The Chemical Educator

Exact solutions of the quantum double square well potential

Enrique Peacock-Lopez1, ∗

1Department of ChemistryWilliams College

Williamstown, MA 01267(Dated: June 13, 2006)

For a symmetrical quantum double square well potential, we find analytical expressions satisfied bythe quantized energies. Graphical or numerical solutions of the former relations allow us to calculatenormalization constants and construct the first eight solutions of the Schrodinger equation. Withthese exact solutions, we analyze quantum tunneling across a potential barrier and compare ourresults with the experimental data for ammonia.

INTRODUCTION

Most chemistry textbooks [1]-[15] in their quantummechanics sections discuss, to different levels of analysis,the one-dimensional particle in a box (PIB) as a relevantone-dimensional quantum system. A natural applicationof the PIB considers delocalized π electrons in hydro-carbons or other organic compounds. In contrast withphysicists [16]-[22], who usually consider other relevantone-dimensional potential related to scattering, chemistsrarely consider other simple but chemically relevant one-dimensional potentials. In some textbooks [5]-[15] wefind a qualitative discussion of quantum tunneling andbarrier penetration but rarely a quantitative analysis .

Some advanced texts [5]-[15] may include a qualitativediscussion of barrier penetration in two minima or doublewell potentials, but again no quantitative analysis. In achemical system, the two minima potential corresponds,usually, to two equilibrium positions or molecular con-formations. In the case of ammonia its so-called periodicinversion is a well documented example of quantum tun-neling.

The energy level splitting resulting from barrier pene-tration is an important quantum mechanical effect thatoccurs whether the double well potential is symmetricor not. The internal rotation in CH3CH3 from onestaggered configuration to another through an eclipsedpositions at the top of the barrier is a good exampleof a symmetric potential. In contrast, in the cases ofthe asymmetric H2O2 hindered rotor or beryllium di-cyclopentadienyl we encounter asymmetric double wellpotentials,. Ammonia (NH3), cyanamide (NH2CN),PH3, and AsH3 are chemical examples of symmetricaldouble well potentials where atoms can tunnel throughthe barrier. In the case of ammonia the double minimarepresents the possible nitrogen positions with respect toplane defined by the three hydrogens.

In the case of ammonia and similar molecules the vi-brational level ting due to barrier penetration has beenobserved through infrared (IR) spectroscopy. The am-monia double well potential can be approximated usingthe Manning potential [23] where the the depth or dis-

sociation energy is estimated to be about 5 eV and theheight of the central barrier to be 0.25 eV. The quanti-tative analysis of the Manning [23] or similar potential[24] is quite difficult and only numerically tractable. Asan alternative, some authors [7, 15, 18, 19, 21] consider aone-dimensional discontinuous infinite depth double wellpotential to emphasize the symmetry of the wave func-tion and quantum tunneling, but none consider a quan-titative analysis or a finite depth well..

In this paper we consider a double well finite depthpotential as an approximation of the ammonia poten-tial. In the second section we consider the discontinu-ous potential and its more general analytical solutions,as well as the continuity and smoothness conditions im-pose on the solutions of the Schrodinger equation by thequantum mechanical postulates. In the third section, weconsider the boundary conditions that yield the allowedquantized energies, and we calculate step by step the re-sultant transcendental equation for the case of infinitedepth and energy greater that the barrier hight, All ofthe algebraic manipulations are straight forward and ac-cessible to chemistry juniors with the typical mathemat-ical background. In section four and five we calculatenumerically the allowed energies and construct the firsteight normalized wave function. In section six we dis-cuss quantum tunneling. Direct comparison between ourresults and experimental data from ammonia is considerin section seven. Finally we summarize and offer somesuggestion of how to use this paper in the classroom.

GENERAL SOLUTION

By dividing into five spatial regions, we set general dif-ferential equations derived from the Schrodinger equation(SE) and construct general solutions. The Double SquareWell Potential (DSWP) is a simplified model of poten-tials found in molecular chemical systems where possibledifferent conformations are separated by an energy bar-rier. Although the DSWP is a discontinuous function ofposition, it gives an appropriate and tractable descrip-tion of the continuous double minima potentials. TheDSWP is defined by the following piecewise function:

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V (x) =

VD (L + a

2 ) ≤ x < ∞0 a

2 < x < (L + a2 )

Vo −a2 ≤ x ≤ a

20 −(L + a

2 ) < x < −a2

VD −∞ < x ≤ −(L + a2 )

(1)

where VD is the depth or dissociation energy and Vo theheight of the central barrier separating two equivalentspatial regions. Also notice that the potential is an evenfunction of position, i.e., V (−x) = V (x). Due to thissymmetry, the solutions of the SE are either odd or evenfunctions of position.

-3 -2 -1 0 1 2 3

1

2

3

4

5E�epb vs x�L

L RIII II I

FIG. 1: The Double Well Square Potential diagram, whereregions L, R, I, II, III are defined

Since the potential rises to the dissociation energy, VD,for |x| ≥ L + a/2 the SE,

− ~2

2m

d2 ΨRL

d x2+ VD ΨRL = E ΨRL (2a)

for E < VD can be reduced to

d2 ΨRL

d x2= κ2 ΨRL (2b)

with

κ =

√8mπ2(VD − E)

h2(2c)

Now we consider the three internal spatial regions.First, we set the SE in each of the three spatial regions.

In region I, defined by a/2 < x < L + a/2, we have

− ~2

2m

d2 ΨI

d x2= E ΨI (3a)

In region II, |x| ≤ a/2, we need to include the constantpotential Vo,

− ~2

2m

d2 ΨII

d x2+ Vo ΨII = E ΨII (3b)

Finally, in region III, −(L + a/2) < x < −a/2,

− ~2

2m

d2 ΨIII

d x2= E ΨIII (3c)

For regions I and III, we can reduce the SE to a simplergeneral equation

d2 Ψd x2

= − α2 Ψ (4a)

where we have defined

α =

√8mEπ2

h2(4b)

In region II we have two cases. In the first case the SEyields the following general equation:

d2 Ψ<II

d x2= β< 2 Ψ<

II (5a)

where E < Vo,, and we have defined

β< ≡√

8mπ2(Vo − E)h2

(5b)

In the second case E > Vo and we get

d2 Ψ>II

d x2= − β> 2 Ψ>

II (6a)

with

β> ≡√

8mπ2(E − Vo)h2

(6b)

Since α, β> and β< are positive real numbers, the solu-tions to eqs 4a, 6a are the simple and well known sineand cosine functions. In the case of eq 5a, the solutionsare the hyperbolic sine and cosine, i.e., sinh and cosh.

The most general solution of the SE in each regionis a linear combination of sine and cosine or sinh andcosh functions. Consequently we have six unknown co-efficients that must be determined. The main problemin many of the one-dimensional quantum potentials isnot the solution of the differential equations but find-ing coefficients that satisfy the required continuity andsmoothness conditions at the regions’ boundaries. In

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other words, quantum mechanics requires that the piece-wise wave function and its first derivative must be contin-uos. In our case, these conditions reduce to the followingequations:

ΨII(a/2) = ΨI(a/2) (7a)

d ΨII

d x

∣∣∣∣x=a/2

=d ΨI

d x

∣∣∣∣x=a/2

(7b)

ΨIII(−a/2) = ΨII(−a/2) (7c)

d ΨIII

d x

∣∣∣∣x=−a/2

=d ΨII

d x

∣∣∣∣x=−a/2

(7d)

The continuity conditions at the edges of the potentialregion yield transcendental relations, whose solutions de-fine the quantized energy levels for the DSWP

ΨI(L + a/2) = ΨR (L + a/2) (8a)

d ΨI

d x

∣∣∣∣x=L+a/2

=d ΨR

d x

∣∣∣∣x=L+a/2

(8b)

ΨIII(−(L + a/2)) = ΨL (−(L + a/2)) (8c)

d ΨIII

d x

∣∣∣∣x=−(L+a/2)

=d ΨL

d x

∣∣∣∣x=−(L+a/2)

(8d)

Finally, for E < VD, we have the following limiting con-ditions

limx→ ∞

ΨR(x) = 0 (9a)

limx→− ∞

ΨL(x) = 0 (9b)

These eight conditions determine the ten coefficients ofthe general solution of the SE for the DSWP. But wecan considerably reduce the algebraic manipulations byrecalling the potential symmetry and its implication tothe solutions of the SE.

From eqs 2b, 9 we notice that the only possible solu-tions for |x| ≥ L + a/2 are simple exponential functions

ΨR(x) = DR exp (− κ x) (10a)

ΨL(x) = DL exp ( κ x) (10b)

Since the potential is an even function of position, thesolutions of the SE are either even or odd functions ofposition,

Ψ(−x) = ±Ψ(x) . (11)

As a consequence for |x| ≥ L + a/2 we are required thatDR and DL differ only by a sign, DL = ±DR dependingon the symmetry of the wave function. In region II weare required to pick either an even or an odd function.In region III we pick the same coefficients as in region I.In other words, our general solution for for a/2 < x <L + a/2 and E > Vo reduces to

ΨI(x) = A cos(αx) + B sin(αx). (12a)

For −a/2 ≤ x ≤ a/2 we have two choices, the symmetricor even function,

ΨSII(x) = CS cos(β> x) , (12b)

and the asymmetric or odd solution,

ΨAII(x) = CA sin(β> x). (12c)

Finally for −(L + a/2) < x < −a/2, we have

ΨAIII = ± ΨI(−x). (12d)

where we pick the positive sign for the even solution andthe negative sign for the odd solution. Notice that byconsidering the potential symmetry, we have reduced thenumber of coefficients and the chosen solution clearlyshows the required symmetry.

In the case where E < Vo, we write the solution inregion I as a/2 < x < L + a/2 and get

ΨI(x) = A sin(αx) + B cos(αx). (13a)

where we have used over-barred coefficients to distinguishthe E < Vo case from the E > Vo case.

For −a/2 ≤ x ≤ a/2 we substitute the sine and cosinefunctions in eqs 12b,12c by the corresponding hyperbolicfunction. In the even case we get

ΨSII(x) = CS cosh(β< x) , (13b)

and in the odd case we get

ΨAII(x) = CA sinh(β< x). (13c)

Finally for −(L + a/2) < x < −a/2, we have

ΨAIII = ± ΨI(−x). (13d)

So far we have dealt with general results that have to becomplemented by the boundary conditions to find par-ticular solutions.

BOUNDARY CONDITIONS

In the previous section we found the most general so-lutions of the SE for the DSWP, but remember thatquantum systems allow only a discrete set of energies,

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{En}. These En are consistent with the the continuityand smoothness condition at the potential boundaries.

The most familiar boundary condition requires, as inthe case of the particle in a box, that the wave func-tion vanishes at the edges of the “box”, and, given thesymmetry of the potential, we only have to consider oneedge,

ΨI(L + a/2) = ΨR (L + a/2) (14)

Using eq 10a, 12a we get

A cos(α(L + a/2)) + B sin(α(L + a/2)) =DR exp (− κ (L + a/2))

(15a)

The slopes of the Ψs yield a second condition

A sin(α(L + a/2)) − B cos(α(L + a/2)) =κ

αDR exp (− κ (L + a/2))

(15b)

Now we use eq 15a in eq 15b and rearrange the

A

[1 +

α

κ

B

A

]cos (α(L + a/2)) =

− B

[1− α

κ

A

B

]sin (α(L + a/2))

(16)

Equation 14 is satisfied in all possible cases, and wecan recast it as

tan (α (L + a/2)) = − A

BF

(A,B,

α

κ

). (17)

where we define

F(A,B,

α

κ

)≡

[1 + α

κBA

][1− α

κAB

] (18)

Equation 17 is a transcendental equation that we needto solve for the quantized energies, which are a conse-quence of the spatial constraint posed on the system.The exact solutions of eq 17 can only be obtained nu-merically, and we can do so in many different ways. Inour approach we first rewrite eq 17 as

α (L+a/2) = n π + arctan

(−A

BF

(A,B,

α

κ

))(19)

As we will show, once we fix the values of the mass,the length of the box, the thickness and the height ofthe barrier, and the depth of the potential, α, κ, A andB depend on the energy, E, and only a finite number ofenergies satisfy the relation expressed by equation eq 19.

Before proceeding with algebraic manipulations andfinding expressions for the coefficients A and B, we rescalethe spatial length and energy as follows:

α x =

√8mL2 Vo

h2

√E

Voπ

x

L≡ γ c π X (20a)

κ x ≡ γ√

D2 − c2 π X (20b)

D ≡√

VD

Vo(20c)

β< x = γ√

1− c2 π X (20d)

β> x = γ√

c2 − 1 π X (20e)

r ≡ a

2 L(20f)

In eq 20a, γ2 represent the height of the barrier in unitsof epb defined as

epb ≡ h2

8mL2, (21)

which is determined by the the mass of the particle andthe length of the box. Also c2 represents the ratio of thesystem’s energy to the height of the barrier, and finallyX is the distance in units of L.

Now we are ready to consider the even and odd solu-tions for E > Vo and E < Vo and to analyze the boundaryconditions expressed by eqs 7. First we consider E > Vo

or c > 1, and we find that the even solution at x = a/2satisfies the following continuity condition:

AS cos (γ c π r) + BS sin (γ c π r)

= CS cos(γ

√c2 − 1 π r

)(22a)

In the case of the smoothness condition we get

AS sin (γ c π r)−BS cos (γ c π r)

= CS

√1− 1

c2sin

(γ

√c2 − 1 π r

)(22b)

We can solve for A if we first multiply eq 22a by cos(γcπr)and eq 22b by sin(γcπr). Second we add the resultingequations, and since sin2 + cos2 = 1 we find that As isgiven by the following expression:

AS = CS[cos(γcπr) cos

(γ√

c2 − 1 π r)

+√1− 1

c2sin(γcπr) sin

(γ√

c2 − 1 π r)] (23a)

Now we can multiply Eq.(22a) by sin and we substracteq 22b multiplied by cos. So we get

BS = CS[sin(γcπr) cos

(γ√

c2 − 1 π r)−√

1− 1c2

cos(γcπr) sin(γ√

c2 − 1 π r)] (23b)

Notice that both A and B are expressed as function ofCs that can be calculated numerically using the required