Consider a^6Li atom (mass of 1.0 times 10^-26 kg) prepared in a 2-dimensional in

Question

Consider a^6Li atom (mass of 1.0 times 10^-26 kg) prepared in a 2-dimensional infinite square well such that the Hamiltonian governing its center of mass motion is given by: H = -h^2/2m (partial differential^2/partial differential x^2 + partial differential^2/partial differential y^2) + V (x, y) where x and y are Cartesian coordinates and the potential is given by: V(x, y) = {0, if 0 lessthanorequalto x lessthanorequalto a and 0 lessthanorequalto y lessthanorequalto b, infinity, otherwise where a = 1mu m and b = 2 mu m. Show that you can separate variables such that the solution to the time independent Schrodinger equation H_psi (x, y) = E psi(x, y) can be written as a product state psi(x, y) = psi (x) chi (y) where phi(x) is a function of only the x coordinate and chi (y) a function of only the y coordinate. Give the lowest energy solution to the time independent Schrodinger equation and the corresponding energy eigenvalue.

Explanation / Answer

Part-1: The solution of time independent schrodinger equation in 2-D is given as,

(x, y) = (x)X(y)

Time independent schrodinger equation for 2-D infinte well is given as,

(d2/dx2 + d2/dy2) (x,y) + (2mE/2) (x,y) = 0

(d2/dx2 + d2/dy2) (x)X(y) + (2mE/2) (x)X(y) = 0

[ X(y) (d2/dx2) (x) + (x) (d2/dy2) X(y) ]+ (2mE/2) (x)X(y) = 0

Divide the above equation by (x)X(y),

[ (1/(x)) (d2/dx2) (x) + (1/X(y)) (d2/dy2) X(y) ]+ (2mE/2) = 0

[ (1/(x)) (d2/dx2) (x) ] + (2mE/2) = - [ (1/X(y)) (d2/dy2) X(y) ] = constant = Ky2

Now both sides of the equation are mutually independent so they can be equivalent to some constant = Ky2.

[ (1/(x)) (d2/dx2) (x) ] + (2mE/2) ] = Ky2....................(1)

- [ (1/X(y)) (d2/dy2) X(y) ] = Ky2 has solution,

X(y) = A sin (Ky y) + B cos (KY y)

using the boundary conditions, at y = 0 we get B = 0 , and at y = b we get,

A sin (Ky b) = 0 which gives, Ky = ny / b here ny = 1,2,3.......

Now solving equation (1) ,

[ (1/(x)) (d2/dx2) (x) ] + (2mE/2) ] = Ky2

[ (1/(x)) (d2/dx2) (x) ] = (2mE/2) - Ky2 = Kx2 has the solution,

(x) = C sin(Kx x) + D cos(Kx x)

Similarly applying the boundary conditions we get, Kx = nx / a here nx = 1,2,3.......

The complete solution is gievn as,

(x, y) = (x)X(y) = C sin (nx x / a) * A sin (ny y / b)

from the condition of normalization we get, A = (2/b)1/2 and C = (2/a)1/2

Hence, (x, y) = (2/a)1/2 (2/b)1/2 sin (nx x / a) * A sin (ny y / b)

PART-2: (2mE/2) - Ky2 = Kx2

(2mE/2) = Kx2 + Ky2

E = (2/2m) ( nx22 / a2 + ny22 / b2 )

The lowest energy solution will be for nx = 1 and ny = 1

E = (22/2m) ( 1/ a2 + 1/ b2 )

On putting all the values we get,

E = 6.8*10-30J = 4.25*10-11 eV

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