# Tunneling in Double Barriers

## Transcription

Tunneling in Double Barriers

Northern Arizona University, Vol. I, No. 1, 1-9 Fall 2014 Tunneling in Double Barriers Congwei Wu12 * Abstract In this research project, the main focus was on the rectangular symmetric double barrier system. There exists an analytic solution for this system hence it is possible to verify numerical results. We applied multiple methods to compute the numerical values of transmission coefficients for different initial energies. As a result, a comparison between analytic solution and numerical solutions is given. Keywords Double Barriers Tunneling, Monte Carlo, Transfer matrix, WKB approximation 1 Department of Physics and Astronomy, Northern Arizona University, Flagstaff, AZ of Mathematics and Statistics, Northern Arizona University, Flagstaff, AZ *Corresponding author: Congwei Wu, [email protected] 2 Department Contents 1 Introduction 1 Physics Model 1 1.1 Quantum Tunneling . . . . . . . . . . . . . . . . . . . 1 Challenge to classical physics • Quantum rectangular potential barrier 1.2 Symmetric Double Barriers . . . . . . . . . . . . . 2 Derivation • Tunneling coefficient • Analytical solution 2 Methods 3 2.1 Method of Solving Non-Linear Equations . . 3 Separation • Metropolis Monte Carlo 2.2 Method of Transfer Matrix . . . . . . . . . . . . . . 4 Background • Transfer matrix • Transmission coefficient • Probability density function f (x) 2.3 WKB Method with Numerov Scheme . . . . . 6 WKB method • Scheme of Numerov’s method • Algorithm 3 Results and Discussion 7 3.1 Transmission Coefficient . . . . . . . . . . . . . . . 7 devices, spontaneous DNA mutation, and Radioactive decay. For the application of semiconductor devices, tunnel diode is one example that uses the quantum tunneling. For a more complicated structure such as double barriers, it is almost impossible to derive analytically solutions, and many numerical methods are being used. For instance, the Wentzel-kramers-Brillouin (WKB) method and Numerov methods are two approximations to solving the Schr¨odinger equation. Meanwhile, new methods such as transfer matrix for solutions of wave functions, and Metropolis Monte Carlo method for estimating ground-state energy have been developed rapidly. Hence there is a tendency for computational statistical methods to be involved in the research of quantum physics. In addition, a circumstance of physicists who become Quantum Bayesianisms also indicates evidence of combination, although there is still a divergence between Frequentists and Bayesians in the area of statistics. Simulation settings • Results and comparison 3.2 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . 8 4 1. Physics Model Conclusion 9 1.1 Quantum Tunneling Appendix 9 1.1.1 Challenge to classical physics References 9 Introduction Quantum tunneling was developed in the 20th Century and it has widely areas of applications, such as Scanning tunneling microscope (STM), semiconductor In classical physics, an object can never be passing through a physical barrier. It is also hard to imagine that someone would pass through a wall with some probability. However, in the world of quantum mechanics, there is some probability for particles to transmit through a barrier, given that the kinetic energy of those particles is lower than the required potential energy. So this type Tunneling in Double Barriers — 2/9 of randomness gives us difficulties not only in physics interpretation, but also in the methods involving partial differential equations (PDE). 1.1.2 Quantum rectangular potential barrier In quantum mechanics, the rectangular potential barrier is a one-dimensional problem which consists of the tunneling phenomenon. This type of problem is required for solving the one-dimensional time-independent Schr¨odinger equation. In addition, under a multiple barriers situation, an analytic solution to the Schr¨odinger equation is not be able to obtained. In this study we are focusing on the double-barrier problem. 1.2 Symmetric Double Barriers In this research study, we are focusing on a symmetric structure of double barrier problem, and are proposing statistical methods to quantum mechanics. As introduced before, solutions to many problems require physicists to solve the Sch¨odinger equation in order to obtain a closeform of the wave function. The wave function consists of the fundamentals of the physical system. d 2 ψ(x) 2m + 2 [E −V (x)]ψ(x) = 0 dx2 h¯ 1.2.2 Tunneling coefficient If E < V0 , then ψI = A1 eik0 x + A2 e−ik0 x , −∞ < x < 0 ψII = B1 ek1 x + B2 e−k1 x , 0 < x < a ψIII = C1 eik0 x +C2 e−ik0 x , a < x < b k1 x ψIV = D1 e ψV = F1 e To avoid tedious calculations, the derivations of transmission coefficient is given as follows. Since the xcoordinate does not have an impact on the derivations, we constructed the double barriers question as: The bar- (1) where m is the mass of the particles, h¯ is the Planck constant, E is the total energy, V (x) is the potential energy, and ψ(x) is the wave function. Due to the conservation of energy, the potential V (x) is constant for all regions. The Schr¨odinger equation can be written as five different wave functions depending on different regions. ik0 x 1.2.1 Derivation −k1 x + D2 e (2) , b<x<c , c<x<∞ Also, the wave numbers k are related to the energy via q k = 2m(V − E)/¯h2 , x ∈ (0, a) ∪ (b, c) 1 0 q k= k = 2mE/¯h2 , otherwise 0 V0 I So the time-independent Schr¨odinger equation for the wave function ψ(x) is (3) II III IV a c b Figure 1. General View of Double Barriers 0 V x riers (in yellow) are positioned between x = 0 and x = a, and between x = b and x = c, respectively. The barriers divide the one dimensional space in five parts. Denote them as: Table 1. Table of Regions Name of Region Region I Region II Region III Region IV Region V Range of x −∞ < x < 0 0<x<a a<x<b b<x<c c<x<∞ The index on coefficients A, B,C, D and F denotes the direction of the velocity vector, where 1 represents the left and 2 represents the right. When E < V0 , k1 becomes imaginary and the wave function is exponentially decaying within the barrier. To determine the coefficients A, B,C, D and F, we applied the boundary conditions of the wave function at x = 0, x = a, x = b and x = c. Also note the derivative of the wave function has to be continuous everywhere. ψI (0) = ψII (0) ⇒ ψI0 (0) = ψII0 (0) 0 ψII (a) = ψIII (a) ⇒ ψII0 (a) = ψIII (a) 0 0 ψIII (b) = ψIV (b) ⇒ ψIII (b) = ψIV (b) (4) 0 ψIV (c) = ψV (c) ⇒ ψIV (c) = ψV0 (c) Substitute all ψ’s by the corresponding wave functions, the boundary conditions yield the restriction on the coef- Tunneling in Double Barriers — 3/9 2.1 Method of Solving Non-Linear Equations ficients 2.1.1 Separation A1 + A2 = B1 + B2 B1 e k1 a −k1 a + B2 e ik0 a = C1 e +C2 e A numerical solution can be given by solving the system of non-linear equations (5) with Monte Carlo simulation. The Euler’s formula yields: −ik0 a C1 eik0 b +C2 e−ik0 b = D1 ek1 b + D2 e−k1 b eix = cos(x) + i sin(x) D1 ek1 c + D2 e−k1 c = Feik0 c A1 (ik0 ) − A2 (ik0 ) = B1 (k1 ) − B2 (k1 ) B1 k1 ek1 a − B2 k1 e−k1 a = C1 ik0 eik0 a −C2 ik0 e−ik0 a C1 ik0 eik0 b −C2 ik0 e−ik0 b = D1 k1 ek1 b − D2 k1 e−k1 b D1 k1 ek1 c − D2 k1 e−k1 c = Fik0 eik0 c . (5) 1.2.3 Analytical solution To simplify this type of problem, we treated the system as consisting of two symmetric double barriers. That is, widths of the two barriers are the same, namely a − 0 = a = c − b. By Burstein and Lundqvist [1], the closed-form of transmission coefficient can be written as: 2 1 Ttrans = cosh(γl) + iα sinh(γl) · e−i2k0 L 2 −2 1 2 + β sinh2 (γl)−2 , (6) 4 (8) One important property of the Euler’s formula is that it separates the imaginary parts from the real parts. So we use the Euler’s formula rewrite (5) in terms of a linear combination of a real part and an imaginary part. That is, express (5) in terms of the form of (Creal ) + (Cimaginary )i. For each part of this combination, we solve for the coefficient independently for Creal and Cimaginary . So (5) can be written as: (i) for all imaginary parts A2 k0 = 0 (C1 −C2 )k0 cos(k0 a) = 0 (C1 +C2 ) sin(k0 a) = 0 (C1 −C2 )k0 cos(k0 b) = 0 (9) (C1 +C2 ) sin(k0 b) = 0 F sin(k0 c) = 0 Fk0 cos(k0 c) = 0, and (ii) for all real parts where A2 − (B1 + B2 ) = 0 l = a−0 = a = c−b (B1 − B2 )k1 = 0 L = b−a k0 γ α= − k0 γ γ k0 β= + k0 γ r 2m γ= (V0 − E). h¯ 2 k1 (B1 ek1 a − B2 e−k1 a ) + (C1 +C2 )k0 sin(k0 a) = 0 (7) Therefore, we obtained the closed-form solution of the transmission coefficient, which can be applied for verifying our numerical results. Note that the analytic solution is generally different for a different quantum tunneling problem, but numerical methods are basically the same. Hence this analytical solution for double barrier problem would help us to determine the best method among various numerical methods. 2. Methods B1 ek1 a + B2 e−k1 a − (C1 +C2 ) cos(k0 a) = 0 k1 (D1 ek1 b − D2 e−k1 b ) + (C1 +C2 )k0 sin(k0 a) = 0 D1 ek1 b + D2 e−k1 b − (C1 +C2 ) cos(k0 a) = 0 D1 ek1 c + D2 e−k2 c − F cos(k0 c) = 0 D1 k1 ek1 c − D2 k1 e−k1 c + Fk0 sin(k0 c) = 0. (10) So now our question is converted into a problem that solves two sytsems of non-linear equations, using the Monte Carlo simulation. In other words, solving a system of non-linear equations as follows f1 (x1 , x2 , ..., xn ) = 0 f2 (x1 , x2 , ..., xn ) = 0 (11) .. . fn (x1 , x2 , ..., xn ) = 0, Tunneling in Double Barriers — 4/9 with a algorithm. We assume the true value of xi , i = 1, 2, ..., n is in some interval (ai , bi ). Each interval does not necessarily be identical but for simplicity they would be set as the same. By evaluating a large number of points from this interval, we will eventually obtain a sequence of points xi∗ , i = 1, 2, ..., n, which would satisfy our equations within a given error ε. The error ε is usually very small, and in this case we picked ε = 0.01. 2.1.2 Metropolis Monte Carlo In statistics, the Metropolis-Hastings algorithm is a Markov chain Monte Carlo (MCMC) method for obtaining a sequence of random samples from our interested probability distribution [2], but it is difficult to sample directly. Statistically speaking, Markov chain is a stochastic process in which every state of the system only depends upon the previous state. Namely, to obtain the information of n-th state, all we need is the information of the n − 1-th state of the system. Therefore, the sequence obtained by MCMC can be used to estimate some statistics that we are interested in, such as expected value of some variable. For our purpose, we could also apply MCMC to sample from the probability distribution (or equivalently square of the wave function) and compute the ground state energy [3]. Since this study focused mainly on tunneling, the ground state energy would not be computed. For each xi , there is a corresponding interval I = 0 (xi − α, xi0 + α) that covers the true value xi∗ . Then we let Xi ∼ Unif(xi0 − α, xi0 + α) for the i-th variable. An algorithm is given below: (for our case, n = 6 since we set A1 = 1 and F2 = 0) 1. Define a detector function Ξ as s n Ξ= ∑ [ fi (x1 , x2 , ..., xn )]2 . (12) 4. If Ξnew < Ξold , then this trial is meaningful and we let Ξold = Ξnew and let xi0 = xij . Because the trail value xij is closer to the true value compared with xi0 . Otherwise, go back to step 3 and select another point xij+1 . 5. Since the trial is good, and we obtain a better Ξ, move to the next variable xi+1 . That is, repeat step ∗ . If i = n, then set i = 1. 3 but for the point xi+1 Therefore, we applied this algorithm to either equation (9) or (10), then we obtained a numerical values of the coefficients A2 , B1 , B2 ,C1 ,C2 , D1 , D2 , and F, given that E,V0 , a, b, c. 2.2 Method of Transfer Matrix 2.2.1 Background The method of transfer matrix has been developed rapidly over the last decades. As Ricco and Azbel mentioned [4], transfer matrix method can be applied to any types of barrier case and extremely useful when it is hard to obtain an analytical solution. 2.2.2 Transfer matrix The Schr¨odinger equation for one dimensional time independent particle with energy V0 is d 2 ψ(x) 2m + 2 (E −V (x))ψ(x) = 0 dx2 h¯ (13) where the m represents mass of the particle. We divided any barrier into N number of intervals, and each interval the solution of the Schr¨odinger equation can be written as ψn (x) = An eikn x + Bn e−ikn x (14) i=1 Note for the sequence of solutions {xi∗ }, we should have Ξ = 0. So a value of Ξ detects whether a sequence of {xi } is our solution. 2. Pick a starting point xi0 for each i, i = 1, 2, ..., n. Let α be some relative big number, for example, 10000. Calculate Ξ, denote as Ξold . Define an interval (xi0 − α, xi0 + α) for which we assume that the true sequence of solutions {xi∗ } is in this ndimensional space. 3. For the j-th sampling, we randomly select a point xij from (xi0 − α, xi0 + α), then plug it into (11). So (11) gives a new value of Ξ. Denote as Ξnew . where xn−1 < x < xn and kn = Denote q 2m(E −V (xn ))/¯h2 . Knxn = eikn xn Kn∗xn = e−ikn xn xn Kn+1 = eikn+1 xn ∗xn Kn+1 = e−ikn+1 xn , (15) For every two consecutive intervals, namely (xn , xn+1 ) and (xn+1 , xn+2 ), the left-hand side of wave functions Tunneling in Double Barriers — 5/9 and their derivatives along with the wave functions is ψn (xn ) = An Knxn + Bn Kn∗xn ψn0 (xn ) = An ikn Knxn − Bn ikn Kn∗xn ∗x x n+1 n+1 ψn+1 (xn+1 ) = An+1 Kn+1 + Bn+1 Kn+1 x ∗x 0 n+1 n+1 ψn+1 (xn+1 ) = An+1 ikn+1 Kn+1 − Bn+1 ikn+1 Kn+1 (16) By the boundary conditions at x = xn+1 , we obtain ψn (xn+1 ) ψn+1 (xn+1 ) ψn (xn+1 − δ ) = = 0 (x ψn0 (xn+1 ) ψn+1 ψn0 (xn+1 − δ ) n+1 ) (17) where the width of interval is defined as δ = xn+1 − xn . Denote ik x e n+1 n+1 eikn+1 xn+1 Λn (xn+1 ) = ikn eikn xn+1 −ikn e−ikn xn+1 ik x e n+1 n+1 eikn+1 xn+1 Λn+1 (xn+1 ) = ikn+1 eikn+1 xn+1 −ikn+1 e−ikn+1 xn+1 (18) Then ψn (xn+1 ) A = Λn (xn+1 ) n ψn0 (xn+1 ) Bn ψn+1 (xn+1 ) An+1 = Λn+1 (xn+1 ) 0 (x ψn+1 Bn+1 n+1 ) So we obtain ψn (xn+1 ) ψn+1 (xn+1 ) = (Tn+1 ) 0 ψn0 (xn+1 ) ψn+1 (xn+1 ) (19) So (20) and (22) together yield ψ0 (x0 ) ψN (xN ) = T1 T2 ...TN 0 ψ00 (x0 ) ψN (xN ) where T j,k , j, k ∈ {1, 2} are elements of the transfer matrix T . So we rewrite (23) and thus ψ0 (x0 ) ψN (xN ) Ta Tb ψN (xN ) =T = ψ00 (x0 ) ψN0 (xN ) Tc Td ψN0 (xN ) (25) which gives ψN (xN ) −1 ψ0 (x0 ) =T ψN0 (xN ) ψ00 (x0 ) (26) 2.2.3 Transmission coefficient By the definition of transfer matrix T , we can cite find the transfer coefficient. Note that the wave functions for region I and region V are defined as: ψ0 (x0 ) = A1 eik0 x0 + A0 e−ik0 x0 (27) ψN (xN ) = AN eikN xN (28) Then by (27), (28) and (25) we have ik x A1 e 0 0 + A0 e−ik0 x0 Ta Tb ψN (xN ) = ψ00 (x0 ) Tc Td ψN0 (xN ) (29) After tedious matrices calculation, it yields that (20) where Tn+1 is derived by Tn+1 = [Λn (xn+1 )][Λn (xn )−1 ], namely 1 cos(kn+1 δ ) − kn+1 sin(kn+1 δ ) Tn+1 = kn+1 sin(kn+1 δ ) cos(kn+1 δ ) (21) So by (21) we define the matrix Tn as cos(kn δ ) − k1n sin(kn δ ) Tn = kn sin(kn δ ) cos(kn δ ) Then define the transfer matrix as 1 N sin(kn+1 δ ) cos(kn+1 δ ) − kn+1 T=∏ cos(kn+1 δ ) n=1 kn+1 sin(kn+1 δ ) (24) AN = 2ik0 ei(k0 x0 −kN xN ) ik0 (Ta + ikN Tb ) + Tc + ikN Td 2 2ik0 ei(k0 x0 −kN xN ) Ttrans = |AN | = ik0 (Ta + ikN Tb ) + Tc + ikN Td 2 (31) Rre f = 1 − Ttrans . (22) (23) (30) (32) 2.2.4 Probability density function f (x) In order to derive the general probability density function for each n state, then we divide the total length of x into N of lengths. By (20), let n + 1 = N then we have ψN−1 (xN−1 ) ψN (xN ) = TN 0 , (33) 0 ψN−2 (xN−2 ) ψN (xN ) Tunneling in Double Barriers — 6/9 Obviously, (23), (24) and (25) indicate that the matrix sequence of [ψn (xn )ψn0 (xn )]T is a geometric matrix sequence, which can be written as: ψn (xn ) AN eikN xN n ψN (xN ) = T = (34) ψn0 (xn ) ψN0 (xN ) ikN AN eikN xN where 1 cos(kn+1 δ ) − kn+1 sin(kn+1 δ ) T =∏ , cos(kn+1 δ ) n=0 kn+1 sin(kn+1 δ ) n Ta Tbn n . (35) T = n Tc Tdn N n So the wave function for each state of n is ψn (xn ) = Tan An eikN xN + ikN Tbn AN eikN xN . (36) Then we obtain the probability density function f (xn ) for each n state, for which the wave function is normalized |ψn (xn )|2 . f (xn ) = R ∞ 2 −∞ |ψn (xn )| dxn (37) R 2.3 WKB Method with Numerov Scheme 2.3.1 WKB method WKB approximation is a half-classical computational method that can be used to solve a linear partial differential equations with varying coefficients [3]. The idea is to use some exponential functions to replace the wave functions in the Schr¨odinger’s equation, then find an asymptotic series as an approximation of the exponential function. Finally we take the limit of the asymptotic series, and obtain an approximation of the wave function. Since we focus on the transmission coefficient in this study, the WKB method was applied to find the transmission coefficient instead of the wave functions. In general, for a differential equation dn y dn−1 y dy + a x + ... + an−1 + an y = 0, (38) 1 n n−1 dx dx dx where ε > 0 is a very small number. We assume that the solution can be expressed as an asymptotic series, namely y(x) ≈ exp 1 ∞ n δ Sn (x) . ∑ δ n=0 φ10 = φ2 2m φ20 = 2 (V (x) − E)φ1 . h¯ (40) (41) We write the wave function as ∞ Since the presence of −∞ |ψn (xn )|2 dxn (constant), the probability density function f (xn ) is normalized, which has a range from 0 to 1. ε n Then we substitute exp δ1 ∑∞ n=0 δ Sn (x) into the differential equation. Let δ → 0, then starting with n = 0 solve for every single Sn (x). Note most cases the series is not convergent, so when n is sufficiently large, δ n Sn (x) would increase. Therefore, once δ n Sn (x) increases, we stop substituting n, and hence an approximate solution yields. For our symmetric double barrier, the differential equation is the Schr¨odinger equation (1), where our variable of interest is the wave function ψ(x). To apply the WKB method, first we write the Schr¨odinger equation (1) as a system of two first-order linear ordinary differential equations with φ1 = ψ and φ2 = dψ/dx. That is, (39) ψ = A(x)eiφ (x) . (42) Assume the potential is changing slowly, then d2 φ /dx2 → 0, the solution of the Schr¨odinger equation is R 1 ψ = A p e±i/¯h |p|dx |p| p k = 2m[E −V0 ]. (43) (44) So the asymptotic forms is a power series ∞ A= ∑ h¯ n An (x). (45) n=0 2.3.2 Scheme of Numerov’s method Numerov’s method is a numerical method to solve ordinary differential equations [3], which is especially useful for the one-dimensional time-independent Schr¨odinger equation. Combined together with the WKB method, the error of our approximation could be reduced to very small. There also exists less accurate methods for solving the Schr¨odinger equation, such as Runge-Kutta method. However, the basic algorithm for both methods is the same. As long as a minimization scheme of the Numerov’s method is well designed, then it should give a relatively close or useful numerical solution to the double-barrier system. In our design scheme, the Numerov’s method is mainly used to reduce the error of the last term of the asymptotic series within the WKB approximation. Tunneling in Double Barriers — 7/9 From the Taylor expansion of g(x) about a point x0 : N∈N g(x) = g(x0 ) + ∑ k=1 (x − x0 )k g(k) (x0 ) k! (46) Then let the distance between x to x0 be h = x − x0 , and then x = h + x0 . So we rewrite the Taylor expansion of g(x) as: So (52) yields the Numerov’s method, and the reminder O(h6 ) is ignored in calculation. In other words, the error in each time step is O(h6 ), which decreases with steps, and the first term of this sequence of error is O(h4 ). Therefore, we derived the Numerov formula for determining ψn , for n = 2, 3, ..., given ψ0 and ψ1 as two initial values. 2.3.3 Algorithm g(x0 + h) = g(x0 ) + N∈N k (k) h g (x ∑ 0) (47) k! k=1 So (47) means the point x0 of the Taylor expansion is taking a step forward by an amount of h. That is, the original expansion point of center is x0 , and we shift this point to x0 + h. Likewise, we can also derive an expansion of backward by taking a step of h back: N∈N g(x0 − h) = g(x0 ) + ∑ k=1 (−h)k g(k) (x0 ) k! (48) For computational purpose, the length h is small, so each time the expansion point is shifted slightly, forward or backward. Note gn+1 = g(xn + h) = g(xn ) + N∈N k (k) h g (xn ) ∑ k=1 N∈N gn−1 = g(xn − h) = g(xn ) + ∑ k=1 k! (49) (−h)k g(k) (xn ) . k! (50) The sum of (49) and (50) gives 00 gn−1 + gn+1 = 2gn + h2 gn + h4 4 g + O(h6 ) (51) 12 n where O(h6 ) denotes terms with order 6 or higher than order 6. Namely, O(h6 ) denotes the reminder of the Taylor expansion. Then we obtain a differential equation as follows: Applying the Numerov’s method to the double-barrier problem requires an adjusted scheme, since each boundary condition contains complex numbers. Therefore, a scheme designed for double-barrier is listed as follows: 1. In order to separate a complex parameter A, we let 2 2 A = Ar + iAi , given that a particle energy E = h¯2mk . Then we guess two initial values ψ0 and ψ1 . 2. The transmission coefficient can be written as [5] T3 (C4 −C3 ) C3 +C4 2 Ttrans = + (54) 4 T3 1 T1 (55) C3 = ( − )eiT2 T1 4 1 T1 (56) C4 = ( + )e−iT2 4 T1 Z x4 |k| T1 = exp(− dx) (57) ¯ x3 h Z x3 |k| dx (58) T2 = ¯ x2 h Z x2 |k| T3 = exp(− dx). (59) ¯ x1 h Therefore, we obtained a numerical result of the transmission coefficient. 3. Results and Discussion 3.1 Transmission Coefficient 3.1.1 Simulation settings Since our study project is focusing on the tunneling in double barrier, it is necessary to obtain the transmission h4 coefficient for our model. Note there are three numerical h2 an gn = 2gn − gn−1 − gn+1 + g4n + O(h6 ) methods, so we set up various numerical systems such 12 (52) that every method can be evaluated under different circumstances. Additionally, the closed-form solution for The solution of (52) is the symmetric double barrier system is obtained, so our numerical simulated environment would be also built on 2 h2 (2 − 5h6 an )gn − (1 + 12 an−1 )gn+1 gn+1 = + O(h6 ) this symmetric system. From (7) we knew that l reph2 1 + 12 an+1 resents the width of each barrier, and L represents the (53) distance between two barriers. Tunneling in Double Barriers — 8/9 Table 2. Table of Simulation Settings Circumstances V0 , in eV l, in m L, in m I: Basic 1 10−10 10−10 II: Huge Barriers 1 10−9 10−10 III: Long Distance 1 10−10 20−9 IV: High Potential 2 10−10 10−10 reach the same level as in the basic model. For example, at V0 = 1eV , it demands E = 0.20eV for a Ttrans ≈ 33%, while at V0 = 2eV , it demands E = 0.40eV for the Ttrans be able to pass 36%. Second we applied the each numerical method to the setting I, the basic model: Table 4. Table of Ttrans from Different Methods There are four settings that base on the symmetric system, which were consisted of four distinct circumstances. The basic circumstance is a possibly realistic model in real world, and the other three are extremal situations in different ways. The huge barriers model has two large barriers with widths of 10−9 meters and the distance between two barriers for the long distance model is large, also reaching 2 × 10−9 meters. Note that the total width of the systems for both models are the same, so it is reasonable that the transmission coefficients for both models be the same. But due to the existence of the double barriers, the transmission coefficients might be somehow different. The last model is called the high potential model, which has a relative high potential energy for both barriers. Thus if given same kinetic energy, the transmission coefficient for this model would be likely lower than the basic model, because it requires higher energy for particles to be tunneling. 3.1.2 Results and comparison First we calculated the exact values for all four settings via (7). The results are as follows: Table 3. Table of Exact Ttrans for All Settings Results E 0.10 0.20 0.30 0.40 0.50 I 0.01068 0.33145 0.41192 0.82572 0.73648 Settings II III 0.00032 0.01052 0.17954 0.33029 0.25698 0.40974 0.56232 0.82003 0.51041 0.72966 IV 0.00120 0.08061 0.20568 0.36557 0.48706 It is obvious that the long distance model has a higher transmission coefficient than the huge barriers model. Although the total widths of both models are the same, the width of barrier has the main effect on the transmission coefficient. Compared the data between basic model and long distance model, it is apparent that the Ttrans of long distance model is very close to the Ttrans of the basic model. So it indicates that the distance between two barriers does not have a big impact on the Ttrans . Finally, for the high potential model, E must be higher to be able to Results E 0.10 0.20 0.30 0.40 0.50 Methods for Setting I Monte Carlo Transfer Matrix 0.01564 0.01395 0.33149 0.35142 0.40190 0.42007 0.83555 0.83134 0.74647 0.75164 WKB 0.00002 0.13415 0.36287 0.52334 0.39787 As we expected, the Monte Carlo method is the most accurate method among all three numerical methods. For each energy level, the Monte Carlo method gives almost the same result as the true value. In addition, the transfer matrix method also yields relative close result, compared to the true values. Note that as long as the simulation time sufficiently large, both Monte Carlo and transfer matrix methods could be very close to the true values, within error less than 10−10 . In our study, we constricted the computing time to about 120 minutes, so both methods did not converge with a very small error. Through this comparison, it is obvious that the convergent speed of Monte Carlo is a little faster than the transfer matrix’s. The reason is probably for the optimization scheme for the Monte Carlo method, and hence this method has a higher efficiency. However, the WKB method seems to be the worst among all numerical methods. Compared to the true values, the WKB method is not only underestimated the Ttrans , but also it has a relative big error, which is over 40% of the estimate. Even though the Numerov’s method came with the WKB to eliminate the error, the results are still not acceptable. The reason for such poor performance is expected to be the divergent asymptotic series. As mentioned before, usually the asymptotic series we chose to approximate is not convergent, and thus it would give a relative large error because of the divergence. 3.2 Discussion Overall, the Monte Carlo method with the scheme of optimization is the best method among all numerical methods. A visualized figure is provided as follows: Tunneling in Double Barriers — 9/9 Figure 2. Figure of data Note that the estimated line of the Monte Carlo is almost the same as the true value line, so is the line of transfer matrix. Therefore, these two methods are actually both good, even though the Monte Carlo method is slightly better in estimation. In addition, it is obvious that the WKB method is off by a large error. References [1] E. Burstein and S. Lundqvist. Tunneling phenomena in solids. Plenum Press, 1:9–12, 1969. [2] R.E. Salvino and F.A. Buot. Self-consistent monte carlo particle transport with model quantum tunneling dynamics: Application to the intrinsic bistability of a symmetric double-barrier structure. Journal of Applied Physics, 72:5975–5981, 1992. [3] T. Pang. An introduction to computational physics. Cambridge University Press, 2:110–115, 2006. [4] B. Ricco and M.Ya. Azbel. Physics of resonant unneling. the one-dimensional double-barrier case. Physical Review Letter B, 29:2, 1984. [5] A. Dutt and S. Kar. Smooth double barriers in quantum mechanics. American Journal of Physics, 78:1352–1360, 2010. 4. Conclusion In this study, for the symmetric rectangular double barrier system, we derived the analytical solution, and also introduced three numerical methods to obtain the transmission coefficient. A simulation study is given and the Monte Carlo and transfer matrix are the two best numerical methods. In the future study, the goal is to develop a better scheme for the WKB method in order to obtain a less error estimation. As a classical method, WKB still plays an important role in the area of quantum mechanics so it is necessary to have a better performance in estimation. Appendix