viii. magnetic properties of ideal fermi-dirac systems ... · viii. magnetic properties of ideal...
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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