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*Introduction *

Suppose we consider the similar particle as in chapter but this time it is constrained to shift in a rectangular box of dimensions a, b and c in length. Within the box (i.e. between x= 0 and a; y = 0 and b and z = 0 and c), the potential energy is zero. At the walls and everywhere outside the box, the potential is.

*Definition of Particle in a Three-Dimensional Box *

The Schrodinger wave equation for 3 dimensional (3D) box is

δ^{2}φ/δx^{2} + δ^{2}φ/δy^{2} + δ^{2}φ/δz^{2} + 8 π^{2}m/h^{2}(E-V) φ =0

where and V are (x,y,z).

since V = 0 inside the box, then the equation becomes

δ^{2}φ/δx^{2} + δ^{2}φ/δy^{2} + δ^{2}φ/δz^{2} + 8 π^{2}m/h^{2}Eφ=0

Equation may be solved by writing the wave function as the product of 3 functions, each depending on one coordinate.

φ(x, y, z) = X(x)Y(y)Z(z)

differentiating equation gives

δφ/δx = Y (y) Z (z) δx/δx

δ^{2}φ/δx^{2}= Y (y) Z(z) δ^{2}x/δx^{2}

and by a similar reasoning

δ^{2}φ/δy^{2} = X (x)Z(z) δ^{2}y/δy^{2}

δ^{2}φ/δz^{2} = X(x)Y(y) δ^{2}z/δz^{2}

Y(y)Z(z)δ^{2}x/δx^{2} + X(x)Z(z) ) δ^{2}y/δy^{2} + X(x)Y(y) δ^{2}z/δz^{2} + 8 π^{2}m/h^{2} EX(x)Y(y)Z(z) = 0 dividing all through by Y(y)X(x)Z(z) one obtains

** ** -h

We can write the energy as the sum of three contributions associated with the coordinates.

E = E_{x} + E_{y }+ E_{z}

using equation, we can divide the expression attained into 3 equations.

-h^{2}/8 π^{2 }m *1/x δ^{2}x/δx^{2 }= E_{x}

-h^{2}/8 π^{2 }m *1/y δ^{2}y/δy^{2} = E_{y }

-h^{2}/8 π^{2 }m *1/z δ^{2}z/δz^{2} = E_{z}

Each of the last three equations is similar to the expression for the particle in a one-dimensional box discussed in chapter. Hence their solutions are respectively:

X = [√(^{2}/a)]sin nπx/a

En,x = n^{2}_{x}h^{2}/8ma^{2}

Y = [√(2/b)]sin nπy/b

E_{n,y }= n^{2}_{y }h^{2}/ 8mb^{2}

X = [√(^{2}/c)] sin nπx/c

E_{n,z }= n^{2}_{z }h^{2}/ 8mc^{2}

Where a, b, c are length in x, y, z directions, correspondingly. Also, nx, ny, nz are the quantum numbers correspondingly.

Since φ (x, y, z) = X(x)Y(y)Z(z) and E = E_{x} + E_{y} + E_{z}, then

ψ (x,y,z) = √a/v sin(nπx/a) sin(nπy/b) sin(nπz/c)

where V is the volume of the box.

E_{x,y,z }= h^{2}/δm * n^{2}_{x}/a^{2} + n^{2}_{y}/b^{2} [+ n^{2}_{z}/c^{2}]

Whenever the three dimensional box has geometrical symmetry, more interesting results are often obtained. For example, in a cubic box, a = b =c, thus equation becomes

E = h^{2}/ φm(n^{2}_{x} + n^{2}_{y }+ n^{2}_{z})

Suppose n_{x} =3, n_{y} = n_{z} =2, then

Ψ(x,y,z) = [/(8/v)] sin3πx/a* sin2πy/a*sin2πz/a]

E = h^{2}/8ma^{2}(3^{2} + 2^{2}+ 2^{2}) = 17h^{2}/8ma^{2}

Assuming we have another set of values n_{x} =2, n_{y} =3, n_{z} =2, then

Ψ(x,y,z) = [/(8/v )]sin 2πx/a sin3πy/a sin2πz/a

E = h^{2}/8ma^{2} (2^{2} + 3^{2}+ 2^{2}) = 17h^{2}/8ma^{2}

Suppose n_{x} =2, n_{y }=2, n_{z} =3, then

Ψ(x,y,z) = [/( 8/v)]sin2πx/a sin2πy/a sin3πz/a

E = h^{2}/8ma^{2}(2^{2} + 2^{2}+ 3^{2}) = 17h^{2}/8ma^{2}

Even though such states are different, the values of the energies are the same. The three states are said to be degenerate since they have equal energy.

For a situation where nx = ny = nz = 1, it corresponds to only one state of the system. The same is true of nx = ny = nz = 2, but situations such as nx ≠ ny = nz three degenerate states are obtained as shown in the Figure below.

Fig: Quantized Energy Levels of a Particle in a Cubic Box

Suppose we wish to calculate the transition energy between the levels E2,2.2 and E3,2,1 then the energy difference is calculated as

ΔE = hν = 14h^{2}/8ma^{2} - 12h^{2}/8ma^{2} = 2h^{2}/8ma^{2} = h^{2}/4ma^{2}

Given suitable data, it should be possible for us to calculate the frequency (ν) of the transition between the two states.

*Zero point energy *

According the old quantum theory, the energy level of a harmonic oscillator is E = hv.

The lowest energy level via n = 0 would have zero energy. Depend on the wave treatment of the system, the energy level corresponds to the state with quantum numbers n_{x }= n_{y} = n_{z} =1. The dissimilarity between such 2 values is termed the zero point energy.

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