Energy Levels in a Symmetric Triple Well

Energy levels in a symmetric triple well

Author: Ferran Torra Clotet Facultat de F´ısica, Universitat de Barcelona, Diagonal 645, 08028 Barcelona, Spain.∗

Advisor: Josep Taron

Abstract: We show, by analogy with the double square well, the energy levels of the triple square well. More specifically, it is interesting the analysis of the lowest energy levels when E < V0 because the three lowest energy levels are non-degenerate in contrast with the classical solution, and the transition between different wells is possible. As a result, we develop the corresponding energy formulae which are confirmed numerically and the tunnel effect is also described.

I. INTRODUCTION

Classically, in a symmetric double well, for E < V0, there are two ground states of equal energy. In contrast, the quantum solution provides a splitting of the two lowest energy levels and the probabilities of the corresponding wave functions are nonzero in classically forbidden regions. In particular, it is shown in [2] with a square Figure 1: The triple square well potential. well potential. As an extension of the results presented in [2], we are going to develop the triple square well case and the tun- to the movement of a particle in each single well. The nelling processes involved in this potential. In addition, quantum possibility of a particle to tunnel through the we will verify properties that hold for all one dimensional barriers produces a splitting of these levels into three symmetric potentials (see [3]): non-degenerate energy levels. In addition, observe that there are two semi-infinite • bound energy levels in one-dimensional potentials square wells and one finite square well. This fact will are non-degenerate, be relevant for the structure of the energy levels as it is going to be demonstrated in this work. • the wave function of the ground state is symmetric and the eigenstates are alternately symmetric and antisymmetric with respect to the centre of sym- metry, x = 0. II. WAVE FUNCTIONS This document, initially, provides an analysis of the two types of possible solutions of the one-dimensional Since the potential is symmetric with respect to the Schrödingerequation in a symmetric potential and then origin, the solutions of the time-independent Schrödinger we determine the corresponding energy levels. These re- equation will be wave functions of definite parity. In ad- sults are compared with the numerical calculations in dition, we have to take into account the boundary condi- (− 3 − ) = ( 3 + ) = order to validate our approximations. Finally, the tunnel tions, ψ 2 L w ψ 2 L w 0. Therefore, the eigen- effect is discussed with the consideration of interesting functions of the Hamiltonian for E < V0 can be written wave functions. as one of the following forms, We consider the symmetric triple well potential (represented in Fig. 1) of the form, • symmetric solutions,

⎧ L ⎪ ∞, SxS > 3 + w, ⎧ [ ( + 3 + )] ⎪ 2 ⎪ D sin k w 2 L x I, V (x) = ⎨ 0, regions I,III,V, (1) ⎪ ⎪ ⎪ ⎩⎪ V0, regions II,IV, ⎪ B sinh[α( w + L + x)]+ ⎪ 2 2 II, ⎪ +C cosh[α( w + L + x)] ⎪ 2 2 where L is the width of a single square well, w is the width ⎪ of the potential barrier and V is the barrier height. ⎪ 0 ψ(x) = ⎨ A cos kx III, (2) Our interest is in the case E < V → ∞ where classi- ⎪ 0 ⎪ cally there are three ground energy levels corresponding ⎪ ⎪ B sinh[α( w + L − x)]+ ⎪ 2 2 IV, ⎪ + [ ( w + L − )] ⎪ C cosh α 2 2 x ⎪ ⎪ ⎪ [ ( + 3 − )] ∗Electronic address: [email protected] ⎩ D sin k w 2 L x V, Energy levels in a symmetric triple well Ferran Torra Clotet

• antisymmetric solutions Observe that, when = 0, this equation becomes 2ηζ2 = 0 which has only one root ζ = 0 (of multiplic- ⎪⎧ −D sin[k(w + 3 L + x)] I, ⎪ 2 ity 2) because η ≠ 0. Thus, there are two roots of the ⎪ ⎪ perturbed equation near ζ0 = 0. Assuming the expansion ⎪ −B sinh[α( w + L + x)]− ( ) = + + 2 + ⎪ 2 2 II, ζ ζ0 ζ1 ζ2 ..., we get the solutions, ⎪ −C cosh[α( w + L + x)] ⎪ 2 2 » ⎪ ζ = 3 ± 9 − 8η2 + o(2). (11) ⎪ 4η ψ(x) = ⎨ A sin kx III, (3) ⎪ ⎪ Recovering the original notation, this condition is ⎪ ⎪ B sinh[α( w + L − x)]+ ¼ ⎪ 2 2 IV, kL k 1 ⎪ + [ ( w + L − )] ≃ ± − 2( ) ⎪ C cosh α 2 2 x cot 3 9 8 tanh αw . (12) ⎪ 2 α 4 tanh αw ⎪ ⎪ [ ( + 3 − )] ⎩ D sin k w 2 L x V, B. Antisymmetric solutions √ » 2mE 2m(V0−E) where k = ̵ and α = ̵ . h h = L In both cases, the solutions must satisfy the continuity The conditions at x 2 in matricial form are, conditions for the wave function and its derivative at the L L sin k L sinh αw cosh αw B points x = ± and x = ±( + w). Note that, since the A 2 = 2 2 . (13) 2 2 L −α cosh αw −α sinh αw C eigenfunctions have definite parity, it suffices to analyze k cos k 2 2 2 the conditions at the points x = L and x = L + w. 2 2 Inverting the matrix, we can isolate the terms B and C, B − sinh αw − 1 cosh αw sin k L = A 2 α 2 2 . (14) A. Symmetric solutions C αw 1 αw L cosh 2 α sinh 2 k cos k 2 = L The conditions at x = L + w in matricial form are, On the one hand, the conditions at the point x 2 in 2 matricial form are, sin kL − sinh αw cosh αw B D = 2 2 . (15) cos k L sinh αw cosh αw B −k cos kL −α cosh αw α sinh αw C A 2 = 2 2 . (4) 2 2 −k sin k L −α cosh αw −α sinh αw C 2 2 2 Thus, using (14), we get the following relationship, Inverting the matrix, we can isolate the terms B and C, sin kL cosh αw 1 sinh αw sin k L D =A α 2 . B − sinh αw − 1 cosh αw cos k L −k cos kL α sinh αw cosh αw k cos k L = A 2 α 2 2 . (5) 2 C αw 1 αw − L (16) cosh 2 α sinh 2 k sin k 2 Dividing the two rows of (16), it leads to the condition On the other hand, the conditions at the point x = L + w 2 ⎡ k L ⎤ k ⎢1 + tanh(αw) cot k ⎥ in matricial form are, tan kL = − ⎢ α 2 ⎥ . (17) ⎢ k L ⎥ α ⎣ tanh(αw) + cot k ⎦ sin kL − sinh αw cosh αw B α 2 = 2 2 D − − αw αw . (6) This expression can also be written as k cos kL α cosh 2 α sinh 2 C 2ζ 1 + ηζ Thus, using (5), we get the following relationship, = − , (18) ζ2 − 1 η + ζ sin kL cosh αw 1 sinh αw cos k L D =A α 2 . where ζ ≡ cot k L , ≡ k and η ≡ tanh(αw). And finally, −k cos kL − L 2 α α sinh αw cosh αw k sin k 2 (7) 2ηζ3 + 3ζ2 + (2 − 2)ηζ − = 0. (19) Dividing the two rows of (7), it leads to the condition Note that, when = 0, this equation becomes 2ηζ = 0 ⎡ L k ⎤ k ⎢cot k − tanh(αw)⎥ which has only one root ζ = 0 (of multiplicity 1) be- tan kL = − ⎢ 2 α ⎥ . (8) α ⎢ ( ) L − k ⎥ cause η ≠ 0. Thus, there is one root of the perturbed ⎣tanh αw cot k 2 α ⎦ equation near ζ0 = 0. Assuming the expansion 2 This expression can also be written as ζ() = ζ0 + ζ1 + ζ2 + ..., we get the solution, 2ζ ζ − η ζ = + o(2). (20) = − , (9) 2η ζ2 − 1 ηζ − Recovering the original notation, this condition is where ζ ≡ cot k L , ≡ k and η ≡ tanh(αw). And finally, 2 α kL k 1 cot ≃ . (21) ζ3 + (2 − 2)ηζ2 − 3ζ + 2η = 0. (10) 2 α 2 tanh αw

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′ = L III. ENERGY LEVELS ζ tan k 2 , which leads to the following respective equations,

Let kG, kF and kS be the three lowest values of k 2η(ζ′)3 − 3η(ζ′)2 + (2 − 2)ηζ′ + = 0, (27) corresponding to the ground state (ψG), the first excited ′ 3 2 ′ 2 ′ 2 state (ψF ) and the second excited state (ψS). −(ζ ) + (2 − )η(ζ ) + 3ζ + η = 0. (28) These three quantities are obtained from (12) and (21) and they are located around kL ∼ π. Our assumption is In a similar way as in (19), the symmetric case has only to consider the case where the height V0 of the potential one solution located in the vicinity of kL ∼ 2nπ for all n ∈ . Likewise, the antisymmetric case has two solu- barrier is huge compared√ with the minimum energy level N ∼ 2mV0 = ≫ tions around each kL ∼ 2nπ for all n ∈ N. Thus, the E, therefore α h̵ α0 k. Moreover, we consider th th ≫ (3(2n + 1)) excited state and the (3(2n + 1) + 2) ex- that the width of the barrier is such that α0w 1. th Consequently, these three values appear as the inter- cited state are antisymmetric and the (3(2n + 1) + 1) kL kL excited state is symmetric for all n ∈ . sections of y = cot with the straight lines y = εG , N 2 2 With this point of view, the energy levels are non- y = ε kL and y = ε kL near kL ∼ π, where the constants F 2 S 2 degenerate as expected since we are dealing with states are of one-dimensional potential. Moreover, it is interesting ⎡ ¼ ⎤ ⎢ 2 ⎥ to note that the energy spectrum consists of alternating coth(α w) ⎢3 + 9 − 8 tanh (α0w)⎥ = 0 ⎢ ⎥ symmetric and antisymmetric states where the ground εG ⎢ ⎥ , (22) α0L ⎢ 2 ⎥ state is always symmetric, the first excited state is anti- ⎣ ⎦ symmetric, and so on. coth(α0w) εF = , (23) Furthermore, it is interesting to analyze the gap be- α0L tween these energy levels. ⎡ ¼ ⎤ ⎢ 2 ⎥ On the one hand, between the ground state and the coth(α w) ⎢3 − 9 − 8 tanh (α0w)⎥ = 0 ⎢ ⎥ first excited state, it is given by, εS ⎢ ⎥ . (24) α0L ⎢ 2 ⎥ ⎣ ⎦ h̵ 2π2 1 1 E − E ≃ − ≃ (29) F G 2 2 2 Thus, since k ∼ π~L, we find the following approximate 2mL (1 + εF ) (1 + εG) values, h̵ 2π2 ≃ (ε − ε ) = (30) π π π mL2 G F kG ≃ , kF ≃ , kS ≃ .(25) ⎡ ¼ ⎤ L(1 + εG) L(1 + εF ) L(1 + εS) 2 2 ⎢ 2 ⎥ h̵ π coth(α w) ⎢1 9 − 8 tanh (α0w)⎥ = 0 ⎢ + ⎥ . 2 ⎢ ⎥ Note that, in our range of parameters, the three constants mL α0L ⎢2 2 ⎥ satisfy ε < ε < ε ≪ 1. Hence, the three quantities are ⎣ ⎦ S F G (31) slightly smaller than π~L which corresponds to the lowest value of the wave number in an individual infinite well According to the assumption that α w ≫ 1, we have of width L. In addition, these three quantities satisfy 0 coth(α w) ≃ 1 + 2e−2α0w and tanh(α w) ≃ 1 − 2e−2α0w. < < 0 0 kG kF kS and the respective energies are Therefore, h̵ 2k2 h̵ 2k2 h̵ 2k2 E = G ,E = F ,E = S , (26) h̵ 2π2 (1 + 2e−2α0w)(1 + 8e−2α0w) G F S E − E ≃ ≃ (32) 2m 2m 2m F G 2 mL α0L where EG is the ground state energy, EF is the first ex- h̵ 2π2 1 ≃ . (33) cited state energy and ES is the second excited state 2 mL α0L energy such that EG < EF < ES. Observe that the wave functions of the ground state We see that this gap decreases as 1~α0 when α0 (i.e. the and the second excited state are symmetric, and the wave height of the potential barrier, V0) increases. function of the first excited state is antisymmetric, as we On the other hand, the split between the first excited expected. state and the second one is We notice that the conditions of (12) and (21) are valid for any group of energy levels of odd order (i.e. h̵ 2π2 1 1 E − E ≃ − ≃ (34) S F 2 2 2 with kL in the vicinity of a odd multiple of π) because 2mL (1 + εS) (1 + εF ) the assumption was that cot k L = 0, which is valid for 2 h̵ 2π2 L th k = (2n + 1)π with n ∈ . Therefore, the (6n) excited ≃ (εF − εS) = (35) 2 Z mL2 state and the (6n + 2)th excited state are symmetric and ⎡ ¼ ⎤ 2 2 ⎢ 2 ⎥ the (6n+1)th excited state is antisymmetric for all n ∈ . h̵ π coth(α w) ⎢ 1 9 − 8 tanh (α0w)⎥ N = 0 ⎢− + ⎥ . Moreover, for any group of energy levels of even order 2 ⎢ ⎥ mL α0L ⎢ 2 2 ⎥ (i.e. with kL in the vicinity of a even multiple of π), we ⎣ ⎦ should analyze the solutions from (8) and (17) around (36)

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Applying the same assumption used in the other gap, Note that these three energy levels are not equidistant, as we expected due to the central well is different of the h̵ 2π2 (1 + 2e−2α0w)(8e−2α0w) other ones. E − E ≃ ≃ (37) S F 2 mL α0L Furthermore, we can obtain the parameters of sym- h̵ 2π2 8e−2α0w metric solutions (2) (resp. antisymmetric solutions (3)) ≃ . (38) 2 by solving numerically (4) and (5) (resp. (13) and (14)); mL α0L and, of course, using the normalization condition. These Note that, in this case, the gap decreases exponentially coefficients are given in Table III for the lowest three en- when α0 (i.e. the height of the potential barrier, V0) ergy levels with w = 0.02L and α0 = 100~L. increases. Thus, the ground state and the first excited state are A B C D more separated than the first and the second excited −2 −2 −1 states. ψG 1.327 1.01 ⋅ 10 2.14 ⋅ 10 3.13 ⋅ 10 −3 −2 −2 −1 ψF 8.52 ⋅ 10 −1.00 ⋅ 10 1.32 ⋅ 10 9.95 ⋅ 10 −1 −2 −3 −1 ψS 4.46 ⋅ 10 1.40 ⋅ 10 −6.60 ⋅ 10 −9.44 ⋅ 10 IV. NUMERICAL SOLUTION Table III: Coefficients of the eigenstates ψG, ψF and ψS of the Hamiltonian with w = 0.02L and α = 100~L. To verify our equations, we solve (8) and (17) with nu- 0 merical methods using [4]. We show the results of the values of k corresponding to the three eigenstates of low- These wave functions can be represented graphically est energy. In this section, the unit of distance will be (Fig. 2). L and, consequently, the unit of wave number k and α0 will be L−1. We present Table I, where the assumptions mentioned above are clearly satisfied. Therefore, our results are consistent with the numerical solutions.

−1 α0[L ] 100 300 500 −1 kG[L ] 3.07381 3.1207869 3.12907627 num −1 kG [L ] 3.07563 3.1207863 3.12907635 −1 kF [L ] 3.10933 3.1311553 3.1353219685 num −1 kF [L ] 3.10934 3.1311551 3.13532200954 −1 kS [L ] 3.11335 3.1311556 3.1353219686 num −1 kS [L ] 3.11149 3.1311557 3.13532200959

Table I: The numerically calculated quantities of k corre- Figure 2: Wave functions for the lowest three energy levels sponding to the lowest three energy levels against those pre- with w = 0.02L and α0 = 100~L. Orange line corresponds to dicted using our equations for w = 0.02L and diﬀerent values the ground state (ψG), blue line to the ﬁrst excited state (ψF ) of α . 0 and green line to the second excited state (ψS ).

The results with a small height of the potential barrier (Table II) are also consistent with what we might predict based on equations above. V. THE TUNNEL EFFECT w[L] 0.2 0.35 0.5 −1 kG[L ] 2.58683 2.61640 2.61791 Classically, in our configuration, there are three ground num −1 kG [L ] 2.56951 2.61025 2.61273 states of equal energy, one for each single square well. −1 kF [L ] 2.84634 2.85552 2.85597 However, the three lowest energy quantum levels are non- num −1 degenerate and the probability density of these eigen- kF [L ] 2.84243 2.85179 2.85231 − states in the regions II and IV is nonzero, whereas these k [L 1] 2.86448 2.85646 2.85602 S regions are classically forbidden. knum[L−1] 2.87891 2.85408 2.85244 S We can combine the wave functions to get the following Table II: The numerically calculated quantities of k corre- configurations sponding to the lowest three energy levels against those pre- dicted using our equations for α0 = 10~L and different values 1 of w . ψC (x) = (−DSψG(x) + DGψS(x)), (39) NC

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» 2 2 where NC = DG + DS, and permits a major probability density in the right conﬁg- uration because the quantum tunnelling again. Unfortu- 1 2 2 nately, this evolution cannot be expressed with a exact ψL(x) = ASψG(x) − AGψS(x)+ (40) NL frequency because there are two exponentials terms,

(AGDS − ASDG) + ψ (x) , (41) 1 EGt ES t F 2 −i h̵ 2 −i h̵ DF ψ2(x, t) = ASψG(x)e − AGψS(x)e + (46) NL 1 2 2 ψ (x) = A ψ (x) − A ψ (x)− (42) (AGDS − ASDG) EF t R S G G S −i h̵ NL + ψF (x)e = (47) DF (AGDS − ASDG) − ( ) −i EGt ψF x , (43) e h̵ DF 2 2 −iωS t = ASψG(x) − AGψS(x)e + (48) ½ NL 2 2 2 (AGDS −AS DG) (A D − A D ) = + + G S S G −iωF t where NL AG AS D2 . + ψ (x)e , (49) F D F These conﬁgurations can be interpreted as the classical F conﬁgurations because they are practically localised in a where hω̵ = E − E and hω̵ = E − E . single square well as shown in Fig. 3. F F G S S G

Figure 4: Probability density of ψ1(x, t) after a time t = π~ω. Figure 3: Probability densities of the classical conﬁgurations with the parameters w = 0.02L and α0 = 100~L; red line corresponds to ψL, green line to ψC and blue line to ψR. VI. CONCLUSIONS

More interesting is to consider a wave function ψ1(x, t) which initially is equal to ψC , i.e., located in the central In this essay, we have shown that the three lowest en- square well. Its time evolution can be expressed, ergy levels are non-degenerate but they are not equidistant. Our formulae have been conﬁrmed with numerical 1 ES t EGt −i ̵ −i ̵ calculations and they show an example of the quantum ψ1(x, t) = DGψS(x)e h − DSψG(x)e h = NC tunnelling phenomena. (44) Our results are consistent with the symmetric triple EGt well using instanton methods which is explained in [1]. −i ̵ e h −iωt = −DSψG(x) + DGψS(x)e , (45) As a ﬁnal comment, we expect that the N lowest en- NC ergy levels in N square wells are non-degenerate and they ̵ tend to create an energy band. where hω = ES − EG. Observe that the probability density varies periodically with the frequency ω. Indeed, the particle can be dis- placed to the other square wells (with major probability after a time t = π~ω) because of quantum tunnelling (see Acknowledgments Fig. 4) and then it is turned back to the center square well . I would like to thank my advisor Josep Taron for all Moreover, the same idea can be applied for a wave useful discussions and corrections. I would also like to function ψ2(x, t) located at the left square well at t = 0. thank the support received from my friends and family, This wave function can describe a time evolution which especially my parents.

[1] H. A. Alhendi and E. I. Lashin, Mod.Phys.Lett. A19, [3] A. Messiah, Quantum mechanics, Vol. 1 (North-Holland, 2103-2112 (2004). Amsterdam, 1961). [2] J. L. Basdevant and J. Dalibard, Quantum mechanics, [4] Wolfram Research, Inc., Mathematica, Version 10.4 2nd. ed. (Springer, Berlin, 2006). (Champaign, 2016).

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