VIII. MAGNETIC PROPERTIES OF IDEAL FERMI-DIRAC SYSTEMS

VIII.1. The Landau’s problem.

In order to study the magnetic properties of an ideal Fermi-Dirac gas, it is necessary to

study the quantum mechanical dynamics of a charged point particle of mass m and charge

−e , e > 0 – an electron – in the presence of a uniform (i.e. constant and homogeneous)

magnetic field: the solution to this problem [ Leon Davidovic Landau (1930): Z. Phys. 64,

629 ] is provided by the celebrated Landau’s levels and degeneracy, which will represent the

starting point for the thermodynamical analysis of the magnetization of an ideal electron

gas.

Owing to its spin, an electron is carrying an intrinsic magnetic moment. According

to relativistic Dirac’s theory - disregarding radiative corrections - the intrinsic magnetic

moment of the electron is given by

~µ =−e2mc

g~s =−eh2mc

~σ = −µ~σ , (1.1)

where we have set the Lande’s g–factor equal to two, according to Dirac’s equation, whereas

~σ = (σx, σy, σz) are the Pauli’s matrices and we have introduced the Bohr’s magneton

µ ≡ (eh/4πmc) = 9.273× 10−21 erg G−1.

Let us suppose that the uniform magnetic field is along the positive Oz axis and let

us choose the asymmetric Landau’s gauge: namely

Ax = −By , Ay = Az = 0 , B = |B| . (1.2)

285

Then the Schrodinger–Pauli hamiltonian operator can be written as

H =1

2m

p +

e

cA(r)

2

− ~µ ·B

=1

2m

(px −

e

cBy

)2

+1

2m(p2

y + p2z

)+ µBσ3 ,

(1.3)

in which px ≡ −ih∂x , et cetera. Since we have[H, σ3

]= 0, we can write the 2–

components Schrodinger–Pauli spinor in the form

ψ(r) =∣∣∣∣ψ+(r)ψ−(r)

∣∣∣∣ ,in such a way that

σ3ψ±(r) = ±ψ±(r) .

Then the eigenvalue equation can be written as

1

2m

(px −

e

cBy

)2

+1

2m(p2

y + p2z

)± µB − E

ψ±,E(r) = 0 . (1.4)

Now, taking into account that[H, px

]=

[H, pz

]= 0, we can set

ψ±,E(r) ≡ 12π

expi

h(xpx + zpz)

Y±,E(y) , (1.5)

and in so doing the reduced eigenfunctions Y±,E(y) fulfill the differential equation

h2

2mY ′′±,E(y) +

E ∓ µB − p2

z

2m− 1

2mω2

c (y − y0)2

Y±,E(y) = 0 , (1.6)

in which we have set

ωc ≡eB

mc, y0 ≡

px

mωc, (1.7)

ωc being the classical cyclotron angular frequency. It can be immediately realized that

eq. (1.6) is nothing but the eigenvalue equation for a one dimensional harmonic oscillator in

286

the Oy direction, with equilibrium position given by y0 and energy ε± = E∓µB−(p2z/2m).

It follows therefrom that the eigenvalues of the original problem are provided by

E±,n,pz= hωc

(n+

12

)+

p2z

2m± µB ≥ 0 , n+ 1 ∈ N , pz ∈ R , (1.8)

which represent the celebrated Landau’s spectrum. It turns out that the spectrum is con-

tinuous and, within the present asymmetric gauge choice, the degeneracy of each eigenvalue

is labelled by px ∈ R, i.e. a continuous infinity1. The corresponding eigenfunctions turn

out to be

ψn,pz,px(x, y, z) =1

2πhexp

i

h(xpx + zpz)

Yn

[√mωc

h

(y − px

mωc

) ], (1.9a)

Yn

[√mωc

h

(y − px

mωc

) ]=

(mωc

πh

)1/4 1√n!2n

×

exp

−mωc

2h

(y − px

mωc

)2Hn

[√mωc

h

(y − px

mωc

) ], (1.9b)

in which Hn denotes the n-th Hermite’s polynomial. The above improper degenerate

eigenfunctions, which do realize a complete orthonormal set, are normalized according to∫ +∞

−∞dx

∫ +∞

−∞dy

∫ +∞

−∞dz ψn,pz,px

(x, y, z)ψ∗m,qz,qx(x, y, z)

= δn,m δ (pz − qz) δ (px − qx) .

(1.10)

Although the degeneracy of each eigenvalue of the continuous Landau’s spectrum

results to be infinite, it turns out that the number of degenerate states per unit area is

finite. As a matter of fact, let us suppose to consider a very large rectangular box, centered

at the origin in the Oxy plane, of sides 2Lx and 2Ly respectively. The plane-wave part of

1 Notice that, after solving the problem in the symmetric gauge Ax = − 12By, Ay =

12Bx, Az = 0, the degeneracy turns out to be infinite although numerable: consequently,

although the spectrum is gauge invariant, the eigenvalues degeneracy appears to be gauge

dependent.

287

the eigenfunction (1.9) along the Ox direction can be seen as the continuous limit, when Lx

becomes very large, of the eigenfunctions in the presence of periodic boundary conditions

at ±Lx: namely,

ψ±,n,pz(x = −Lx, y, z; px) = ψ±,n,pz

(x = Lx, y, z; px) , (1.11)

that means

px(N) =πNh

Lx, N ∈ Z . (1.12)

It follows therefrom that, if we require

|y0| =|px|mωc

=πh|N |mωcLx

≤ Ly , (1.13)

we eventually find

|N |h2mωc

≤ LxLy ⇔ |N | ≤ mωc

2h4LxLy . (1.14)

The above equation means that we can understand the quantity

∆L ≡mωc

h=eB

hc, (1.15)

as the number of degenerate states per unit area, as anticipated: this is the so called

Landau’s degeneracy. It is worthwhile to remark that the quantity φ0 ≡ hc/e ' 4.136 ×

10−7 G cm2 is the unit of quantum flux. Consequently, for a standard laboratory magnetic

field of one Tesla, we have approximately 2.5×1010 energy eigenstates per cm2 in each

Landau’s band.

To sum up, the number of the degenerate eigenstates within an infinitesimal interval

of the continuous spectrum between pz and pz + dpz is

∆Γ =2Lzdpz

h

eB

hc4LxLy = V dpz

eB

h2c, (1.16)

288

if we consider a symmetric parallelepiped of volume V = 8LxLyLz.

We are ready now to compute the one-electron partition function density: actually we

easily find

Z(β;B) =∫ +∞

−∞dpz

eB

h2c

∑±

∞∑n=0

exp−β

[hωc

(n+

12

)+

p2z

2m± µB

]

=(

2πmh2β

)3/2

2βµB cothβµB ,(β ≡ 1

kT

) (1.17)

from which all the magnetic properties of an ideal electron gas will be extracted as we shall

see below.

289

VIII.2. The density of the states.

According to § VII.1., from eq. (1.17) we can now compute the so called distribution

of the states τ(εF ;B), i.e., the number of the one-electron energy eigenstates per unit

volume up to the Fermi energy εF in the presence of the uniform magnetic field B. To this

purpose, we have to evaluate the following inverse Laplace transform: namely,

τ(εF ;B) =(

2πmh2

)3/2µB

2πi

∫ γ+i∞

γ−i∞ds 2 exp sεF s−3/2 cothµBs . (2.1)

The integrand can be rewritten as

I(s; εF , B) ≡ 4 exp s (εF − µB) coshµBss√s (1− exp−2µBs)

;

it has a branch point at s = 0 and simple poles at µBs = ±rπi, r ∈ N and, consequently,

we have to choose γ > 0.

Let us now consider the contour (ABCDEFA) ≡ C in the complex s-plane (see Fig.

VIII.1), such that it does not pass through any of the poles of the integrand. The residues

at the poles µBs = ±rπi are given by

2 exp± (εF − µB)

πir

µB

cosh±rπi(µB)1/2(±rπi)−3/2 =

∓ 2i√µB

(1πr

)3/2

exp±i

(rπεFµB

− π

4

)and after summation over ± we obtain

4õB

(1πr

)3/2

sinπr

εFµB

− π

4

,

whence we can write(2πmh2

)3/2

µB

∮C

ds

2πi2 exp sεF s−3/2 cothµBs

=32π

(µB

εF

)3/2 83π

(2mεFh2

)3/2 ∞∑r=1

r−3/2 sinπr

εFµB

− π

4

.

(2.2)

290

Now, the integrals over the arcs BC and AF do vanish as the radius of the arcs goes

to infinity: for instance, if s ∈ BC ⇒ s = Reiθ, π/2 < θ ≤ π, then we get∫BC

ds I(s; εF , B) = 2∫ π

π/2

dθ

2πexp − iθ/2√

R

× exp RεF (cos θ + i sin θ) cothµBReiθ

,

which goes to zero as R→∞ owing to cos θ ≤ 0.

Let us then consider the contour CDEF ≡ σ in the limit when the radius r of the

small circle goes to zero; from the expansion

cothxx√x

=1

x2√x

+1

3√x− x

√x

45+ . . . ,

it follows that∫σ

ds

2πiexpsεF (µBs)−3/2 cothµBs =∫

σ

ds

2πiexpsεF

(µBs)−5/2 +

13

(µBs)−1/2

−∫

σ

ds

2πiexpsεF

(µBs)−5/2 +

13

(µBs)−1/2 − (µBs)−3/2 cothµBs.

(2.3)

The first two terms in the RHS of the above equation just correspond to the Hankel’s

representation of the inverse of the Euler’s gamma function: namely,

1Γ(z)

= −∫

σ

ds

2πiess−z , (2.4)

whilst the last contour integral in the RHS of eq. (2.3) becomes a real integral as the small

circle contribution goes to zero when r → 0, the argument of s being (−π) on EF and

(+π) on CD. Collecting all together we eventually find

τ(εF ;B) =83π

(2mεFh2

)3/2

×

×

1 +

14x2

F +32π

x3/2F

∞∑r=1

r−3/2 sin(πr

xF− π

4

)− 3

4x

3/2F

∫ ∞

0

dy√πy

exp− y

xF

(13

+1y2− coth y

y

),

(2.5)

291

where we have set xF ≡ (µB/εF ).

The density of the states, i.e., the number of the one-electron energy eigenstates per

unit volume with energies between E and E + dE, can be obtained in this case from the

very definition and a direct calculation – quite similar to that one outlined above – yields

%(E;B) =(

2πmh2

)3/2µB

πi

∫ γ+i∞

γ−i∞

ds√s

exp sE cothµBs

= 4π√E

(2mh2

)3/2

1 +∞∑

r=1

õB

r2Ecos

(πrE

µB− π

4

)

+

õB

πE

∫ ∞

0

dy√y

exp− yEµB

(1y− coth y

).

(2.6)

It should be noticed that, even if we set εF ≡ (p2F /2m), what is still consistent with the fact

that the Landau’s spectrum is continuous, the Fermi surface in the present case turns out

to be geometrically represented by a quite complicated manifold in the three-dimensional

pF space.

From the expression (2.5) of the distribution of the states, i.e., the number of

one-electron energy eigenstates per unit volume up to the energy E, from the basic

eq. (VII.1.22a) and after setting z ≡ expβεF = expTF /T, x ≡ (µB/kT ), xF ≡

(µB/kTF ), we can write

βλ3

T

2VΩ(z, T ;B) = f5/2(z) +

13x2f1/2(z)

+ 2∞∑

r=1

( x

πr

)3/2∫ ∞

0

dtsin πr t/x− π/4

1 + z−1et

+x3/2

π

∫ ∞

0

dy

(y−3/2 coth y − y−5/2 − 1

3y−1/2

)×

∫ ∞

0

dtexp − t y/x

1 + z−1et.

(2.7)

292

Now, the third term in the RHS of the above equation can be rewritten in the form

2∞∑

r=1

( x

πr

)3/2

=m

exp(iπr

xF− iπ

4

) ∫ ∞

−βεF

dtexp (−t+ iπr t/x)

1 + e−t

= −2∞∑

r=1

( x

πr

)3/2

cos(πr

xF− π

4

)csch

(π2r

x

)+O (1/z) ,

(2.8)

where we used the asymptotic result when z 1: namely,∫ ∞

−βεF

dtexp (−t+ iπr t/x)

1 + e−t=∫ +∞

−∞

exp (−t+ iπr t/x)1 + e−t

+O

(1z

)= π cscπ

(1− i

πr

x

)+O (1/z) .

(2.9)

As a consequence, we can eventually cast eq. (2.7) in the simpler form

λ3T

2VlnZ(z, T ;B) = f5/2(z) +

13x2f1/2(z)

− 2π∞∑

r=1

( x

πr

)3/2

cos(πr

xF− π

4

)csch

(π2r

x

)+R

(x,TF

T

)+O (1/z) ,

(2.10)

in which we have set

R(x,TF

T

)≡ x3/2

∫ ∞

0

dy

πexp

−y TF

xT

(y−3/2 coth y − y−5/2 − 1

3y−1/2

)×

∫ ∞

−(TF /T )

dtexp− ty/x

1 + et.

(2.11)

It is possible to show – see Appendix – that the above expression can be neglected in the

high degeneracy case.

To sum up, in the highly degenerate regime z 1 we can write

βλ3

T

2VΩ(z, x, xF )

z1≈ f5/2(z) +

13x2f1/2(z)

− 2π∞∑

r=1

( x

πr

)3/2

cos(πr

xF− π

4

)csch

(π2r

x

),

(2.12)

Starting from the above basic equation (2.12), we shall be able to discuss in details the

different regimes of physical interest, i.e., the high-degeneracy case in the weak magnetic

field limit and for small or large x respectively. As we shall see below, this study will

beautifully unravel the diamagnetic and paramagnetic properties of the metals.

293

VIII.3. Magnetic properties.

We start the discussion of the magnetic properties of an ideal electrons gas from

the low degenerate case. From the general form of the state equation (VII.2.5), we can

immediately write the lowest order contibution as

βΩ(z, β, V ;B)z1≈ zV Z(β;B) =

V

λ3T

2zx cothx ,(x ≡ µB

kT

), (2.13a)

nz1≈ 2

λ3T

zx cothx . (2.13b)

On the other hand, the mean magnetic moment per unit volume, or magnetization, or even

magnetic polarizability, of the electron gas is defined by

M ≡ 1V

(∂Ω∂B

)z,T,V

. (2.14)

Notice that in the C. G. S. gaussian system of units the magnetization has the same

dimension of an electromagnetic field, i.e. G or dine/ues. Now, up to the same lowest

order approximation we find

Mz1≈ µn

tanhx−

(cothx− 1

x

), (2.15)

where the first term in the RHS represents the paramegnetic contribution, whereas the

second one the (negative) diamagnetic amount to the total magnetization.

In the weak field and/or high temperature limit, i.e. x 1, we obtain

Mx1≈ 2

3nµ2B

kT, (2.16)

that means the Curie’s law with a magnetic susceptibility given by

χm ≡(∂M

∂B

)x1≈ 2nµ2

3kT. (2.17)

294

At room temperature and for ordinary Alkali metals the magnetic susceptibility, which

turns out to be dimensionless in the C. G. S. gaussian system of units, turns out to be

of the order 10−5 ÷ 10−4, taking into account that an effective (smaller) electron’s mass

is suitably involved to correctly estimate the diamagnetic contribution due to the orbital

motion.

For strong applied magnetic fields, i.e. x 1, eq. (2.15) yields M ' 0, which means

that the para- and dia-magnetic contributions exactly compensate for the low degenerate

electrons gas in the presence of strong magnetic fields.

Let us now turn to the more interesting case of a highly degenerate electrons gas, that

means z 1 or, equivalently, T TF . In order to carefully treat this regime, we have to

start from the corresponding expression of eq. (2.12): namely,

Ωλ3T

2V kTz1≈ f5/2(z) +

13x2f1/2(z)

− 2π∞∑

r=1

( x

πr

)3/2

cos(πr

xF− π

4

)csch

(π2r

x

),

from which, taking the definition of eq. (2.14) into account, we can easily obtain

Mλ3T

2µ=

23xf1/2(z)− 2

√πx

xF

∞∑r=1

r−1/2 sin(πr

xF− π

4

)csch

(π2r

x

)

− 2π√π

x

∞∑r=1

r−1/2 cos(πr

xF− π

4

)csch

(π2r

x

)coth

(π2r

x

)

− 3√x

π

∞∑r=1

r−3/2 cos(πr

xF− π

4

)csch

(π2r

x

)+ o

(T

TF

).

(2.18)

Now, we notice that in the highly degenerate case T TF and taking into account that

xF 1 because (µ/εF ) ' 1.84 × 10−9 G−1, we can retain the leading terms within the

two regimes:

(i) weak fields x 1, in which only the first term in the RHS of the above equation is

relevant, owing to the presence of csch(π2r/x);

295

(ii) strong fields x > 1, in which only the first two terms of the RHS of the above equation

are the leading ones.

Moreover, from eq. (VII.2.10) we readily get the leading terms in the asymptotic expansion

fs(z)TTF≈ (TF /T )s

Γ(s+ 1)

1 + s(s− 1)

π2

6

(T

TF

)2

+O

(T

TF

)4

. (2.19)

As a consequence, according to the previous remarks, we can rewrite eq. (2.18) keeping

only the leading terms into account: namely

MTTF≈ B

16π3h3

m3/2µ2√

2εF

×

1− 3

2πkT

µB

√εFµB

∞∑r=1

r−1/2 sin(πr

xF− π

4

)csch

(π2r

x

),

(2.20)

and taking one more derivative with respect to the magnetic field strength we finally come

to the magnetic susceptibility, whose leading terms for xF 1 and x > 1 read

χm ≡(∂M/∂B

) TTF≈ 16π3h3

m3/2µ2√

2εF

×

1 +

3π2

2xx−3/2F

∞∑r=1

√r cos

(πr

xF− π

4

)csch

(π2r

x

).

(2.21)

In order to compare the above expression for the susceptibility with the experimental

data on metals, we have to notice that the effect of the weak binding of the electrons to

the crystal lattice of the metal can be represented, as is done in many other branches of

the theory of metals, by the introduction of an effective mass m∗ for the electron, i.e.,

typically m∗ = 0.98me . In so doing, the Bohr’s magneton due to the orbital motion –

which is responsible for the diamagnetism – has to be replaced by µ∗ = (eh/2m∗c), whilst

the spin magnetic moment – which is responsible for the paramagnetism – however, is still

µ whatever the effective mass of the electron may be, so that the one–electron partition

function density of eq. (1.17) is replaced by

Z(β;B) =(

2πm∗h2β

)3/2 2βµ∗Bsinhβµ∗B

coshβµB . (2.22)

296

Furthermore, to the lowest order in the high degeneracy case, from eq. (2.5) we can write

n ' 83π

(2m∗εFh2

)3/2

, (2.23)

in such a way that it is a simple exercise to show that the magnetic susceptibility, up to

the leading terms, can be cast in the form

χmTTF≈ χ0 + χosc =

12nµ2∗εF

3m2∗

m2− 1 + 3π2 kT

µ∗B

(εFµ∗B

)3/2

×∞∑

r=1

(−1)r√r cos

(rπm∗m

)cos

(πrεFµ∗B

− π

4

)csch

(rπ2 kT

µ∗B

).

(2.24)

(i) x 1: in this case it is the first line in the RHS to be the dominant one and to

give rise to the steady susceptibility of the metals, viz.

χ0 =nµ2

∗2εF

(3m2∗

m2− 1

), (2.25)

which is temperature independent and of the order 10−7, in a reasonable accordance

with experimental data keeping in mind the crudeness of the present approximation.

For instance we have, after multiplication by 107, the experimental (theoretical) values

of the susceptibilities for the following metals in the IA group of the Mendeleev’s

periodic table: 5.8 (4.38) for Sodium (Na), 5.1 (3.40) for Potassium (K), 0.6 (3.26)

for Rubidium (Rb), –0.5 (3.02) for Cesium (Cs). The disagreements, especially for

Rb and Cs – in which also the sign is opposite – are basically due to the fact that

the assumption that the valence electrons are free particles is too a rough one and

corrections are mandatory. Notice that the term (3m2∗/m

2) is the Pauli’s steady

paramagnetism - the spin contribution - whereas the (−1) is the Landau’s steady

diamagnetism. It is also worthwhile to stress that the Curie’s susceptibility - see

eq. (2.16) - at room temperature is such that χCurie(T = 300 K) ' 1.33×102 χ0, what

297

endorses that at room temperature the valence electrons in metals can be thought,

in a first approximation, to behave like a highly degenerate ideal Fermi-Dirac gas of

quasi–free particles.

(ii) x > 1: in this case, a further oscillatory contribution to the magnetism arises, which

is known as the de Haas - van Alphen effect: namely,

χosc '3nµ∗2B3/2

√εFµ∗

cos(π εFµ∗B

− π

4

).

This effect, which appears at sufficiently low temperatures as T/B < µ∗/π2k ' 6.8×

10−6 K G−1, is quite important since it allows an experimental determination of the

Fermi energy: for ordinary metals the latter turns out to be of the order εF ' 3.14

eV= 5.03×10−12 erg, which corresponds to a Fermi temperature TF ' 3.64×104 K.

Bibliography

1. L.D. Landau, E.M. Lifchitz (1967): Physique Statistique, MIR, Mosca.

2. R.K. Pathria 1972): Statistical Mechanics, Pergamon Press, Oxford.

3. K. Huang (1987): Statistical Mechanics, Wyley, New York.

298

Appendix D: evaluation of integral representations.

Here we want to evaluate the double integral

R (x, z) ≡ x3/2

π

∫ ∞

0

dy

∫ ∞

0

dte−ty/x

1 + z−1 ety−3/2

(coth y − 1

y− y

3

). (A1)

To this aim, let us first change the integration variable y = wx/t so that

R (x, z) ≡ x

π

∫ ∞

0

dt

∫ ∞

0

dwt1/2w−3/2e−w

1 + z−1 et

coth

(wxt

)− t

wx− wx

3t

(A2)

and then consider the following change of variables in the double integral: namely,

ξ ≡ wx

tπ, η ≡ t+ w , 0 ≤ ξ ≤ 1 , η ≥ 0 , (A3)

the inversion of which reads

w =xη

x+ πξ, t =

πξη

x+ πξ, (A4)

whereas the jacobian turns out to be∣∣∣∣∂(w, t)∂(ξ, η)

∣∣∣∣ =πxη

(x+ πξ)2. (A5)

Then we have

R (x, z) =x5/2

π√π

∫ ∞

0

dη

∫ 1

0

dξξ−3/2

x+ πξexp

− πξη

x+ πξ

×

(cothπξ − 1

πξ− πξ

3

) (1 + z−1 exp

xη

x+ πξ

)−1

.

(A6)

Now we can expand the hyperbolic cotangent according to [ I. S. Gradshteyn and I. M.

Ryzhik: Table of Integrals, Series, and Products, Fifth Edition, Academic Press, San Diego

(1994); Eq. 1.4118. pg. 42 ] and integrating by series we finally obtain

R (x, z) =(xπ

)5/2 ∞∑k=2

(2π)2kB2k

(2k)!

∫ 1

0

dξξ2k−5/2

x+ πξ

∫ ∞

0

dηz e−η

1 + z exp −xη/(x+ πξ),

(A7)

299

where B2k denote the Bernoulli’s numbers. Here we write the first values:

B1 =16, B2 =

130

, B3 =142

, B4 =130

. (A8)

Let us first consider the case of weak magnetic fields, i.e. x 1: then the leading term

clearly reads

R (x, z) x1∼ x5/2 z

1 + zπ−7/2

∞∑k=2

(2π)2kB2k

(2k)!(2k − 5/2). (A9)

It follows therefrom that in the highly degenerate regime z →∞ and for weak fields we

definitely find

R (x, z) = O(x5/2) , ln z 1 , x 1 , (A10)

which is manifestly negligible with respect to the second addendum in the right hand side

of eq. (2.10).

On the other hand, in the case of strong magnetic fields x 1 it is convenient to

set η = t(1+πξ/x) so that the double integral of eq. (A7) can be rewritten in the form

R (x, z) =x3/2

π5/2

∞∑k=2

(2π)2kB2k

(2k)!

∫ 1

0

dξ ξ2k−5/2

∫ ∞

0

dtz e−t

1 + z e−texp −tπξ/x . (A11)

Now, in the limit of strong magnetic fields the dominant contribution reads

R (x, z) x1∼ x3/2∞∑

k=2

(2π)2kB2k

(2k)!(2k − 3/2)π−5/2 ln(1 + z) (A12)

and turns out to be manifestly negligible with respect to the second addendum of the right

hand side of eq. (2.10), as well as with respect to the third addendum of the right hand side

of eq. (2.10), the behaviour of which is O(x5/2) for large magnetic fields. In conclusion,

the leading behaviour of eq. (2.12) does indeed hold true.

300

As a final remark, it is interesting to notice that the state distribution per unit volume

can be expressed in terms of the Riemann’s zeta function. As a matter of fact, according

to eq.s (2.1) and (2.2), the states distribution can be written as follows

τ(E;B) =(

2πmh2

)3/2µB

2πi

∫ γ+i∞

γ−i∞ds 2 exp sE s−3/2 cothµBs

= −(

2πmh2

)3/2

µB

∫σ

ds

2πi2 exp sE s−3/2 cothµBs

+ 4(

2mµBh2

)3/2 ∞∑r=1

r−3/2 sinπr

E

µB− π

4

,

(A13)

where E > 0 and for a given ε, 0 < ε < 1, the oriented contour σ consists of the half-line

<es ≤ −ε, arg(s) = π covered from the left to the right, of the circle s = εeiθ; −π ≤

θ ≤ π covered clockwise, and of the half-line <es ≤ −ε, arg(s) = −π covered from the

right to the left. The above integral can be expressed in terms of the Riemann’s ζ-function

[ I. S. Gradshteyn and I. M. Ryzhik: Table of Integrals, Series, and Products, Fifth Edition,

Academic Press, San Diego (1994); Eq. 3.5513. pg. 403 ]. To this aim, let us change

the integration variable 2s = −w to rewrite the integral in the second line of eq. (A1):

namely,

τ(q;B) =(

4πmµBh2

)3/2 ∫ (0+)

∞

dw

2πi(−w)−3/2 exp −wq+ exp −w(1 + q)

1− exp −w

+ 4(

2mµBh2

)3/2 ∞∑r=1

r−3/2 sin

2πrq − π

4

= − 4π

√2

(2mµBh2

)3/2 ζ

(−1

2; q

)+ ζ

(−1

2; q + 1

)+ 4

(2mµBh2

)3/2 ∞∑r=1

r−3/2 sin

2πrq − π

4

= 8π

√2

(2mµBh2

)3/2 √q

2− ζ

(−1

2; q

)+ 4

(2mµBh2

)3/2 ∞∑r=1

r−3/2 sin

2πrq − π

4

,

(A14)

301

where q ≡ E/2µB whereas the path integration now consists of the half-line <ew ≥

ε, arg(w) = π covered from the right to the left, of the circle w = εeiθ; −π ≤ θ ≤ π

covered counterclockwise, and of the half-line <ew ≥ ε, arg(w) = −π covered from the

left to the right. To sum up we eventually have

τ(E;B) =4π2

(mµB

h2

)3/2√

E

8µB− ζ

(−1

2;E

2µB

)+

∞∑r=1

sin [ (πrE/µB)− (π/4) ]π(2r)3/2

.

Ferro .... Tf = 1808 K ......... Tc = 1043 K

Nichel .... Tf = 1726 K ......... Tc = 631 K

Cobalto ..... Tf = 1768 K .......... Tc = 1295 K

Tf temperatura di fusione Tc temperatura di Curie

Magnetic susceptibilities at room temperature in units of 10−7

Bismuth .... –1700

Copper ......... –100

Silver ............ –190

Alluminium .... 220

Platinum ..... 3600

302

PROBLEMS

Problema VIII.1. Un sistema meccanico e costituito da N elettroni identici puntiformi

di massa m e carica −e , immersi in un campo magnetico omogeneo e indipendente dal

tempo B = ∇ × A(r) . Se indichiamo con (ri, pi) , i = 1, 2, . . . , N , le coordinate e

l’operatore momento canonico del i-esimo elettrone, l’operatore di Hamilton del sistema si

scrive

H =N∑

i=1

1

2m

[pi +

e

cA(ri)

]2

+gµ

hSi ·B

+ UCoulomb

dove µ = eh/2mc = 0.927 × 10−20 erg/Gauss e il magnetone di Bohr, g = 2.0023 e il

fattore di Lande dell’elettrone, Si rappresenta l’operatore di spin dell’i-esimo elettrone,

mentre UCoulomb rappresenta l’energia coulombiana tra gli elettroni stessi e gli elettroni e

i nuclei. Il sistema e posto in contatto termico con un termostato a temperatura assoluta

T . Si calcoli la suscettivita diamagnetica del sistema.

Soluzione

Scegliamo la gauge simmetrica

A(ri) =12

B× ri , i = 1, 2, . . . , N

in modo tale che

[ pj , A(rk) ] = −ih δjk∇ ·A(rj) ≡ 0

Tenendo conto di questa circostanza l’hamiltoniano si puo mettere nella forma

H = H0 +e

mc

N∑i=1

A(ri) · pi +e2

8mc2

N∑i=1

(B× ri)2 +

gµ

hS ·B

= H0 +µ

h

(L + 2S

)·B +

e2

8mc2

N∑i=1

(B× ri)2

303

dove S :=∑N

i=1 Si e l’operatore dello spin totale elettronico, L :=∑N

i=1 ri×pi e l’operatore

del momento angolare totale elettronico, mentre

H0 :=N∑

i=1

p2i

2m+ UCoulomb

denota l’hamiltoniano in assenza del campo magnetico.

Problem VIII.2. Calculate the Green’s function of the Schrodinger-Pauli hamiltonian

operator (1.3) in the asymmetric Landau’s gauge Ax = −By , Ay = Az = 0 , B > 0 .

Solution. By definition the Green’s function of the hamiltonian self-adjoint operator is the

integral kernel of the resolvent operator R(E) ≡ (E − H)−1 when the complex energy

variable is in the upper or lower half planes. Thus we have

G+(E; r, r′) ≡ 〈 r|(E − H

)−1

|r′ 〉 , =m(E) > 0 .

From eq.s (1.8) and (1.9) we obtain the spectral resolution [ ωc = eB/mc ]

G+(E; r, r′) =∞∑

n=0

∫ +∞

−∞dpz

(E − nhωc −

hωc

2− p2

z

2m∓ µB

)−1

×∫ +∞

−∞dpx ψn,pz,px

(x, y, z)ψ∗n,pz,px(x′, y′, z′)

=(mωc

πh

)1/2 (4π2h2n! 2n

)−1∞∑

n=0

∫ +∞

−∞dpz

×(E − nhωc −

hωc

2− p2

z

2m∓ µB

)−1

×∫ +∞

−∞dpx exp

i

h[ px(x− x′) + pz(z − z′) ]

× exp

−mωc

2h

(y − px

mωc

)2

− mωc

2h

(y′ − px

mωc

)2

×Hn

[√mωc

h

(y − px

mωc

) ]Hn

[√mωc

h

(y′ − px

mωc

) ].

304

Let us first consider an integral representation for the Green’s function which is valid for

<e(E) < 0: from the identity

(E − E±,n,pz)−1 = −

∫ ∞

0

dt exptE − thωc

2− tnhωc −

tp2z

2m∓ tµB

after a suitable rearrangement of the factorized integrals and series we easily obtain

G(E; r, r′) = −√mωc

πhexp

−mωc

2h(y2 + y′2

)×

∫ ∞

0

dt expt

(E − hωc

2∓ µB

)×

∫ +∞

−∞

dpz

2πhexp

− tp

2z

2m+ipz

h(z − z′)

×

∫ +∞

−∞

dpx

2πhexp

− p2

x

mhωc− ipx

h[ (x− x′) + i(y + y′) ]

×

∞∑n=0

1n!

(exp−thωc

2

)n

×Hn

[√mωc

h

(y − px

mωc

) ]Hn

[√mωc

h

(y′ − px

mωc

) ].

The gaussian integral over pz can be easily performed, whilst the series can also be summed

thanks to the formula [ |u| < 1/2 ]

∞∑n=0

un

n!Hn(ξ)Hn(ξ′) = (1− 4u2)−1/2 exp

[4uξξ′ − 4u2(ξ2 + ξ′ 2)

1− 4u2

]and the result is

G(E; r, r′) = −√mωc

πhexp

−mωc

2h(y2 + y′2

)×

∫ ∞

0

dt t−1/2 expt

(E − hωc

2∓ µB

)×

√m

2πh2 exp− m

2h2t(z − z′)2

×

∫ +∞

−∞

dpx

2πhexp

− p2

x

mhωc− ipx

h[ (x− x′) + i(y + y′) ]

× expthωc/2√

2 sinh(thωc)exp

mωc

h sinh(thωc)

(yy′ − px

y + y′

mωc+

p2x

m2ω2c

)× exp

−mωc

2h[ coth(thωc)− 1 ]

(y2 + y′2 − 2px

y + y′

mωc+

2p2x

m2ω2c

).

305

The gaussian integral over px gives∫ +∞

−∞

dpx

2πhexp

− p2

x

mhωc+ipx

h[ (x− x′) + i(y + y′) ]

× exp

mωc

h sinh(thωc)

(pxy + y′

mωc+

p2x

m2ω2c

)× exp

−mωc

2h[ coth(thωc)− 1 ]

(2px

y + y′

mωc+

2p2x

m2ω2c

)=

√mωc

4πhcoth

(thωc

2

)× exp

−mωc

4h

[(x− x′)2 coth

(thωc

2

)− (y + y′)2 tanh

(thωc

2

)+ 2i(x− x′)(y + y′)

]so that we finally obtain for <e(E) < 0

G(E; r, r′) = − mωc

4πh2

√m

2πexp −imωc (x− x′)(y + y′)/2h

×∫ ∞

0

dt t−1/2

sinh(thωc/2)exp

Et∓ tµB − m

2h2t(z − z′)2

× exp

−mωc

4h[(x− x′)2 + (y − y′)2

]coth

(thωc

2

).

Notice that the canonical dimensions [G] of the Green’s function in the C.G.S. Gaussian

units system are [G] = erg−1 cm−3 = cm−5 g−1 s2 . It is now convenient to change the

integration variable and to introduce the dimensionless integration variable α ≡ thωc/2 :

G(E; r, r′) = − mωc

4πh2

√m

πhωcexp −imωc (x− x′)(y + y′)/2h

×∫ ∞

0

dα√α sinhα

exp

2αE ∓ µB

hωc− mωc

4hα(z − z′)2

× exp

−mωc

4h[(x− x′)2 + (y − y′)2

]cothα

.

Analytic continuation to the upper half-plane of the complex energy =m(E) > 0 can

be obtained under the change of variable α = iθ that gives

G+(E; r, r′) =mωc

4πh2

√m

πhωcexp

−i mωc

2h(x− x′)(y + y′) +

3πi4

×

∫ ∞

0

dθ√θ sin θ

exp

2iθE ∓ µB

hωc+ i

mωc

4hθ(z − z′)2

× exp

imωc

4h[(x− x′)2 + (y − y′)2

]cot θ

.

306

Problem VIII.3. Calculate the gauge transformation which connects the asymmetric

Landau’s gauge Ax = −By , Ay = Az = 0 to the symmetric gauge A′x = −By/2 , A′y =

Bx/2 , A′z = 0 , with B > 0 .

Solution. In classical Electrodynamics the time independent gauge transformation on the

vector potentials relating two different gauge choices is given by

A′(r) = A(r) +∇f(r) .

Notice that the canonical dimensions of the gauge function f(t, r) are G cm2, i.e. the

gauge function is a magnetic field flux. In Quantum Mechanics the gauge transformations

on the wave functions are defined as follows: under the above gauge transformation on the

vector potentials, the wave function must undergo a local phase transformation

ψ′(t, r) = ψ(t, r) exp−iφ(r) ,

such that [p− e

cA′(r)

]ψ′(t, r) = exp−iφ(r)

[p− e

cA(r)

]ψ(t, r) ,

that means, the wave function must transform covariantly under the local phase transfor-

mation. It is very easy to find the phase function φ(r) in terms of the gauge function f(r).

As a matter of fact we have[p− e

cA′(r)

]ψ′(t, r) =[

p− e

cA(r)− e

c∇f(r)

]ψ(t, r) exp−iφ(r) =

exp−iφ(r)[p− e

cA(r)

]ψ(t, r)− e

c∇f(r)ψ′(t, r)− hψ′(t, r)∇φ(r) ,

so that we must require

φ(r) = − e

hcf(r) + constant .

307

Notice that the phase function is dimensionless, as it does, owing to the presence of the so

called quantum flux hc/e = 4.136× 10−7 G cm2. Now we have evidently

f(x, y) =B

2xy , φ(x, y) = −ef

hc,

so that the gauge transformation of the Schrodinger-Pauli’s spinor reads

ψ′(t, r) = expiπeB

hcxy

ψ(t, r) .

As a consequence, taking eq. (1.9) into account, it follows that the degenerate eigenfunc-

tions of the Landau’s problem in the symmetric gauge turn out to be

ψn,pz,px(x, y, z) =1

2πhexp

i

h(xpx + zpz) + iπ

eB

hcxy

×

(mωc

πh

)1/4 1√n!2n

exp

−mωc

2h

(y +

px

mωc

)2Hn

[√mωc

h

(y +

px

mωc

) ],

in which Hn denotes the n-th Hermite’s polynomial. The above improper degenerate

eigenfunctions are normalized according to

∫ +∞

−∞

∫ +∞

−∞

∫ +∞

−∞dxdydz ψn,pz,px(x, y, z)ψ∗m,qz,qx

(x, y, z) = δnm δ (pz − qz) δ (px − qx) .

As a matter of fact, it is not difficult to show that the stationary Schrodinger-Pauli’s

equation transforms covariantly under the local phase change of the spinor: actually,[p− e

cA′(r)

]2

ψ′(t, r) =[p− e

cA′(r)

]exp−iφ(r)

[p− e

cA(r)

]ψ(t, r) =

exp−iφ(r)[p− e

cA(r)

]2

ψ(t, r) .

It immediately follows therefrom

H ′ψ′(t, r) =[

12m

p− e

cA′(r)

2

− ~µ ·B]ψ′(t, r) = exp−iφ(r) H ψ(t, r)

308

which manifestly implies that the spectrum of the hamiltonian is gauge invariant, whereas

the eigenfunctions are related through an overall multiplicative local phase factor.

Problem VIII.4. Calculate the electric current density carried on by the Landau’s bands.

Solution. In Quantum Mechanics the probability current density vector j(t, r) is defined

to be [ cfr. e.g. E. Merzbacher (1970): Quantum Mechanics, John Wiley & Sons, New

York, eq. (4.5a) p. 37 ]

j(t, r) =h

2mi[ψ∗∇ψ − (∇ψ∗)ψ ] .

Notice that, iff the wave functions have the canonical dimension [ ψ ]=cm−3/2, then the

canonical dimensions of the probability current density are [ j ] = cm−2 s−1 as it does.

However, in the presence of the electromagnetic field, in order to keep the property of

the gauge invariance of the probability current density vector, it must be generalized as

follows: namely,

j(t, r) =h

2mi[ψ∗Dψ − ψ(Dψ)∗ ] ,

where we have introduced the so called covariant derivative

Dx := ∂x −ie

hcAx(x, y, z) , et cetera .

Then we have

D′xψ′(r) =

∂x −

ie

hc[Ax(r) + ∂xf(r) ]

exp

ie

hcf(r)

ψ(r)

= expie

hcf(r)

Dx ψ(r) , et cetera ,

so that the probability current density vector turns out to be manifestly gauge invariant.

In the asymmetric Landau’s gauge, the Landau’s bands are spread out by the continuous

309

infinity of the degenerate eigenfunctions

Ψn,pz,px(x, y, z) = (2πh)−1 exp

i

h(xpx + zpz)

√mhωc ×(mωc

πh

)1/4 1√n!2n

exp

−mωc

2h

(y +

px

mωc

)2Hn

[√mωc

h

(y +

px

mωc

) ],

where px is the degeneracy quantum number, whereas the pairs (n, pz) uniquely identify

a specific Landau’s band. Notice that the above eigenfunctions differ from those ones of

eq. (1.9) by an overall factor√mhωc to guarantee [ Ψ ] = cm−3/2. Then we readily

find

jx(n, pz, px; r) =(px

m+ ωcy

)|Ψn,pz,px

(x, y, z)|2 =ωc

4π2h(px +mωcy)

(mωc

πh

)1/2 2−n

n!

× exp

−mωc

h

(y +

px

mωc

)2H2

n

[√mωc

h

(y +

px

mωc

) ].

For any Landau’s band, it is convenient to label its continuous infinite degeneracy in terms

of the dimensionless real parameter η := px(mhωc)−1/2 . Then the probability current

density along the Ox-direction carried on by each Landau’s band becomes

jbx =∫ +∞

−∞

dpx√mhωc

ωc

4π2h(px +mωcy)

(mωc

πh

)1/2 2−n

n!

× exp

−mωc

h

(y +

px

mωc

)2H2

n

[√mωc

h

(y +

px

mωc

) ]and after setting υ := y

√mωc/h we can write

jbx =mω2

c

4π2h

∫ +∞

−∞dη

η + υ

n!2n√π

exp−(υ + η)2

H2

n(υ + η) = 0 .

Concerning the Oy-direction we find instead

DyΨn,pz,px= ∂yΨn,pz,px

(x, y, z) = −mωc

h

(y +

px

mωc

)Ψn,pz,px

(x, y, z)

+√mωc

h(2πh)−1 exp

i

h(xpx + zpz)

(mωc

πh

)1/4√mhωc

n!2n

× exp

−mωc

2h

(y +

px

mωc

)2

2nHn−1

[√mωc

h

(y +

px

mωc

) ].

310

From the recursion formula of the Hermite’s polynomials

2nHn−1(ξ) = 2ξHn(ξ)−Hn+1(ξ) ,

we get

DyΨn,pz,px(x, y, z) =

mωc

h

(y +

px

mωc

)Ψn,pz,px

(x, y, z)

−√mωc

h(2πh)−1 exp

i

h(xpx + zpz)

(mωc

πh

)1/4√mhωc

n!2n

× exp

−mωc

2h

(y +

px

mωc

)2Hn+1

[√mωc

h

(y +

px

mωc

) ]that means eventually

DyΨn,pz,px(x, y, z) =

mωc

h

(y +

px

mωc

)Ψn,pz,px

(x, y, z)−√

2mωc

h(n+ 1) Ψn+1,pz,px

(x, y, z)

whence we finally obtain

jy(n, pz, px;x, y, z) = 0 .

Along the Oz-direction we have

jz(n, pz, px;x, y, z) =pz

m|Ψn,pz,px

(x, y, z)|2 ,

so that the resulting current intensities are

Ix = Iy = 0 , dIz = −evzeB

hcA dpz

h= −evz

N

ν

dpz

h,

where A is the area of a section in the Oxy−plane, N is the average number of electrons

whereas ν ≡ hcN/eBA is the so called filling fraction of the Landau’s bands.

311