SSPD_Chapter 1_Part 8_continued_AN ELECTRON IN A FINITE HEIGHT 1-D POTENTIAL WELL.

Figure 1

Fig(1.28) Electron in first quantum state in a finite potential well.

Electron total energy E is less than V₁ hence classically it cannot come out of the well.

Inside the potential well for 0 < z < W:

∂ 2 ψ /∂z 2 + (2mE/ћ 2 )ψ = 0 1.64

Outside the potential well for z < 0 and W < z :

∂ 2 ψ /∂z 2 + (2m(E-qV 1 )/ћ 2 )ψ = 0 1.65

The wave vector inside the potential well is :

k₁ and k₁*= ±i[√(2mE)]/ћ;

The wave vector outside the potential well is:

k₂ = -[√(2m(qV₁-E)]/ћ;

Inside the potential well we have harmonic solution:

That is ψ(z,t) = BSin[k₁(z+δ)]Exp[iωt]

Where k₁= 2π/λ₁ and by inspection λ₁= 2(W+2 δ) and

the solution is a standing wave with nodes at z = - δ and at z = (W+ δ);

Outside the potential well we have exponentially decaying solution:

For z < 0, we have ψ(z,t) = AExp[k 2 z].Exp[i ωt]

For W < z, we have ψ(z,t) = AExp[-k 2 (z-W)].Exp[i ωt] 1.66

To satisfy the boundary condition:

At x=0, A= B Sin[π δ/(W+2 δ)]

At x=W, A= B Sin[π (W+δ)/(W+2 δ)] 1.67

Since π - π (W+δ)/(W+2 δ) = π δ/(W+2 δ)

Therefore the two boundary conditions are:

A= BSin(π-Ө);

A = BSin(Ө)

where Ө= π (W+δ)/(W+2 δ);

This a consistent boundary condition therefore the solution is a valid solution .

Fig(1.29) depicts an electron in second quantum state n= 2.

Figure 2

Fig(1.29) Electron Matter Wave in second quantum state n=2;

Thus here we get a counter-intuitive result. Classically electron should remain confined to the potential well because it has insufficient energy to climb over the potential barrier at the two nodes but we find that electron has exponentially decaying probability of being found inside the Potenial Barrier region as it penetrates into the barrier. As a consequence we donot find the node of the standing wave pattern at the edges of the well. Infact it has shifted inside the barrier as shown in the Figure 1.29.

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Last edited by Bijay_Kumar Sharma on Jan 23, 2010 1:43 am -0600.