The quantum Gaussian wellSaikat Nandia�

Tata Institute of Fundamental Research, Mumbai-400005, India

�Received 9 April 2010; accepted 13 July 2010�

Different features of a potential in the form of a Gaussian well have been discussed extensively.Although the details of the calculation are involved, the general approach uses a variational methodand WKB approximation, techniques that should be familiar to advanced undergraduates. Anumerical solution of the Schrödinger equation through diagonalization has been developed in aself-contained way, and physical applications of the potential are mentioned. © 2010 AmericanAssociation of Physics Teachers.

�DOI: 10.1119/1.3474665�

I. INTRODUCTION

Potentials such as the infinite square well, the harmonicoscillator, the delta function well, and the finite square wellare frequently discussed in textbooks1,2 as examples thathave bound states. In this paper, we consider solutions of thetime independent Schrödinger equation,1

−�2

2m

d2

dx2��x� + V�x���x� = E��x� , �1�

for the Gaussian well given by

V�x� = − V0e−�x2�− � � x � + �� , �2�

where V0�0 and ��0. In Eq. �1�, m is the mass of theparticle and ��x� is the eigenfunction. We will obtain anestimate of the ground state energy from a simple variationalmethod and determine that there are a finite number of boundstates using the WKB approximation. We also formulate Eq.�1� as a matrix eigenvalue problem, which can be used forany Hamiltonian. For simplicity, we choose units such that�=m=1 in the following.

II. BOUND STATE CRITERION

Consider the Hamiltonian H for a particle in one dimen-sion subjected to a potential V�x�,

H = −1

2

d2

dx2 + V�x� . �3�

To demonstrate that H possesses a bound state, it is sufficientto construct a real normalized trial function ��x� such thatthe expectation value of H for this trial function is negative,that is,

�H� � �−�

+�

�H�dx � 0. �4�

This sufficiency condition becomes

�H� = �−�

+� 1

2d�

dx�2

+ V�2�dx . �5�

Defining L to be the width �or, more precisely, the length

scale� of ��x�, we can write � in the form

1341 Am. J. Phys. 78 �12�, December 2010 http://aapt.org/ajp

��x� =1 L

� x

L� . �6�

We substitute this form for � in Eq. �5� and obtain

�H� =a1

2L2 +a2

L, �7�

where a1=�−�+��d� /dx��2dx� �with x�=x /L� is a dimension-

less positive number and a2=�−�+�V�2dx. If L is sufficiently

large, the second term on the right-hand side of Eq. �7� domi-nates the first term, and ���x /L��2 becomes ����0��2 �apositive constant� so that it can be taken out of the integral.Hence, the sign of a2 depends only on the sign of �−�

+�Vdx.For both a1 and a2 to be positive, it is not possible to obtaina positive value of L for which �H� is negative. Thus, thecondition on V�x� to have at least one bound state is

�−�

+�

Vdx � 0, �8�

which depends on the shape of the potential, not on itsstrength.

Equation �8� is a sufficient condition for a potential tohave a bound state. It is not a necessary condition as theexample of the harmonic oscillator potential shows. Notethat this general criterion for the existence of the bound statefollows from a simple argument involving dimensionalanalysis and the variational principle.

III. GROUND STATE ENERGY FROM THEVARIATIONAL METHOD

In practice, we can obtain various upper bounds of theground state energy by calculating �H� for suitably chosentrial functions. There is a tradeoff between improving theapproximate ground state energy using a complex trial wavefunction and the ease of calculation. Here we choose a nor-malized Gaussian trial wave function ��x� with adjustablewidth,

��x� = 2b

�1/4

e−bx2, �9�

where b is related to the width L. The expectation value of

the Hamiltonian H becomes

1341© 2010 American Association of Physics Teachers

�H� =b

2− V0 2b

2b + �. �10�

From Eq. �10�, the condition for a minimum is d�H� /db=0or

b�2b + ��3 = 2V02�2. �11�

Equation �11� determines the variational parameter b forgiven values of V0 and �. The variational method groundstate energies for different values of V0 and � are given inTable I. As expected, these estimates are greater than theexact �numerically obtained� ground state energies but areremarkably close.

IV. FINITE NUMBER OF BOUND STATES

The wave function �n�x� corresponding to the eigenvalueEn of a discrete spectrum has n nodes. The nature of thenumber of bound states is best demonstrated by the WKBapproximation. States belonging to the discrete energy spec-trum are semiclassical only for large values of n. In onedimension the WKB integral for the energy E is given by2

�x1

x2 2�E − V�x��dx = n −1

2�, for n = 1,2,3, . . . ,

�12�

where x1 and x2 are the turning points of the classical motion.Because the discrete spectrum lies in the range of energyvalues for which the particle cannot move to infinity, theenergy must be less than the limiting values V�x= ��. Forthe Gaussian potential, this condition implies that E�0.Thus, the number of discrete levels N is obtained by settingE=0 in Eq. �12� so that n=N is close to the quantum numberfor the last bound state. For the classical turning points x= �, Eq. �12� becomes

2V0�−�

+�

e−�x2/2dx = N −1

2� , �13�

which gives

N =2

V0

�+

1

2. �14�

For V0 and � to be finite, N is also finite. Therefore, thenumber of bound states is finite for the Gaussian well. Thevalue of N calculated from Eq. �14� compared to the numeri-

Table I. Comparison of the ground state energy �H� calculated from thevariational principle and the numerical solution Enum. �H� is within 2% ofthe corresponding numerically evaluated value.

V0 � b �H� Enum

1.0 1.0 0.3742 0.4671 0.47742.5 0.5 0.6113 1.8005 1.80383.0 1.0 0.8717 1.9557 1.96373.0 0.1 0.3504 2.6312 2.6316

cal solution is in Table II.

1342 Am. J. Phys., Vol. 78, No. 12, December 2010

Equation �14� implies that the number of bound states is afunction of the ratio V0 /�, which is a measure of the lengthscale L of the wave functions.3 Because V�x�� � � /�e−�x2

,V�x� becomes a delta function in the limit �→� for whichEq. �13� gives N=1 /2, implying that there exists exactly onebound state for the delta function well.1 The reason for thehalf-quantum number is that the ground state has no nodes. IfV0→�, then N→�, which is the case for an infinite squarewell.1

V. NUMERICAL SOLUTION OF THESCHRÖDINGER EQUATION

The one-dimensional Schrödinger equation has few ana-lytic solutions, and most problems must be solved numeri-cally. We first use a Taylor’s series to obtain a discretizationof derivatives for a function f�x� and write the second de-rivative of the function f�x� in the three point central differ-ence form,

f��x� =f�x + �� − 2f�x� + f�x − ��

�2 + O��2� , �15�

where � is the step size. If p is the number of mesh pointsand xmax and xmin are the maximum and minimum values ofthe variable x, the step size � is

� =xmax − xmin

p. �16�

The solution of Eq. �1� is accomplished by discretizing itusing Eq. �15� and evaluating the functions and derivatives atxk=xmin+k� �for k=1,2 , . . . , p−1�,

−��xk + �� − 2��xk� + ��xk − ��

2�2 + V�xk���xk� = E��xk� ,

�17�

Because the solution is expected to decay exponentially out-side xmax and xmin, we solve Eq. �17� in the interval�xmax,xmin�. These cut-offs and the step size need to be ad-justed to obtain the desired accuracy.

Equation �17� is equivalent to the tight-binding approxi-mation applied to a chain of atoms with spacing � and oneorbital per atom.4 We can develop a matrix representation ofthe Schrödinger equation, with −�−2 /2 as the subdiagonaland superdiagonal matrix elements and the diagonal ele-ments are �−2+V�xk�. Equation �17� can then be written as a

Table II. Number of bound state given by the WKB approximation �N� andthe numerical method �Nnum�.

V0 /� N Nnum

0.5 1.3 11.0 1.6 1

10.0 4.1 4100.0 11.8 11

matrix equation as

1342Saikat Nandi

��−2 + V�x1� �− �−2/2� 0 0 . . . 0 0

�− �−2/2� �−2 + V�x2� �− �−2/2� 0 . . . 0 0

. . . . . . . . . . . . . . . . . . . . .

0 . . . . . . . . . . . . �−2 + V�xp−2� �− �−2/2�0 . . . . . . . . . . . . �− �−2/2� �−2 + V�xp−1�

����x1���x2�]

��xp−2���xp−1�

� = E���x1���x2�]

��xp−2���xp−1�

� . �18�

Equation �18� is a matrix eigenvalue problem with a tridi-agonal matrix of dimension �p−1�� �p−1�; thus, there are�p−1� eigenvalues. Because all the matrix elements are realand the transpose of the matrix is equal to the matrix itself, itis a Hermitian matrix, and hence all the eigenvalues are real.An efficient way of diagonalizing a tridiagonal matrix is touse the standard LAPACK routine DSTEVX,5 which storesthe symmetric tridiagonal matrix in two one-dimensional ar-rays, one of length �p−1� containing the diagonal elementsand one of length �p−2� containing the off-diagonal ele-ments, and returns the eigenvalues along with the eigenfunc-tions of the matrix.

The eigenvalues and corresponding eigenfunctions for theGaussian well with V0=3.0 and �=0.1 are shown in Fig. 1.The ground state is symmetric with respect to the center ofthe potential �even parity�, and the first excited state is anti-symmetric �odd parity�. The wave functions resemble thoseof the harmonic oscillator1 because for x�1, the dominantterm in the expansion of e−�x2

is proportional to x2.The virtue of the matrix method is that it can be applied to

any potential for which the Hamiltonian can be brought intoa symmetric tridiagonal or bidiagonal matrix. However, forsystems with periodic boundary conditions, the Hamiltonianis no longer tridiagonal, and we cannot use the simple matrixmethod.

VI. TUNNELING IN A GAUSSIAN BARRIER

A Gaussian barrier can be constructed by changing thesign of V0 in Eq. �2�. Because the condition in Eq. �8� is not

-3

-2.5

-2

-1.5

-1

-0.5

0

-10 -5 0 5 10

Ene

rgy

(a.u

.)

Position (a.u.)

Fig. 1. The energy eigenfunctions and the corresponding eigenvalues for

V0=3.0 and �=0.1. Note the unequal spacing between different levels.

1343 Am. J. Phys., Vol. 78, No. 12, December 2010

satisfied, there is no bound state. The barrier formed in thisway is an interesting example of tunneling. We start with thetime energy uncertainty principle,

�E�t � 12 . �19�

Denote the energy of the incident particle by E. The uncer-tainty in the energy is �E and for sufficiently small �t, theenergy of the particle E+�E is greater than the height of thebarrier V0. Tunneling will take place if in the time �t theparticle can traverse the barrier. We take �−1/2 as the width ofthe barrier and write

�t ��−1/2

2�E + �E − V0�. �20�

From Eqs. �19� and �20�, we find �E to satisfy the equation

��E�2 −�

2�E +

�

2�V0 − E� = 0. �21�

The condition for �E to be real implies that

�

8� �V0 − E� , �22�

which is the condition for tunneling. The left-hand side ofEq. �22� is the kinetic energy of the particle obtained fromthe position-momentum uncertainty relation, with an uncer-tainty of �x��−1/2 in the particle’s position. We see that fortunneling to occur, the kinetic energy of the particle must begreater than the difference between the height of the barrierV0 and the total energy E.

We now investigate tunneling using the WKB approxima-tion. Consider the classically inaccessible region, E�V�x�,as a�x�b so that we can write the transmission coefficientas2

T � exp− 2�a

b

��x�dx� , �23�

where ��x�= 2�V�x�−E�. We define the opacity of the bar-rier by

� = exp�a

b

��x�dx� , �24�

so that Eq. �23� becomes T��−2. No attempt been made toobtain an analytical solution to the integral in Eq. �23�. Butwe can find an approximate solution of the integral that cor-rectly predicts the transmission coefficient T.

We introduce the dimensionless quantity �=V0 /E�1 andwrite the classical turning points as a=− ln � /� and b= ln � /�. The change in variables, x=y ln � /�, changes

Eq. �24� to

1343Saikat Nandi

� = exp 2V0 ln �

��

−1

+1 �1 − �1 − �−y2+ �−1��dy� .

�25�

The term �1−�−y2+�−1� is always less than unity. If we as-

sume that this term is approximately equal to unity,6 then thebinomial expansion of �1− �1−�−y2

+�−1��1/2 to first order is

� � exp 2V0 ln �

��

−1

+1 1

2+

1

2�−y2

−1

2��dy� , �26�

which can be further expressed as

� � exp 2V0

� ln � +

2erf� ln �� −

ln �

��� ,

�27�

where erf�x� is the error function of x. Equation �27� impliesthat T depends only on the ratio �=V0 /E, a common featureof barrier tunneling. The dependence of T on � can be seenfrom Fig. 2. It is evident that the transmission coefficientdecreases as � increases. As expected for �=1, that is, E=V�x=0�, T=1, implying no reflection from the barrier.

-1.4

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

-4 -3 -2 -1 0 1 2 3 4

Ene

rgy

(a.u

.)

Position (a.u.)

Fig. 3. The energy levels �ground state and the first excited state� along withthe wave functions for U�x� with V0=3.0 and �=1.0. We can see the tun-

0

0.2

0.4

0.6

0.8

1

1 1.5 2 2.5 3

Tra

nsm

issi

onC

oeffi

cien

t(T

)

V0/E

T

Fig. 2. Dependence of the transmission coefficient T on �=V0 /E. As ex-pected, T decreases as � increases.

neling of the wave function through the barrier.

1344 Am. J. Phys., Vol. 78, No. 12, December 2010

VII. A DOUBLE GAUSSIAN WELL

An interesting double well potential may be formed bymultiplying V�x� in Eq. �2� by a factor of x2. Because the

potential U�x�=−V0x2e−�x2is symmetric, the Hamiltonian

can be brought into a symmetric tridiagonal form. The nu-merical solution for this potential is given in Fig. 3. Becausethe barrier is finite, the wave function extends into bothwells. The ground state wave function is small but nonzeroinside the barrier. Assume that the particle starts in the rightwell. From Fig. 3, we see that initially �=�1+�2 �where �1and �2 are the ground state and first excited state, respec-tively� is canceled on the left. As a consequence, the particlewill oscillate between the wells with the period �=2� / �E2−E1�. Thus the tunneling rate depends on the en-ergy difference, �E=E2−E1, and a double well with a highor wide barrier will have a smaller �E than one with a low ornarrow barrier. Also, �E will become larger as the energyincreases �that is, as �V0−E� decreases�.

The double Gaussian well is a good model for a two levelsystem. An example is a quantum well laser based on thetransition that an electron makes between the ground andfirst excited state of a double Gaussian well. By choosing �and V0, we can tune the wavelength of the light emitted.Although the double Gaussian well has multiple energy lev-els, it can be seen from Fig. 4 that the spacing between theground and first excited state is very small compared to thatbetween the first and second excited states. Therefore, thelowest two bound states can be effectively decoupled fromthe other states by choosing appropriate values of V0 and �.

ACKNOWLEDGMENTS

The author would like to thank Sayan Chakraborti, Swas-tik Bhattacharya, and Shamashis Sengupta for many helpfulsuggestions. The author is grateful to Professor Deepak Dharfor help in understanding the different aspects of quantummechanics. Thanks are also due to the anonymous refereesfor their constructive reviews.

a�Electronic mail: saikat@tifr.res.in1

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

-4 -3 -2 -1 0 1 2 3 4

Ene

rgy

(a.u

.)

Position (a.u.)

Fig. 4. The wave functions and the corresponding energy eigenvalues ofU�x� for V0=10.0 and �=1.0. The proximity between the lowest two statesis evident.

D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed. �Prentice-

1344Saikat Nandi

Hall, Englewood Cliff, NJ, 2005�, Chap. 2.2E. Merzbacher, Quantum Mechanics, 3rd ed. �Wiley, New York, 2005�,Chap. 7.

3The ratio 1 /� is the length scale for the potential. For fixed V0 /�, thelength scale of the wavefunction can be identified as L��1 / �� / �V0 /��= � /V0.

4

T. B. Boykin, “Tight-binding-like expressions for the continuous-space

1345 Am. J. Phys., Vol. 78, No. 12, December 2010

electromagnetic coupling Hamiltonian,” Am. J. Phys. 69, 793–798�2001�.

5E. Anderson et al., LAPACK Users’ Guide, 3rd ed. �SIAM, Philadelphia,1999�, p. 338.

6We can add more terms in the binomial expansion to obtain more accu-rate values of T. However, the essential physical behavior of T will re-

main the same.

1345Saikat Nandi