Lecture 5 Motion of a Charged Particle in a Magnetic Field

Home , Charged particle, Magnetic flux, Orbital motion (quantum), Scalar potential

Lecture 5 Motion of a charged particle in a magnetic ﬁeld Charged particle in a magnetic ﬁeld: Outline

1 Canonical quantization: lessons from classical dynamics

2 Quantum mechanics of a particle in a ﬁeld

3 Atomic hydrogen in a uniform ﬁeld: Normal Zeeman eﬀect

4 Gauge invariance and the Aharonov-Bohm eﬀect

5 Free electrons in a magnetic ﬁeld: Landau levels

6 Integer Quantum Hall eﬀect Lorentz force

What is eﬀect of a static electromagnetic ﬁeld on a charged particle?

Classically, in electric and magnetic ﬁeld, particles experience a Lorentz force:

F = q (E + v B) ×

q denotes charge (notation: q = e for electron). − Velocity-dependent force qv B very diﬀerent from that derived from scalar potential, and programme× for transferring from classical to quantum mechanics has to be carried out with more care.

As preparation, helpful to revise(?) how the Lorentz force law arises classically from Lagrangian formulation. Analytical dynamics: a short primer

For a system with m degrees of freedom speciﬁed by coordinates q , q , classical action determined from Lagrangian L(q , q˙ ) by 1 ··· m i i

S[qi ]= dt L(qi , q˙ i ) ! For conservative forces (those which conserve mechanical energy), L = T V , with T the kinetic and V the potential energy. − Hamilton’s extremal principle: trajectories qi (t) that minimize action specify classical (Euler-Lagrange) equations of motion,

d (∂ L(q , q˙ )) ∂ L(q , q˙ )=0 dt q˙ i i i − qi i i mq˙ 2 e.g. for a particle in a potential V (q), L(q, q˙ )= V (q) and 2 − from Euler-Lagrange equations, mq¨ = ∂ V (q) − q Analytical dynamics: a short primer

To determine the classical Hamiltonian H from the Lagrangian, ﬁrst

obtain the canonical momentum pi = ∂q˙ i L and then set

H(q , p )= p q˙ L(q , q˙ ) i i i i − i i "i mq˙ 2 e.g. for L(q, q˙ )= 2 V (q), p = ∂q˙ L = mq˙ , and − 2 2 H = pq˙ L = p p ( p V (q)) = p + V (q). − m − 2m − 2m In Hamiltonian formulation, minimization of classical action

S = dt pi q˙ i H(qi , pi ) , leads to Hamilton’s equations: # − $ ! "i q˙ = ∂ H, p˙ = ∂ H i pi i − qi

i.e. if Hamiltonian is independent of qi , corresponding momentum pi is conserved, i.e. pi is a constant of the motion. Analytical dynamics: Lorentz force

As Lorentz force F = qv B is velocity dependent, it can not be expressed as gradient of some× potential – nevertheless, classical equations of motion still specifed by principle of least action. With electric and magnetic ﬁelds written in terms of scalar and vector potential, B = A, E = ϕ ∂ A, Lagrangian: ∇× −∇ − t 1 L = mv2 qϕ + qv A 2 − ·

q x =(x , x , x ) andq ˙ v = (x ˙ , x˙ , x˙ ) i ≡ i 1 2 3 i ≡ i 1 2 3 N.B. form of Lagrangian more natural in relativistic formulation: qv µA = qϕ + qv A where v µ =(c, v) and Aµ =(ϕ/c, A) − µ − ·

Canonical momentum: pi = ∂x˙i L = mvi + qAi no longer given by mass velocity – there is an extra term! × Analytical dynamics: Lorentz force

From H(q , p )= p q˙ L(q , q˙ ), Hamiltonian given by: i i i i − i i "i 1 1 H = (mv + qA ) v mv2 qϕ + qv A = mv2 + qϕ i i i − 2 − · 2 i ) * " = pi = L(˙qi , qi ) % &' ( % &' ( To determine classical equations of motion, H must be expressed solely in terms of coordinates and canonical momenta, p = mv + qA

1 H = (p qA(x, t))2 + qϕ(x, t) 2m −

Then, from classical equations of motionx ˙i = ∂pi H and p˙ = ∂ H, and a little algebra, we recover Lorentz force law i − xi

mx¨ = F = q (E + v B) × Lessons from classical dynamics

So, in summary, the classical Hamiltonian for a charged particle in an electromagnetic ﬁeld is given by

1 H = (p qA(x, t))2 + qϕ(x, t) 2m −

This is all that you need to recall – its ﬁrst principles derivation from the Lagrangian formulation is not formally examinable!

Using this result as a platform, we can now turn to the quantum mechanical formulation. Quantum mechanics of particle in a ﬁeld

Canonical quantization: promote conjugate variables to operators, p pˆ = i! , x xˆ with commutation relations [ˆpi , xˆj ]= i!δij → − ∇ → − 1 Hˆ = (pˆ qA(x, t))2 + qϕ(x, t) 2m −

Gauge freedom: Note that the vector potential, A, is speciﬁed only up to some gauge: For a given vector potential A(x, t), the gauge transformation

A A! = A + Λ,ϕ ϕ! = ϕ ∂ Λ &→ ∇ &→ − t with Λ(x, t) an arbitrary (scalar) function, leads to the same physical magnetic and electric ﬁeld, B = A, and E = ϕ ∂ A. ∇× −∇ − t

In the following, we will adopt the Coulomb gauge condition, ( A) = 0. ∇· Quantum mechanics of particle in a ﬁeld

1 Hˆ = (pˆ qA(x, t))2 + qϕ(x, t) 2m −

Expanding the Hamiltonian in A, we can identify two types of contribution: the cross-term (known as the paramagnetic term),

q iq! iq! (pˆ A + A pˆ)= ( A + A )= A −2m · · 2m ∇· ·∇ m ·∇ where equality follows from Coulomb gauge condition, ( A) = 0. ∇· q2 And the diagonal term (known as the diamagnetic term) A2. 2m Together, they lead to the expansion

!2 iq! q2 Hˆ = 2 + A + A2 + qϕ −2m ∇ m ·∇ 2m Quantum mechanics of particle in a uniform ﬁeld

!2 iq! q2 Hˆ = 2 + A + A2 + qϕ −2m ∇ m ·∇ 2m

For a stationary uniform magnetic ﬁeld, A(x)= 1 x B (known as − 2 × the symmetric gauge), the paramagnetic component of Hˆ given by,

iq! iq! iq! q A = (x B) = (x ) B = Lˆ B m ·∇ − 2m × ·∇ 2m ×∇ · −2m ·

where Lˆ = x ( i! ) denotes the angular momentum operator. × − ∇ For ﬁeld, B = Beˆz oriented along z, diamagnetic term,

q2 q2 q2 q2B2 A2 = (x B)2 = x2B2 (x B)2 = (x 2 + y 2) 2m 8m × 8m − · 8m + , Quantum mechanics of particle in a uniform ﬁeld

!2 qB q2B2 Hˆ = 2 Lˆ + (x 2 + y 2)+qϕ −2m ∇ − 2m z 8m

In the following, we will address two examples of electron (q = e) − motion in a uniform magnetic ﬁeld, B = Beˆz :

1 Atomic hydrogen: where electron is bound to a proton by the Coulomb potential,

1 e2 V (r)=qϕ(r)= −4π%0 r

2 Free electrons: where the electron is unbound, ϕ = 0. In the first case, we will see that the diamagnetic term has a negligible role whereas, in the second, both terms contribute significantly to the dynamics. Atomic hydrogen in uniform field

2 2 2 2 ! 2 eB e B 2 2 1 e Hˆ = + Lˆz + (x + y ) −2m ∇ 2m 8m − 4π%0 r

2 2 2 With x + y a0, where a0 is Bohr radius, and Lz !, ' () ratio of paramagnetic and diamagnetic' () terms,

2 2 2 2 (e /8me ) x + y B e 2 6 ' ( = a0B 10− B/T (e/2me ) Lz B 4! ) ' ( i.e. for bound electrons, diamagnetic term is negligible. not so for unbound electrons or on neutron stars! When compared with Coulomb energy scale,

(e/2m)!B e! 5 2 2 = 2 B 10− B/T me c α /2 (me cα) )

2 where α = e 1 1 denotes fine structure constant, 4π"0 !c 137 paramagnetic) term effects only a small perturbation. Atomic hydrogen in uniform field

!2 e 1 e2 Hˆ 2 + B Lˆ )−2m ∇ 2m · − 4π%0 r

In general, term linear in B deﬁnes magnetic dipole moment µ: HˆM = µ B. Result above shows that orbital degrees of freedom of the electron− · lead to a magnetic moment, e µ = Lˆ −2me cf. classical result: for an electron in a circular orbit around a proton, I = e/τ = ev/2πr. With angular momentum L = m vr, − − e ev 2 e e µ = IA = πr = me vr = L −2πr −2me −2me In the next lecture, we will see that there is an additional instrinsic contribution to the magnetic moment of the electron which derives from quantum mechanical spin.

Since Lˆ !, scale of µ set by the Bohr magneton, ' (∼ e! 5 µB = =5.79 10− eV/T 2me × Atomic hydrogen: Normal Zeeman eﬀect

So, in a uniform magnetic ﬁeld, B = Beˆz , the electron Hamiltonian for atomic hydrogen is given by,

e pˆ2 1 e2 Hˆ = Hˆ0 + BLˆz , Hˆ0 = 2m 2m − 4π%0 r

Since [Hˆ0, Lz ] = 0, eigenstates of unperturbed Hamiltonian, Hˆ0, deﬁned by ψn#m(x), remain eigenstates of Hˆ , with eigenvalues,

1 En#m = Ry + !ωLm −n2 eB where ω = denotes the Larmor frequency. L 2m (Without spin contribution) uniform magnetic ﬁeld ! splitting of (2* + 1)-fold degeneracy with multiplets separated by !ωL. Normal Zeeman eﬀect: experiment

Experiment shows Zeeman splitting of spectral lines...

e.g. Splitting of Sodium D lines (involving 3p to 3s transitions)

P. Zeeman, Nature 55, 347 (1897). ...but the reality is made more complicated by the existence of spin and relativistic corrections – see later in the course. Gauge invariance

1 Hˆ = (pˆ qA(x, t))2 + qϕ(x, t) 2m −

Hamiltonian of charged particle depends on vector potential, A. Since A deﬁned only up to some gauge choice wavefunction is not a gauge invariant object. ⇒ To explore gauge freedom, consider eﬀect of gauge transformation

A A! = A + Λ,ϕ ϕ! = ϕ ∂ Λ &→ ∇ &→ − t where Λ(x, t) denotes arbitrary scalar function. ˆ ˆ Under gauge transformation: i!∂t ψ = H[A]ψ i!∂t ψ! = H[A!]ψ! where wavefunction acquires additional phase,&→

q ψ!(x, t) = exp i Λ(x, t) ψ(x, t) ! - . 2 2 but probability density, ψ!(x, t) = ψ(x, t) is conserved. | | | | Gauge invariance

q ψ!(x, t) = exp i Λ(x, t) ψ(x, t) ! - . Proof: using the identity q q (pˆ qA q Λ) exp i Λ = exp i Λ (pˆ qA) − − ∇ ! ! − - . - .

1 2 q Hˆ [A!]ψ! = (pˆ qA q Λ) + qϕ q∂t Λ exp i Λ ψ 2m − − ∇ − ! / 0 - . q 1 = exp i Λ (pˆ qA)2 + qϕ q∂ Λ ψ 2m − − t ! / 0 - q . = exp i Λ Hˆ [A] q∂t Λ ψ ! − Similarly - .- . q i!∂t ψ! = exp i Λ (i!∂t q∂t Λ) ψ ! − - . ˆ ˆ Therefore, if i!∂t ψ = H[A]ψ, we have i!∂t ψ! = H[A!]ψ!. Gauge invariance: physical consequences

q ψ!(x, t) = exp i Λ(x, t) ψ(x, t) ! - . Consider particle (charge q) travelling along path, P, in which the magnetic ﬁeld, B = 0. However, B =0 A = 0: any Λ(x) such,⇒ that A = Λleads to B = 0. ∇ In traversing path, wavefunction acquires phase q φ = A dx. · ! !P If we consider two separate paths P and P! with same initial and ﬁnal points, relative phase of the wavefunction, q q q q ∆φ = A dx A dx = A dx Stokes= B d 2x ! P · − ! P · ! · ! A · ! ! ! 1 !

where A runs over area enclosed by loop formed from P and P’ 2 Gauge invariance: physical consequences

q ∆φ = B d 2x · ! !A

i.e. for paths P and P!, wavefunction components acquire relative phase diﬀerence,

q ∆φ = magnetic ﬂux through area ! ×

If paths enclose region of non-vanishing field, even if B identically zero on paths P and P!, ψ(x) acquires non-vanishing relative phase. This phenomenon, known as the Aharonov-Bohm effect, leads to quantum interference which can influence observable properties. Aharanov-Bohm effect: example I

Influence of quantum interference effects is visible in transport properties of low-dimensional semiconductor devices. When electrons are forced to detour around a potential barrier, conductance through the device shows Aharonov-Bohm oscillations. Aharanov-Bohm effect: example II

But can we demonstrate interference eﬀects when electrons transverse region where B is truly zero? Deﬁnitive experimental proof provided in 1986 by Tonomura:

If a superconductor completely encloses a toroidal magnet, ﬂux through superconducting loop quantized in units of h/2e. Electrons which pass inside or outside the loop therefore acquire a relative phase e h diﬀerence of ϕ = n = nπ. −! × 2e If n is even, there is no phase shift, while if n Tonomura et al., 1986 is odd, there is a phase shift of π.

Experiment confirms both Aharonov-Bohm effect and the phenomenon of flux quantization in a superconductor! Summary: charged particle in a field

Starting from the classical Lagrangian for a particle moving in a a static electromagnetic ﬁeld,

1 L = mv2 qϕ + qv A 2 − ·

we derived the quantum Hamiltonian,

1 H = (p qA(x, t))2 + qϕ(x, t) 2m −

An expansion in A leads to a paramagnetic and diamagnetic contribution which, in the Coulomb gauge ( A = 0), is given by ∇·

!2 iq! q2 Hˆ = 2 + A + A2 + qϕ −2m ∇ m ·∇ 2m Summary: charged particle in a ﬁeld

Applied to atomic hydrogen, a uniform magnetic ﬁeld, B = Beˆz leads to the Hamiltonian

ˆ2 2 ˆ 1 2 L ˆ 1 2 2 2 2 1 e H = pˆr + 2 + eBLz + e B (x + y ) 2m 3 r 4 4 − 4π%0 r

For weak fields, diamagnetic contribution is negligible in comparison with paramagnetic and can be dropped. Therefore, (continuing to ignore electron spin), magnetic field splits orbital degeneracy leading to normal Zeeman effect,

1 E = Ry + µ Bm n#m −n2 B

However, when diamagnetic term O(B2n3) competes with Coulomb Ry energy scale n2 classical dynamics becomes irregular and system enters “quantum− chaotic regime”. Summary: charged particle in a ﬁeld

Gauge invariance of electromagnetic ﬁeld wavefunction not gauge invariant. ⇒

Under gauge transformation, A A! = A + Λ, &→ ∇ ϕ ϕ! = ϕ ∂ Λ, wavefunction acquires additional phase, &→ − t q ψ!(x, t) = exp i Λ(x, t) ψ(x, t) ! - . ! Aharanov-Bohm effect: (even if no orbital effect) particles encircling magnetic flux acquire relative phase, ∆φ = q B d 2x. ! A · 2 i.e. for ∆φ =2πn expect constructive 1 h interference n e oscillations. ⇒ Lecture 5: continuedFree electrons in a magnetic field: Landau levels

But what happens when free (i.e. unbound) charged particles experience a magnetic ﬁeld which inﬂuences orbital motion? e.g. electrons in a metal.

1 Hˆ = (pˆ qA(x, t))2 + qϕ(x, t), q = e 2m − −

In this case, classical orbits can be macroscopic in extent, and there is no reason to neglect the diamagnetic contribution. Here it is convenient (but not essential – see PS1) to adopt Landau gauge, A(x)=( By, 0, 0), B = A = Beˆ , where − ∇× z 1 Hˆ = (ˆp eBy)2 +ˆp2 +ˆp2 2m x − y z 5 6 Free electrons in a magnetic ﬁeld: Landau levels

1 Hˆ ψ(x)= (ˆp eBy)2 +ˆp2 +ˆp2 ψ(x)=Eψ(x) 2m x − y z 5 6 Since [Hˆ , pˆx ] = [Hˆ , pˆz ] = 0, both px and pz conserved, i.e. ψ(x)=ei(px x+pz z)/!χ(y) with

pˆ 2 1 p2 y + mω2(y y )2 χ(y)= E z χ(y) 2m 2 − 0 − 2m 3 4 ) * p eB where y = x and ω = is classical cyclotron frequency 0 eB m px deﬁnes centre of harmonic oscillator in y with frequency ω, i.e.

2 pz En,pz =(n +1/2)!ω + 2m

The quantum numbers, n, index inﬁnite set of Landau levels. Free electrons in a magnetic ﬁeld: Landau levels

2 pz En,pz =(n +1/2)!ω + 2m

Taking pz = 0 (for simplicity), for lowest Landau level, n = 0, !ω E0 = 2 ; what is level degeneracy?

Consider periodic rectangular geometry of area A = Lx Ly . Centre px × of oscillator wavefunction, y0 = eB , lies in [0, Ly ]. With periodic boundary conditions eipx Lx /! = 1, p =2πn ! , i.e. y x Lx 0 set by evenly-spaced discrete values separated by ∆y = ∆px = h . 0 eB eBLx degeneracy of lowest Landau level N = Ly = Ly = BA , ∴ ∆y0 h/eBLx Φ0 e | N | 14 2 1 where Φ = denotes “ﬂux quantum”,( 10 m− T− ). 0 h BA ) Quantum Hall eﬀect

The existence of Landau levels leads to the remarkable phenomenon of the Quantum Hall Eﬀect, discovered in 1980 by von Kiltzing, Dorda and Pepper (formerly of the Cavendish).

Experiment: linear increase in ρxy with B punctuated by plateaus at which ρxx = 0 – dissipationless ﬂow!

Classically, in a crossed electric E = eˆ and magnetic ﬁeld E y B = Beˆz , electron drifts in direction eˆ with speed v = /B. x E F = q(E + v B) ×

With current density jx = nev, Hall resistivity, −

Ey B ρxy = = E = − jx nev en where n is charge density. Quantum Hall Eﬀect

Origin of phenomenon lies in Landau level quantization: For a state of the lowest Landau level,

ipx x/ ! 1/4 mω 2 e mω (y y0) 2! ψpx (y)= e− − √Lx π! 7 8 1 current jx = 2m (ψ∗(ˆpx + eAx )ψ + ψ((ˆpx + eAx )ψ)∗), i.e. 1 j (y)= (ψ∗ (ˆp eBy)ψ + ψ ((ˆp eBy)ψ∗)) x 2m px x − px x −

mω 2 mω 1 px eBy (y y0) = − e− ! − π L m 9 ! x eB (y y) m 0− % &' ( is non-vanishing. (Note that current operator also gauge invarant.) However, if we compute total current along x by integrating along Ly y, sum vanishes, Ix = 0 dy jx (y) = 0. 2 Quantum Hall Eﬀect

If electric ﬁeld now imposed along y, eϕ(y)= e y, symmetry is broken; but wavefunction still harmonic− oscillator-like,− E

2 pˆy 1 2 2 + mω (y y0) e y χ(y)=Eχ(y) 3 2m 2 − − E 4

px m but now centered around y0 = eB + eBE2 . However, the current is still given by

mω 2 mω 1 px eBy eB m (y y0) j (y)= − y y E e− ! − x π L m m 0 − − eB2 9 ! x ) * Integrating, we now obtain a non-vanishing current ﬂow

Ly e I = dy j (y)= E x x −BL !0 x Quantum Hall Eﬀect

Ly e I = dy j (y)= E x x −BL !0 x

To obtain total current ﬂow from all electrons, we must multiply Ix by the total number of occupied states. If Fermi energy lies between two Landau levels with n occupied, eB e e2 Itot = nN Ix = n Lx Ly E = n Ly × − h × BLx − h E

With V = Ly , voltage drop across y, Hall conductance (equal to conductivity−E in two-dimensions),

I e2 σ = tot = n xy − V h Since no current flow in direction of applied field, longitudinal conductivity σyy vanishes. Quantum Hall Effect

Since there is no potential drop in the direction of current ﬂow, the longitudinal resistivity ρxx also vanishes, while

1 h ρ = yx n e2

Experimental measurements of these values provides the best determination of fundamental ratio e2/h, better than 1 part in 108.