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XI Discrete Variable Representation

The goal of this section is to introduce a generic discrete variable representation (DVR) method, introduced by Colbert and

Miller [J. Chem. Phys.(1992)96:1982-1991] to solve the time-independent Schrodinger equation,

HCj CjEj = 0, (80)

and obtain the eigenvaluesEj corresponding to the eigenstatesj(x)in a grid-based representation: (x) =

jCj(x

xj). The DVR method simply evaluates the Hamiltonian matrix H in a discrete (grid-based) representation and obtains

the eigenstates and eigenvalues by using standard numerical diagonalization methods e.g., TRED2, TQLI and EIGSRT,

as described in Numerical Recipes (Ch. 11, Numerical Recipes), or Lanczos-type (iterative linear algebra methods) that

exploit the sparsity of H.

The rest of this section shows that the Hamiltonian matrix elements are:

H(j, j) = V(xj)jj+

2

2m(1)jj

jj

2

3 + (1 jj)

2

(jj)2

, (81)

when the delta functions are placed on a grid xj = j that extends over the interval x = (,) withj = 1, 2,....

Furthermore, we show that for the particular case of a radial coordinate defined in the interval x= (0,), the Hamiltonian

matrix elements are:

H(j, j) = V(xj)jj+

2

2m(1)jj

jj

2

3

1

2j2

+ (1 jj )

2

(jj)2

2

(j+j)2

. (82)

In order to derive Eq. (81) and Eq. (82), we consider the Hamiltonian,

H= T+ V(x), (83)

whereV(x)and T = p2

2m are the potential and kinetic energy operators. In the representation of delta functions (x xj),

equally spaced at coordinates,

xj =xmin+j, (84)

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with = (xmax xmin)/N, the potential energy matrixV() is diagonal with matrix elements defined as follows:

V()(j, k) = j|V(x)|k= dx

(x xj)V(x)(x xk),

=V(xk)j,k.

(85)

In order to obtain the kinetic energy matrix T() in the same grid-based representation, we first obtain the kinetic energy

matrixT() in the representation of eigenstatesn(x)of the particle in the box x= (xmin, xmax). Then, we rotateT() to

the representation of delta functions by using the following similarity transformation:

T() =C1T()C, (86)

whereCis the transformation matrix defined by the linear combinations,

k(x) =j

C(j, k)(x xj), (87)

where

C(j, k) = k(xj). (88)

The eigenstates of the particle in the box are:

k(x) =

2

xmax xminSin

k (x xmin)

(xmax xmin)

, (89)

withk(xmin) = 0andk(xmax) = 0. Therefore,

Tk(x) =(k)2

2m k(x), (90)

andT()

is diagonal with matrix elements,

T()(j, k) = j |T|k=(k)2

2m

2

(xmax xmin)2jk . (91)

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Therefore, substituting Eq. (91) and Eq. (88) into Eq. (86) we obtain,

T()(i, i) =

j,k=1

C1(i, j)T()(j, k)C(k, i) =

j,k=1

C(j, i)T()(j, k)C(k, i),

= 22

(xmax xmin)2

j,k=1

j(xi)(k)2

2m jkk(x

i) = 22

(xmax xmin)2

k=1

k(xi)(k)2

2m k(x

i),

= 222

2m(xmax xmin)22

(xmax xmin)

k=1

k2Sin

k

(xi xmin)

(xmax xmin)

Sin

k

(xi xmin)

(xmax xmin)

.

(92)

Finally, substituting Eq. (84) into Eq. (92) we obtain:

T()(j, j) = 22

2m(xmax xmin)2

2

N

k=1

k2SinkjNSinkj

N .

(93)

In order to calculate the infinite sum analytically, we note that

2Sin

kj

N

Sin

kj

N

=Cos

k(jj

N

Cos

k(j+j)

N

,

=Re

Exp

ik(jj

N

Exp

ik(j+j)

N

.

(94)

Therefore, Eq. (93) can be written as follows:

T()(j, j) = 22

2m(xmax xmin)22

N

Re

k=1

k2Exp

ik(jj

N

Re

k=1

k2Exp

ik(j+j)

N

. (95)

Evaluating the sum overk analytically, we obtain:

T()(j, j) = 2(1)jj

2m(xmax xmin)22

2

1

Sin2[(jj)/(2N)]

1

Sin2[(j+j)/(2N)]

, (96)

forj =j and

T()(j, j) = 2

2m(xmax xmin)22

2

(2N2 + 1)

3

1

Sin2[j/N]

. (97)

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These equations are obtained by considering the geometric series

k=0 xk = 1/(1 x)and noting that:

xx

k=0

xk =

k=0

kxk,

x2 2

x2

k=0

xk =

k=0

k2xk k=0

kxk.

(98)

Therefore,

k=1

k2xk =k=0

k2xk,

=x2 2

x2

1

1 x+ x

x

1

1 x ,

= 2x2

(1 x)3+

x

(1 x)2.

(99)

Equation (81) is obtained from Eq. (96) and Eq. (97), by taking the limit xmin , xmax , at finite . This

requiresN . Furthermore, since (j + j ) = xj +xj 2xmin and(j j ) = xj xj , this limit implies

(j+j) while(j+j)remains finite.

Equation (82) is obtained from Eq. (96) and Eq. (97), by making xmin = 0, and taking the limit xmax , at finite

. This requiresN . In this case, (j+j) =xj+ xj and(jj) =xj xj , and therefore both(j+j

)and

(j+j

)remain finite.

Computational Problem 15 (Due November 16, 2008):

15.1Write a program to solve the time independent Schr odinger equation by using the DVR method and apply it to find the

first 10 eigenvalues and eigenfunctions of the Harmonic oscillator introduced by Eq. (10) with m = 1 and = 1. Verify

that the eigenvalues areE() = (1/2 + ),= 010.

15.2Change the potential of the code written in 15.1 to that of a Morse oscillatorV(x) = De(1 exp(a(xxe)))2, with

xe = 0,De = 8, anda= k/(2De), wherek = m2, and recompute the eigenvalues and eigenfunctions.

15.3Generalize the program developed in 15.1 to solve the 2-dimensional Harmonic oscillatorV(x, y) = 1/2m2(x2+y2)

and apply it to find the first 10 eigenvalues and eigenfunctions of the Harmonic oscillator introduced by Eq. (10) with m= 1

and = 1. Verify that the eigenvalues are E() = (1 + 1+ 2).

15.3 Change the potential of the code written in 15.3 to that of a 2-dimensional Morse oscillator V(x,y) = De(1

exp(a(x xe)))2 +De(1 exp(a(y xe)))2, withxe = 0, De = 8, anda =k/(2De), wherek = m2, and

recompute the eigenvalues and eigenfunctions.

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