Question

Show that sin(x) is an eigenfunction of the free particle (no potential energy) Hamiltonian operator. Please be explicit.

Hsin(x) = Esin(x)

p²x = − iℏ ∂/∂x = ℏi ∂/∂x

H = p²/2m

Answer 1

The simplest system in quantum mechanics has the potential energy V equal to zero everywhere. This is called a free particle since it has no forces acting on it. We consider the one-dimensional case, with motion only in the x-direction, giving the Schr¨odinger equation

− ¯h ² /2m . d ²ψ(x) dx² = Eψ(x)............... (1)

Total derivatives can be used since there is but one independent variable. The equation simplifies to ψ ⁿ(x) + k ² ψ(x) = 0.............(2)

with the definition, k ² ≡ 2mE/¯h ²..............(3)

Possible solutions of Eq (2) are

ψ(x) = const ( sin kx ,cos kx, e^±ikx).........(4)

There is no restriction on the value of k. Thus a free particle, even in quantum mechanics, can have any non-negative value of the energy

E = ¯h ²k ² /2m ≥ 0.............(5)

The energy levels in this case are not quantized and correspond to the same continuum of kinetic energy shown by a classical particle

It is of interest also to consider the x-component of linear momentum for the free-particle solutions (4).the eigenvalue equation for momentum should read

pˆ_xψ(x) = −i¯h dψ(x)/ dx = p ψ(x)

where we have denoted the momentum eigenvalue as p. It is easily shown that neither of the functions sin kx or cos kx from (4) is an eigenfunction of ˆpx. But e^±ikx are both eigenfunctions with eigenvalues p = ±¯hk, respectively. Evidently the momentum p can take on any real value between −∞ and +∞. The kinetic energy, equal to E = p ²/2m, can correspondingly have any value between 0 and +∞

The functions sin kx and cos kx, while not eigenfunctions of ˆpx, are each superpositions of the two eigenfunctions e^±ikx, by virtue of the trigonometric identities

cos kx = 1/ 2 (e ^ikx + e^−ikx) and sin kx = 1/ 2i (e ^ikx − e^−ikx)

The eigenfunction e ikx for k > 0 represents the particle moving from left to right on the x-axis, with momentum p > 0. Correspondingly, e−ikx represents motion from right to left with p < 0. The functions sin kx and cos kx represent standing waves, obtained by superposition of opposing wave motions. Although these latter two are not eigenfunctions of ˆpx but are eigenfunctions of ˆp 2 x , hence of the Hamiltonian Hˆ .