−Ωr)t

2 , which can be expanded and used to find b.

a = ei∆t2

(K+ cos

(ΩRt

2

)+ iK− sin

(ΩRt

2

))

b = e−i∆t2

[(2∆K+

ω1+

2ΩRK−ω1

)cos

(ΩRt

2

)+ i

(2∆K−ω1

+2ΩRK+

ω1

)sin

(ΩRt

2

)]Where, for the initial states a(0) = ao and b(0) = bo then K+ = ao and K− = 1

ΩR

(ω1bo

2 −∆ao). If

ao = 1 and bo = 0 then on resonance ∆ = 0 and ΩR = ω1

2 :

a = ei∆t2 cos

(ΩRt

2

)

b = e−i∆t2 sin

(ΩRt

2

)The probability for the system can be found as can the expectation values.

〈Sz〉 = − h2

cos(ΩRt)

This is clearly different to the normal precession where 〈Sz〉 = 0 and so it therefore states that theoscillating magnetic field is solely causing the value of 〈Sz〉. It can be seen that that 〈Sz〉 is maximallyuncertain when the spin has a definite value in only the x−y plane. When the system is not in resonancethe oscillations still occur, but not with maximum certainty.

11

3

Quantum Cryptography

3.1 Cryptographic Keys

A cryptographic key is a long string of bits. The key can be used to encrypt and decrypt messages. Theproblem with keys is that repeated use of them are able to be broken. One way of improving this is byallowing only one person to have the key. Online cryptography allows a person to send encrypted dataand the other person to open it by first sending the encryption key privately. This is normally done byproduct of prime numbers.

3.1.1 Sending Keys

If Alice and Bob want to exchange confidential information then they could use a private channel. Aprivate channel is able to send keys because it is secure. Otherwise a public channel can be used, butanyone can access a public channel. The private channel can be hacked, and so this needs to be detectedfor sending secure data.

A B

Hack

Private Channel

Public Channel

Figure 3.1: Sending Data

The private channel can represented as an electron where the preserved spin state represents a bit. Whenthe spin state is known then the system acts as the non-quantum system would. If both Alice and Bobhave a Sx and a Sy filter then Alice can randomly chose which filter to use for each bit. The possiblestates for the electron in the private channel is:

|1x〉 =1√2

(11

)|0x〉 =

1√2

(1−1

)

|1z〉 =

(10

)|0z〉 =

(01

)

12

If Bob randomly measures Sx or Sz then Alice and Bob tell each other which orientation was used for eachbit and any incorrect components are discarded. If the private channel is hacked then the probabilitieschange in a way that can be detected.

Table 3.1: Probability of Determining Bits

Alice Hacker Bob Probability

Z 1Z 1 50%X 0 25%X 1 25%

Z 1Z 1

Z1 50%

X 0 0 or 1 25%X 1 0 or 1 25%

When the hacker measures Sx then the Sz orientation becomes maximally uncertain and so when thebits are compared by Alice and Bob if 25% of the bits are incorrect then it is known the private channelhas been hacked. The practical problem with this method is that electrons cannot travel very far withoutthe spin being altered. If polarised light is used then this can be rectified over long distances using fibreoptic cables.

13

4

Quantum Dynamics

4.1 Wavepackets

Space is continuous so that the probability of a particle being precisely at a position r˜ is vanishingly

small. This means that the probability density must be defined in a volume element P(r˜)δ3r˜. If a systemis unbound then the wavefunction may not converge, so there would still be a relative probability, butno absolute scale.

4.1.1 Charged Particle in an Electromagnetic Field

Classically the Hamiltonian for this system is H = 12m

∣∣∣p˜+ qA˜∣∣∣2 − qϕ where A is the electromagnetic

vector potential and ϕ is the scalar potential, both of which are derived from E˜ = −∇ϕ −∂A

∂t and

B˜ = ∇ × A˜ . Converting this into the quantum Hamiltonian operator gives H = 12

∣∣−ih∇+ qA˜∣∣2 − qϕ.In classical mechanics a particle only responds to a non-zero region of field, but in quantum mechanicsreal effects, such as the Aharonov-Bohm effect, can be seen even when the particle cannot pass througha non-zero region.

4.2 Time Evolution

If a particle has a mass m and is subject to a potential V then the time dependent Schrodinger Equationis:

ih∂ψ

∂t= − h2

2m∇2ψ + V ψ

These waves are dispersive and so can change in shape as well as position. A probability current can bedefined as the rate at which probability flows from a region.

j˜ = − ih

2m

(ψ†∇ψ − ψ∇ψ†

)This must satisfy the continuity equation ∂P

∂t +∇ · j˜ = 0. This is the reason why the first derivative of

the wavefunction must be smooth and continuous.

4.3 Heisenberg’s Uncertainty Principle

A wavefunction can be created that has both finite ∆r˜ and ∆p˜. If the momentum is defined by ψ = eikox

and the position is localised by a Gaussian function ψ = e−(x−xo)2

4`2 then the state can be defined:

ψ = Ceikoxe−(x−xo)2

4`2

The normalisation function of this system is C = (2π`2)−14 .

14

Figure 4.1: Wavepacket with Finite Uncertainty in Position and Momentum

The expectation of the position will occur at the central point of the wavepacket (〈x〉 = xo) and theexpectation of the momentum will be the wavelength (〈p〉 = hko).

4.3.1 Uncertainties

Uncertainties are given by ∆j2 =

⟨(j − 〈j〉

)2⟩

=⟨j2⟩−⟨j⟩2

. By calculating 〈x〉,⟨x2⟩, 〈p〉 and

⟨p2⟩

the uncertainty of the wavepacket can be calculated.

∆x =√〈x2〉 − 〈x〉2 = `

∆p =√〈p2〉 − 〈p〉2 =

h

2`

∆p∆x =h

2

This shows that this particular wavefunction has the minimum possible uncertainty product. The meankinetic energy can be calculated as: ⟨

K⟩

=

⟨p2

2m

⟩=h2k2

0

2m+

h2

8m`2

This contains the motion of the wavepacket and energy from the confinement.

4.3.2 Time Evolution

When this wavefunction evolves over time the probability spreads out and oscillations occurs. For free

particles V = 0 and so the Hamiltonian is H = p2

2m . Ehrenfest’s Theorem shows that:

⟨x2⟩

=i

h

⟨[p2

2m, x2

]⟩= 0

And so ∆p is conserved. As ∆x∆p = h2 at t = 0 then ∆x must increase with time.

δ

δt〈xp+ px〉 =

i

h

⟨[p2

2m, px+ xp

]⟩=

2⟨p2⟩

m

This shows that the uncertainty in position is time-dependent:

∆x(t) = ∆x(0)2 +∆2pt

2

m

If the particle is initially very well localised then the uncertainty is larger and so the uncertainty inposition grows quickly.

15

5

Quantum Systems

5.1 Particle in an Infinite Well

L0

V =∞V = 0V =∞

Figure 5.1: Inifinite Potential Well

The potential is V = 0 from 0 ≤ x ≤ L and V = ∞ from 0 < x and x > L and so the Schrodingerequation is:

Eψ = − h2

2m

∂2ψ

∂x2+ V ψ

5.1.1 Outside the Well

Because the particle cannot have infinite energy the value of the wavefunction must be 0. The particle istherefore bounded by the potential well.

5.1.2 Inside the Well

The potential is V = 0 so the Schrodinger equation is:

− h2

2m

∂2ψ

∂x= Eψ

With solutions sin(kx), cos(kx) and eikx. The general solution is:

ψ = A′(eikx − e−ikx

)Boundary Conditions

The boundary conditions state that ψ(0) = 0 and ψ(L) = 0, where ψ and ∂ψ∂x must be smooth with no

abrupt changers. Energy is quantised because A sin(kL) = 0 and it is given by:

En =h2π2n2

2mL2

The solutions are reminiscent of standing waves on a vibrating string where the number of nodes increaseswith n.

16

Figure 5.2: Wavefunctions of the Inifinite Potential Well

The lowest possible energy is greater than zero. If the particle is confined it has an energy of E1 = h2k2

2mL2 .This energy is called zero point energy.

5.1.3 Finite Well

If the well does not have an infinite potential on one side then there is a probability that the particle willtunnel into the classical forbidden region.

Figure 5.3: Wavefunction of the Finite Potential Well

The potential is V = Vo and so the Schrodinger equation is:

− h2

2m

∂2ψ

∂x2= (E − Vo)ψ

In an unbound system the wavelength of the wavefunction increases and the amplitude decreases. In abound system the wavefunction decreases exponentially.

5.2 Rectangular Potential Barrier

Figure 5.4: Finite Barrier

A particle with mass m, momentum hk and energy E = h2k2

2m moves in the positive x direction. Thepotential barrier has a height V = Vo and begins at x.

17

x < 0

− h2

2m

∂2ψ

∂x2= Eψ

This gives the general solution ψ = CIeikx + CRe

−ikx, where |CI |2 is the intensity of the incident wavein the positive direction and |CR|2 is the intensity of the reflected wave in the negative direction.

0 ≤ x ≤ L

In the unbound case the solution to the Schrodinger equation is ψ = Beikx + B′e−ikx which is theoscillatory form. In the bound case then the wavefunction decays exponentially ψ = De−αx. This wouldbe classically forbidden.

x > L

The Schrodinger equation is the same as before the barrier with an intensity |CT |2 and wavefunctionψ = CT e

ikx.

5.2.1 Complex Coefficients

By calculating the value of ψ(0), ∂ψ(0)∂x , ψ(L) and ∂ψ(L)

∂x then complex coefficients can be found. Theprobability of finding the reflected wave at x = 0 is:

R =|CR|2

|CI |2

Similarly the probability of the particle being transmitted through the barrier and emerging at x > L is:

T =|CT |2

|CI |2

5.2.2 Quantum Tunneling

T =|CT |2

|CI |2=

16α2k2e−2αL

(k2 + α2)

There is a finite chance of a particle tunnelling through into the classically forbidden region. This can

be expressed in terms of the energy parameters by substituting in k2 = 2mEh2 and α2 = 2m(Vo−E)

h2 . Thisgives:

T = 16

(E

Vo

)(1− E

Vo

)e−

2√

2m(Vo−E)L

h

Tunnelling Electrons

Electrons can tunnel from a surface across a region of low potential, such as air, and be found on anothersurface. The fractional change in the tunnelling probability is given by ∆T

T ≈ −2α∆L. This gives a 2%change in probability for a distance of about 10−12m. This is used in scanning tunnelling microscopy.

Nuclear Fusion

The core of the sun reaches temperatures of 107K. Given that the protons energy is approximately kTthen this temperature equates to 1keV . The protons need to be within rn ≈ 2× 10−15m to be attractedby the strong nuclear force. The Coulomb repulsion between the two protons must be overcome to reachthis distance. The amount of energy needed by the proton to fuse is:

U =e2

4πεorN≈ 1MeV

The distance of closest approach of a proton with an energy of 1keV is rC ≈ 1.4× 10−12m. The protonsdo not have enough energy to get over the potential barrier so they must tunnel through to fuse. Theprobability of a proton tunnelling and arriving is T ≈ 3× 10−10, so only one proton in every three billionwill undergo fusion. This accounts for the large timescale of stars.

18

Alpha Particle Decay

In the opposite way to protons fusing together, alpha particles can escape the Coulomb barrier to leavethe nuclei of larger atoms. If the alpha particles have a kinetic energy much lower than the height of thepotential barrier then the alpha particle must tunnel through. The probability of radioactive decay isgiven by the decay constant λ = fT , where f is the frequency of collisions. Typically f ≈ 1021s−1, buta decay can take around a thousand years to occur. This shows how unlikely the probability is, but it isstill finite.

5.3 Quantum Harmonic Oscillators

The Hamiltonian of the classical harmonic oscillator is H = p2

2m + 12kx

2 = p2

2m + 12mω

2x2 so that thequantum harmonic oscillator is:

H = − h2

2m

∂2

∂x2+

1

2mω2x2

If this is placed into the Schrodinger equation then it can be seen that there are an infinite number ofsolutions En, ψ(x) where n is the quantum number.

Figure 5.5: Energy Levels and Wavefunctions of the Quantum Harmonic Oscillator

The energy levels are equally spaced where the energy is given by En =(n+ 1

2

)hω. There is zero point

energy which arises from the Heisenberg Uncertainty Principle. The nodes of the wavefunction increase

with n and the ground state is centred on x = 0. As the natural scale length is a =√

hmω with a natural

unit of energy hω. It is useful to define E = εhω and x = q√

hmω so that the Schrodinger equation

becomes:

− h2

∂2ψ(q)

∂q2+h

2ωq2ψ(q) = Eψ(q)

When the energy is substituted in then this becomes:(− ∂2

∂q2+ q2

)ψ(q) = 2εψ(q)

This can be written in two ways: (q − ∂

∂q

)(q +

∂

∂q

)ψ = (2ε− 1)ψ

Or: (q +

∂

∂q

)(q − ∂

∂q

)ψ = (2ε+ 1)ψ

This defines the raising and lowering operators.

19

5.3.1 Raising and Lowering Operators

The raising operator is

A† =1√2

(q − ∂

∂q

)When applied to the ground state then the first excited state is generated. Higher states can further beobtained by successively applying the operator. The lowering operator is

A =1√2

(q +

∂

∂q

)It generates ψn−1 with εn−1 = εn−1. The raising and lowering operators can be called Ladder operatorsor creation and annihilation operators.

Hermite Polynomials

Successively applying the raising operator n times gives:(A†)n

ψ(q) = C ′Hnψ(q)

Where Hn is a hermit polynomial.

5.3.2 Ground State

If ε = − 12 and

(q − ∂

∂q

)ψ = 0 then

(q + ∂

∂q

)(q − ∂

∂q

)ψ = (2ε + 1)ψ is satisfied and the differential

equation has the solution ψ = Ceq2

2 . This is physically inadmissible as the wavefunction is not integrable

square. If ε = 12 and

(q − ∂

∂q

)ψ = 0 then

(q − ∂

∂q

)(q + ∂

∂q

)ψ = (2ε − 1)ψ is satisfied and has the

solution ψ = Ce−q2

2 . This is integrable square and therefore a solution, which is the ground state of the

quantum harmonic oscillator. E = 12 hω and n = 0 so ψ = Ce−

mωx2

2h .

5.3.3 Probability

The state is given by ψ =∑n cnun which is a linear superposition of the quantum harmonic oscillator

eigenfunctions. The probability of an energy level occurring is Pn = c†ncn. As particular observationsare randomly distributed, individual events cannot be predicted. A series of data can be predicted byfinding the mean energy weighted by the individual energies. The expectation of the value of the positionis 〈x〉 = 〈ψ|x|ψ〉 which can be expressed in natural units to give:

〈q〉 =a

2cos(ωt)

Which is the oscillation inside the quantum harmonic oscillator.

5.4 Hydrogen Atom

p+

r

e−

Figure 5.6: Model of the Hydrogen Atom System

The potential energy of the system is the electrostatic attraction of Coulombs law V = −Ze24πεor˜ where Z

is the nuclear charge, which for hydrogen is 1. The electron moves in three dimensions and its mass is

20

assumed to be mp me so the nucleus can be made the centre of mass. The classical Hamiltonian is

H = p2

2m + V and so the quantum Hamiltonian Operator is:

H = − h2

2m∇2 − e2

4πεor

As the system has spherical symmetry it is more convent to write the Hamiltonian in polar form where:

∇2 =∂2

∂r+

2

r

∂

∂r+

1

r2

(∂2

∂θ2+ cot θ

∂

∂θ+

1

sin2 θ

∂2

∂ϕ2

)

The term inside the brackets is −L2 and so the Hamiltonian is:

H = − h2

2me

(∂2

∂r2+

2

r

∂

∂r− L2

r2

)− e2

4πεo

The Schrodinger Equation can be found where it is known that there must be solutions which satisfy L2

and the potential because[L2, V

]= 0. This means there can be separation of variables.

HR(r)Y m`

` (θ, ϕ) = ER(r)Y m`

` (θ, ϕ)

Where Y m`

` are spherical harmonics with quantum numbersm` and `. The radial equation of the hydrogenatom is:

− h2

2me

(∂2

∂r2+

2

r

∂

∂r

)+

(h2`(`+ 1)

2mer2− e2

4πεor

)R(r) = ER(r)

The first term is the kinetic energy associated with the radial energy and the second term is the effectivepotential energy. This is made up of the Coulomb Potential and the quantum mechanical form of thecentrifugal potential energy.

Veff

` = 3` = 2

` = 1` = 0

r

Bohr Radius

Figure 5.7: Effective Potential vs. Radius

5.4.1 Solution

When E > 0 the electron is not bound and so can travel anywhere so that there is a continuum of states.When E < 0 the electron is bound to the nucleus and so the solutions of the radial equations are definedin terms of quantum numbers where nr is the radial quantum number and ` is the angular momentumnumber. There is no dependence on m` so the energy of the bound state is:

Enr,` = − ER(nr + `+ 1)2

Where ER = me

2h2

(e2

4πεo

)2

= 13.6eV is the Rydberg Energy.

21

−ER

−ER

22

−ER

32

−ER

42

nr = 0

nr = 1

nr = 2

nr = 3

` = 0

nr = 0

nr = 1

nr = 2

` = 1

nr = 0

nr = 1

` = 2

nr = 0

` = 3

Figure 5.8: Degenerate States

There is degeneracy of the states with the same value nr + `. This can be written as n = nr + `+ 1 whichis the principle quantum number.

En = −ERn2

5.4.2 Eigenfunctions

For large r an approximate solution is e−rrB . In fact the ground state has n = 1 and so nr = 0 and ` = 0has the form:

Rn,`(r) = R10(r) = Ce− r

rB

R10 is a solution of the Schrodinger equation when 2h2

2merrB= e2

4πεo. In this case the energy is:

E10 = − h2

2mer2B

= −me

2h2

(e2

4πεo

)2

= −ER

More generally the solution to the radial equation is of the form:

Rn,`(r) = Cn,`e− r

nrB

(r

rB

)`L2`+1n−`−1

(2r

nrB

)Where Cn,` is a normalisation constant and L2`+1

n−`−1 is an associated Laguerre Polynomaial.

Lqnr

(2r

nrB

)= ao + a1

(2r

nrB

)+ a2

(2r

nrB

)2

+ a3

(2r

nrB

)3

+ ...+ aq

(2r

nrB

)nr

The polynomial is of order nr and so it defines the number of oscillations. For n = 1 then ` = 0 is theonly possibility. For n = 2 then nr = 0, 1 and so ` = 0 or ` = 1. The general rule is ` = 0, 1, ..., (n− 1).The complete wavefunction of the Hydrogen atoms are:

ψ(r, θ, ϕ) = Rn`(r)Ym`

` (θ, ϕ)

Table 5.1: State of the Hydrogen Atoms

n ` m` State1 0 0 1s2 0 0 2s2 1 −`,0,` 2p3 0 0 3s3 1 −`,0,` 3p3 2 −2`,−`,0,`,2` 3d

` = 0 is labelled the s state. ` = 1 is labelled the p state. ` = 2 is labelled d state. ` = 3 is labelled fstate. ` = 4 is labelled the g state. The nth energy level possesses n different ` values, ranging from 0 ton − 1. Each of them possesses (2` + 1) different values of m`. All of these have the same energy. Thedegeneracy of the nth shell is therefore:

gn =

n1∑`=0

(2`+ 1) = n2

22

5.4.3 Probability Distribution

The probability of finding the electron in a volume element δV centred at (r, θ, ϕ) is P = |ψn,`,m`|2δV

where δV = r2 sin θδrδθδϕ. The probability can be separated into its different variables to give:

P = |Rn,`(r)|2r2δr|P |m`|` (cos θ)|2 sin θδθδϕ

For the ground state (1s) of the hydrogen atom ` = 0 and m` = 0. As the integrals over θ and ϕ isY m`

` = 1√4π

then the normalisation of the wavefunction can be found. This gives the wavefunction as:

ψ(r, θ, ϕ) =2

r32

B

1√4πe− r

rB

The expectation of the position and momentum can be calculated, where the root mean squared value ofthe position is: √

〈r2〉 =√

3rB

The probability density of the radial equation gives wavelike characteristics. For s states (` = 0) theradial equation is finite at r = 0, but the probability of finding a particle there is zero. The number ofnodes in the radial equation increases with energy eigenvalue and is equal to n− ` = nr + 1.

5.4.4 Reduced Mass Effect

If it is not assumed that mp me then the centre of mass is shifted and the proton would also orbit thecentre of the system. In the centre of mass frame of reference the mass can be written as:

M =memp

me +mp

The Rydberg energy must now be corrected:

E′R =M

2π2

(e2

4πεo

)2

=M

meER

This leads to small corrections in the energies of the hydrogen atom.

E(H)n = −0.999456

ERn2

For Deuterium then the mass of the mass must also be taken into account and so E(D)n = −0.999728ER

n2 .

23

6

Appendix

6.1 Gaussian Integrals

If an integral is of the form∫∞−∞ e−αx

2

δx then is decays quickly and must be a well defined number. Thiscan be found by squaring the original integral and converting to polar coordinates. This can be donearbitrarily as the original integral is over an infinite plane.

I2 =

∫ ∞−∞

e−αx2

δx×∫ ∞−∞

e−αy2

δy =

∫ ∞0

e−αr2

rδr

∫ 2π

0

δθ

If u = αr2 is substituted in then:

I2 = 2π

∫ ∞0

e−u

2αδu =

π

α=

∫ ∞0

e−uδu =π

α

This means that:

I =

√π

α

6.2 Quantum Harmonic Expectation Value

The expectation value 〈q〉 = a2

2 〈ψ|q|ψ〉 = a2

2

∫∞−∞ ψ†qψδq

ψ =1√2

(√1

a√πe−

q2

2 eiEot

h +

√2

a√πqe−

q2

2 eiE1t

h

)

〈q〉 =a

2

(e−

i(E1−E0)th + e

i(E1−E0)th

)〈q〉 =

a

2cos

((E1 − E0)t

h

)As the energy is E0 = hω

2 and E1 = 3hω2 so E1 − E0 = hω then:

〈q〉 =a

2cosωt

24