Reexamination on the problem of the finite square well in quantum mechanics

Young-Sea Huang∗ and Kuan-Wei Lee†Department of Physics, Soochow University, Shih-Lin, Taipei 111, Taiwan

AbstractRecently, the problem of the infinite square was reexamined by using the self-adjointness of the

Hamiltonian operator and the momentum operator, instead of postulating boundary conditions.The solutions so obtained are free from those peculiarities in the standard solution of the infinitesquare well. Also, the solutions provide theoretical justification for the periodic boundary conditionsapplied to the infinite square well. Here, the same tact is used to study the problem of the finitesquare well. Boundary conditions allowed for the finite square well are found by the self-adjointnessof the Hamiltonian operator. Solutions for bound states of the finite square well are classified intotwo kinds: (1) the particle is completely confined inside the well, (2) the particle has possibility toleak out of the well. For the first kind, solutions are those of the infinite square well, as obtainedpreviously, for which their energies are less than the potential of the finite square well. For thesecond kind, there are infinitely many families of solutions, each of which is characterized by theallowed boundary conditions. The well-known standard solution of the finite square well is just oneof these families of solutions.

PACS numbers: Quantum mechanics, 03.65.-w; Foundations of quantum mechanics, 03.65.TaKeywords: Quantum mechanics, Schrodinger wave equation, Finite potential square well, Infinite potentialsquare well, Boundary conditions, Hermitian operator, Self-adjoint operator, Physical observable.

I. INTRODUCTION

Peculiarities in the standard solution of the infinite square well were pointed out as originated from the boundarycondition — the continuity of wave functions at boundaries [1]. The problem of the infinite square well was solved byusing the self-adjointness of the Hamiltonian operator and the momentum operator. The solutions obtained are freefrom those peculiarities in the standard solution of the infinite square well. Only two types of boundary conditions areallowed, the periodic boundary conditions and the anti-periodic boundary conditions. The results provide theoreticaljustification for the periodic boundary conditions applied to the infinite square well which is often used as the modelfor spatially confined physical systems in solid state physics and quantum statistical physics. Herein, we will use thesame tact to solve the problem of the finite square well.

First, we recapitulate the usual way of solving the problem of the finite square well [2, 3]. Consider a particle in anone-dimensional finite square well of width L = 2a and the potential

V (x) ={

0 , −a ≤ x ≤ aV0 , otherwise (1)

where V0 is constant. By the time-independent Schrodinger wave equation in quantum mechanics,

− ~2

2md2

dx2ψ(x) + V (x)ψ(x) = E ψ(x), (2)

inside the well, V (x) = 0, we have

d2ψ(x)dx2

+ k2ψ(x) = 0, (3)

where k =√

2mE/~ and E is the total energy of the particle. The possible solutions of Eq. (3) are eikx and e−ikx. Bylinear combination of the two solutions, sin(kx) and cos(kx) are also possible solutions. Outside the well, V (x) = V0,we have

d2ψ(x)dx2

− q2ψ(x) = 0, (4)

∗Electronic address: yshuang@mail.scu.edu.tw†Electronic address: eirc74@gmail.com

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where q =√

2m(V0 − E)/~ and E < V0 for bound states. The possible solutions of Eq. (4) are e−qx for a < x < ∞,and eqx for −∞ < x < −a. Then, the possible solution of even-parity is

ψ(x) =

Aeqx , −∞ < x < −aB cos(kx) , −a ≤ x ≤ aA e−qx , a < x <∞

(5)

where A and B are constants to be determined by boundary conditions and normalization requirement. The possiblesolution of odd-parity is

ψ(x) =

−Aeqx , −∞ < x < −aB sin(kx) , −a ≤ x ≤ aA e−qx , a < x < ∞

(6)

Apply the conventional boundary condition — both ψ(x) and ψ′(x) are continuous at boundaries x = ±a. To thesolution of even-parity Eq. (5), we get

k tan(ka) = q. (7)

To the solution of odd-parity Eq. (6), we get

−k cot(ka) = q. (8)

Because k and q are functions of the energy E of the particle, the allowed eigen-energies can be found by solving Eqs. (7)and (8). By substituting these eigen-energies into the tentative wave functions Eqs. (5) and (6), and normalizing them,eigen-functions of the allowed bound states can be obtained.

However, the derivative ψ′′(x) in the Schrodinger wave equation Eq. (2) is undefined at the boundaries of thefinite square well where there is abrupt potential jump. The postulation that ψ(x) and ψ′(x) are continuous at theboundaries is not justified theoretically [4]. To circumvent this difficulty, ψ(x) and its derivatives at the boundaries canbe treated by the one-sided limit, for example, at the boundary x = a, using the left-hand limit ψ(a−) = limε→0ψ(a−ε)inside the well, and the right-hand limit ψ(a+) = limε→0ψ(a + ε) outside the well, where ε > 0.

II. BOUNDARY CONDITIONS CONSTRAINED BY SELF-ADJOINT OPERATORS H AND p

With the one-sided limit, we investigate possible boundary conditions constrained by the self-adjointness of theHamiltonian H and the momentum operator p. For the finite square well, let us consider a Hilbert space which is asubset of L2(−∞,∞) consisting of square-integrable functions defined on the interval (−∞,∞), and the HamiltonianH = p2/2m + V (x) is a self-adjoint operator on the Hilbert space. For any functions ψ(x) and φ(x) in the Hilbertspace, using the integration by part, we have

〈ψ(x)| H |φ(x)〉 =−~2

2m

(K(x)

∣∣∣−a−

−∞+K(x)

∣∣∣a−

−a++K(x)

∣∣∣∞

a+

)+ (〈φ(x)| H |ψ(x)〉)∗, (9)

where

K(x) ≡ ψ∗(x)dφ(x)

dx− dψ∗(x)

dxφ(x). (10)

Because H is a self-adjoint operator on the Hilbert space, 〈ψ(x)| H |φ(x)〉 = (〈φ(x)| H |ψ(x)〉)∗. Therefore, anyfunctions ψ(x) and φ(x) in the Hilbert space satisfy the requirement

K(x)∣∣∣−a−

−∞+K(x)

∣∣∣a−

−a++K(x)

∣∣∣∞

a+= 0. (11)

For the momentum operator p = −i~ ddx , by using the integration by part, we have

〈ψ(x)| p |φ(x)〉 = −i~(χ(x)

∣∣∣−a−

−∞+ χ(x)

∣∣∣a−

−a++ χ(x)

∣∣∣∞

a+

)+ (〈φ(x)| p |ψ(x)〉)∗, (12)

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where χ(x) ≡ ψ∗(x)φ(x). Suppose that the momentum operator is also a self-adjoint operator on the Hilbert space,i.e., 〈ψ(x)| p |φ(x)〉 = (〈φ(x)| p |ψ(x)〉)∗, then any functions ψ(x) and φ(x) in the Hilbert space satisfy the additionalrequirement

χ(x)∣∣∣−a−

−∞+ χ(x)

∣∣∣a−

−a++ χ(x)

∣∣∣∞

a+= 0. (13)

There are infinitely many possible boundary conditions satisfying the requirements Eqs. (11) and (13); the conven-tional boundary condition is just one of them [4]. The choice of boundary conditions should be determined by relevantphysical conditions [5]. The common reason to impose the conventional boundary condition is that this boundarycondition ensures both the probability density and the probability current being continuous at the boundaries. Nev-ertheless, we will show that the conventional boundary condition is not the only one to ensure the continuity of theprobability density and the probability current.

III. SOLUTIONS OF BOUND STATES OF THE FINITE SQUARE WELL

On the basis of classical physics, particles are completely confined inside the well, provided that their kineticenergies are less than the potential energy of the well. In contrast, according to quantum mechanics, it is possiblethat particles leak out of the well, even if their kinetic energies are less than the potential energy of the well — theso-called tunneling effect which has been confirmed experimentally. However, the tunneling effect is insufficient to ruleout the possibility that the particle can be completely confined inside the well under certain circumstances. Actually,ψ(x) = 0 is also a possible solution of Eq. (4) outside the well. Namely, it is possible that a particle can be completelyconfined inside the well. Therefore, bound states allowed for the finite square well are classified into two kinds: (1)the particle is completely confined inside the well, and (2) there is possibility that the particle leaks out of the well.

A. The particle is completely confined inside the well

For the case that the particle is completely confined inside the finite square well, the problem can be solved exactlyby the method used previously in solving the infinite square well [1]. Consequently, there are two sets of solutions forbound states of the finite square well:

Set 1:Energies of the allowed bound states are

En =n2π2~2

2ma2=

(2n)2π2~2

2mL2, (14)

where n = 0, 1, 2, 3, · · ·. It should be noted that the allowed values of n are limited by En < V0; that is, theallowed energies En must be less than the potential energy V0. Corresponding to each eigen-energy En, there are twoindependent eigen-functions,

φ+n (x) =

{ √1L Exp(iknx) , −a ≤ x ≤ a

0 , otherwise(15)

and

φ−n (x) =

{ √1L Exp(−iknx) , −a ≤ x ≤ a

0 , otherwise(16)

where kn = nπ/a. The two eigen-functions, φ+n (x) and φ−

n (x), are also eigen-states of the momentum operator pcorresponding to the quantized momentum ±℘n, respectively, where ℘n = ~kn = nπ~/a.

It is easy to show that these eigen-functions φ±n (x) satisfy the requirements Eq. (11) and Eq. (13), by substituting

them into Eq. (11) and Eq. (13). For instance, let ψ(x) = φ+n (x) and φ(x) = φ−

m(x) in Eq. (10). Outside the

well, φ+n (x) = 0 and φ−

m(x) = 0, thus K(x)∣∣∞a+ = 0 and K(x)

∣∣−a−−∞ = 0. Inside the well, K(x) = φ+

n (x)∗ d φ−m(x)dx −

d φ+n (x)∗

dxφ−m(x) = i

L(kn − km) Exp[−i(kn + km)x]. Thus, K(−a+) = i

L(kn − km) Exp[i(n + m)π] at the boundary

x = −a, and K(a−) = iL (kn − km) Exp[−i(n +m)π] at the boundary x = a. Because Exp[−i(n+m)π] = Exp[i(n +

4

m)π] for any integers n and m, K(x)∣∣a−−a+ = 0. Therefore, the requirement Eq. (11) is fulfilled by these eigen-

functions. Because φ+n (x) = 0 and φ−

m(x) = 0 outside the well, χ(x)∣∣−a−−∞ = 0 and χ(x)

∣∣∞a+ = 0. Inside the well,

χ(x) ≡ φ+n (x)∗ φ−

m(x) = 1L

Exp[−i(kn + km)x]. Then, χ(−a+) = 1L

Exp[i(n +m)a] and χ(a−) = 1L

Exp[−i(n +m)a];consequently, χ(−a+) = χ(a−). Therefore, the requirement Eq. (13) is fulfilled by these eigen-functions.

In addition, φ+n (x) and φ−

n (x) obey the periodic boundary conditions at the boundaries x = ±a, i.e., φ±n (−a+) =

φ±n (a−) and dφ±

n (−a+)/dx = dφ±n (a−)/dx. Let H be the linear space spanned by the complete set of basis eigen-

functions φ±n (x). Then, any function ψ(x) in the Hilbert space H not only satisfies the requirements Eqs. (11) and

(13), but also obeys the periodic boundary conditions

ψ(−a+) = ψ(a−) and ψ′(−a+) = ψ′(a−). (17)

From the eigen-functions φ±n (x), we have (n 6= 0)

ψen(x) = 1√

2(φ+

n (x) + φ−n (x) )

=

{ √2L cos(knx) , −a ≤ x ≤ a

0 , otherwise

(18)

and

ψon(x) = −i√

2(φ+

n (x) − φ−n (x) )

=

{ √2L

sin(knx) , −a ≤ x ≤ a

0 , otherwise

(19)

The parity of eigen-functions ψen(x) is even, and the parity of eigen-functions ψo

n(x) is odd. Although, theseeigen-functions are eigen-states of the Hamiltonian H, they are not eigen-states of the momentum operator p.Corresponding to each non-zero eigen-energy En (n 6= 0), there are two eigen-states. Yet, for n = 0, there is only oneground state φ0(x) =

√1/L of zero energy.

Set 2:Energies of the allowed bound states are

En =(n+ 1

2)2π2~2

2ma2=

(2n+ 1)2π2~2

2mL2, (20)

where n = 0, 1, 2, 3, · · ·, such that n is limited by En < V0. Corresponding to each eigen-energy En, there are twoindependent eigen-functions:

φ+n (x) =

{ √1L Exp(iknx) , −a ≤ x ≤ a

0 , otherwise(21)

and

φ−n (x) =

{ √1L Exp(−iknx) , −a ≤ x ≤ a

0 , otherwise(22)

where kn = (n+ 12)π/a . The two eigen-functions, φ+

n (x) and φ−n (x), are also eigen-states of the momentum operator

p corresponding to the quantized momentum ±℘n, respectively, where ℘n = ~kn = (n + 12)π/a .

These eigen-functions satisfy the requirements Eq. (11) and Eq. (13). Furthermore, φ+n (x) and φ−

n (x) obey theanti-periodic boundary conditions at the boundaries x = ±a, i.e., φ±

n (−a+) = −φ±n (a−) and dφ±

n (−a+)/dx =−dφ±

n (a−)/dx. Let H be the linear space spanned by the complete set of basis eigen-functions φ±n (x). Then, any

function ψ(x) in the Hilbert space H not only satisfies the requirements Eqs. (11) and (13), but also obeys theanti-periodic boundary conditions

ψ(−a+) = −ψ(a−) and ψ′(−a+) = −ψ′(a−). (23)

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From the eigen-functions φ±n (x), we have

ψen(x) = 1√

2( φ+

n (x) + φ−n (x) )

=

{ √2L

cos(knx) , −a ≤ x ≤ a

0 , otherwise

(24)

and

ψon(x) = −i√

2( φ+

n (x) − φ−n (x) )

=

{ √2L

sin(knx) , −a ≤ x ≤ a

0 , otherwise

(25)

The parity of eigen-functions ψen(x) is even, and the parity of eigen-functions ψo

n(x) is odd.The two sets of solutions are those eigen-states of the infinite square well, as obtained previously [1], for which

their energies are less than the potential energy. As a numerical example, consider the finite square well of V0 =50 (~2/2ma2). Fig. 1 illustrates schematic plots of the eigen-functions of Eqs. (18) and (19), whose wave vectorsare labeled as kn, as well as eigen-functions of Eqs. (24) and (25), whose wave vectors are labeled as kn. To eacheigen-energy, there are two eigen-functions, one is symmetric and the other one is anti-symmetric. The ground stateφ0(x) of wave vector k0 = 0 is not shown in this figure.

B. There is possibility that the particle leaks out of the well

In the case that the particle has a nonzero probability of leaking out of the well, outside the well, the possiblesolutions of Eq. (4) are e−qx for a < x < ∞, and eqx for −∞ < x < −a. To find eigen-states of even-parity, try thewave function

ψ =

Aeqx , −∞ < x < −acos(kx) , −a ≤ x ≤ aA e−qx , a < x < ∞

(26)

where A is to be determined. From physical point of view, the probability density ρ(x) = ψ(x)∗ψ(x) is continuous atthe boundaries. Thus, it is reasonable to postulate that the tentative wave function is continuous at the boundaries.Then, we get A = cos(ka) eqa, and the tentative wave function becomes

ψek(x) =

cos(ka) eq(x+a) , −∞ < x < −acos(kx) , −a ≤ x ≤ a

cos(ka) e−q(x−a) , a < x < ∞(27)

Similarly, to find eigen-states of odd-parity, the tentative wave function is

ψok(x) =

−sin(ka) eq(x+a) , −∞ < x < −asin(kx) , −a ≤ x ≤ a

sin(ka) e−q(x−a) , a < x < ∞(28)

One more boundary condition is needed to find the allowed eigen-states. As mentioned in Sec. II, because theHamiltonian H is a self-adjoint operator on the Hilbert space, any functions ψ(x) and φ(x) in the Hilbert space mustobey the requirement Eq. (11). With these tentative wave functions Eqs. (27) and (28), K(−∞) = 0 and K(∞) = 0,because e−qx → 0 as x → ∞ and eqx → 0 as x→ −∞. If one of them is of even-parity and the other is of odd-parity,then K(−x) = K(x). For such case, the requirement Eq. (11) is automatically fulfilled. If both of them are ofeven-parity, or odd-parity, then K(−x) = −K(x). Therefore, the requirement Eq. (11) is simplified to

K(a−) = K(a+) and K(−a−) = K(−a+). (29)

Because K(−a−) = K(−a+) at the boundary x = −a is equivalent to K(a−) = K(a+) at the boundary x = a,we consider only the boundary condition at x = a. For arbitrary tentative wave functions of even-parity ψe

k1(x) and

ψek2

(x), we have, at the boundary x = a,

K(a−) = k1 sin(k1a) cos(k2 a) − k2 cos(k1a) sin(k2 a), (30)

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and

K(a+) = (q1 − q2) cos(k1a) cos(k2 a). (31)

Substituting Eqs. (30) and (31) into Eq. (29) yields

k1 tan(k1a) − q1 = k2 tan(k2 a) − q2. (32)

Since q1 is a function of k1, the left-hand side of Eq. (32) depends on k1 only. Similarly, the right-hand side of Eq. (32)depends on k2 only. Therefore, we have

k tan(k a) − q = α, (33)

where α is a real constant. By Eq. (33), the boundary condition on ψek(x) is: at the boundary x = a,

d ψek(a+)dx = −q cos(k a)

= −(k tan(k a) − α) cos(k a)= −k sin(k a) + α cos(k a)= d ψe

k(a−)dx

+ αψek(a

−)

(34)

Similarly, for arbitrary tentative wave functions of odd-parity ψok1

(x) and ψok2

(x),

k1 cot(k1a) + q1 = k2 cot(k2 a) + q2. (35)

Then, we have

−k cot(k a) − q = β, (36)

where β is a real constant. By Eq. (36), the boundary condition on ψok(x) is: at the boundary x = a,

d ψok(a+)dx = −q sin(k a)

= (k cot(k a) + β) sin(k a)= k cos(k a) + β sin(k a)= d ψo

k(a−)dx + β ψo

k(a−)

(37)

Let α = β, then the boundary conditions allowed for the wave function of even-parity Eq. (27) and the wave functionof odd-parity Eq. (28) are

ψ(a+) = ψ(a−) and ψ′(a+) = ψ′(a−) + αψ(a−). (38)

The allowed eigen-states can be obtained by solving the Schrodinger wave equation Eq. (2) with the boundaryconditions Eq. (38). Let Hα be the Hilbert space spanned by those eigen-states as obtained. For any real number α,there is a Hilbert space Hα which is characterized by the boundary conditions Eq. (38). The Hilbert space H0 (α = 0)is just the standard solution which is obtained by imposing the conventional boundary condition — both ψ(x) andψ′(x) are continuous at boundaries [2, 3].

From physical point of view on the second kind solution, both the probability density and the probability currentare required to be continuous at boundaries. The first condition in Eq. (38), ψ(a+) = ψ(a−), is assumed to ensurethe probability density is continuous at the boundaries. The probability current of wave function Ψ(x, t) is

j(x, t) ≡ i~2m

(Ψ(x, t)

∂Ψ∗(x, t)∂ x

− Ψ∗(x, t)∂Ψ(x, t)∂ x

). (39)

Letting φ(x) = ψ(x) in Eq. (10) yields

J(x) ≡ ψ∗(x)dψ(x)

dx− ψ(x)

dψ∗(x)dx

. (40)

Because Ψ(x, t) = ψ(x) Exp[−i εt/~] for wave function ψ(x) of an eigen-energy ε, J(a+) = J(a−) implies j(a+, t) =j(a−, t) for all time. In addition, J(a+) = J(a−) for wave functions ψ(x) which obey the boundary conditionsEq. (38). Thus, the probability current of wave functions, which obey the boundary conditions Eq. (38), is continuousat boundaries. Consequently, for wave functions which obey the boundary conditions Eq. (38), their probabilitydensity and probability current are continuous at boundaries. The conventional boundary condition is not the onlyone to ensure the continuity of the probability density and the probability current at boundaries.

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IV. NUMERICAL SOLUTIONS OF BOUND STATES OF THE FINITE SQUARE WELL

To show a numerical example for the second kind solution, let us consider V0 = 50 (in units of ~2/2ma2) so as tocompare with the numerical example given for the first kind solution. Define z = k a and z0 =

√2mV0a2/~. Then,

q a =√z20 − z2. Re-write Eq. (33) as

tan(z) =

√(z0z

)2

− 1 +αa

z. (41)

Also, Eq. (36) is re-written as

− cot(z) =

√(z0z

)2

− 1 +αa

z. (42)

First, we solve the case of α = 0/a. Solutions of the transcendental equations Eqs. (41) and (42) are illustratedgraphically in Fig. 2. The five intersection points of the curves tan(z) and − cot(z) with the curve

√(z0/z)2 − 1

correspond to five allowed eigen-energies. From the numerical solutions of Eqs. (41) and (42), numerical values ofquantized wave vectors kn and quantized eigen-energies En = ~2 k2

n/2m are obtained. Then, eigen-functions of theallowed bound states are obtained by substituting these values of kn into tentative wave functions ψe

k(x) and ψok(x),

and normalizing them. Schematic plots of these eigen-functions are shown in Fig. 3. Contrast to the first kind solution,there is only one eigen-state to each eigen-energy. The leaking probability of the particle in each eigen-state can beevaluated from its eigen-function. Relevant properties of the eigen-states such as eigen-energy, eigen-wavevector,parity and leaking probability are summarized in Table I.

Referring to Fig. 2 and by Eq. (41), no matter how small the potential energy V0 is, there is a solution of Eq. (41).Namely, there exists at least one bound state of vanishingly small energy, even if V0 ≈ 0. This is consistent with theexistence of the ground state of zero energy in the first kind solution.

For the case of α = 10/a, graphical solution of the transcendental equations Eqs. (41) and (42) is shown in Fig. 4.Only four eigen-states are allowed for this case, since the energy of the state n = 5 becomes larger than V0 as α = 10/a;this state is no longer bound by the potential. Schematic plots of the eigen-functions of allowed bound states are shownin Fig. 5. Eigen-energy, eigen-wavevector, parity and leaking probability of the eigen-states are given in Table II.

Referring to Table I and Table II, to each eigen-state, its energy is higher for larger α, whereas its leaking probabilityis smaller for larger α. According to Eqs. (41) and (42), by increasing α to infinity, eigen-functions of states n = 1and n = 3, which are symmetric, respectively become like those symmetric ones of eigen-states of k0 = π/2 andk1 = 3π/2 in the first kind solution (refer to Figs. 1–5). Meanwhile, eigen-functions of states n = 2 and n = 4, whichare anti-symmetric, respectively become like those anti-symmetric ones of eigen-states of k1 = π and k2 = 2π in thefirst kind solution.

For the case of α = −5/a, graphical solution of the transcendental equations Eqs. (41) and (42) is shown in Fig. 6.Schematic plots of the eigen-functions of allowed bound states are shown in Fig. 7. Eigen-energy, eigen-wavevector,parity and leaking probability of the eigen-states are given in Table III.

According to Eqs. (41) and (42), by decreasing α → −∞, eigen-functions of states n = 3 and n = 5, respectively,become like those symmetric ones of eigen-states of k0 = π/2 and k1 = 3π/2 in the first kind solution. Yet, thestate n = 1 ceases to exist. Meanwhile, the eigen-function of the state n = 4 becomes like the anti-symmetric one ofeigen-states of k1 = π in the first kind solution. Yet, the eigen-function of the state n = 2, which is anti-symmetric,does not become like the ground state of k0 = 0, which is symmetric, in the first kind solution.

To more clearly show how eigen-states vary with the parameter α, consider states n = 3 and n = 2, for example.Schematic plots of the eigen-functions of the state n = 3 for various values of α such as 100/a, 10/a, 0/a, -10/a,-100/a are shown in Fig. 8. Eigen-energy, eigen-wavevector and leaking probability of the state n = 3 for these valuesof α are given in Table IV.

By increasing α→ ∞, the wave vector of the state n = 3 approaches 3π/2 and its eigen-energy approaches (3π/2)2.Yet, its leaking probability is decreased to 0. By decreasing α → −∞, its wave vector approaches π/2 and its eigen-energy approaches (π/2)2. Also, its leaking probability is decreased to 0. The leaking probability is decreased to 0,as the wave vector of the state approaches either 3π/2, or π/2. The leaking probability is dictated by wave vectors ofeigen-states, rather than their energies.

By increasing α → ∞, the wave vector of the state n = 2 approaches π and its eigen-energy approaches π2. Itsleaking probability is decreased to 0. Yet, the wave vector of the state n = 2 can not approach 0, by decreasingα → −∞. If α < −(1 + 5

√2)/a ≈ −8.071/a, then Eq. (42) has no solution in the region 0 ≤ z ≤ π. For instance

α = −10/a, the curves − cot(z) and√

(z0/z)2 − 1 + −10z

do not intersect in the region 0 ≤ z ≤ π, as shown in Fig. 9.Thus, the state n = 2 ceases to exist, if α < −(1 + 5

√2)/a. This is consistent with there is only one ground state of

8

zero energy in the first kind solution. Fig. 10 shows schematic plots of the eigen-functions of the state n = 2, for α =100/a, 0/a, -5/a, -7/a and -8/a. Eigen-energy, eigen-wavevector and leaking probability of the state n = 2 for thesevalues of α are given in Table V.

Finally, when the potential energy V0 of the finite square well becomes infinite, then ψ(x) = 0 is the only solutionof Eq. (4) outside the well, since q = ∞. Namely, the particle is not allowed to be outside the infinite square well.Consequently, we retain all the solutions of the infinite square well as obtained previously [1].

V. CONCLUSION

Two kinds of solutions are found for the finite square well, (1) the particle is completely confined inside the well, and(2) there is possibility that the particle leaks out of the well. The solutions of the first kind are those eigen-states of theinfinite square well [1], for which their eigen-energies are less than the potential energy. For the second kind solution,there are infinitely many families of solutions, each of which is characterized by the allowed boundary conditions. Thestandard solution of the finite square well is just one of these families of solutions. Nevertheless, the standard solutionhas the advantage over other families of solutions, because it seems impossible to set up discontinuous jump of finitepotential in real physical world. In this respect, only the standard solution can be considered as the approximatesolution for the finite ”square” well whose potential raises continuously with extremely large slope at boundaries.

Acknowledgments

We gratefully acknowledge Dr. C.M.L. Leonard and Prof. Chyi-Lung Lin for valuable comments during thepreparation of this paper.

[1] Young-Sea Huang, ”Reexamination on the problem of the infinite square well in quantum mechanics”, DOI:10.13140/RG.2.1.1277.5843.

[2] D.H. McIntyre, QUANTUM MECHANICS, A Paradigms Approach (Pearson Education, Inc., 2012).[3] D.J. Griffiths, Introduction to quantum mechanics, 2nd ed., (Pearson Education, Inc., 2005).[4] D. Branson, ”Continuity condition on Schrodinger wave functions at discontinuities of the potential”, Am. J. Phys. 47,

1000-1003 (1979).[5] A.Z. Capri, ”Self-adjointness and spontaneously broken symmetry”, Am. J. Phys. 45, 823-825 (1977).

Tables and Figures:

n Parity kn En Leaking Prob.

1 even 1.37508 1.8909 0.4765 %2 odd 2.74319 7.5251 2.0021 %3 even 4.09478 16.7672 4.9572 %4 odd 5.41164 29.2859 10.5510 %5 even 6.63586 44.0346 25.5836 %

TABLE I: Eigen-energy, eigen-wavevector, parity and leaking probability of eigen-states are given for the case α = 0. Eigen-wavevector kn is in units of 1/a. Eigen-energy En is in units of ~2/2ma2. Leaking Prob. is the probability of the particleoccurring outside the well.

9

n Parity kn En Leaking Prob.

1 even 1.48332 2.2002 0.10417 %2 odd 2.96306 8.7797 0.46169 %3 even 4.43391 19.6596 1.27815 %4 odd 5.88337 34.6140 3.51297 %

TABLE II: Eigen-energy, eigen-wavevector, parity and leaking probability of eigen-states are given for the case α = 10/a.

n Parity kn En Leaking Prob.

1 even 1.07519 1.1560 2.2766 %2 odd 2.22621 4.9560 7.1450 %3 even 3.46553 12.0099 11.8264 %4 odd 4.76038 22.6612 15.8895 %5 even 6.06245 36.7532 21.3303 %

TABLE III: Eigen-energy, eigen-wavevector, parity and leaking probability of eigen-states are given for the case α = −5/a.

α k3 E3 Leaking Prob.

100/a 4.66809 21.7911 0.0366 %10/a 4.43391 19.6596 1.2782 %0/a 4.09478 16.7672 4.9572 %

-10/a 2.15397 4.6396 5.4142 %-100/a 1.58785 2.5213 0.0055 %

TABLE IV: Eigen-energy, eigen-wavevector and leaking probability of the state n = 3 are given for various values of α.

α k2 E2 Leaking Prob.

100/a 3.11234 9.6866 0.0134 %0/a 2.74319 7.5251 2.0021 %

-5/a 2.22621 4.9560 7.1450 %-7/a 1.50993 2.2799 13.0637 %-8/a 0.41734 0.1742 17.1884 %

TABLE V: Eigen-energy, eigen-wavevector and leaking probability of the state n = 2 are given for various values of α.

k�

0 = Π�2k1 = Π

k�

1 = 3Π�2

k2 = 2 Π

-2 -1 1 2x

10

20

30

40

50

E

FIG. 1: (Color online) Schematic plots of the eigen-functions of bound states of the finite square well of V0 = 50 (~2/2ma2).Energy E is in units of ~2/2ma2. Position x is in units of a. Wave vector k is in units of 1/a. Eigen-functions of even-parityand odd-parity are, respectively, in solid blue line and dashed red line.

10

0 Π2

Π 3 Π2

2 Π 5 Π2

-5

0

5

10

15

zfHzL

FIG. 2: (Color online) Graphical solution of the transcendental equations Eqs. (41) and (42) for the case α = 0/a. Curves of

tan(z), − cot(z) and g(z) =√

(z0/z)2 − 1 are in solid red line, dashed blue line, and dot-dashed black line, respectively. Theintersection points of the curves represent solutions of the transcendental equations.

n = 1

n = 2

n = 3

n = 4

n = 5

-2 -1 1 2x

10

20

30

40

50

E

FIG. 3: (Color online) Schematic plots of the eigen-functions of bound states of the finite square well for the case α = 0/a.Energy E is in units of ~2/2ma2. Position x is in units of a.

0 Π2

Π 3 Π2

2 Π 5 Π2

-5

0

5

10

15

z

fHzL

FIG. 4: (Color online) Graphical solution of the transcendental equations Eqs. (41) and (42) for the case α = 10/a. Curves of

tan(z), − cot(z) and g(z) =√

(z0/z)2 − 1 + 10z

are in solid red line, dashed blue line, and dot-dashed black line, respectively.

11

n = 1

n = 2

n = 3

n = 4

-2 -1 1 2x

10

20

30

40

50

E

FIG. 5: (Color online) Schematic plots of the eigen-functions of bound states of the finite square well for the case α = 10/a.

0 Π2

Π 3 Π2

2 Π 5 Π2

-2

-1

0

1

2

3

z

fHzL

FIG. 6: (Color online) Graphical solution of the transcendental equations Eqs. (41) and (42) for the case α = −5/a. Curves of

tan(z), − cot(z) and g(z) =√

(z0/z)2 − 1 + −5z

are in solid red line, dashed blue line, and dot-dashed black line, respectively.

n = 1n = 2

n = 3

n = 4

n = 5

-2 -1 1 2x

10

20

30

40

50

E

FIG. 7: (Color online) Schematic plots of the eigen-functions of bound states of the finite square well for the case α = −5/a.

12

Α = -100�aΑ = -10�aΑ = 0�aΑ = 10�aΑ = 100�a

FIG. 8: (Color online) Schematic plots of the wave functions of the state n = 3 for α = 100/a, 10/a, 0/a, −10/a, −100/a.

0 Π4

Π2

3 Π4

Π-50

-40

-30

-20

-10

0

10

20

z

fHzL

FIG. 9: (Color online) For the case α = −10/a, graphical solution of the transcendental equations Eqs. (41) and (42) in the

region 0 ≤ x ≤ π. Curves of tan(z), − cot(z) and g(z) =√

(z0/z)2 − 1 + −10z

are in solid red line, dashed blue line, anddot-dashed black line, respectively.

Α = -8�aΑ = -7�aΑ = -5�aΑ = 0�aΑ = 100�a

FIG. 10: (Color online) Schematic plots of the wave functions of the state n = 2 for α = 100/a, 0/a, −5/a, −7/a, −8/a.