VIII. MAGNETIC PROPERTIES OF IDEAL FERMI-DIRAC SYSTEMS VIII.1. The Landau’s problem.

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 = −e

2mc g~ s = −e¯ h

2mc ~ σ = −µ~ σ , (1.1)

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

~ σ = (σ _x , σ _y , σ _z ) 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

A _x = −By , A _y = A _z = 0 , B = |B| . (1.2)

Then the Schr¨ odinger–Pauli hamiltonian operator can be written as

H = b 1 2m

n p + b e

c A(r) o 2

− ~ µ · B

= 1 2m

 ˆ p x − e

c By  2

+ 1

2m p ˆ ² _y + ˆ p ² _z
+ µBσ 3 ,

(1.3)

in which ˆ p _x ≡ −i¯ h∂ _x , et cetera. Since we have h ˆ H, σ ₃ i

= 0, we can write the 2–

components Schr¨ odinger–Pauli spinor in the form

ψ(r) =    

ψ + (r) ψ ₋ (r)    

,

in such a way that

σ 3 ψ _± (r) = ±ψ _± (r) .

Then the eigenvalue equation can be written as

 1 2m

 ˆ p x − e

c By  ² + 1

2m p ˆ ² _y + ˆ p ² _z
± µB − E



ψ _±,E (r) = 0 . (1.4)

Now, taking into account that h ˆ H, ˆ p x

i

= h ˆ H, ˆ p z

i

= 0, we can set

ψ ±,E (r) ≡ 1

2π exp  i

¯

h (xp x + zp z )



Y ±,E (y) , (1.5)

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

¯ h ²

2m Y _±,E ⁰⁰ (y) +



E ∓ µB − p ² _z 2m − 1

2 mω _c ² (y − y 0 ) ²



Y ±,E (y) = 0 , (1.6)

in which we have set

ω c ≡ eB

mc , y 0 ≡ p x

mω _c , (1.7)

ω c 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

the Oy direction, with equilibrium position given by y ₀ and energy ε _± = E ∓µB −(p ² _z /2m).

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

E ±,n,p

z

= ¯ hω c

 n + 1

2

 + p ² _z

2m ± µB ≥ 0 , n + 1 ∈ N , p z ∈ 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 p x ∈ R, i.e. a continuous infinity ¹ . The corresponding eigenfunctions turn out to be

ψ n,p

_z

,p

_x

(x, y, z) = 1

2π¯ h exp  i

¯

h (xp x + zp z )

 Y n

 r mω _c

¯ h



y − p x

mω _c

 

, (1.9a) Y n

 r mω _c

¯ h



y − p x

mω _c

 

=  mω c

π¯ h

 ^1/4 1

√

n!2 ⁿ × exp

(

− mω _c 2¯ h



y − p _x mω c

 2 )

H _n  r mω _c

¯ h



y − p _x mω c

 

, (1.9b)

in which H n denotes the n-th Hermite’s polynomial. The above improper degenerate eigenfunctions, which do realize a complete orthonormal set, are normalized according to

Z +∞

−∞

dx Z +∞

−∞

dy Z +∞

−∞

dz ψ n,p

z

,p

x

(x, y, z) ψ ^∗ _m,q

_z

_,q

_x

(x, y, z)

= δ _n,m δ (p _z − q _z ) δ (p _x − q _x ) .

(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 2L x and 2L y respectively. The plane-wave part of

1 Notice that, after solving the problem in the symmetric gauge A x = − ¹ ₂ By, A y =

1

2 Bx, A _z = 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.

the eigenfunction (1.9) along the Ox direction can be seen as the continuous limit, when L _x becomes very large, of the eigenfunctions in the presence of periodic boundary conditions at ±L x : namely,

ψ _±,n,p

_z

(x = −L _x , y, z; p _x ) = ψ _±,n,p

_z

(x = L _x , y, z; p _x ) , (1.11)

that means

p x (N ) = πN ¯ h

L _x , N ∈ Z . (1.12)

It follows therefrom that, if we require

|y 0 | = |p x |

mω _c = π¯ h|N |

mω _c L _x ≤ L y , (1.13)

we eventually find

|N |h 2mω c

≤ L x L y ⇔ |N | ≤ mω c

2h 4L x L y . (1.14)

The above equation means that we can understand the quantity

∆ L ≡ mω c

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 φ ₀ ≡ hc/e ' 4.136 × 10 ⁻⁷ G cm ² is the unit of quantum flux. Consequently, for a standard laboratory magnetic field of one Tesla, we have approximately 2.5×10 ¹⁰ energy eigenstates per cm ² in each Landau’s band.

To sum up, the number of the degenerate eigenstates within an infinitesimal interval of the continuous spectrum between p _z and p _z + dp _z is

∆Γ = 2L _z dp _z h

eB

hc 4L _x L _y = V dp _z eB

h ² c , (1.16)

if we consider a symmetric parallelepiped of volume V = 8L _x L _y L _z .

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

Z(β; B) = Z +∞

−∞

dp _z eB h ² c

X

±

∞

X

n=0

exp



−β



¯ hω _c

 n + 1

2

 + p ² _z

2m ± µB



=  2πm h ² β

 3/2

2βµB coth{βµB} ,



β ≡ 1 kT



(1.17)

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

see below.

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 ( _F ; B), i.e., the number of the one-electron energy eigenstates per unit volume up to the Fermi energy  F in the presence of the uniform magnetic field B. To this purpose, we have to evaluate the following inverse Laplace transform: namely,

τ ( _F ; B) =  2πm h ²

 3/2

µB 2πi

Z γ+i∞

γ−i∞

ds 2 exp {s _F } s ^−3/2 coth{µBs} . (2.1)

The integrand can be rewritten as

I(s;  F , B) ≡ 4 exp {s ( F − µB)} cosh{µBs}

s √

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 (ABCDEF A) ≡ 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



± ( F − µB) πir µB



cosh{±rπi}(µB) ^1/2 (±rπi) ^−3/2 =

∓ 2i p

µB  1 πr

 3/2

exp



±i  rπ _F µB − π

4



and after summation over ± we obtain

4 p

µB  1 πr

 3/2

sin

 πr  F

µB − π 4

 ,

whence we can write

 2πm h ²

 3/2

µB I

C

ds

2πi 2 exp {s F } s ^−3/2 coth{µBs}

= 3 2π

 µB

 _F

 3/2

8

3 π  2m _F h ²

 3/2 ∞

X

r=1

r ^−3/2 sin

 πr  F

µB − π 4

 .

(2.2)

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 = Re ^iθ , π/2 < θ ≤ π, then we get

Z

BC

ds I(s;  _F , B) = 2 Z π

π/2

dθ 2π

exp {− iθ/2}

√ R

× exp {R _F (cos θ + i sin θ)} coth µBRe ^iθ , 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

coth x x √

x = 1

x ² √

x + 1 3 √

x − x √ x

45 + . . . , it follows that

Z

σ

ds

2πi exp{s F }(µBs) ^−3/2 coth{µBs} = Z

σ

ds

2πi exp{s F }



(µBs) ^−5/2 + 1

3 (µBs) ^−1/2



− Z

σ

ds

2πi exp{s _F }



(µBs) ^−5/2 + 1

3 (µBs) ^−1/2 − (µBs) ^−3/2 coth{µBs}

 .

(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,

1

Γ(z) = − Z

σ

ds

2πi e ^s s ^−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 ( F ; B) = 8

3 π  2m _F h ²

 3/2

×

× (

1 + 1

4 x ² _F + 3 2π x ^3/2 _F

∞

X

r=1

r ^−3/2 sin  πr x F

− π 4



− 3 4 x ^3/2 _F

Z ∞ 0

√ dy

πy exp



− y x _F

  1 3 + 1

y ² − coth y y



,

(2.5)

where we have set x _F ≡ (µB/ _F ).

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) =  2πm h ²

 3/2

µB πi

Z γ+i∞

γ−i∞

√ ds

s exp {sE} coth{µBs}

= 4π √

E  2m h ²

 3/2 ( 1 +

∞

X

r=1

r µB

r ² E cos  πrE µB − π

4



+ r µB

πE Z ∞

0

√ dy y exp



− yE µB

  1

y − coth y

 ) .

(2.6)

It should be noticed that, even if we set  _F ≡ (p ² _F /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 p F 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 ≡ exp{β _F } = exp{T _F /T }, x ≡ (µB/kT ), x _F ≡ (µB/kT F ), we can write

β λ ³ _T

2V Ω(z, T ; B) = f _5/2 (z) + 1

3 x ² f _1/2 (z) + 2

∞

X

r=1

 x πr

 3/2 Z ∞ 0

dt sin {πr t/x − π/4}

1 + z ⁻¹ e ^t + x ^3/2

π Z ∞

0

dy



y ^−3/2 coth y − y ^−5/2 − 1 3 y ^−1/2



× Z ∞

0

dt exp {− t y/x}

1 + z ⁻¹ e ^t .

(2.7)

Now, the third term in the RHS of the above equation can be rewritten in the form 2

∞

X

r=1

 x πr

 3/2

=m



exp  iπr x F

− iπ 4

 Z ∞

−β

F

dt exp (−t + iπr t/x) 1 + e ^−t



= −2

∞

X

r=1

 x πr

 ^3/2

cos  πr x _F − π

4



csch  π ² r x



+ O (1/z) ,

(2.8)

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

−β

F

dt exp (−t + iπr t/x) 1 + e ^−t = Z +∞

−∞

exp (−t + iπr t/x)

1 + e ^−t + O  1 z



= π csc π 

1 − i πr x



+ O (1/z) .

(2.9)

As a consequence, we can eventually cast eq. (2.7) in the simpler form λ ³ _T

2V ln Z(z, T ; B) = f _5/2 (z) + 1

3 x ² f _1/2 (z)

− 2π

∞

X

r=1

 x πr

 3/2

cos  πr x _F − π

4



csch  π ² r x



+ R

 x, T F

T



+ O (1/z) ,

(2.10)

in which we have set R

 x, T F

T



≡ x ^3/2 Z ∞

0

dy π exp



− y T F

xT

 

y ^−3/2 coth y − y ^−5/2 − 1 3 y ^−1/2



× Z ∞

−(T

F

/T )

dt exp{− ty/x}

1 + e ^t .

(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 β λ ³ _T

2V Ω(z, x, x F ) ^z1 ≈ f _5/2 (z) + 1

3 x ² f _1/2 (z)

− 2π

∞

X

r=1

 x πr

 3/2

cos  πr x _F − π

4



csch  π ² r 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.

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

λ ³ _T 2zx coth x ,



x ≡ µB kT



, (2.13a) n ^z1 ≈ 2

λ ³ _T zx coth x . (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 ≡ 1 V

 ∂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

M ^z1 ≈ µ n



tanh x −



coth x − 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

M ^x1 ≈ 2

3 n µ ² B

kT , (2.16)

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

χ _m ≡  ∂M

∂B

 x1

≈ 2nµ ²

3kT . (2.17)

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  T _F . In order to carefully treat this regime, we have to start from the corresponding expression of eq. (2.12): namely,

Ω λ ³ _T 2V kT

z1 ≈ f _5/2 (z) + 1

3 x ² f _1/2 (z)

− 2π

∞

X

r=1

 x πr

 3/2

cos  πr x _F − π

4



csch  π ² r x

 ,

from which, taking the definition of eq. (2.14) into account, we can easily obtain M λ ³ _T

2µ = 2

3 xf _1/2 (z) − 2

√ πx x _F

∞

X

r=1

r ^−1/2 sin  πr x _F − π

4



csch  π ² r x



− 2π r π x

∞

X

r=1

r ^−1/2 cos  πr x F

− π 4



csch  π ² r x



coth  π ² r x



− 3 r x π

∞

X

r=1

r ^−3/2 cos  πr x _F − π

4



csch  π ² r x



+ o  T T _F

 .

(2.18)

Now, we notice that in the highly degenerate case T  T _F and taking into account that x F  1 because (µ/ F ) ' 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(π ² r/x);

(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

f s (z) ^{T T} ≈

^F

(T _F /T ) ^s Γ(s + 1)

(

1 + s(s − 1) π ² 6

 T T F

 2

+ O  T T F

 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

M ^{T T} ≈

^F

B 16π

3h ³ m ^3/2 µ ² √ 2 F

× (

1 − 3 2 π kT

µB r  F

µB

∞

X

r=1

r ^−1/2 sin  πr x F

− π 4



csch  π ² r 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 x _F  1 and x > 1 read

χ _m ≡ ∂M /∂B
^{T T}

^F

≈ 16π

3h ³ m ^3/2 µ ² √ 2 _F

× (

1 + 3π ² 2 x x ^−3/2 _F

∞

X

r=1

√ r cos  πr x _F − π

4



csch  π ² r 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 _∗ = 0.98 m _e . In so doing, the Bohr’s magneton due to the orbital motion – which is responsible for the diamagnetism – has to be replaced by µ ∗ = (e¯ h/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 _∗ h ² β

 3/2

2βµ _∗ B

sinh{βµ _∗ B} cosh{βµB} . (2.22)

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

n ' 8

3 π  2m _∗  _F h ²

 ^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

χ _m ^{T T} ≈

^F

χ ₀ + χ _osc = 1 2 n µ ² _∗

 F

( 3 m ² _∗

m ² − 1 + 3π ² kT µ ∗ B

  _F µ ∗ B

 3/2

×

∞

X

r=1

(−1) ^r √

r cos  rπm _∗ m



cos  πr _F µ ∗ B − π

4

 csch



rπ ² 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.

χ ₀ = nµ ² _∗ 2 _F

 3 m ² _∗

m ² − 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 10 ⁷ , 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 (3m ² _∗ /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 χ _Curie (T = 300 ^◦ K) ' 1.33×10 ² χ ₀ , what

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,

χ osc ' 3nµ ∗

2B ^3/2 r  F

µ _∗ cos  π  _F µ _∗ B − π

4

 .

This effect, which appears at sufficiently low temperatures as T /B < µ _∗ /π ² k ' 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  F ' 3.14 eV= 5.03 × 10 ⁻¹² erg, which corresponds to a Fermi temperature T _F ' 3.64 × 10 ^{4 ◦} 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.

Appendix D: evaluation of integral representations.

Here we want to evaluate the double integral

R (x, z) ≡ x ^3/2 π

Z ∞ 0

dy Z ∞

0

dt e ^−ty/x

1 + z ⁻¹ e ^t y ^−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

Z ∞ 0

dt Z ∞

0

dw t ^1/2 w ^−3/2 e ^−w 1 + z ⁻¹ e ^t



coth  wx t

 − 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) = x ^5/2 π √

π Z ∞

0

dη Z 1

0

dξ ξ ^−3/2 x + πξ exp



− πξη x + πξ



×



coth πξ − 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) =  x π

 ^5/2 X ^∞

k=2

(2π) ^2k B 2k

(2k)!

Z 1 0

dξ ξ ^2k−5/2 x + πξ

Z ∞ 0

dη z e ^−η

1 + z exp {−xη/(x + πξ)} ,

(A7)

where B _2k denote the Bernoulli’s numbers. Here we write the first values:

B 1 = 1

6 , B 2 = 1

30 , B 3 = 1

42 , B 4 = 1

30 . (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 ∼ x ^5/2 z

1 + z π ^−7/2

∞

X

k=2

(2π) ^2k B 2k

(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(x ^5/2 ) , 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) = x ^3/2 π ^5/2

∞

X

k=2

(2π) ^2k B 2k

(2k)!

Z 1 0

dξ ξ ^2k−5/2 Z ∞

0

dt z e ^−t

1 + z e ^−t exp {−tπξ/x} . (A11) Now, in the limit of strong magnetic fields the dominant contribution reads

R (x, z) ^x1 ∼ x ^3/2

∞

X

k=2

(2π) ^2k B 2k

(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(x ^5/2 ) for large magnetic fields. In conclusion,

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

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) =  2πm h ²

 ^3/2 µB 2πi

Z γ+i∞

γ−i∞

ds 2 exp {sE} s ^−3/2 coth{µBs}

= −  2πm h ²

 ^3/2 µB

Z

σ

ds

2πi 2 exp {sE} s ^−3/2 coth{µBs}

+ 4  2mµB h ²

 ^{3/2 ∞} X

r=1

r ^−3/2 sin

 πr 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 = εe ^iθ ; −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) =  4πmµB h ²

 3/2 Z (0+)

∞

dw

2πi (−w) ^−3/2 exp {−wq} + exp {−w(1 + q)}

1 − exp {−w}

+ 4  2mµB h ²

 3/2 ∞

X

r=1

r ^−3/2 sin n

2πrq − π 4

o

= − 4π √

2  2mµB h ²

 3/2  ζ



− 1 2 ; q

 + ζ



− 1 2 ; q + 1



+ 4  2mµB h ²

 3/2 ∞

X

r=1

r ^−3/2 sin n

2πrq − π 4

o

= 8π √

2  2mµB h ²

 3/2

 √q 2 − ζ



− 1 2 ; q



+ 4  2mµB h ²

 3/2 ∞

X

r=1

r ^−3/2 sin n

2πrq − π 4

o ,

(A14)

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 = εe ^iθ ; −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) = 4 π ²

 mµB

¯ h ²

 3/2 (s E 8µB − ζ



− 1 2 ; E

2µB

 +

∞

X

r=1

sin [ (πrE/µB) − (π/4) ] π(2r) ^3/2

) .

Ferro .... T f = 1808 ^◦ K ... T c = 1043 ^◦ K Nichel .... T _f = 1726 ^◦ K ... T _c = 631 ^◦ K Cobalto ... T f = 1768 ^◦ K ... T c = 1295 ^◦ K T _f temperatura di fusione T _c temperatura di Curie

Magnetic susceptibilities at room temperature in units of 10 ⁻⁷

Bismuth .... –1700

Copper ... –100

Silver ... –190

Alluminium .... 220

Platinum ... 3600

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 (r i , p b i ) , i = 1, 2, . . . , N , le coordinate e l’operatore momento canonico del i-esimo elettrone, l’operatore di Hamilton del sistema si scrive

H = b

N

X

i=1

 1 2m

h

p b _i + e

c A(r _i ) i 2

+ gµ

¯

h S b _i · B



+ U _Coulomb

dove µ = e¯ h/2mc = 0.927 × 10 ⁻²⁰ erg/Gauss ´ e il magnetone di Bohr, g = 2.0023 ´ e il fattore di Land´ e dell’elettrone, b S i rappresenta l’operatore di spin dell’i-esimo elettrone, mentre U _Coulomb 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 suscettivit´ a diamagnetica del sistema.

Soluzione

Scegliamo la gauge simmetrica

A(r i ) = 1

2 B × r i , i = 1, 2, . . . , N

in modo tale che

[ p b j , A(r k ) ] = −i¯ h δ jk ∇ · A(r j ) ≡ 0

Tenendo conto di questa circostanza l’hamiltoniano si pu´ o mettere nella forma

H = b b H 0 + e mc

N

X

i=1

A(r i ) · p b i + e ² 8mc ²

N

X

i=1

(B × r i ) ² + gµ

¯ h S · B b

= b H 0 + µ

¯ h



L + 2b b S



· B + e ² 8mc ²

N

X

i=1

(B × r i ) ²

dove S := P N

i=1 b S _i ´ e l’operatore dello spin totale elettronico, L := P N

i=1 r _i ×p _i ´ e l’operatore del momento angolare totale elettronico, mentre

H b 0 :=

N

X

i=1

p b ² _i

2m + U Coulomb

denota l’hamiltoniano in assenza del campo magnetico.

Problem VIII.2. Calculate the Green’s function of the Schr¨ odinger-Pauli hamiltonian operator (1.3) in the asymmetric Landau’s gauge A _x = −By , A _y = A _z = 0 , B > 0 .

Solution. By definition the Green’s function of the hamiltonian self-adjoint operator is the integral kernel of the resolvent operator b R(E) ≡ (E − b H) ⁻¹ when the complex energy variable is in the upper or lower half planes. Thus we have

G ⁺ (E; r, r ⁰ ) ≡ h r| 

E − b H  −1

|r ⁰ i , =m(E) > 0 .

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

G ⁺ (E; r, r ⁰ ) =

∞

X

n=0

Z +∞

−∞

dp _z



E − n¯ hω _c − ¯ hω _c 2 − p ² _z

2m ∓ µB

 ⁻¹

× Z +∞

−∞

dp x ψ n,p

_z

,p

_x

(x, y, z) ψ _n,p ^∗

_z

_,p

_x

(x ⁰ , y ⁰ , z ⁰ )

=

 mω _c π¯ h

 1/2

4π ² ¯ h ² n! 2 ^n
−1

∞

X

n=0

Z +∞

−∞

dp z

×



E − n¯ hω c − ¯ hω c

2 − p ² _z

2m ∓ µB

 −1

× Z +∞

−∞

dp x exp  i

¯

h [ p x (x − x ⁰ ) + p z (z − z ⁰ ) ]



× exp (

− mω c

2¯ h



y − p x

mω c

 ²

− mω c

2¯ h



y ⁰ − p x

mω c

 ² )

× H _n  r mω _c

¯ h



y − p _x mω c

 

H _n  r mω _c

¯ h



y ⁰ − p _x mω c

 

.

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,p

_z

) ⁻¹ = −

Z ∞ 0

dt exp



tE − t¯ hω _c

2 − tn¯ hω _c − tp ² _z

2m ∓ tµB



after a suitable rearrangement of the factorized integrals and series we easily obtain G(E; r, r ⁰ ) = − r mω c

π¯ h exp n

− mω c

2¯ h y ² + y ⁰²
o

× Z ∞

0

dt exp

 t



E − ¯ hω _c 2 ∓ µB



× Z +∞

−∞

dp _z 2π¯ h exp



− tp ² _z

2m + ip _z

¯

h (z − z ⁰ )



× Z +∞

−∞

dp x

2π¯ h exp



− p ² _x

m¯ hω _c − ip x

¯

h [ (x − x ⁰ ) + i(y + y ⁰ ) ]



×

∞

X

n=0

1 n!

 exp{−t¯ hω c } 2

 n

× H _n  r mω _c

¯ h



y − p _x mω c

 

H _n  r mω _c

¯ h



y ⁰ − p _x mω c

  .

The gaussian integral over p z can be easily performed, whilst the series can also be summed thanks to the formula [ |u| < 1/2 ]

∞

X

n=0

u ⁿ

n! H n (ξ) H n (ξ ⁰ ) = (1 − 4u ² ) ^−1/2 exp  4uξξ ⁰ − 4u ² (ξ ² + ξ ^{0 2} ) 1 − 4u ²



and the result is

G(E; r, r ⁰ ) = − r mω _c

π¯ h exp n

− mω _c

2¯ h y ² + y ⁰²
o

× Z ∞

0

dt t ^−1/2 exp

 t



E − ¯ hω c

2 ∓ µB



×

r m

2π¯ h ² exp



− m

2¯ h ² t (z − z ⁰ ) ²



× Z +∞

−∞

dp _x 2π¯ h exp



− p ² _x m¯ hω c

− ip _x

¯

h [ (x − x ⁰ ) + i(y + y ⁰ ) ]



× exp{t¯ hω _c /2}

p2 sinh(t¯hω c ) exp

 mω _c

¯

h sinh(t¯ hω c )



yy ⁰ − p _x y + y ⁰ mω c

+ p ² _x m ² ω _c ²



× exp



− mω _c

2¯ h [ coth(t¯ hω _c ) − 1 ]



y ² + y ⁰² − 2p _x y + y ⁰ mω c

+ 2p ² _x m ² ω ² _c



.

The gaussian integral over p _x gives Z +∞

−∞

dp x

2π¯ h exp



− p ² _x

m¯ hω _c + ip x

¯

h [ (x − x ⁰ ) + i(y + y ⁰ ) ]



× exp

 mω c

¯

h sinh(t¯ hω c )

 p x

y + y ⁰ mω c

+ p ² _x m ² ω _c ²



× exp



− mω _c

2¯ h [ coth(t¯ hω c ) − 1 ]

 2p x

y + y ⁰ mω c

+ 2p ² _x m ² ω ² _c



= s

mω _c

4π¯ h coth  t¯ hω _c 2



× exp



− mω _c 4¯ h



(x − x ⁰ ) ² coth  t¯ hω _c 2



− (y + y ⁰ ) ² tanh  t¯ hω _c 2



+ 2i(x − x ⁰ )(y + y ⁰ )



so that we finally obtain for <e(E) < 0 G(E; r, r ⁰ ) = − mω _c

4π¯ h ² r m

2π exp {−imω _c (x − x ⁰ )(y + y ⁰ )/2¯ h}

× Z ∞

0

dt t ^−1/2

sinh(t¯ hω c /2) exp



Et ∓ tµB − m

2¯ h ² t (z − z ⁰ ) ²



× exp



− mω c

4¯ h  (x − x ⁰ ) ² + (y − y ⁰ ) ²  coth  t¯ hω c

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 s ² . It is now convenient to change the integration variable and to introduce the dimensionless integration variable α ≡ t¯ hω c /2 :

G(E; r, r ⁰ ) = − mω c

4π¯ h ²

r m

π¯ hω _c exp {−imω _c (x − x ⁰ )(y + y ⁰ )/2¯ h}

× Z ∞

0

√ dα

α sinh α exp



2α E ∓ µB

¯ hω c

− mω c

4¯ hα (z − z ⁰ ) ²



× exp n

− mω _c

4¯ h  (x − x ⁰ ) ² + (y − y ⁰ ) ²  coth α o .

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 ⁰ ) = mω c

4π¯ h ²

r m

π¯ hω _c exp



−i mω c

2¯ h (x − x ⁰ )(y + y ⁰ ) + 3πi 4



× Z ∞

0

√ dθ

θ sin θ exp



2iθ E ∓ µB

¯ hω c

+ i mω _c

4¯ hθ (z − z ⁰ ) ²



× exp n i mω _c

4¯ h  (x − x ⁰ ) ² + (y − y ⁰ ) ²  cot θ o

.

Problem VIII.3. Calculate the gauge transformation which connects the asymmetric Landau’s gauge A x = −By , A y = A z = 0 to the symmetric gauge A ⁰ _x = −By/2 , A ⁰ _y = Bx/2 , A ⁰ _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 ⁰ (r) = A(r) + ∇f (r) .

Notice that the canonical dimensions of the gauge function f (t, r) are G cm ² , 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 ⁰ (t, r) = ψ(t, r) exp{−iφ(r)} ,

such that h

b p − e

c A ⁰ (r) i

ψ ⁰ (t, r) = exp{−iφ(r)} h p − b e

c A(r) i

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 h

p − b e

c A ⁰ (r) i

ψ ⁰ (t, r) = h

p − b e

c A(r) − e

c ∇f (r) i

ψ(t, r) exp{−iφ(r)} = exp{−iφ(r)}

h p − b e

c A(r) i

ψ(t, r) − e

c ∇f (r)ψ ⁰ (t, r) − ¯ hψ ⁰ (t, r)∇φ(r) , so that we must require

φ(r) = − e

¯

hc f (r) + constant .

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 cm ² . Now we have evidently

f (x, y) = B

2 xy , φ(x, y) = − ef

¯ hc , so that the gauge transformation of the Schr¨ odinger-Pauli’s spinor reads

ψ ⁰ (t, r) = exp

 iπ eB

hc xy



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

ψ n,p

z

,p

x

(x, y, z) = 1

2π¯ h exp  i

¯

h (xp x + zp z ) + iπ eB hc xy



×

 mω _c π¯ h

 1/4 1

√ n!2 ⁿ exp (

− mω _c 2¯ h



y + p _x mω c

 ² )

H _n  r mω _c

¯ h



y + p _x mω c

  ,

in which H n denotes the n-th Hermite’s polynomial. The above improper degenerate eigenfunctions are normalized according to

Z +∞

−∞

Z +∞

−∞

Z +∞

−∞

dxdydz ψ n,p

z

,p

x

(x, y, z) ψ ^∗ _m,q

_z

_,q

_x

(x, y, z) = δ nm δ (p z − q z ) δ (p x − q x ) .

As a matter of fact, it is not difficult to show that the stationary Schr¨ odinger-Pauli’s equation transforms covariantly under the local phase change of the spinor: actually,

h p − b e

c A ⁰ (r) i 2

ψ ⁰ (t, r) = h

p − b e

c A ⁰ (r) i

exp{−iφ(r)} h p − b e

c A(r) i

ψ(t, r) = exp{−iφ(r)}

h p − b e

c A(r) i 2

ψ(t, r) . It immediately follows therefrom

H b ⁰ ψ ⁰ (t, r) =

 1 2m

n p − b e

c A ⁰ (r) o 2

− ~ µ · B



ψ ⁰ (t, r) = exp{−iφ(r)} b H ψ(t, r)

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

D x := ∂ x − ie

¯

hc A x (x, y, z) , et cetera . Then we have

D ⁰ _x ψ ⁰ (r) =



∂ _x − ie

¯

hc [ A _x (r) + ∂ _x f (r) ]



exp  ie

¯ hc f (r)

 ψ(r)

= exp  ie

¯ hc f (r)



D _x 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

infinity of the degenerate eigenfunctions Ψ n,p

_z

,p

_x

(x, y, z) = (2π¯ h) ⁻¹ exp  i

¯

h (xp x + zp z )



p m¯ hω c ×

 mω _c π¯ h

 1/4 1

√ n!2 ⁿ exp (

− mω _c 2¯ h



y + p _x mω c

 2 ) H n

 r mω _c

¯ h



y + p _x mω c

  ,

where p x is the degeneracy quantum number, whereas the pairs (n, p z ) uniquely identify a specific Landau’s band. Notice that the above eigenfunctions differ from those ones of eq. (1.9) by an overall factor √

m¯ hω c to guarantee [ Ψ ] = cm ^−3/2 . Then we readily find

j x (n, p z , p x ; r) =  p x

m + ω c y 

|Ψ n,p

_z

,p

_x

(x, y, z)| ² = ω c

4π ² ¯ h (p x + mω c y)  mω c

π¯ h

 ^1/2 2 ⁻ⁿ n!

× exp (

− mω c

¯ h



y + p x

mω c

 2 )

H _n ²  r mω _c

¯ h



y + p x

mω c

  .

For any Landau’s band, it is convenient to label its continuous infinite degeneracy in terms of the dimensionless real parameter η := p _x (m¯ hω _c ) ^−1/2 . Then the probability current density along the Ox-direction carried on by each Landau’s band becomes

j _x ^b = Z +∞

−∞

dp _x

√ m¯ hω c

ω _c

4π ² ¯ h (p _x + mω _c y)  mω _c π¯ h

 1/2 2 ⁻ⁿ n!

× exp (

− mω c

¯ h



y + p x

mω _c

 2 )

H _n ²  r mω _c

¯ h



y + p x

mω _c

 

and after setting υ := ypmω c /¯ h we can write

j _x ^b = mω _c ² 4π ² ¯ h

Z +∞

−∞

dη η + υ n!2 ⁿ √

π exp −(υ + η) ² H _n ² (υ + η) = 0 . Concerning the Oy-direction we find instead

D _y Ψ _n,p

_z

_,p

_x

= ∂ _y Ψ _n,p

_z

_,p

_x

(x, y, z) = − mω _c

¯ h



y + p _x mω c



Ψ _n,p

_z

_,p

_x

(x, y, z)

+ r mω _c

¯

h (2π¯ h) ⁻¹ exp  i

¯

h (xp x + zp z )

  mω _c π¯ h

 1/4 r m¯ hω _c

n!2 ⁿ

× exp (

− mω c

2¯ h



y + p x

mω _c

 ² )

2nH _n−1  r mω c

¯ h



y + p x

mω _c

 

.

From the recursion formula of the Hermite’s polynomials

2nH n−1 (ξ) = 2ξH n (ξ) − H n+1 (ξ) ,

we get

D y Ψ n,p

_z

,p

_x

(x, y, z) = mω c

¯ h



y + p x

mω _c



Ψ n,p

_z

,p

_x

(x, y, z)

− r mω c

¯

h (2π¯ h) ⁻¹ exp  i

¯

h (xp _x + zp _z )

  mω c

π¯ h

 1/4 r m¯ hω c

n!2 ⁿ

× exp (

− mω c

2¯ h



y + p x

mω c

 2 ) H n+1

 r mω _c

¯ h



y + p x

mω c

 

that means eventually

D y Ψ n,p

_z

,p

_x

(x, y, z) = mω c

¯ h



y + p x

mω c



Ψ n,p

z

,p

x

(x, y, z) −

r 2mω c

¯

h (n + 1) Ψ n+1,p

z

,p

x

(x, y, z) whence we finally obtain

j y (n, p z , p x ; x, y, z) = 0 . Along the Oz-direction we have

j z (n, p z , p x ; x, y, z) = p z

m |Ψ n,p

_z

,p

_x

(x, y, z)| ² , so that the resulting current intensities are

I x = I y = 0 , dI z = −ev z

eB hc A dp z

h = −ev z

N ν

dp z

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.