The Schr¨ odinger Equation in One Dimension
2.2 One-Dimensional Harmonic Oscillator
"
1 + V2sin2p8m(E − V )a/~
4E(E − V )
#−1/2
. (2.27)
(b) Show that for E < V , one need simply interchange E and V and let sin → sinh.
(c) Find the values of E − V that lead to 100% transmission.
Problem 2.14 Total transmission. An electron encounters a barrier of height 10 eV and width 1A (= 10◦ −10m). What is the lowest energy E > V (answer in electron volts, or eV) that the electron must have to pass the barrier without reflection?
Problem 2.15 Partial transmission. Using part (b) from Problem 2.13, find what percentage of a beam of 5 eV electrons incident on a barrier of 25 volts in height and width a = 5.2 × 10−11 m will be transmitted.
2.2 One-Dimensional Harmonic Oscillator
While potential barriers and wells that are rectangular are much easier to solve than more complicated shapes, the real world is rarely well-approximated by these simple shapes. Another potential distribution, that of a parabolic po-tential well, which is characteristic of the harmonic oscillator problem, is an excellent fit to many real problems, especially for their lowest energy states.
Furthermore, for a great many real potentials, the very bottom of the poten-tial well may be approximated by a quadratic, since if we expand a general potential V (x) about the lowest point, x0, then
V (x) = V (x0) + V0(x0)(x − x0) +12V00(x0)(x − x0)2+ · · · ' V (x0) +12V00(x0)(x − x0)2,
since at the bottom (by definition), V0(x0) = 0. For the lowest energy states in a broad variety of problems, the harmonic oscillator problem provides a good first approximation. When additional terms in the series are required, perturbation theory, treated in Chapter 5, allows us to extend the accuracy.
The Schr¨odinger Equation in One Dimension 57 2.2.1 The Schr¨odinger Equation
For a force of the type F (x) = −kx, the potential energy is given by V (x) =
−Rx
F (x0) dx0= 12kx2 so the Hamiltonian is
H = −ˆ ~2 2m
d2
dx2+12kx2. (2.28)
For the time-independent Schr¨odinger equation, then, we may write
−~2 2m
d2ψ
dx2 +12kx2ψ = Eψ . (2.29) In order to solve this equation, we shall use the general procedure:
1. Express the equation in dimensionless form. This is for convenience, so that there will only be one constant in the equation.
2. Determine the asymptotic behavior of ψ(x) as |x| → ∞. Since the wave function must be normalizable, the solution must vanish as |x| tends toward infinity. This procedure generally allows one to eliminate one possible solution which is unbounded.
3. Change to a new dependent variable that has better asymptotic behavior by factoring out the asymptotic behavior. Without this step, a power se-ries solution typically has a three- or four-term recursion formula which makes the analysis of the convergence of the power series almost im-possible. Factoring out this asymptotic behavior usually results in a simpler differential equation that can be solved by a power series with a two-term recursion formula, permitting the analysis of the convergence of the series.
4. Solve the equation for this new dependent variable by the power series method. This is usually straightforward, leading to a two-term recursion formula if the previous steps are properly carried out.
5. Determine what values of energy will lead to well-behaved solutions.
Typically, the power series will diverge for arbitrary values of the re-maining dimensionless constant. The constant must be chosen so that the overall solution is convergent, typically by truncating the series to a polynomial.
6. Write the complete solution in terms of the original constants. Re-turning to the original constants in the wave function and the energy completes the solution of the differential equation and prepares one for the interpretation of the results.
2.2.2 Changing to Dimensionless Variables The first step is to change variables to
ξ = αx ,
(or ξ = α(x − x0) if the origin is offset) where α is yet to be determined.
Denoting the derivative with respect to the argument (ξ in this case) by ψ0, we have
−~2α2
2m ψ00+ kξ2 2α2 − E
ψ = 0 , (2.30)
and dividing by the leading constant term, we obtain ψ00+ 2mE
~2α2 − km
~2α4ξ2
ψ = 0 . (2.31)
The equation can now be simplified by choosing α4≡km
~2
and λ = E
~pk/m, so that the equation in dimensionless form is
ψ00+ (2λ − ξ2)ψ = 0 . (2.32) In terms of λ, we find E = λ~ω since ω =pk/m for the classical harmonic oscillator.
2.2.3 The Asymptotic Form As |ξ| → ∞, Equation (2.32) takes the form
ψ00≈ ξ2ψ , (2.33)
which if ξ in the coefficient were constant would have solutions of the form ψ → e±ξ·ξ = e±ξ2.
Since ξ is not a constant, we try a solution of the form ψ ≈ e±cξ2 where c is a constant to be found. Then from Equation (2.33) we find that
4c2ξ2e±cξ2± 2c e±cξ2− ξ2e±cξ2 ≈ 0 ,
so that we must choose c = 1/2 for the dominant terms as |ξ| → ∞ to cancel.
This leaves us with the asymptotic form ψ ≈ e±ξ2/2,
but we must reject the positive sign since we require the solution to be bounded.
The Schr¨odinger Equation in One Dimension 59 2.2.4 Factoring out the Asymptotic Behavior
The approximate solution may have the correct asymptotic form, but we require an exact solution that has the same asymptotic form, so we assume a solution of the form
ψ(ξ) = N e−ξ2/2H(ξ) , (2.34) where N is a normalization constant. Substituting this solution into Equation (2.32) and factoring out the common exponential factor, we find that H(ξ) satisfies
H00− 2ξH0+ (2λ − 1)H = 0 . (2.35) Problem 2.16 Hermite equation. Show that H(ξ) as defined by Equation (2.34) satisfies Equation (2.35) if ψ(ξ) satisfies Equation (2.32). This equation is known as the Hermite equation.
2.2.5 Finding the Power Series Solution
We assume Equation (2.35) has a power series solution of the form
H(ξ) =
∞
X
`=0
a`ξ`. (2.36)
Substituting this into Equation (2.35), this leads to
∞
X
`=0
[`(` − 1)a`ξ`−2− 2`a`ξ`+ (2λ − 1)a`ξ`] = 0 . (2.37)
Since this must be satisfied for any value of ξ, the coefficient of each power of ξ must vanish separately. (If even one coefficient did not vanish, the sum would not vanish for arbitrary values of ξ.) Allowing no negative powers (to avoid a singularity at the origin), we set ` = s + 2 in the first term of Equation (2.37) and ` = s in the next two terms to obtain
∞
X
s=0
[(s + 2)(s + 1)as+2− 2sas+ (2λ − 1)as]ξs= 0 , (2.38)
so setting each coefficient to zero gives the recursion relation, as+2
as
= 2s − (2λ − 1)
(s + 2)(s + 1). (2.39)
Since for large s, as+2/as→ 2/s → 0, the series will be convergent, but we need to examine its form. If we examine the series expression for eξ2, we find that it also has the ratio for successive terms as+2/as → 2/s → 0, so it seems that H(ξ) → eξ2 and hence ψ(ξ) → eξ2/2, which is again unbounded
as |ξ| → ∞. The only way to prevent this from happening is to truncate the series at some value s = n by choosing λ so that the numerator of Equation (2.39) vanishes when s = n. We thus arrive at the necessary relation that
λn= n +12, (2.40)
where n is a nonnegative integer.
This procedure has several implications we should note. First, it leads to the result that Hn(ξ) is a polynomial of order n (we introduce the subscript n to denote a polynomial of order n). Second, these polynomials must be either even or odd, since the recursion relation relates terms of every other order, and the condition on λn can only truncate either an even series or an odd series, but not both simultaneously.
The first few polynomials may be found by noting that the lowest value of λn is λ0 = 12 so the lowest order polynomial is a constant (a2 = 0, and we choose a1= 0 to eliminate all of the odd terms), and we choose a0to be unity.
Hence H0(ξ) = 1. Setting n = 1 so that λ1= 32, a3= 0 and we choose a0= 0 to eliminate the even terms, so only a16= 0 and we choose to define a1= 2 so H1(ξ) = 2ξ. We choose the arbitrary first coefficient to make the polynomials agree with the conventional Hermite polynomials [the highest order term of Hn(ξ) is always (2ξ)n]. These polynomials may be conveniently represented by the expression
Hn(ξ) = (−1)neξ2 dn
dξne−ξ2. (2.41)
Problem 2.17 Hermite polynomials. Expand Equation (2.41) for n = 1 through n = 5 and show that the highest order term is (2ξ)n for each n.
2.2.6 The Energy Eigenvalues
Since only certain values of λn are allowed, it follows from the definition of λn that there are only certain values for En. From
λn= En
~ω , it follows that
En= λn~ω = ~ω(n +12) . (2.42) Using the classical relation ω =pk/m, it is useful to note that
α2= mω
~
. (2.43)
2.2.7 Normalized Wave Functions Writing the normalized wave functions as
ψn(x) = Nne−α2x2/2Hn(αx) , (2.44)
The Schr¨odinger Equation in One Dimension 61 the normalization constant is shown in Appendix B.1 to be
Nn =
α
√π2nn!
1/2
. (2.45)
The first few normalized wave functions are:
ψ0(x) = α1/2
π1/4e−α2x2/2 (2.46)
ψ1(x) = α1/2
√2π1/42αxe−α2x2/2 (2.47)
ψ2(x) = α1/2
√8π1/4(4α2x2− 2)e−α2x2/2 (2.48)
ψ3(x) = α1/2 4√
3π1/4(8α3x3− 12αx)e−α2x2/2. (2.49) There are several features of the harmonic oscillator wave functions that can most easily be illustrated in a series of graphs. In Figure 2.6, the first six wave functions are illustrated, but each one has the zero reference shifted by En so they can be plotted along with the potential function (the argument is z = αx). With this vertical shift, one can see that at the value of the argument where En− V (z) changes sign, each wave function changes from oscillatory (inside) to exponential (outside). The alternating of even and odd functions with even and odd n is also apparent.
In Figure 2.7, the absolute magnitudes |ψ(z)|2 are shown for several wave functions, this time on the same plot. For the n = 3 case, it begins to be apparent that the outermost amplitude peaks are largest, indicating that the particle is most likely to be found there as the classical velocity vanishes at the turning point and moves quickly through the middle of the potential well where the velocity peaks. Figure 2.8 is a plot of the probability density for n = 15 and is compared to the classical probability density. Here we begin to see how the quantum mechanical probability density begins to correspond to the classical probability density (shown as a dotted curve), which is proportional to the time a particle spends between x and x + dx, so that it is maximum at the end points.
We may note that the wave functions tabulated in Equations (2.46) through (2.49) are final solutions back in ordinary variables since α is not dimensionless but carries a combination of the original constants through Equation (2.43).
Problem 2.18 Nonclassical behavior.
(a) Find the classical limit of the x-motion for an oscillator whose energy is
1 2~ω.
(b) Find the probability that the particle will be found outside this limit.
[Hint: Use ψ0. The integral cannot be evaluated exactly, but leads to the error function, erf(x), which is a tabulated function.]
−4 −2 0 2 4
Wave functions (not normalized) for n = 0 through n = 5 compared to the classical density.
(c) Find the probability that the particle will be found more than twice this far from the origin. (See hint above.)
Problem 2.19 Higher order wave functions.
(a) Find ψ4(x) and ψ5(x).
(b) Show by direct integration that ψ4(x) is normalized.
Problem 2.20 Mean speed. A harmonic oscillator is in a state ψ = c0ψ0+
The Schr¨odinger Equation in One Dimension 63
−4 −3 −2 −1 0 1 2 3 4
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
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FIGURE 2.7
Probability densities for the first four harmonic oscillator wave functions.