4.7: Simple Harmonic Oscillator
- Last updated
-
Apr 1, 2025
-
Save as PDF
-
The classical Hamiltonian of a simple harmonic oscillator is H=p22m+12Kx2,
where K>0 is the so-called force constant of the oscillator. Assuming that the quantum mechanical Hamiltonian has the same form as the classical Hamiltonian, the time-independent Schrödinger equation for a particle of mass m and energy E moving in a simple harmonic potential becomes d2ψdx2=2mℏ2(12Kx2−E)ψ.
Let ω=√K/m, where ω is the oscillator’s classical angular frequency of oscillation. Furthermore, let y=√mωℏx,
and ϵ=2Eℏω.
Equation ([e5.90]) reduces to d2ψdy2−(y2−ϵ)ψ=0.
We need to find solutions to the previous equation which are bounded at infinity: that is, solutions which satisfy the boundary condition ψ→0 as |y|→∞.
Consider the behavior of the solution to Equation ([e5.93]) in the limit |y|≫1. As is easily seen, in this limit the equation simplifies somewhat to give d2ψdy2−y2ψ≃0.
The approximate solutions to the previous equation are ψ(y)≃A(y)e±y2/2,
where A(y) is a relatively slowly varying function of y. Clearly, if ψ(y) is to remain bounded as |y|→∞ then we must chose the exponentially decaying solution. This suggests that we should write ψ(y)=h(y)e−y2/2,
where we would expect h(y) to be an algebraic, rather than an exponential, function of y.
Substituting Equation ([e5.96]) into Equation ([e5.93]), we obtain d2hdy2−2ydhdy+(ϵ−1)h=0.
Let us attempt a power-law solution of the form h(y)=∑i=0,∞ciyi.
Inserting this test solution into Equation ([e5.97]), and equating the coefficients of yi, we obtain the recursion relation ci+2=(2i−ϵ+1)(i+1)(i+2)ci.
Consider the behavior of h(y) in the limit |y|→∞. The previous recursion relation simplifies to ci+2≃2ici.
Hence, at large |y|, when the higher powers of y dominate, we have h(y)∼C∑jy2jj!∼Cey2.
It follows that ψ(y)=h(y)exp(−y2/2) varies as exp(y2/2) as |y|→∞. This behavior is unacceptable, because it does not satisfy the boundary condition ψ→0 as |y|→∞. The only way in which we can prevent ψ from blowing up as |y|→∞ is to demand that the power series ([e5.98]) terminate at some finite value of i. This implies, from the recursion relation ([e5.99]), that ϵ=2n+1,
where n is a non-negative integer. Note that the number of terms in the power series ([e5.98]) is n+1. Finally, using Equation ([e5.92]), we obtain E=(n+1/2)ℏω,
for n=0,1,2,⋯.
Hence, we conclude that a particle moving in a harmonic potential has quantized energy levels that are equally spaced. The spacing between successive energy levels is ℏω, where ω is the classical oscillation frequency. Furthermore, the lowest energy state (n=0) possesses the finite energy (1/2)ℏω. This is sometimes called zero-point energy. It is easily demonstrated that the (normalized) wavefunction of the lowest energy state takes the form ψ0(x)=e−x2/2d2π1/4√d
Let ψn(x) be an energy eigenstate of the harmonic oscillator corresponding to the eigenvalue En=(n+1/2)ℏω.
Assuming that the ψn are properly normalized (and real), we have ∫∞−∞ψnψmdx=δnm.
Now, Equation ([e5.93]) can be written (−d2dy2+y2)ψn=(2n+1)ψn,
where x=dy, and d=√ℏ/mω. It is helpful to define the operators a±=1√2(∓ddy+y).
As is easily demonstrated, these operators satisfy the commutation relation [a+,a−]=−1.
Using these operators, Equation ([e5.108]) can also be written in the forms a+a−ψn=nψn,
or a−a+ψn=(n+1)ψn.
The previous two equations imply that a+ψn=√n+1ψn+1,a−ψn=√nψn−1.
We conclude that a+ and a− are raising and lowering operators, respectively, for the harmonic oscillator: that is, operating on the wavefunction with a+ causes the quantum number n to increase by unity, and vice versa. The Hamiltonian for the harmonic oscillator can be written in the form H=ℏω(a+a−+12),
from which the result Hψn=(n+1/2)ℏωψn=Enψn
is readily deduced. Finally, Equations ([e5.107]), ([e5.113]), and ([e5.114]) yield the useful expression ∫∞−∞ψmxψndx=d√2∫∞−∞ψm(a++a−)ψndx=√ℏ2mω(√mδm,n+1+√nδm,n−1).
Contributors and Attributions