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SSPD_Chapter1_Part 8continued_Electron in an Infinite Potential Well_Solution of Schrodinger Equation.

Module by: Bijay_Kumar Sharma. E-mail the author

Summary: This is the continuation of Chapter 1_Part 8. This gives the solution of Schrodinger Equation for an electron in an 1-D Infinite Potential Well.This is analogous to the analysis of an electron in the ground state of hydrogen atom except that for a hydrogen atom the Schrodinger Equation will have to be solved for spherical coordinates. The problem of Hydrogen Atom will be taken up in a latter chapter.


Figure 1
Figure 1 (Picture 3.png)

Fig(1.24) Electron in 1-D infinite potential well. Probability Amplitude ψ(z,t) and probability density |ψ(z,t)|^ 2 are respectively plotted along the spatial axis-z axis.

Writing the Schrodinger Equation[ Appendix XXIX] for an electron in an infinite potential well where V(z)=0 we get:

2 /2m) ∂ 2 ψ /∂z 2 + E ψ = 0 1.48

2 ψ /∂z 2 + (2mE/ћ 2 )ψ = 0

According to Linear Operator Theory, second order linear differential equation has two solutions: the complementary solution also known as transient solution or the natural solution and the second solution is the particular solution also known as the steady state solution or the forced solution.

The total solution = transient solution + steady state solution.

The transient solution depends on the characteristic equation of the system. A second order linear system has second degree characteristic equation which has two roots and two arbitrary constants which need two boundary conditions.

The steady state solution depends on the forcing function or the driving function. If the forcing function is a harmonic function then the steady state solution is a sinusoidal solution. If the forcing function is a constant the steady state output is also a constant.

In Eq.(1.48) Right Hand Side is zero hence steady state solution is zero.

The characteristic equation is:

D2 + 2mE/ћ2 = 0

Where D is the differential operator.

The roots are imaginary:

D1= -i[√(2mE)]/ ћ and D2 = +i[√(2mE)]/ ћ

Therefore the solution is harmonic. If the roots were real then the solution would be hyperbolic or exponential.

The solution of Eq.(1.48) is:

ψ = A.Exp[+i{(√(2mE))/ ћ}x] + B.Exp[-i{(√(2mE))/ ћ}x];

From two boundary conditions the two arbitrary constants are determined.

At x = 0, ψ(0) = 0 and at x = W ( the width of the potential well), ψ(W) = 0.

From these two boundary conditions we obtain two simultaneous equations.

A + B=0;

A.Exp[+i{(√(2mE))/ ћ}W] + B.Exp[-i{(√(2mE))/ ћ}W] = 0;

The determinant is:

Δ = 1 1

Exp[+i{(√(2mE))/ ћ}W] Exp[-i{(√(2mE))/ ћ}W]

Therefore Δ = Exp[-i{(√(2mE))/ ћ}W] - Exp[+i{(√(2mE))/ ћ}W]

If these two simultaneous equations are to have non-trivial solutions then the determinant of the set of simultaneous equations should be zero.

Therefore Δ = Exp[-i{(√(2mE))/ ћ}W] - Exp[+i{(√(2mE))/ ћ}W] =0

Therefore Exp[-i{(√(2mE))/ ћ}W] = Exp[+i{(√(2mE))/ ћ}W]

Therefore Cos[{(√(2mE))/ ћ}W] –i Sin[{(√(2mE))/ ћ}W]

= Cos[{(√(2mE))/ ћ}W] +i Sin[{(√(2mE))/ ћ}W]

Or –i Sin[{(√(2mE))/ ћ}W] = i Sin[{(√(2mE))/ ћ}W]

This relationship is satisfied only when Sin[{(√(2mE))/ ћ}W]=0;

This requires {(√(2mE))/ ћ}W = 0 radians or

{(√(2mE))/ ћ}W = π radians or

{(√(2mE))/ ћ}W = n π radians where n= 0,1,2,3,4…….

Squaring both sides we get:

{(2mE)/ ћ2 }W2 = (n π)2

or E = n 2 h 2 /(8mW 2 ) 1.49

Eq.(1.49) implies that the permissible energy states of an electron in an infinite potential well are quantized.

Why electrons cannot occupy a continuum energy states as they do in free space? The answer is the following:

We had assumed at the beginning of the analysis that V(x) = 0. This implies that potential energy is zero and electron possesses only Kinetic Energy.

Therefore total energy E = p2/(2m) = n2h2/(8mW2)

Therefore p = (nh)/(2W) 1.50

From de Broglie postulate: λ = h/p = h/[(nh)/(2W)]

To satisfy the standing wave condition in bounded space which an infinite 1-D potential well is, following boundary condition must be satisfied

W = n λ/2 1.51

Eq.(1.51) is the necessary condition for Standing Wave pattern. This standing wave pattern requirement causes the quantization of energy states.

Here we digress briefly to the chapter of light to fully understand the behavior of matter wave.

Figure 2
Figure 2 (Picture 5.png)

Fig.(1.25) A plane wavefront light ray perpendicularly incident on an interface of two optical mediums.

Whenever light travels from one optical medium of refractive index n1 to the other optical medium of refractive index n2 , the incident wave Transverse Electromagnetic Wave (TEM) experiences partial reflection at the interface of the two media and partial transmission into the second medium.

Let us assume that medium 1 is absolute vacuum hence its refractive index = n1 = 1 and medium 2 is a solid medium of refractive index n2 = n. The mathematical form of the incident wave, reflected wave and the transmitted wave is given in Fig(1.25).

Wave vector in medium 1 is k1 = 2π/λ1 and wave vector in medium 2 is k2 = 2π/λ2;

And ν λ1= c, ν λ2 = v; 1.52

Therefore c/v = λ1 / λ2 = n/1;

Therefore λ1 = n. λ2 1.53

We know that if the second medium is metal, the incident light is totally reflected and the reflected light experiences a phase change of 180°.

Also at the interface Eincident + Ereflected = 0;

But if the second medium is dielectric then we have partial reflection and partial transmission and at the interface we have: Eincident + Ereflected = Etransmitted ;

The incident wave is the forward wave:

E(z,t)= E xoincident Exp[j(k z1 z – ω.t)];

The reflected wave is the backward wave:

E(z,t)= E xoreflected Exp[j(k z1 z + ω.t)];

The transmitted wave is also moving in forward direction therefore it is:

E(z,t)= E xotransmitted Exp[j(k z2 z – ω.t)]; 1.54

Wave vectors have been defined in Eq.(1.52).and Eq.(1.53).

The incident forward and reflected backward wave interfere to form Standing Wave as they do on a mismatched transmission line. If a transmission line is not terminated in Characteristic Impedance then partial reflection takes place at the load and a partial standing wave pattern is formed on the transmission line. Standing Wave implies there is no transmission of energy. In case of metal there is total reflection. Hence we have 100% standing wave in medium 1 and there is no penetration of light into the metallic medium 2. Hence no transmission of light energy. For dielectric medium 2 , we have partial reflection hence only partial standing wave.

The abrupt change in the refractive index or the dielectric constant is referred to as step index change. In the same way abrupt voltage change is referred to as step voltage change.

Matter wave behaves in an analogous fashion at a step voltage change as a light wave behaves at step index change.

Figure 3
Figure 3 (Picture 7.png)

Fig(1.26) Reflection and Transmission of electron matter wave across step voltage change from V 1 to V 2 .

For matter wave,

wave vector k=2π/λ = 2πp/h = [√{2m(E-V)}]/ћ;

Therefore the two wave vectors are:

k1= [√{2m(E-V1)}]/ћ and k2 = [√{2m(E-V2)}]/ћ;

The incident forward electron matter wave:

ψ(z,t) incident = ψ 0incident Exp[j(ωt- k z1 .z)]

The reflected backward electron wave:

ψ(z,t) reflected = ψ 0reflected Exp[j(ωt+ k z1 .z)] 1.55

The transmitted forward electron wave:

ψ(z,t) transmitted = ψ 0transmitted Exp[j(ωt- k z2 .z)] 1.56

At the interface or at the step,

ψ 0incident - ψ 0reflected = ψ 0transmitted 1.57

Eq.(1.57) is the interface boundary condition.

Now we return to the original infinite potential well case.

Inside the potential well we have both the wave vectors ±i[√{2mE}]/ћ] hence we have both forward and backward traveling wave components.

Outside the well Schrodinger Equation is of the form:

2 ψ /∂x2 + (2m(E-V)/ћ2)ψ = 0

But outside the well, V = ∞ therefore the wave equation reduces to:

2 ψ /∂x2 - (2m∞/ћ2)ψ = 0

The characteristic equation has real roots hence the solution is hyperbolic.

Therefore ψ(z,t) = [AExp(-k 2 .z) + BExp(+k 2 .z)]Exp(iωt) 1.58

Where k2 = [√{2m∞}]/ћ]

In Eq(1.58), the term Exp(+k2.z) is not admissible as it is exponentially growing function. Hence B=0 and the solution outside the well is:

ψ(z,t) = [AExp(-k 2 .z)]Exp(iωt) 1.59

But since k2 = infinity implies the wave function does not exist outside the well. The transmitted wave outside the well almost instantaneously attenuates to zero since k2 is the attenuation coefficient of the exponentially decaying wave and this coefficient is infinite.

Inside the potential well there are two waves one forward and the second backward and both have equal amplitude but opposite sign so that at the two boundaries of the potential well the sum of the two produce nodes.

ψ 0incident - ψ 0reflected = 0 at the boundaries 1.60


ψ(z,t) incident = A Exp[j(ωt- k z1 .z)]

ψ(z,t) reflected = -AExp[j(ωt+ k z1 .z)] 1.61

The superposition of the two waves is:

ψ(z,t) = AExp[j(ωt)] [ Exp(- kz1.z)- Exp(+ kz1.z)]

ψ(z,t) = AExp[j(ωt)][Cos(kz1.z)-iSin(kz1.z) -Cos(kz1.z)-iSin(kz1.z)]

ψ(z,t) = -2iAExp[j(ωt)][Sin(kz1.z)]

or ψ(z,t) = BExp[j(ωt)][Sin(kz1.z)] 1.62

Eq.(1.62) is the mathematical formulation of the standing wave.

According to boundary conditions, this standing wave is suppose to have two nodes :

ψ(0,t)= ψ(W,t) =0

therefore kz1.W = (2π/λn)W = n π where n = 1, 2, 3,………….

Therefore W=n. λn/2 1.63

This is the same restriction as was imposed in Eq.(1.51). This Eq(1.63) sheds a very important light on the quantization of energy levels permitted for an electron in an infinite potential well.

Inside the potential well there is the superposition of incident and reflected wave. There are only certain cases for which constructive interference takes place for the remaining cases there is destructive interference.

The cases for which the superposition leads to node formation at the boundaries z = 0 and W, only for those cases electron has an existence. For all the remaining cases electron goes out of existence by destructive interference.

Therefore we assert that in an infinite potential well , the electron can be in existence only for the energy levels:

E= E0 , 4 E0 , 9 E0 , 16 E0 ………

Where E0 = h2/(8mW2)

The above result comes from Eq.(1.50) and its previous line.

An identical situation prevails in a hydrogen atom. The orbits which allow constructive interference to take place only those orbits are permitted. Remaining are forbidden. The orbital radii which support constructive interference are the radii where electron can exist as standing wave.

These are the radii exactly predicted by Bohr’s Law:

Orbital Angular Momentum = Integral Multiple of (h/(2π)) as already seen in Chapter 1_part 6.

Figure 4
Figure 4 (Picture 8.png)

Fig(1.27) An electron in 3 quantum states n=1,2,3 in an infinite potential well.

An electron in first three permissible quantum states in an infinite potential well are shown in Fig.(1.27

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