**3.1.1 Classical statistical mechanics of ideal gas in magnetic field**

The Hamiltonian of an electron in a magnetic field has the form:

$$k = \left(1/2m\right)\left(\mathbf{p} + e\mathbf{A}\right)^2\tag{21}$$

where *m* and *e* are the mass and the charge of an electron, **p** is momentum, and **A** is vector potential of the magnetic field:

$$\mathbf{A} = \frac{1}{2} [\mathbf{H} \mathbf{r}] = \frac{1}{2} (-yH\_{\prime} \mathbf{x}H\_{\prime} \mathbf{0}) \,. \tag{22}$$

This Hamiltonian does not have the translation symmetry. This symmetry, seemingly, should be, if the magnetic field is uniform at an unlimited plane. But a uniform magnetic field at unlimited plane is impossible because an electrical current that generates it according to Maxwell equation should envelope a part of this plane. It is asserted (Landau, & Lifshitz, E.M. 1980b; Vagner et al., 2006) that the Hamiltonian (21) with the vector potential (22) would be converted by gauge transformation **A A** *f x*, , *y z* . If the function *f eHxy* 2 , then Hamiltonian will be in the Landau form :

$$\mathcal{R}\_{\perp} = \left(\mathbf{1}/2m\right) \left[ \left(p\_x - eHy\right)^2 + p\_y^2\right]\_{\perp} \tag{23}$$

and will have the translation symmetry in the direction of the axis **X** in return for axial symmetry. That is strange assertion because the symmetry is the physical property of the system rather than of a method of it description. In fact the transformation to the Hamiltonian (23) in classical mechanics is result of the canonical transformation of the Hamilton variables with the generating function 2 *x y px p y eHxy* . Then *p p x x* 1 2 , *eHy p p y y* 1 2 , *eHx* , *x xy y* . Therefore the , *<sup>x</sup> <sup>y</sup> p p* (in fact , *<sup>x</sup> <sup>y</sup> p p* ) in the Landau Hamiltonian (23) are not the momentum components in the Cartesian coordinates and the absence of the *x* coordinate in this Hamiltonian does not lead to the momentum *x* component conservation. In quantum mechanics the unitary transformation with operator exp 2 *ieHxy* is equivalent to this canonical transformation. The boundary is created by the line of intersection of the plane with a solenoid that generates the magnetic field. Electrons, orbits of which transverse this boundary, will be extruded from the area, and the gas will evaporate. The more realistic problem is gas in the area with a reflecting boundary.

An isolated electron has three motion integrals. Those are the angular momentum relative to the centre of area and two coordinates of the centre electron orbit:

$$I\_z = \mathbf{x}p\_y - \mathbf{y}p\_{x'} \quad X = -\frac{p\_y}{eH} + \frac{\mathbf{x}}{2} \quad Y = \frac{p\_x}{eH} + \frac{\mathbf{y}}{2} \tag{24}$$

Two motion integrals that have the physical importance would be created from it: energy *E* and squared centre electron orbit distance from the centre of area <sup>2</sup> *R* :

Statistical Mechanics That Takes into Account Angular

function of the total system (gas and background) is:

 

distribution for the gas it is needed to integrate the function

2 ! *<sup>g</sup>*

*zg g <sup>i</sup> l L* and the Eqs. (8). Then it is obtained:

The integration over *<sup>z</sup>*<sup>1</sup> *l* leads to change 1 <sup>2</sup>

mentioned at the Introduction. The magnetic moment

The Hamiltonian H*g g* **R** would be represented in the form:

*N*

 2 2 22 2

1

3 2 1 1 11 <sup>Z</sup> exp <sup>d</sup> 2 ! 2 8 *<sup>g</sup> N*

neutralize it. The model that regards this interaction will be considered below.

**3.1.2 Quantum problem of electron in magnetic field at bounded area** 

space of the background. Then

as expH*gg B* **R** *k T* . Then:

1 *N*

Momentum Conservation Law - Theory and Application 457

momentum should be equal to zero. But then it should be nonuniform as is evident from the foregoing consideration. It should be regarded more comprehensively. Let us go to classical statistic mechanics for a gas of charged particles in magnetic field. The characteristic

**RR R R R R** *g b gg bb gg bb* H +H E

Here the indexes *g* and *b* denote the quantities those relating to the gas and to the background, H is a Hamiltonian, E is the value of the total energy. To provide of Gibbs

H*g g* **R** can be factored from integral by the method Krutkov (Krutkov, 1933; Zubarev,1974)

<sup>3</sup> <sup>1</sup> <sup>Z</sup> exp <sup>d</sup>

*<sup>N</sup> <sup>N</sup> Lg g g g <sup>B</sup> <sup>g</sup> k T*

**R R**

2 1 1 1

 *mr m m* 

*N N N*

*g g i ri zi i i i i*

Let us substitute Hamiltonian (30) to the formula (29) and take into account formula

2 2 22

*N N ri g g B i i l eHr p L <sup>N</sup> kT m r m*

form to the standard appearance the Hamiltonian of ideal gas that is collection of 1 *N* particles in harmonic potential field is obtained. Obviously, that taking into account conservation of the zero value of the angular momentum eliminates the paradoxes that were

is confined by the magnetic field. But that confinement provides to the inconsistency of the model that neglects of the electrostatic interaction because the uniform background cannot

The quasiclassical description of an electron in a magnetic field would not give the correct picture of the probability density distribution and the current density. It would not also describe the alternation of the energy spectrum when a perturbation does the classical

2 82

2

*N*

*zi i*

<sup>H</sup> **<sup>R</sup>** h . (30)

*l e H r eH p l*

*zi i*

(31)

*<sup>z</sup> H kT B N* ln Z 0 . The gas

*<sup>z</sup> zi <sup>i</sup> l l* and after reduction of the quadratic

 

*Lg g* **R** is factored out from integral. The function from

*L L* (28)

H . (29)

**R R** *<sup>g</sup> <sup>b</sup>* over the phase

$$\begin{aligned} \text{R}^2(\mathbf{p}, \mathbf{r}) &= X^2 + Y^2 = \frac{1}{e^2 H^2} \left[ \left( p\_x + \frac{eHy}{2} \right)^2 + \left( p\_y - \frac{eHx}{2} \right)^2 \right]; \\ \text{c}(\mathbf{p}, \mathbf{r}) &= \frac{e^2 H^2}{2m} \left( R^2 + \frac{2}{eH} l\_z \right) = \frac{e^2 H^2}{2m} \rho^2 = k \left( \mathbf{p}, \mathbf{r} \right). \end{aligned} \tag{25}$$

Here is the radius of the electron orbit. The motion integral <sup>2</sup> *R* is proportional to the Hamiltonian with opposite direction of the magnetic field. The values of <sup>2</sup> *R* and should be discrete by the rules of the quasiclassical quantization:

$$R\_k^2 = 2\lambda^2 \left(k + \frac{1}{2}\right), \quad \varepsilon\_n = \hbar o\_\varepsilon \left(n + \frac{1}{2}\right); \quad \lambda = \sqrt{\frac{\hbar}{eH}} = \sqrt{\frac{\hbar}{m o\_\varepsilon}}, \quad o\_\varepsilon = \frac{eH}{m}.\tag{26}$$

Here is magnetic length and *<sup>c</sup>* is cyclotron frequency. Every energy level *En* is degenerated because the integer *k* that determines the position of the orbit can take any values from zero to 1 *k* , where 1 *k* is determined by the condition that electron orbit does not have common points with the boundary. That condition would be expressed by formula when the boundary is a circle with radius R . Then max *Rk n* R or 2 1 21 *k n* <sup>1</sup> R . Those are the "ordinary" states. When 1 2 *k kk* , where 2 1 21 *k n* <sup>2</sup> R , the states are nominated "near-boundary state ". Their energies are not described by the formula (26). The instant magnetic moment is determined by the formula:

$$
\mu\_z = -\frac{e}{2} \left( \mathbf{x} \boldsymbol{\upsilon}\_y - \mathbf{y} \boldsymbol{\upsilon}\_x \right) = -\frac{\partial}{\partial H} \boldsymbol{k} \left( \mathbf{p}\_\prime \mathbf{r} \right) \tag{27}
$$

The first determination is valid for any negative charged particle with and without an external magnetic field. The second equality is valid when the vector potential has the form (22), or when the equality *v p i i* h is transformed by canonical transformation. For the ordinary states the averaged over the orbit magnetic moment is *z n H* . The trajectories of the near-boundary states are composed from arcs and envelop the all area. Their magnetic moment is positive. If in the area exists a potential field *U r* the orbit centers of the ordinary states also move along equipotential lines and the energy values depend on *k* . The degeneration of the energy levels goes off. The angular momentum does not represent the electron motion, but rather the position of the electron orbit. From Eqs. (25- 26) it follows that 2 2 2 *zl eH R n k* . Then *lz* 0 if the orbit envelops the centre, and *lz* 0 if the area centre locates out the orbit. If *<sup>n</sup>* R a large share of the states with energy *<sup>n</sup>* have angular momentum *lz* 0 .

Going to consideration of the ideal gas with electron-elektron collisions, let us suppose that the interaction does by a central force. Then total energy and angular momentum are conserved. It is generally believed that the area is filled by uniform and motionless positive charged background that neutralizes the electrostatic interaction. This assumption is inconsistently. If electron gas is in equilibrium with motionless background, its angular

, ; 2 2

is the radius of the electron orbit. The motion integral <sup>2</sup> *R* is proportional to the

 

*x y*

  2 2

*c*

*eH m m*

(26)

*<sup>c</sup>* is cyclotron frequency. Every energy level *En* is

. Then *lz* 0 if the orbit envelops the

h **p r** (27)

*<sup>n</sup>* R a large share of the states

 *z n H* . The

(25)

should

R or

**p r** h **p r**

 

Hamiltonian with opposite direction of the magnetic field. The values of <sup>2</sup> *R* and

*e H e H R l m eH m*

<sup>2</sup> , , . 2 2

*eHy eHx R XY p p e H*

*z*

2 2

2 2 1 1 2 , ; , . 2 2 *<sup>k</sup> n c <sup>c</sup>*

degenerated because the integer *k* that determines the position of the orbit can take any values from zero to 1 *k* , where 1 *k* is determined by the condition that electron orbit does not have common points with the boundary. That condition would be expressed by

2 1 21 *k n* <sup>1</sup> R . Those are the "ordinary" states. When 1 2 *k kk* , where

 2 1 21 *k n* <sup>2</sup> R , the states are nominated "near-boundary state ". Their energies are not described by the formula (26). The instant magnetic moment is determined by the

, <sup>2</sup> *z yx*

The first determination is valid for any negative charged particle with and without an external magnetic field. The second equality is valid when the vector potential has the form (22), or when the equality *v p i i* h is transformed by canonical transformation. For the

trajectories of the near-boundary states are composed from arcs and envelop the all area. Their magnetic moment is positive. If in the area exists a potential field *U r* the orbit centers of the ordinary states also move along equipotential lines and the energy values depend on *k* . The degeneration of the energy levels goes off. The angular momentum does not represent the electron motion, but rather the position of the electron orbit. From Eqs. (25-

Going to consideration of the ideal gas with electron-elektron collisions, let us suppose that the interaction does by a central force. Then total energy and angular momentum are conserved. It is generally believed that the area is filled by uniform and motionless positive charged background that neutralizes the electrostatic interaction. This assumption is inconsistently. If electron gas is in equilibrium with motionless background, its angular

*xv yv <sup>H</sup>*

formula when the boundary is a circle with radius R . Then max *Rk n*

*eH Rk n*

2 2 2 2 2 2

1

**p r**

is magnetic length and

Here 

Here 

formula:

with energy

2 22

be discrete by the rules of the quasiclassical quantization:

 

*e*

ordinary states the averaged over the orbit magnetic moment is

26) it follows that 2 2 2 *zl eH R n k*

centre, and *lz* 0 if the area centre locates out the orbit. If

*<sup>n</sup>* have angular momentum *lz* 0 .

momentum should be equal to zero. But then it should be nonuniform as is evident from the foregoing consideration. It should be regarded more comprehensively. Let us go to classical statistic mechanics for a gas of charged particles in magnetic field. The characteristic function of the total system (gas and background) is:

$$\operatorname{\mathcal{O}}\left(\mathbf{R}\_{\mathcal{S}} + \mathbf{R}\_{b}\right) = \operatorname{\mathcal{S}}\left(\operatorname{\mathcal{H}}\_{\mathcal{S}}\left(\mathbf{R}\_{\mathcal{S}}\right) + \operatorname{\mathcal{H}}\_{b}\left(\mathbf{R}\_{b}\right) - \operatorname{\mathcal{E}}\right)\operatorname{\mathcal{S}}\left(L\_{\mathcal{S}}\left(\mathbf{R}\_{\mathcal{S}}\right)\right)\operatorname{\mathcal{S}}\left(L\_{b}\left(\mathbf{R}\_{b}\right)\right) \tag{28}$$

Here the indexes *g* and *b* denote the quantities those relating to the gas and to the background, H is a Hamiltonian, E is the value of the total energy. To provide of Gibbs distribution for the gas it is needed to integrate the function **R R** *<sup>g</sup> <sup>b</sup>* over the phase space of the background. Then *Lg g* **R** is factored out from integral. The function from H*g g* **R** can be factored from integral by the method Krutkov (Krutkov, 1933; Zubarev,1974) as expH*gg B* **R** *k T* . Then:

$$\mathbf{Z}\_{N} = \frac{1}{\left(2\pi\hbar\right)^{3N} N!} \int\_{\Gamma\_{\mathcal{S}}} \delta\left(L\_{\mathcal{S}}\left(\mathbf{R}\_{\mathcal{S}}\right)\right) \exp\left(-\mathfrak{K}\_{\mathcal{S}}\left(\mathbf{R}\_{\mathcal{S}}\right)\Big/\hbar\_{\mathcal{S}}T\right) \mathrm{d}\Gamma\_{\mathcal{S}}\,\mathrm{d}.\tag{29}$$

The Hamiltonian H*g g* **R** would be represented in the form:

$$\text{SiC}\_{\text{g}}\left(\text{R}\_{\text{g}}\right) = \sum\_{i=1}^{N} \mathbb{K}\_{i} = \sum\_{i=1}^{N} \left( \frac{1}{2m} \left( p\_{ri}^{2} + \frac{I\_{\text{ai}}^{2}}{r\_{i}^{2}} \right) + \frac{e^{2}H^{2}r\_{i}^{2}}{8m} \right) + \frac{eH}{2m} \sum\_{i=1}^{N} I\_{\text{ai}} \,. \tag{30}$$

Let us substitute Hamiltonian (30) to the formula (29) and take into account formula 1 *N zg g <sup>i</sup> l L* and the Eqs. (8). Then it is obtained:

$$Z\_{\aleph} = \frac{1}{\left(2\pi\hbar\right)^{3N} N!} \int \exp\left\{-\frac{1}{k\_B T} \sum\_{l=1}^{N} \left(\frac{1}{2m} \left(p\_{rl}^2 + \frac{I\_{zl}^2}{r\_i^2}\right) + \frac{e^2 H^2 r\_l^2}{8m}\right) \right\} \delta\left(L\_{\g}\right) d\Gamma\_{\g} \tag{31}$$

The integration over *<sup>z</sup>*<sup>1</sup> *l* leads to change 1 <sup>2</sup> *N <sup>z</sup> zi <sup>i</sup> l l* and after reduction of the quadratic form to the standard appearance the Hamiltonian of ideal gas that is collection of 1 *N* particles in harmonic potential field is obtained. Obviously, that taking into account conservation of the zero value of the angular momentum eliminates the paradoxes that were mentioned at the Introduction. The magnetic moment *<sup>z</sup> H kT B N* ln Z 0 . The gas is confined by the magnetic field. But that confinement provides to the inconsistency of the model that neglects of the electrostatic interaction because the uniform background cannot neutralize it. The model that regards this interaction will be considered below.
