Physlets run in a Java-enabled browser, except Chrome, on the latest Windows & Mac operating systems. If Physlets do not run, click here for help updating Java & setting Java security.
Section 12.2: The Quantum-mechanical Harmonic Oscillator
Please wait for the animation to completely load.
The solution of the one-dimensional quantum harmonic oscillator problem begins with the time-independent Schrödinger equation
[−(ħ2/2m)(d2/dx2) + (1/2)mω2x2] ψ(x) = Eψ(x) , (12.4)
where k/2 = mω2/2. We can put this equation into a more standard form as
[d2/dx2 − m2ω2x2/ħ2 + 2mE/ħ2 ] ψ(x) = 0 . (12.5)
We can further simplify the differential equation by defining the following: β2 = mω/ħ and ξ2 = β2x2, which when substituted into the differential equation, gives
[d2/dξ2 − ξ2 + 2E/(ħω)] ψ(ξ) = 0 . (12.6)
Even though this differential equation looks nothing like our original differential equation, the three terms are just the kinetic, potential, and total energies, respectively. Consider the two special limiting cases of Eq. (12.6):
Case I: −ξ2 + 2E/(ħω) ≈ −ξ2, this situation results when the potential energy at a given position in the well is much greater than the total energy, V >> E. This is forbidden classically. We explicitly solve the differential equation
[d2/dξ2 − ξ2] ψ(ξ) = 0 ,
which for large ξ yields the solutions ψ(ξ) = exp(±ξ2/2).
Case II: −ξ2 + 2E/(ħω) ≈ 2E/(ħω), this corresponds to the case where the total energy is much greater than the potential energy, E >> V, which occurs near x = 0. We explicitly solve the differential equation
[d2/dξ2 +2E/(ħω)]ψ(ξ) = 0 ,
which yields the real solutions ψ(ξ) = Acos(Kξ) + Bsin(Kξ) where K2 = 2E/(ħω).
What does the total solution look like? In the region where E>>V we have an oscillating solution. However, in the classically forbidden region V >> E, we have some leakage of the energy eigenfunction into the classically-forbidden region. The well-behaved solution in this region is exp(−ξ2/2). as it goes to zero for large ξ → ±∞.
Now that we have an idea of what the bound states should look like, we can find the entire solution to to Eq. (12.6). This solution can be written as Hermite polynomials weighted by a factor of exp(−ξ2/2) = exp(−β2x2/2). These solutions can be written in terms of β and x as:
ψn(x) = AnHn(βx) exp(−β2x2/2) , (12.7)
where n = 0,1,2,3,4,…. The term Hn(βx) refers to the Hermite polynomials of order n and the An is the normalization factor which is equal to
An = ((mω)1/2/(2nn!(ħπ)1/2)1/2 . (12.8)
In terms of the argument ξ, the first 4 Hermite polynomials are
H0(ξ) = 1 H1(ξ) = 2ξ H2(ξ) = 4ξ2 − 2 H3(ξ) = 8ξ3 − 12ξ .
In the animations, the energy eigenfunctions for a quantum harmonic oscillator are shown. The animation uses ħ = 2m = 1 and ω = 2. Note that in the well, the energy eigenfunction's amplitude and curviness change with position. These changes are due to the fact that the potential energy function itself changes with position. The energy eigenfunction is curvier and has a smaller amplitude nearer the center of the well as compared to the positions closer to the classical turning point.
By brute force integration we can use these energy eigenfunctions to calculate the expectation values.2 We find that <x>n = 0 and <p>n = 0, as expected, and
<x2>n = (ħ/2mω) (2n + 1) (12.9)
<p2>n = (mħω/2) (2n + 1) , (12.10)
<E>n = <p2>n/2m + mω2<x2>n/2 = (n + 1/2)ħω . (12.11)
This uniformity in the spacing of the harmonic oscillator energies is shown in the "spectrum" animation. To see the other bound states simply click-drag in the energy level diagram on the left to select a level. The selected level will turn red. Since we have chosen ω = 2 and ħ = 2m = 1, the energy spectrum is just En = (2n + 1).
2There is an easier way. This method uses operators, called raising and lowering operators, to write the harmonic oscillator Hamiltonian. This method was pioneered by Dirac and is the easiest way to calculate expectation values.