Particle in one dimensional box, particle in two dimensional box

Particle in one dimensional box

Let us consider one dimensional box having its edge length 'a'. The length of the edge at any point is 'x'. Potential energy of particle inside the box is minimum and let it be zero. The potential energy at the boundary and outside the box is very high. Let it be '∞'. The Schrodinger wave equation for a particle in one dimensional box is

$$\frac{d^2\psi(x)}{dx^2}$$ + $$\frac{8\pi^2m(E-V)\psi(x)}{h^2}$$ = 0.............(i)

(a) Wave function of a particle outside the box, $$\psi$$(x) = 0

Outside the box, v =∞

$$\frac{d^2\psi(x)}{dx^2}$$ + $$\frac{8\pi^2m(E-V)\psi(x)}{h^2}$$ = 0

$$\frac{d^2\psi(x)}{dx^2}$$ - $$\frac{8\pi^2m∞\psi(x)}{h^2}$$ = 0 [since, E-V≈ -V -∞]

$$\frac{8\pi^2m∞\psi(x)}{h^2}$$ =$$\frac{d^2\psi(x)}{dx^2}$$

$$\psi$$(x) = $$\frac{1}{∞}$$$$\frac{8\pi^2m∞\psi(x)}{h^2}$$ $$\frac{h^2}{8\pi^2m}$$

$$\psi(x)$$ = 0..........(2)

(Electron doesnot exist outside the box hence its wave function is zero) $$\psi(x)$$ = 0

b) Wave function inside the box:

Inside the box, P.E = 0

Now, ewuation (i) becomes,$$\frac{d^2\psi(x)}{dx^2}$$ +$$\frac{8\pi^2mE\psi(x)}{h^2}$$ = 0..........(iii)

The kinetic energy of particle within the box is taken constant.

Hence, $$\frac{8\pi^2mE}{h^2}$$ is constant. Let it be $$\alpha$$²

Now, equation(iii) becomes,$$\frac{d^2\psi(x)}{dx^2}$$ + $$\alpha$$²$$\psi(x)$$ = 0..........(iv)

The solution of equation (iv) is ,

$$\psi(x)$$ = A $$\sin$$ $$\alpha$$x + B $$\cos$$ $$\alpha$$x........(v)

Applying the boundary condition, when x=0

Now, equation (v) becomes.

$$\psi(x)$$ = A$$\sin$$ $$\alpha$$ 0 + B $$\cos$$ $$\alpha$$ 0

0 = A$$\sin$$ 0° + B $$\cos$$ 0°

or, 0 = 0 + B

∴ B = 0

Now, equation (v) becomes,

$$\psi(x)$$ = A$$\sin$$ $$\alpha$$ x........(vi)

When x = a

$$\psi(x) = A$$\sin$$ $$\alpha$$ a

0 = A$$\sin$$ $$\alpha$$ a

Either, A = 0 or,$$\sin$$ $$\alpha$$ a = 0

A≠ 0 So,$$\sin$$ $$\alpha$$ a = 0

$$\sin$$ $$\alpha$$ a =$$\sin$$ n$$\pi$$

∴ $$\alpha$$ a = n $$\pi$$..........(vii)

Putting the value of $$\alpha$$ in equation (vi)

$$\psi(x)$$ = A$$\sin$$$$\frac{n\pi x}{a}$$...............(viii)

We know,

$$\alpha$$²=$$\frac{8\pi^2mE\psi(x)}{h^2}$$

or, $$\frac{n^2\pi^2}{a^2}$$ =$$\frac{8\pi^2mE\psi(x)}{h^2}$$

∴E = $$\frac{n^2h^2}{8ma^2}$$........(ix)

h$$\nu$$ = $$\frac{n^2h^2}{8ma^2}$$

Calculation of value of A

The value of A is calculated by normalizing A

i.e$$\int_{a}^{0} $$ $$\psi^*$$ $$\psi$$ dx = 1

or,$$\int_{a}^{0} $$ A $$\sin$$ $$\frac{n\pi x}{a}$$ A $$\sin$$ $$\frac{n\pi x}{a}$$ dx = 1

or,$$\int_{a}^{0} $$ A²$$\sin^2$$ $$\frac{n\pi x}{a}$$ dx = 1

or, A²$$\int_{a}^{0} $$$$\sin^2$$ $$\frac{n\pi x}{a}$$ dx = 1

or, A² $$\frac{a}{2}$$ = 1 [$$\because$$$$\sin^2$$ $$\frac{n\pi x}{a}$$ dx=$$\frac{a}{2}$$]

or, A = $$\frac{2}{a}$$

or, A = $$\frac{\sqrt2}{\sqrt a}$$

Now, equation (8) becomes,

$$\psi(x)$$ =$$\frac{\sqrt2}{\sqrt a}$$$$$$\sin^2$$ $$\frac{n\pi x}{a}$$..........(x)

Particle in three dimensional box

Let us consider a particle in three dimensional box having edge length a,b and c. The potential energy of the particle outside the box is maximum and let it be infinite. Since, the potential energy of the particle within the box is zero, the Schrodinger wave equation for the particle within 3D box is given by,

$$\frac{d^2}{dx^2}$$ $$\psi$$(x, y, z) + $$\frac{d^2}{dy^2}$$ $$\psi$$ $$\psi$$ (x.y, z) +$$\frac{d^2}{dz^2}$$ $$\psi$$ (x. y, z) + $$\frac{8\pi^2 mE}{h^2}$$ $$\psi$$ (x. y, z)= 0.......(1)

The wave function is the product of the wave function along x-axis, y-axis and z-axis

$$\psi$$ (x,y, z) = X_x. Y_y. Z_z...........(2)

X_x= wave function along x-axis

Y_y= wave function along y-axis

Z_z= wace function along z-axis

Putting the value of$$\psi$$ (x,y, z) in equation (1)

$$\frac{d^2}{dx^2}$$X_x. Y_y. Z_z+$$\frac{d^2}{dy^2}$$X_x. Y_y. Z_z+$$\frac{d^2}{dz^2}$$X_x. Y_y. Z_z+$$\frac{8\pi^2 mE}{h^2}$$X_x. Y_y. Z_z= 0

$$\frac{d^2}{dx^2}$$ operates only X_x$$\frac{d^2}{dy^2}$$ operates only Y_yand$$\frac{d^2}{dz^2}$$ operates only Z_z. Now above expression becomes,

Y_y.Z_z$$\frac{d^2}{dx^2}$$ X_x+ X_xZ_z$$\frac{d^2}{dy^2}$$Y_y+ X_x.Y_y$$\frac{d^2}{dz^2}$$ Z_z + $$\frac{8\pi^2 mE}{h^2}$$X_x. Y_y. Z_z= 0

Dividing by X_x. Y_y. Z_z

$$\frac{Y_y Z_z}{X_x Y_Y Z_z}$$$$\frac{d^2}{dx^2}$$ Xx+ $$\frac{X_x Z_z}{X_x Y_Y Z_z}$$$$\frac{d^2}{dy^2}$$Yy + $$\frac{X_x Y_Y}{X_x Y_Y Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z+$$\frac{8\pi^2 mE}{h^2}$$ $$\frac{X_x Y_Y Z_z}{X_x Y_Y Z_z}$$ = 0

or, $$\frac{1}{X_x}$$$$\frac{d^2}{dx^2}$$ Xx + $$\frac{1}{Y_Y}$$$$\frac{d^2}{dy^2}$$Yy + $$\frac{1}{Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z+$$\frac{8\pi^2 mE}{h^2}$$ = 0

or, $$\frac{1}{X_x}$$$$\frac{d^2}{dx^2}$$ Xx + $$\frac{1}{Y_Y}$$$$\frac{d^2}{dy^2}$$Yy + $$\frac{1}{Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z =-$$\frac{8\pi^2 mE}{h^2}$$...........(3)

In these calculation, we carried out the separation of variables. The first term, second term and third term at L.H.S of the above equation only depends upon x, y and z respectively. The sum of first, second and third term is constant.

If we differentiate the above equation at constant Y_Yand Z_zor at constant Z_zand X_xor at constant X_xand Y_Y. The value obtained is same. It shows that the three terms at L.H.S side are independent with each other and each term is independent of corresponding wave function because the value obtained is constant. It shows that the separation of variable was possible.

Ie each term at L.H.S is constant with its component

$$\frac{1}{X_x}$$$$\frac{d^2}{dx^2}$$ Xx =-$$\frac{8\pi^2 mE}{h^2}$$ or$$\frac{1}{X_x}$$$$\frac{d^2}{dx^2}$$ Xx= - $$\alpha_x$$²

$$\frac{1}{Y_Y}$$$$\frac{d^2}{dy^2}$$Y_Y =-$$\frac{8\pi^2 mE_y}{h^2}$$ or $$\frac{1}{Y_Y}$$$$\frac{d^2}{dy^2}$$Yy = -$$\alpha_y$$²

$$\frac{1}{Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z= -$$\frac{8\pi^2 mE_z}{h^2}$$ or $$\frac{1}{Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z= - $$\alpha_z$$²

Now equation (3) becomes,

- $$\alpha_x$$²-$$\alpha_y$$²-$$\alpha_z$$²=-$$\frac{8\pi^2 mE}{h^2}$$

or, $$\alpha_x$$²+$$\alpha_y$$²+$$\alpha_z$$² =$$\frac{8\pi^2 mE}{h^2}$$......(4)

In the above expression, each degree of freedom is expressed as,

Multiplying by X_x,

$$\frac{1}{X_x}$$$$\frac{d^2}{dx^2}$$ Xx +$$\alpha_x$$² = 0

Multiplying by Y_y,

$$\frac{1}{Y_Y}$$$$\frac{d^2}{dy^2}$$Y_Y+ $$\alpha_y$$²= 0

Multiplying by Z_z,

$$\frac{1}{Z_z}$$$$\frac{d^2}{dz^2}$$ Z_z+$$\alpha_z$$²= 0

Equation (5) can also be expressed as,

$$\frac{d^2}{dx^2}$$ Xx +$$\alpha_x$$²X_x= 0

$$\frac{d^2}{dy^2}$$Y_Y+ $$\alpha_y$$²Y_y= 0

$$\frac{d^2}{dz^2}$$ Z_z+$$\alpha_z$$²Z_z= 0

These are equation similar to particle in one dimension box. The normalised wave function in each case is,

X_x= $$\sqrt{\frac 2a}$$ $$\sin$$ $$\frac{n_x \pi_x}{a}$$

Y_y=$$\sqrt{\frac 2b}$$ $$\sin$$ $$\frac{n_y \pi_y}{b}$$

Z_z=$$\sqrt{\frac 2c}$$ $$\sin$$ $$\frac{n_z \pi_z}{c}$$

From equation (2) $$\psi(x,y,z)$$ = X_x. Y_y. Z_z

$$\psi(x,y,z)$$ = $$\sqrt{\frac 2a}$$ $$\sin$$ $$\frac{n_x \pi_x}{a}$$ .$$\sqrt{\frac 2b}$$ $$\sin$$ $$\frac{n_y \pi_y}{b}$$ .$$\sqrt{\frac 2c}$$ $$\sin$$ $$\frac{n_z \pi_z}{c}$$

= $$\frac{8

The energy can be claculated as,

E = E_x+ E_y+ E_z

= $$\frac{n_x^2 h^2}{8ma^2}$$ + $$\frac{n_y^2 h^2}{8mb^2}$$ + $$\frac{n_z^2 h^2}{8mc^2}$$

= $$\frac{h^2}{8m}$$ ($$\frac{n_x^2}{a^2}$$ + $$\frac{n_y^2}{b^2}$$ + $$\frac{n_z^2}{c^2}$$)

For cube, a = b = c

E = $$\frac{h^2(n_x^2 + n_y^2 + n_z^2)}{8ma^2}$$....(10)