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Section 11.2: Finite Potential Energy Wells: Quantitative
check, then drag a slider to see the odd states' TE.
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The finite square well problem is defined by a potential energy function that is zero everywhere except1
V(x) = −|V0| −a < x < a.
Since the potential energy function is finite, quantum mechanically there will be some leakage of the wave function into the classically-forbidden regions specified by x > a and x < −a. We will have three regions in which we must solve the time-independent Schrödinger equation. In Region I (x < −a) and Region III (x > a), for bound-state solutions, E < 0, we can write the time-independent Schrödinger equation as
[d2/dx2 − κ2] ψ(x) = 0 , (11.1)
where2 κ2 ≡ 2m|E|/ħ2. The equation above has the solutions
ψ(x) = Aexp(+κx) + Bexp(−κx). (11.2)
Because the energy eigenfunction must be zero at ±∞ we have
ψI(x) = Aexp(+κx) and ψIII(x) = Dexp(−κx) (11.3)
for the solutions in Region I and III, respectively.
In Region II (−a < x < a), we expect an oscillatory solution since the energy is greater than the potential energy: E > V0 or |V0| > |E|. In this region we can write the time-independent Schrödinger as
[d2/dx2 + k2] ψ(x) = 0 , (11.4)
where k2 ≡ 2m(|V0| − |E|)/ħ2. The above equation has the solutions ψII(x) = Bsin(kx) + Ccos(kx) which are valid solutions for Region II.
Next, we must match the solutions across the boundaries at x = −a and x = a. Matching the energy eigenfunctions across these boundaries means that the energy eigenfunctions and the first derivatives of the energy eigenfunctions must match at each boundary so that we have a continuous and smooth wave function (no jumps or kinks).
Since the potential energy function is symmetric about the origin, there are even and odd parity solutions to the bound-state problem.3 We begin by considering the even (parity) solutions and therefore the ψII(x) = Ccos(kx) solution in Region II.
Matching proceeds much like the scattering cases we considered in Chapter 8. At x = −a we have the conditions
ψI(−a) = ψII(−a) → Aexp(−κa) = Ccos(−ka),
ψ'I(−a) = ψ'II(−a) → Aκ exp(−κa) = Cksin(−ka).
From the symmetry in the problem, we need not consider the boundary at x = a as it yields the exact same condition on energy eigenfunctions. We now divide the resulting two equations to give a condition for the existence of even solutions: κ/k = tan(ka). This is actually a constraint on the allowed energies, as both k and κ involve the energy.
We now consider the following substitutions in terms of dimensionless variables:
ζ ≡ ka = [2m(|V0| − |E|)a2/ħ2]1/2 (11.5)
ζ0 ≡ [2m|V0|a2/ħ2]1/2 (11.6)
where ζ0 > ζ. Using these variables we have
[(ζ0/ζ)2 − 1]1/2 = tan(ζ) . (11.7)
This equation is a transcendental equation for ζ which itself is related via Eq. (11.6) to the energy. In addition, Eq. (11.7) only has solutions for particular values of ζ. We can solve this equation numerically or graphically, and we choose graphically in the animation. The right-hand side of Eq. (11.7) is shown in black and the left-hand side is shown in red. In the animation, ħ = 2m = 1. You may also select the Show the transcendental equation as a function of energy instead link to see the equations as a function of energy. You can change a and |V0| by dragging the sliders to a particular value to see how the left-hand side of Eq. (11.7) changes.
We note that as the potential energy well gets shallower and/or narrower, ζ0 < π/2 and there exists just one bound state. No matter how shallow or narrow the potential energy well, there will always be at least one bound state.
As ζ0 gets larger (meaning larger a and |V0|), the number of bound-state solutions increases. In addition, the intersection of the curves on the graph approaches ζ = nπ/2, with n odd. This means that the energy (as measured from the bottom of the well) approaches that of the infinite square well of length 2a (ζ ≈ nπ /2 yields |V0| − |E| = ħ2k2/2m ≈ n2π2ħ2/2m(2a)2).
The solution for the odd (parity) energy eigenfunctions proceeds like the even-parity case except that we use the sine solution in Region II:
ψI(−a) = ψII(−a) → Aexp(−κa) = Bsin(−ka),
ψ'I(−a) = ψ'II(−a) → Aκ exp(−κa) = Bk cos(−ka).
Again, we need not consider the equations for x = a because by symmetry, they yield the same result. We again divide the two equations to give κ/k = −cot(ka), and using the same substitutions as above yields: [(ζ0/ζ)2 − 1]1/2 = −cot(ζ).
This equation for the odd-parity solutions is shown in the animation by checking the text box and moving the slider. Note that as the potential energy well gets shallower and/or narrower, ζ0 gets smaller, and it is possible for there to be no intersections on the graph which means that there will not be any odd-parity states. No matter how shallow or narrow the symmetric finite potential energy well, there will always be at least one bound state and it is an even-parity state.
As ζ0 gets larger (meaning larger a and |V0|), the number of bound-state solutions increases. In addition, the intersection of the curves on the graph approaches ζ = nπ/2, with n even. Again this means that the energy as measured relative to the bottom of the well approaches that of the infinite square well of length 2a.
In order to find the energy eigenfunctions, we must solve for the constants A, B, C, and D. This requires using the matching equations and then normalizing the wave function. In practice this is time consuming, instead you can view the numerical solution by clicking the Show the energy eigenfunction and well instead link. To see other bound states, simply click-drag in the energy level diagram and select a level. The selected level will turn red.
1Note the differences between the potential energy functions describing the finite well and the infinite well. The width of the finite well is 2a and its walls are at V = 0 while the well is at V = −|V0|. Bound states of the finite well, therefore, have E < 0. With the infinite square well, the width is L and its walls are at V = ∞ while the well is at V = 0. Bound states of the infinite well, therefore, have E > 0.
2Since E < 0, we choose to write E = −|E| to avoid any ambiguity in sign.
3This is due to the fact that for even potential energy functions, the Hamiltonian commutes with the parity operator. As a consequence, there are even states in which ψe(−x) = ψe(x), and odd states in which ψo(−x) = −ψo(x).