Amplitudes of two-channel scattering by a two-dimensional potential barrier with a constant height

Amplitudes of Two-Channel Scattering by a Two-Dimensional Potential Barrier

with a Constant Height D. M. Sedrakiana, E. M. Kazaryanb, and L. R. Sedrakianb

aYerevan State University, Yerevan, Armenia bRussian-Armenian (Slavonic) State University, Yerevan, Armenia

Received November 27, 2009

Abstract⎯The immersing method is used to solve the two-channel scattering problem in the case of concrete potential. In particular, we consider the particle scattering by a two-dimensional barrier which is constant in the scattering direction and is arbitrary in the transverse direction. For this case the scattering amplitudes 1 2 1 ,, , t t r and 2r are determined. Expressions for the transmission and reflection amplitudes are obtained for the case of δ-potential. The behavior of the scattering amplitudes in the limit 2 0k → is studied. It is also shown that the ratio of products of transmission and reflection amplitudes for two channels does not depend on the coordinate of the middle of coordinate.

DOI: 10.3103/S1068337210030035 Key words: two-channel scattering, potential barrier, transmission, reflection, immersing method

1. INTRODUCTION Let us consider the two-channel scattering of an electron by the potential ( ) ( ) ( ), ,V x y V x V y= where

( ), ,

0, ,K a x b

V xa x b≤ ≤⎧

= ⎨ ≥ ≥⎩ (1)

( )V y is an arbitrary function of y, which satisfies the condition ( ) ( )0 ,V V c= =∞ and K is a constant (see the figure).

Geometry of the problem.

In works [1–5] it was shown that for finding the amplitudes of the two-channel scattering one must solve the system of linear equations for the functions ( )1L x and ( )2 :L x

ISSN 1068–3372, Journal of Contemporary Physics (Armenian Academy of Sciences), 2010, Vol. 45, No. 3, pp. 111–117. © Allerton Press, Inc., 2010. Original Russian Text © D.M. Sedrakian, E.M. Kazaryan, L.R. Sedrakian, 2010, published in Izvestiya NAN Armenii, Fizika, 2010, Vol. 45, No. 3, pp. 173–182.

z

c y a b

x

111

SEDRAKIAN et al.

112

JOURNAL OF CONTEMPORARY PHYSICS (ARMENIAN Ac. Sci.) Vol. 45 No. 3 2010

( ) ( ) ( )

( ) ( ) ( )

21 2

1 1 12 22

22 2

2 2 12 12

0,

0,

d L xq L x V L x

dxd L x

q L x V L xdx

+ − =

+ − =

(2)

where

( ) ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 21 1 11 2 2 22

2 211 1 22 2 12 1 2

0 0 0

, ,

, , . c c c

q k V q k V

V K V y y dy V K V y y dy V K y V y y dy

= − = −

= Φ = Φ = Φ Φ∫ ∫ ∫ (3)

Here the functions ( )n yΦ are defined by the formula ( ) ( )2 sin ,i y c n c yΦ = π 1,2 ,i = … and the potentials 11,V 12 ,V and 22V are constant in the region a x b≤ ≤ and are equal to zero at .a x b≥ ≥

Equations (2) are integrated up to the point x b= with the following initial conditions given at the point :x a=

( ) 1 111 1, ,ik a ik a

x a

dLL a e ik edx

−

=

= − = (4a)

( ) 22 0, 0.

x a

dLL adx =

= = (4b)

We denote by 1L and 2L the values of the desired functions ( )1L x and ( )2L x at the point .x b= The two-channel scattering amplitudes are determined by means of the four quantities ( 1 1 2, , ,D D D and 2D ) which are connected with 1L and 2L by the following relations [1]:

( ) ( )

( ) ( )

1 , 1,2,21 , 1,2,2

m

m

ik bm m m

ik bm m m

D b M L e m

D b M L e m

−= + =

= − = (5)

where ( )1 .m m m x bM ik dL dx

==

The formulas, which connect the two-channel scattering amplitudes with the quantities 1 1 2, , ,D D D and 2D are presented in [3]. However, the phases ϕ and ϕ entering formula (35) of that work are as yet indeterminate. By requiring that the dependences of phases of the functions 2T and 2R on the distance of the potential’s center 0z have the form ( )1 2 0k k z− and ( )1 2 0 ,k k z+ respectively, for ϕ and ϕ we get the following values: 0,ϕ = ,ϕ = α (6) where α is the phase of the function 2.D Note also that in [3, 4] the quantities 1 2, T T and 1 2, R R were transmission and reflection amplitudes, whereas 2 2

1 2, T T and 2 21 2, R R are the transmission and

reflection flows in the case when the flow of the incident particle is normalized to unity. Hence, if we denote the scattering amplitudes by 1 2 1, , ,t t r and 2 ,r then they can be expressed in terms of 1 2 1, , ,T T R and 2R by the following formulas:

1 11 1 1 1 2 2 2 2

2 2

, , , .k kt T r R t T r Rk k

= = = = (7)

Using the formulas (7) and (35) from the work [3] and our formula (6), we obtain for the scattering amplitudes

( )1 11 2 2

1 2 1 22 21

, , , ,D DD D Dt t r rD D DD D

∗∗ ∗ ∗

= = = = (8)

where ( )2 2 21 2 1 2 .D D k k D= +

Thus, by integrating equations (2) and determining the values of the functions ( )1L x and ( )2L x at the point x b= we can find amplitudes of the two-channel scattering by the potential barrier of type (1).

AMPLITUDES OF TWO-CHANNEL SCATTERING

113


2. SOLUTION OF THE SYSTEM OF EQUATIONS (2) We find a solution of the system (2) in the form

1 1 2 2, ,i ii x i xi i

i i

L A e L A eχ χ= =∑ ∑ (9)

where iχ are yet indeterminate constant quantities, whereas 1iA and 2iA are determined from algebraic equations which can be obtained by substituting solutions (9) into equations (2):

( )

( )

2 21 1 12 2

2 212 1 2 2

0,

0.

i i i

i i i

q A V A

V A q A

−χ + − =

− + −χ + = (10)

The system of linear equations (10) is homogeneous and, hence, it has nonzero solutions if the determinant of this system equals zero. This condition leads to the equation

( )( )2 2 2 2 21 2 12 0,i iq q Vχ − χ − − = (11)

the roots of which are the quantities .iχ There are four roots: 1Q± and 2Q± , where 1Q and 2Q are defined as

( ) ( )

( ) ( )

12 2 21 2

1

12 2 21 2

2

1 1 ,2 2

1 1 ,2 2

q qQ

q qQ

⎡ ⎤= + χ + − χ⎢ ⎥⎣ ⎦

⎡ ⎤= − χ + + χ⎢ ⎥⎣ ⎦

(12)

where

12 2

122 21 2

21 .Vq q

⎡ ⎤⎛ ⎞⎢ ⎥χ = + ⎜ ⎟−⎢ ⎥⎝ ⎠⎣ ⎦

(13)

If condition (11) is fulfilled then from the first equation of system (10) the amplitudes 2iA are uniquely expressed in terms of 1 :iA

( )2 2

12 1

12

.ii i

qA A

V− χ

= (14)

If we denote unknown amplitudes of solutions in the form 11 12 13 14, , , ,A A A B A C A D= = = = then the solutions to the system of equations (2) can be represented as

( ) ( ) ( )

( ) ( ) ( )

1 11 22

2 2 2 21 1 1 2

2 11 2212 12

,

,

L x L x L x

q Q q QL x L x L xV V

= +

− −= +

(15)

where

( ) ( ) ( ) ( ) ( ) ( )1 1 2 211 22, .iQ x a iQ x a iQ x a iQ x aL x Ae Ce L x Be De− − − − − −= + = +

Initial conditions (4a) and (4b) result in the following equations determining the amplitudes A, B, C, and D:

( )

( )

1

1

2 21 12 21 2

2 21 1 1 1 1

2 22 1 2 1 2

,

.

ik a

ik a

q QB D e A C A Cq Q

Q k Q q QB D e A C A CQ Q Q q Q

−

−+ = − − − = − +

−

⎛ ⎞ −− = − − + = − −⎜ ⎟ −⎝ ⎠

(16)

SEDRAKIAN et al.

114


By solving equations (16) we derive the following expressions for the unknown amplitudes:

1 1

1 1 1 2

1 1 1 2

2 2 2 21 2 1 1 1 12 2 2 2

1 2 1 1 2 2

1 1 , ,1 1

1 , 1 .2 2

ik a ik a

k Q k QC A D Bk Q k Q

q Q k q Q ke eA BQ Q Q Q Q Q

− −

+ += =

− −

⎛ ⎞ ⎛ ⎞− −= − − = −⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

(17)

If we introduce the notations

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( ) ( )

2 2 2 2 2 21 1 1 1 2 2 1 1 2 12

1

1

, ,

2 cos sin , 1,2,

2 cos sin , 1,2,m m m m

m m m m

q Q Q Q q Q V

l x Q x a i k Q Q x a m

q x Q x a i Q k Q x a m

η = − − η = η −

= − − + − =

= − − − =

(18)

then the desired functions ( )1L x and ( )2L x take the form

( ) ( ) ( ) ( )

( ) ( ) ( )

1

1

1 1 1 1 2

2 2 1 2

1 ,2

.2

ik a

ik a

eL x l x l x

eL x l x l x

−

−

= ⎡ + η −η ⎤⎣ ⎦

= η ⎡ − ⎤⎣ ⎦

(19)

Note also that solutions for the functions ( )1M x and ( )2M x take the form

( ) ( ) ( ) ( )( )

( ) ( ) ( )

1

1 1 1 1 2

12 2 1 2

2

,2

.2

ik a

ika

eM x q x q x q x

k eM x q x q xk

−

−

⎡ ⎤= + η −⎣ ⎦

= η ⎡ − ⎤⎣ ⎦

(20)

3. DETERMINATION OF THE QUANTITIES ( )mD b AND ( )mD b

In order to find the desired quantities ( )mD b and ( ) ,mD b we should use equations (5). By substituting the values of the functions ( )mM x and ( )mL x at the point x b= into these equations we get

( ) ( ) ( ) ( )( ) ( ) ( )( )

( ) ( ) ( ) ( )( ) ( ) ( )( )

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) ( ) ( )( )

11

11

12

12

1 1 1 1 1 2 2

1 1 1 1 1 2 2

12 2 1 2 2 1

2

12 2 1 2 1 2

2

1 1 ,2 21 1 ,2 2

1 ,2 2

1 ,2 2

ik aik b

ik aik b

ik aik b

ik aik b

eD b q b l b q b l b e

eD b q b l b q b l b e

k eD b q b q b l b l b ek

k eD b l b l b q b q b ek

−

−−

−

−−

⎡ ⎤= + η − −η −⎣ ⎦

⎡ ⎤= + η + − η +⎣ ⎦

⎡ ⎤= η − + −⎢ ⎥

⎣ ⎦⎡ ⎤

= η − + −⎢ ⎥⎣ ⎦

(21)

where ( )il b and ( )iq b are determined by equations (18). Substituting equations (18) into solutions (21), we have finally

( ) ( )

( )

1

2 21 1

1 1 1 1 1 21 1

2 2 2 21 1 2 1

1 21 1 1 2

2 2 2 2 2 21 1 1 1 1 2

1 1 1 1 21 1 1 1 1 2

cos sin cos cos 2

sin sin ,2 2

sin sin sin2 2 2

ik d Q kD b e Q d i Q d Q d Q dk Q

Q k Q ki Q d i Q dk Q k Q

k Q k Q k QD b i Q d i Q d Q d ek Q k Q k Q

⎧⎡ ⎤+⎪= − + η ⎡ − −⎨⎢ ⎥ ⎣⎪⎣ ⎦⎩

⎫⎤+ + ⎪− + ⎬⎥⎪⎦⎭

⎧ ⎫⎡ ⎤− − −⎪ ⎪= + η −⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

1 02ik x−

(22)


115


for the quantities ( )1D b and ( )1D b and

( ) ( )

( ) ( )

( ) ( )

1 2 1 2 0

1

21 2 1 1 2

2 2 1 2 12 2 1

22 1 2 2

22 2

21 2 1 1 2

2 2 1 2 12 2 1

22 1 2 2

22 2

1 cos cos sin 2 2

sin ,

1 cos cos sin 2 2

sin

i k k d i k k x

i k k

k k Q k kiD b Q d Q d Q dk k Q

Q k k Q d e ek Q

k k Q k kiD b Q d Q d Q dk k Q

Q k k Q d ek Q

+ − −

−

⎧ ⎡+ +⎪= η − − +⎨ ⎢⎪ ⎣⎩

⎫⎤+ ⎪+ ⎬⎥⎪⎦⎭

⎧ ⎡− −⎪= η − − −⎨ ⎢⎪ ⎣⎩

⎫⎤− ⎪− ⎬⎥⎪⎦⎭

( ) ( )2 1 2 0d i k k xe− +

(23)

for the quantities ( )2D b and ( )2 .D b Note that ( ) ( )2 2 2 2 .D k b D k b= − − Here we introduce the notations 02a b x+ = and ,b a d− = where d is the barrier thickness in the x-direction and 0x is the coordinate of

the barrier center on the x-axis. The quantity 1η entering formulas (22) has the form

11 ,2− λ

η =λ

(24)

where λ is defined from formula (13):

( )1

2 2 122 21 2

1 4 , .Vq q

λ = + δ δ =−

(25)

The quantity 2η entering formulas (23) is defined from formula (18):

2 .δη =λ

(26)

By substituting ( ) ( ) ( )1 1 2, , ,D b D b D b and ( )2D b from expressions (22) and (23) into formulas (8) we can determine the scattering amplitudes 1 2 1, , ,t t r and 2r of the problem under consideration.

Let us find the scattering amplitudes for a specific potential which is nonzero only at the point 0 ,x 0y and has the form ( ) ( ) ( )0 0v , .x y P x x y y= δ − δ − (27)

The quantities 1 1 2, , ,D D D and 2D for this problem can be determined from formulas (22) and (23), if we choose the potentials ikV so that they would be finite at 0.d → Other quantities proportional to d, naturally, will tend to zero. To make this transition, one should first replace ( )V y by ( )0y yδ − in formulas (3) and require that

0lim .dk

d K P→→∞

= (28)

Thus, for determining 1 1 2, , ,D D D and 2D for potential (27) it is necessary in formulas (22) and (23) to tend d to zero and K to infinity and to replace everywhere ikd V by vik which are already finite: ( ) ( )0 0v .ik i kP y y= Φ Φ (29)

Before to write the final expressions for 1 1 2, , ,D D D and 2 ,D one should make replacements in formulas (22) and (23) at small values of d: 1 2cos cos 1,Q d Q d≈ ≈ 1 1sin ,Q d Q d≈ and 2 2sin .Q d Q d≈ Then we obtain

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2 0 1 2

2 1 0 1 21 0

22 2 2 21 1 2 2 2

1 2 1 2 2 11 1 2

222 2 2 21 1 2 2

1 2 1 2 2 11 1 2

1 , ,2 2 2

, .2 2 2

i di k k x k k

ii k k x k k dik x

iQ d i iD Q Q d D Q Q dek k k

iQ d i iD Q Q d e D Q Q dek k k

− − + +

− − + −−

η η= − + − = −

⎡ ⎤η η= − + − = −⎢ ⎥⎣ ⎦

(30)

Substituting 1η and 2η from formulas (18) and taking into account that ( )2 21 1 11 110 0

lim lim v ,d d

q d k V d→ →

= − = − ( )( )2 2 2 2 2 2

1 1 1 2 120lim v ,d

q Q q Q d→

− − = − we derive finally

SEDRAKIAN et al.

116


( ) ( )

1 0

1 2 0 1 2 0

211 111 1

1 1

12 122 2

2 2

v v1 , ,2 2

v v, .2 2

ik x

i k k x i k k x

i iD D ek k

i iD e D ek k

−

− − − +

= + =

= = (31)

Substituting expressions (31) into formulas (8), we determine the scattering amplitudes for potential (27).

4. DISCUSSION OF RESULTS As seen from formulas (22) and (23), at 12 0V = the scattering along the second channel is absent 2 2( 0)D D= = and only the scattering along the first channel remains. The formulas for 1D and 1D pass

into the corresponding expressions for these quantities at the one-dimensional scattering [6], however, instead of the one-dimensional potential V, the potential 11V enters these expressions. The value of this potential, naturally, depends on the particle motion along the y-direction. This property of scattering takes place also for potential (27). If the excitation potential 12V is nonzero but small ( )12 11 1V V << then in expressions for 1D and 1D corrections proportional to ( )2

1 12 11~ V Vη appear. In these case the quantities 2D and 2D are proportional to the first power 2 12 11~ .V Vη Using expressions (22) and (23), from formulas (8) one can derive the scattering amplitudes and

coefficients of transmission and reflection for the two-channel scattering both for the scattering amplitudes of potentials (1) and (27). The expressions for the scattering amplitudes of potential (1) are too cumbersome and we do not present them here. However, it is easy to show that for potential (7) the calculation of scattering amplitudes is simple. By substituting expressions (31) into formula (8) we derive for the scattering amplitudes

( ) ( )

1 0

1 2 01 2 0

11 11 111 2 1 2

1 1 1 21 1 1 12 2 22 212 11 12

21

12 1121

22 2 12 2 212 12

v v v1 12 2 2

, ,v v v14 4 4

vv22 , ,

v v4 4

ik x

i k k xi k k x

i ik k i k kk k k

t r e

k

ki ei k e kt r

+−

⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= =⎡ ⎤ ⎛ ⎞ ⎡ ⎤Γ +⎢ ⎥ + Γ +⎜ ⎟ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎝ ⎠

= − = −⎡ ⎤ ⎡ ⎤Γ +⎢ ⎥ Γ +⎢ ⎥⎣ ⎦ ⎣ ⎦

(32)

where ( )2 211 1 1 21 v 4 .k k kΓ = +

For the transmission and reflection coefficients we get the following expressions:

2 22 211 11

1 2 1 222 21 1

1 12 221212

2212 1212

12 2 22 22 22

1212

v v14 4, ,

vv44

vv44 , .

vv44

k k k kk kt r

kk kt r

⎛ ⎞+⎜ ⎟

⎝ ⎠= =⎡ ⎤⎡ ⎤ Γ +Γ + ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

= =⎡ ⎤⎡ ⎤ Γ +Γ + ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦

(33)

Using formulas (33), it is easy to check that

( )2 2 2 211 2 2 2

2

1.kt t t rk

+ + + = (34)

Equation (34) expresses the condition of continuity of the particle flow.


117


As seen from solutions (32) and (33), the scattering amplitudes and coefficients depend only on the potentials 11v and 12v , i.e., the potential 22v does not enter their expressions. The cause of such a behavior of the scattering amplitudes is determined by the fact that scattering potential (27) is localized at the point 0.x The incident flow has the momentum 1k and, hence, the scattering along the first channel must be described by the potential 11v . But the scattering along the second channel is the transition of the particle at the point 0x into the state with the longitudinal momentum 2k and its propagation in two opposite directions of x. Therefore the scattering amplitudes should be proportional only to 12v . However, if the potential has a finite thickness d, as in the case (1), then the potential 22v enters the expression for scattering amplitudes.

It is of interest to consider the behavior of scattering amplitudes in the case of small values of 2.k This condition can be satisfied because

2

2 22 1

0

3 ,k kc

⎛ ⎞π= − ⎜ ⎟

⎝ ⎠ (35)

and at ( )21 03k c→ π we have 2 0.k → In this limit, as seen from formulas (33), for the potential (18) the

coefficients of transmission and reflection along the first channel tend to zero:

2 2

2 21 10 0

lim lim 0.k k

t r→ →

= = (36)

As to the scattering along the second channel, the transmission coefficient 22t is finite, whereas the

reflection coefficient 22r tends to infinity so that

2

2220

1

lim 1.k

k rk→

= (37)

It is easy to show that the scattering amplitudes 2 2 21 1 2, , ,t r t and 2

2r for potential (1) also have such behavior. Thus, the behavior of 2

2r at 2 0k → described by formula (37) does not depend on the form of the potential. The increase in 2 1

2 2~r k − at 2 0k → is connected with the fact that in this limit the particle scattering along the first channel is stopped and, hence, all incident particles pass into the second channel. Since the velocity of their removal in this channel tends to zero, then 2

2r tends to infinity. Note also one more important property of the two-channel scattering. From formulas (8), (22), (23),

and (31) follows that the transmission amplitude 1t does not depend on the position of the middle of potential 0 ,x whereas the reflection amplitude 1r has the phase factor ( )1 0exp 2 .ik x As to the scattering along the second channel, the transmission and reflection amplitudes 2t and 2r have the phase factor

( )( )1 2 0exp i k k x− and ( )( )1 2 0exp ,i k k x+ respectively. Hence it follows that the phase factor for the product 2 2r t equals ( )1 0exp 2ik x and, hence, the expression 2 2 1 1r t r t does not depend on 0.x This property of the two-channel scattering will be used in solution of the problem of electron localization on a quasiperiodical system of potentials.

In conclusion we note that in the regime 2 0k → the interaction of free electrons with an electromagnetic wave with a frequency of the order of 2 2

03 cπ can lead to the amplification of the wave at the expense of the electric current energy. Therefore it is of interest to consider the problem of interaction of the electromagnetic wave with scattering electrons.

REFERENCES 1. Boese, D., Lischka, M., and Reichl, L.E., Phys. Rev. B, 2000, vol. 62, p. 16933. 2. Souma, S. and Suzuki, A., Phys. Rev. B., 2002, vol. 65, p. 115307. 3. Sedrakian, D.M., Kazaryan, E.M., and Sedrakian, L.R., J. Contemp. Phys. (Armenian Ac. Sci.), 2009, vol. 44, p.

257. 4. Sedrakian, D.M., Kazaryan, E.M., and Sedrakian, L.R., Proc. seventh int. conf. Semiconductor micro- and

nanoelectronics, Tsakhcadzor, Armenia, 2009, p. 11. 5. Sedrakian, L.R., Doklady NAN Armenii, 2009, vol. 109, p. 214. 6. Sedrakian, D.M., Khachatrian, A.Zh., Kazaryan, E.M., and Sedrakian, L.R., J. Contemp. Phys. (Armenian Ac.

Sci.), 2009, vol. 44, p. 113.

Documents

Amplitudes of two-channel scattering by a two-dimensional potential barrier with a constant height