Application of schrodinger equation in quantum mechanics
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Consider one dimensional closed box of width L. A particle of mass ‘m’ is moving in a one-dimensional region along X-axis specified by the limits x=0 and x=L as shown in fig. The potential energy of particle inside the box is zero and infinity elsewhere.
I.e Potential energy V(x) is of the form
V(x) = {o; if o<x<L
∞: elsewhere
The one-dimensional time independent Schrodinger wave equation is given by
d2ψ/dx2+ 2m/Ћ2[E-V] ψ=0 (1)
Here we have changed partial derivatives in to exact because equation now contains only one variable i.e x-Co-ordinate. Inside the box V(x) =0
Therefore the Schrodinger equation in this region becomes
d2/ψ/dx2+ 2m/Ћ2Eψ=0
Or d2ψ/dx2+ K2ψ=0 (2)
Where k= 2mE/Ћ2 (3)
K is called the Propagation constant of the wave associated with particle and it has dimensions reciprocal of length.
The general solution of eq (2) is
Ψ=A sin Kx + B cos K x (4)
Where A and B are arbitrary conditions and these will be determined by the boundary conditions.
(ii) Boundary Conditions
The particle will always remain inside the box because of infinite potential barrier at the walls. So the probability of finding the particle outside the box is zero i.e.ψx=0 outside the box.
We know that the wave function must be continuous at the boundaries of potential well at x=0 and x=L, i.e.
Ψ(x)=0 at x=0 (5)
Ψ(x)=0 at x= L (6)
These equations are known as Boundary conditions.
(iii) Determination of Energy of Particle
Apply Boundary condition of eq.(5) to eq.(4)
0=A sin (X*0) +B cos (K*0)
0= 0+B*1
B=0 (7)
Therefore eq.(4) becomes
Ψ(x) = A sin Kx (8)
Applying the boundary condition of eq.(6) to eq.(8) ,we have
0=A sin KL
Sin KL=0
KL=nπ
K=nπ/L (9)
Where n= 1, 2, 3 – – –
A Cannot be zero in eq. (9) because then both A and B would be zero. This will give a zero wave function every where which means particle is not inside the box.
Wave functions. Substitute the value of K from eq. (9) in eq. (8) to get
Ψ(x)=A sin(nπ/Lx)
As the wave function depends on quantum number π so we write it ψn. Thus
Ψn=A sin (nπx/L)0<x<L
This is the wave function or eigen function of the particle in a box.
Ψn=0 outside the box
Energy value or Eigen value of particle in a box: Put this value of K from equation (9) in eq. (3)
nπ/L = 2m E/Ћ2
Squaring both sides
n2π2/L2=2mE/Ћ2
E=n2π2Ћ2/2mL2
Where n= 1, 2, 3… Is called the Quantum number
As E depends on n, we shall denote the energy of particle ar En. Thus
En= n2π2Ћ2/2mL2 (10)
This is the eigen value or energy value of the particle in a box.
I.e Potential energy V(x) is of the form
V(x) = {o; if o<x<L
∞: elsewhere
The one-dimensional time independent Schrodinger wave equation is given by
d2ψ/dx2+ 2m/Ћ2[E-V] ψ=0 (1)
Here we have changed partial derivatives in to exact because equation now contains only one variable i.e x-Co-ordinate. Inside the box V(x) =0
Therefore the Schrodinger equation in this region becomes
d2/ψ/dx2+ 2m/Ћ2Eψ=0
Or d2ψ/dx2+ K2ψ=0 (2)
Where k= 2mE/Ћ2 (3)
K is called the Propagation constant of the wave associated with particle and it has dimensions reciprocal of length.
The general solution of eq (2) is
Ψ=A sin Kx + B cos K x (4)
Where A and B are arbitrary conditions and these will be determined by the boundary conditions.
(ii) Boundary Conditions
The particle will always remain inside the box because of infinite potential barrier at the walls. So the probability of finding the particle outside the box is zero i.e.ψx=0 outside the box.
We know that the wave function must be continuous at the boundaries of potential well at x=0 and x=L, i.e.
Ψ(x)=0 at x=0 (5)
Ψ(x)=0 at x= L (6)
These equations are known as Boundary conditions.
(iii) Determination of Energy of Particle
Apply Boundary condition of eq.(5) to eq.(4)
0=A sin (X*0) +B cos (K*0)
0= 0+B*1
B=0 (7)
Therefore eq.(4) becomes
Ψ(x) = A sin Kx (8)
Applying the boundary condition of eq.(6) to eq.(8) ,we have
0=A sin KL
Sin KL=0
KL=nπ
K=nπ/L (9)
Where n= 1, 2, 3 – – –
A Cannot be zero in eq. (9) because then both A and B would be zero. This will give a zero wave function every where which means particle is not inside the box.
Wave functions. Substitute the value of K from eq. (9) in eq. (8) to get
Ψ(x)=A sin(nπ/Lx)
As the wave function depends on quantum number π so we write it ψn. Thus
Ψn=A sin (nπx/L)0<x<L
This is the wave function or eigen function of the particle in a box.
Ψn=0 outside the box
Energy value or Eigen value of particle in a box: Put this value of K from equation (9) in eq. (3)
nπ/L = 2m E/Ћ2
Squaring both sides
n2π2/L2=2mE/Ћ2
E=n2π2Ћ2/2mL2
Where n= 1, 2, 3… Is called the Quantum number
As E depends on n, we shall denote the energy of particle ar En. Thus
En= n2π2Ћ2/2mL2 (10)
This is the eigen value or energy value of the particle in a box.
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