and if this is to hold true for all values ofxandt, then we must
have~ω=~(^2) k 2
2 m, which is simply an expression of the Newtonian
relationK=p^2 / 2 m, sincekλ= 2πandp=h/λ. Flipping the sign
ofk results in an equally valid solution, and a negativekis how
we would represent a wave traveling to the left.
We have two solutions to the Schrodinger equation corresponding ̈
to the two signs ofk, and because the Schrodinger equation is ̈
linear, it follows that we can make a more general solution of the
form
Aei(k x−ωt)+Bei(−k x−ωt),
whereAandBare any two complex numbers. (We could also try
to elaborate on this theme by allowing for an arbitrary phase angle
δinside the complex exponentials, e.g., changing the argument
of the first exponential toi(k x−ωt+δ). However, this would be
equivalent to changingAtoAeiδ, which is just a change inA’s
phase angle, not a new solution.)
Dispersion of a wave packet example 4
An annoying feature of example 3 is that the wavefunction can-
not be normalized because it extends in all directions to infinity.
This type of infinite plane wave is at best an idealization of the
wavefunction for a realistic example such an electron launched
by a cathode ray tube, or an alpha particle emitted by a nucleus.
As a more realistic example, we might try something like a wave
packet, such as a pulse with a certain shape, traveling in a cer-
tain direction. This works for waves on a string or for electromag-
netic waves: the pulse or packet simply glides along while rigidly
maintaining its shape. To investigate this idea using the time-
dependent Schrodinger equation, it will be convenient to adopt ̈
the frame of reference in which the particle is at rest. In this frame,
we would expect the wave to look frozen, just as an ocean wave
looks frozen in place to a surfer who is riding it. It must therefore
be of the form
Ψ=e−iωtf(x),
wherefis some function specifying the shape of the wave packet.
But thisΨis not a solution to the Schrodinger equation. On ̈
the left-hand side of the Schrodinger equation, evaluating the ̈
time derivative gives~ωΨ, which is just the original wavefunc-
tion multiplied by a constant. If we are to satisfy the Schrodinger ̈
equation, then the right-hand side, which is the second derivative
with respect tox, must also be equal to the original wavefunc-
tion multiplied by a constant. But the only functions for which
(d^2 /dx^2 )(...) = (constant)(...) are exponentials and sine waves.
An exponential shape obviously isn’t a physical realization of a
wave packet. A sine wave works, but it just describes an infinite
plane wave like the one in example 3, not a wave packet that can
be localized and normalized.
968 Chapter 14 Additional Topics in Quantum Physics