Introduction

The particle in one dimensional potential box can be expanded to consider a particle  within a higher dimensions as demonstrated elsewhere for a  Particle in a three dimensional potential box . Now in this perticular article we are going to discuss about solutions of Schrodinger equation,enery eigen value and cubical potential box also degeneracy of energy levels etc.

Particle in a three dimensional potential box

  • Solutions of Schrodinger equation

Consider motion of a particle of mass m and in three dimensional potential box of sides a,b,c.  Hence The potential inside the box is zero.

KE operator in 3D: Now just set up the Schrodinger equation: Schrodinger eq for particle in 3D box.

V(x,y,z) = 0 for o<x<a,  0<y<b , 0<z<c

And infinite outside (elsewhere)

The time independent Schrodinger equation governing the motion of the particle inside the box is,

δ²ψ/δx² +δ²ψ/δy² + δ²ψ/δz² + 2m(E-V)ψ/h² = 0

Because Inside the box V =0

δ²ψ/δx² +δ²ψ/δy² + δ²ψ/δz² + 2m(E)ψ/h² = 0          (1)

Where,

K= √2mE/h²

since, K is a positive constant.

And now, By the method of separation of variable-

ψ( x y z ) = X(x) Y(y) Z(z)

Hence by Putting this value in equation (1) and devide by XYZ hence we get,

1/Xδ²ψ/δx² +1/Yδ²ψ/δy² + 1/Zδ²ψ/δz² + 2m(E)ψ/h² = 0

Or ,

δ²ψ/δx² +δ²ψ/δy² + δ²ψ/δz² = -K²

Therefore These equations can be written as

δ²ψ/δx² = -K₁²

And , δ²ψ/δy² = -K₂²

also, δ²ψ/δz²  = − K₃²

Hence,

δ²ψ/δx² +k₁²=0        (3)

and,δ²ψ/δy² +K₂² =0        (4)

also, δ²ψ/δz² + K₃² =0        (5)

And ,

K₁² +K₂²+ K₃²= 2m(E)/h²        (6)

Hence Solution of this equation is given by-

X(x)=Asin K₁x+ B cos K₂y           (7)

And now Applying boundary condition-

(1)  At x=0 and ψ = 0

since XYZ =0

But YZ are independent hence X=0

(2)  when x=a , X=0

also, B=0 , K₁=nπ/a

and K₁= n₁π/a

Hence, X(x) = A sin (n₁π/a)x

(3)  normalization condition-

∫₀ª X(x) X*(x)dx =1

A= √2/a

hence X(x) = √2/a (sin K₁(x))

Where. K₁ = n₁π/ a

Similarly, Y(y) = √2/b (sin K₂(x))

because K₂ = n₂π/a

Z(z) = √2/c sin K₃z

where K₃ = n₃π /c

Hence the total number of wave function is given by-

ψ = XYZ

and, ψ ( x,y,z) = √8/abc (sin K₁x sin K₂y sinK₃z)

Where √8/abc is called Eigen function.

If a=b=c

ψ ( x,y,z) = √8/a³ (sin K₁x sin K₂y sinK₃z)

Energy Eigen value

We have ,

K₁² +K₂²+ K₃²= 2m(E)/h²

therefore, (n₁π/a)² + (n₂π/a)² + (n₃π /c)²= 2m(E)/h²

Separate the terms and then rearranging we get-

(n₁² + n₂² + n₃²)π²/a²+b²+c²=  2mE/h²

If a = b =c

Hence E = π²h²(n₁² + n₂² + n₃²)/2ma²

Cubical potential Box

Particle in a three dimensional potential box

 

 

The energy of a particle for a cubical potential box (a=b=c) is given by-

E = π²h²(n₁² + n₂² + n₃²)/2ma²

And now Introducing subscript (n₁n₂n₃) it may be written as

En₁n₂n₃ =π²h²(n₁² + n₂² + n₃²)/2ma²

Where n₁,n₂,n₃ ∈ {1,2,3,4,…}

The corresponding eigen functions are-

ψ n₁n₂n₃ =√8/a³ sin (n₁πx/ a) sin(n₂πy/a ) sin(n₃πz /c )

Thus the ground state energy term of particle confined in three dimensional cubical Box is for n₁ = n₂ = n₃ =1

E₁₁₁ = 3(π²h²/2ma²) = 3E₁, where E₁ = (π²h²/2ma²)

and the corresponding Eigen function is given by-

ψ₁₁₁ =√8/a³ sin (πx/ a) sin(πy/a ) sin(πz /c )

In addition The next energy eigen value is for (n₁ n₂ n₃) =(112),(121), (211)

E₁₁₂ = E₁₂₁  = E₂₁₁ =6 ((π²h²/2ma²) = 6E₁

Hence, Three different Eigen function ψ₁₁₂ ,ψ₁₂₁ and ψ₂₁₁ correspond to the same energy level

ψ₁₁₂ =√8/a³ sin (πx/ a) sin(πy/a ) sin(2πz /c )

and, ψ₁₂₁=√8/a³ sin (πx/ a) sin(2πy/a ) sin(πz /c )

also, ψ₂₁₁ =√8/a³ sin (2πx/ a) sin(πy/a ) sin(πz /c )

Hence Continuing in this manner, we will see that the energy eigen value for three dimensional cubical potential Box are

3E₁ , 6E₁ ,9E₁, 11E₁, 12E₁, 14E₁…

Where, E₁ =(π²h²/2ma²). These energy levels are plotted in fig given bellow,

Degeneracy of the energy level

When a certain energy level is associated with several wave functions or states, then as a result we say that there is the energy level.And the order of degeneracy of an energy level is equal to the number of different (independent)wave functions for the given energy.

In addition let us consider the energy eigenvalues for a particle in a three dimensional cubical potential Box of side ‘a’

En₁n₂n₃ =π²h²(n₁² + n₂² + n₃²)/2ma²

Hence ,We see that energy depends only on (n₁² + n₂² + n₃²)≡ ζ².consequently , as a result we can say  that all states corresponding to all Integers n₁n₂,n₃, and it gives the same value for ζ², have the same energy.

The order of degeneracy of an energy level is denoted by g. And This order is equal to the number of different (independent) wave function for the given energy.

For example, let the degeneracy is g=3 for E=9 (π²h²/2ma²) therefore,

Conclusion

So in this perticular article ,Particle in a three dimensional potential box we have gone through many important theories and formulas in easiest way possible.

 

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