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Chapter 10. Particle motion in potential step field

To solve Equation (10.1) we shall assume that there is a superposition of quantum states of incident and reflected particles, i.e. their coordinates are indistinguishable x1 x3 x.

Let 1 (x) and 3 (x) be solutions to Equation (10.1). It can be shown that the superposition of these solutions in the form:

13 x 1 3 − 2 1 3

is also a solution to this equation, if E P12 / 2m P22 / 2m. However, for the coordinates of incident and reflected particles to be indistinguishable along the direction of motion they must be delocalized over the

whole space.	Let us fulfill the relevant transition in solutions for
1 , 3 , 2 ( xi	% &), then we obtain boundary conditions in the form:
	10 30 2		20,
		10 30
	P1( 10 30) P2 20,

which provides obtaining formulae of Problem 1 for R and D from [1], which are independent from the Plank constant and, therefore, do not provide the classical limit when % 0.

D 4P1P2 / (P1 P2)2.

The existence of classical limit for the transmission coefficient D should not certainly depend on the steepness of the potential step boundary. The reason is in the traditional view of a free quantum particle as de Broglie wave.

Reference

1.L.D. Landau, E.M. Lifshitz. Quantum Mechanics. Nonrelativistic Theory. M.: Nauka 1974. P.103.

CHAPTER 11

TUNNELING

The tunneling problem was solved in quasi-hydrodynamic representation in Paragraph 7, when alpha-decay of nuclei was studied. Tunneling was looked at from a potential well with finite motion of particles with a zero translation motion component. We shall consider particle tunneling in a more general case when particles collide with a rectangular wall with some momentum.

Let us consider a particle passing over a rectangular barrier of a finite height U 0 (E ! U 0 ) and finite width а, Fig.11.1.

Fig. 11.1.

As there is no macroscopic particle momentum in the vicinity of the barrier, it is necessary to solve a system of Equations (3.2), (3.6).

div J 0,

(3.2)

(J / )

J 2

2( )2

(3.6)

2m 2

We shall make some simplifying assumptions to solve the system of

equations. We assume that

0 at tunneling in stationary poten-

tial fields, i.e. the energy of tunneling particles remains constant

J 2

( )2 = const.

(11.1)

2m 2

8m 2

Chapter 11. Tunneling

To simplify the formulae further we shall consider one-dimensional spatially localized particles — needle states. If there are transversal components of quantum oscillation energy, they are «pulled» over the barrier without unchanged as they are motion invariants. We shall take into account the transversal components in final formulae.

Then, it follows from (11.1) that J / depends only on coordinate x and

J		2 2	2		2
					(U 0 E)2m.	(11.2)
	2 x 2
			4 2	x

To fulfill this relation it is necessary to accept t (t) x (x). Then, Equation (3.2) can be solved by the method of variable sepa-

ration in quadratures. We obtain

	m	x
		x
J (x,t) exp(t / ) J 0
		0

(x)dx . (11.3)

Here m / P12 , x = 0 is the position of the front wall of the barrier, where the flow density of particle probability equals J0. The solution of Equation (3.2) can be written in a different form:

	.	1		x		1	m	x
J (x,t) J 0	exp 0	1	t m x (x)dx 3, J x (x) J 0				m	x (x)dx. (11.4)
J (x,t) J 0	exp 0		t m x (x)dx 3, J x (x) J 0					x (x)dx. (11.4)
	0			0	J x (x)	3		0
	/			0		2		0

One can make sure that these are similar solutions by differentiating them with respect to x coordinate. It follows from (11.4) that the probability density flow J(x,t) J 0 is transferred in accordance with the law:

x
t m x (x) dx.		(11.5)
0	J x (x)

Having substituted Solution (11.3) into (11.1) we obtain a system of equations:

J 2(x)

E U 0

4m x

2m x2(x) 8m x2

dx 2

(11.6)

J x (x) J 0

x (x)dx.

We can write this system in a different way:

2J x(((J x( (J x(()2 42(J x( )2 52J x2 0, x (x) J x( , m

Chapter 11. Tunneling


where 4 2	8m	(U 0 E), 5		2 4P4	/ 4 , the prime mark indicates x-de-

	2			1
rivatives. The solution J x			has been found in the form:

J x (x) A exp(−x),

where [4 2 / 2 44 / 4 52 ]1/2 , А is constant.

Then, we calculate the tunneling barrier-transmission coefficient D. For this purpose one can use solutions for probability density obtained earlier in areas before the barrier and after passing the barrier. We assign index 1, as previously, to particles incident onto the barrier, and index 3 to reflected particles, index 2 is given to particles tunneling in the vicinity of the barrier, and index 4 — to particles which have passed the barrier. Thus, we have:

1 10 cos2 (tP1 / m x1 x / x1),

2(x,t)

30 cos2( (tP3 / m x3 x / x3) 03),

	t			t
20	exp	x , J 2	(x,t) J 20	exp	x ,

40 cos2( (tP4 / m x4 x / x4) 04).

Laws of energy conservation for free particles can be represented in the following way:

E	P 2	2	2 2	,
	1	10
	2m	8m
	2m	8m
E	P 2	2	2 2	,
	3	30
	2m	30
	2m	8m
E	P 2	2	2 2	.
	4	40
	2m	40
	2m	8m

Boundary conditions can be written, as previously, in this form:

P1 1(t0, x0) P3 3(t0, x0) J 2(t0, x0) J 2(t0, x0 a) P4 4(t0, x0 a).

1(t0, x0) 3(t0, x0) 2(t0, x0) 2(t0, x0 a) 4(t0, x0 a),

We assume further, x0 0,t0 0, as the front wall of the barrier is tied to coordinate х0 = 0 in the solution for tunneling particles.

Solving the written system of equations together with relations for laws of energy conservation of free particles, one can calculate D.

Chapter 11. Tunneling

D	P4 4	(a)	.	(11.7)
	P1 10

It follows from boundary conditions:

4(a) 2(a) 20e a ( 10 30) e a.

Then D equals:

D P4( 10 30) exp( a).

P1 10

At	a % 0, D % 1, 30		%	0 there must be P4 = P1. Then it follows
from	the energy conservation					law: 40 10 . From the solution for
probability density		4( a)		10	cos 2 .
						4

Let us write out final formulae to calculate the tunnel barrier-trans- mission coefficient D.

D	10 30 exp( a ) cos 62.	(11.8)
	10
	4

It can be seen from (11.8) that the particles which have tunneled through the barrier preserve their energy, momentum, and the wave of probability density changes the phase.

From the law of conservation of energy of reflected particles and the definition R P3 30 / P1 10 we can find the expression for 30 :

		1 .											11/ 2
30		1 .		2m(E )				4m2(E )2	27 (1			D)2 P 2 2	11/ 2	, (11.9)
			0										3


	7	2 /										1 10	2
					.						11/ 2
					.						11/ 2
				4	01	/ 2		1 / 4 52	/ 44 3			,
					/					2
								7 / 2.
				42	8m		(U	0 E), 52 4P 4		/ 4.
				42			(U	0 E), 52 4P 4		/ 4.
					2				1

Using Equations (11.8) and (11.9) one can calculate the barrier transmission ratio, if E, P1 , 10 are given. The barrier transmission in the classical limit must be equal to zero, indeed, if we formally fix % 0, we obtain % & and D % 0.

The conventional formula for the barrier under consideration (see [1], p. 104) is as follows:

Chapter 11. Tunneling

		4k 2				P / , (2m(U		E
D		1			, k		0		) / 2 )1 / 2	. (11.10)
D					, k				) / 2 )1 / 2	. (11.10)
0	(k 2	)2 sh2 a	0	4k 2	1	1			k
	1		0	1

P 2

Here Ek 1 is the kinetic energy of a particle. Comparing Equations 2m

(11.10) and (11.8), (11.9), one can see that the tunneling of particles with spatially structured probability density differs from conventional tunneling. The preexponential factor in Equation (11.8) changes from 1 to 2. The exponent index includes parameter which will be written in the form convenient for comparison:

				2
2 2m U		E	k	10	2 2 .
2 2m U		E			2 2 .
	0
				8m

It is evident that 2 0 , however in the formula for there is a parameter , which is equal to

2		E k2
4		4(U 0 E )

and which increases the coefficient to some extent. Thus, barrier-tran- smission coefficients should be compared for particular cases. Let us compare the formulae for the case when a 0 1 as usual. Then, from (11.10) we have:

D	16k 2
		1		e 2 a 0 .	(11.11)

0	(k 2		)2

	1

We shall carry out numerical comparison of transmission coeffi-

cients. We consider needle states, when	0 and 2 / 4 1, then
2 0 and exponent indices coincide in	formulae for D and D0.

The difference in barrier-transmission coefficients will be defined by

preexponential		factors ratio. Let k 2	/ 2	E	k	/ (U	0	E	k	) 1, then
D / D	0,5 1.	1	0
		The situation changes radically with P						k 0 when
	0					1		1

tunneling is possible mainly due to the energy of quantum particle fluctuations which are not considered by the conventional approach, then:

D / D0 20 / 8k12 1.

Thus, tunneling of particles with spatially structured probability density actually always differ from those calculated by means of traditional formulae.

Chapter 11. Tunneling

The time of tunneling of particles with spatially structured probability density is of quite a definite value and, in accordance with formula (11.5) and m / P12 , equals:

t a 0 / P1,

(11.12)

where 0 is the transit time for the а-wide barrier, with the momentum

Р1 0 am.

It can be seen that the process of tunneling can be slow when the momentum of a particle incident onto the barrier is small.

Reference

1.L.D. Landau, E.M. Lifshitz. Quantum Mechanics. Nonrelativistic Theory. M.: Nauka 1974. P.103.

CONCLUSION

Quasi-hydrodynamic representation of infinite motion of quantum particles with physical variables makes it possible to obtain new solutions which change the idea of them. The «cost» of the physical form of the initial system of equations is nonlinearity of one of them. For theoretical physics analytical solutions to quantum problems are invaluable. However, in the era of computer technologies solving quantum problems in quasi-hydrodynamic representation should not be of great difficulty, especially since transport problems in particular quantum devices are solved by means of computers due to their complexity.

Quantum particles in infinite motion beside classical kinetic energy always have quite certain quantum energy, so called, energy of quantum fluctuations. This approach eliminates contradictions of the traditional theory for transport phenomena, i.e. all the new quantum formulae and equations have a classical limit. In traditional quantum mechanics free particles are described by means of wave packets, as they are based on the notion of particle momentum fluctuations. However, the energy of these fluctuations which is transferred by free particles among other things was not taken into account by the energy conservation law.

The formulae for barrier-transmission coefficients have been obtained in the implicit form. Nominally, it is attributed to nonlinearity of one of the quantum motion equations. The process of transmitting barriers is essentially self-consistent for the probability density distribution, the statistic wave field of a particle changes with motion invariants preserved. In the mentioned traditional solutions to quantum problems this circumstance is not taken into account.

Considering the spatial structure of the probability density of free quantum particles gives, in our opinion, more correct and, sometimes, new relations for transport phenomena. Ultimately, using quasi-hydro- dynamic representation is justified if there are new results which are or can be confirmed experimentally. In particular, experimental proof of resonant pumping the quantum component of the energy of infinite particle motion opens a fundamentally new approach to solving some applied problems and will provide an additional evidence of the wave character of infinite motion of quantum particles.

Appendix 1. Deriving quantum motion equations in quasi hydrodynamic representation

APPENDIX 1

DERIVING QUANTUM MOTION EQUATIONS IN QUASI HYDRODYNAMIC REPRESENTATION

Given below are initial Schroedinger equations necessary for further calculations:

				*
i	t	H ,	i	t	H *.	(Ap. 1.1)
	t			t

Hamiltonian operator for a particle with mass m, which moves in an arbitrary potential field U = U(r,t) has a form of:

	2	U (r,t).	(Ap. 1.2)
H
	2m

Multiplying the first Schroedinger equation by * and the other byand subtracting the second equation from the first one, we obtain:

			2		2
i	t	: :		( * *)		div ( * *),
			2m		2m

where * . Now we introduce probability flow density J/m

J i ( * * ).

m2m

We get the law of particles mass conservation in the form:

m div J 0.

Then we calculate the derivative:

/ )

(ln * ln )

1 *

: * *:

(Ap. 1.3)

(Ap. 1.4)

Here Schroedinger equations given above are used. Substituting the explicit form of Hamiltonian operator we obtain:

(J / )		2	* *
	U			.	(Ap. 1.5)
t		4m

Appendix 2. Quantum particles motion in stationary external fields

The expression with wave functions in the right part of this equation should be substituted by expressions depending on the probability density. For this purpose we calculate:

* * 2 *,

	( )2 ( * * )2 ( * * )2 4 *	4J 2
	( )2 ( * * )2 ( * * )2 4 *	2
		2
	Excluding with help of above formulae the expression with
	functions in (Ap. 1.5), we finally get the following equation:

wave

(J / )		2	J 2	2( )2
	U				,	(Ap. 1.6)
t		4m	2m 2	8m 2

which coincides with Equation (3.6). If there is a macroscopic momentum Р at infinite motion, we substitute J / P in Equation (Ap. 6) and obtain Expression (3.5).

P	P 2			2( )2	2
			U			.	(Ap. 1.7)
t				8m 2
		2m			4m

APPENDIX 2

QUANTUM PARTICLES MOTION

IN STATIONARY EXTERNAL FIELDS

Let us consider a one-dimensional case of quantum particle motion in a stationary external field U(x). The quantum particle is in a needle state when transversal components of quantum energy are equal to zero. The problem is to understand to what extend the solution for a free particle can be used in weak-gradient fields. It is necessary to solve the following system of equations:

m		P	0.	(Ap. 2.1)

	t	x

Here Р = Рх(x).

E	P 2	U ( ) const.	(Ap. 2.2)
	2m

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E	P 2	2	2 2	,
	1	10
	2m	8m
	2m	8m
E	P 2	2	2 2	,
	3	30
	2m	30
	2m	8m
E	P 2	2	2 2	.
	4	40
	2m	40
	2m	8m

E	P 2	2	2 2	,
	1	10
	2m	8m
	2m	8m
E	P 2	2	2 2	,
	3	30
	2m	30
	2m	8m
E	P 2	2	2 2	.
	4	40
	2m	40
	2m	8m

E	P 2	2	2 2	,
	1	10
	2m	8m
	2m	8m
E	P 2	2	2 2	,
	3	30
	2m	30
	2m	8m
E	P 2	2	2 2	.
	4	40
	2m	40
	2m	8m