wave along a deterministic trajectory.
Another example that shows the contrast with the classical de-
scription is that if I show you a snapshot of a wave on a string, you
can’t tell which direction it’s going — as with the baseball, you need
to know its initial velocity in addition. But if I show you a snap-
shot of a quantum-mechanical traveling wave, youcan tell which
direction it’s going, because of the complex phase, as shown in fig-
ures u/2 and u/3 on page 914. Note that this mechanism wouldn’t
work if wavefunctions were always real numbers, so wavefunction
fundamentalism implies complex wavefunctions.
Given the wavefunction at some initial time, we can predict its
evolution into the future by making use of the principle thatE=
hf. Suppose for example that we have a sinusoidal plane wave
traveling to the right. Then we expect the value of the wavefunction
at a particular point in space to rotate clockwise about the origin
in the complex plane at the appropriate frequencyf, showing a
time dependencee−iωt(where, as usual,ω= 2πf). Thus the time
derivative of the wavefunction is Ψ′=−iωΨ =−i(E/~)Ψ, so that
EΨ =i~Ψ′. Then to generalize the time-independent Schr ̈odinger
equation to its time-dependent version, the most obvious thing to
try is simply to substitutei~∂Ψ/∂tforEΨ, which gives
i~∂Ψ
∂t=−
~^2
2 m∇^2 Ψ +UΨ.
(In section 14.6.4, p. 988, we will generalize this to cases where the
wavefunction is not expressed in terms of the spatial coordinatesx,
y, andz.) Unlike Newton’s laws of motion, which refer to a second
derivative with respect to time, the Schr ̈odinger equation involves
only a first time derivative. This is why we don’t need initial data
on∂Ψ/∂t, but only Ψ: if we know Ψ, then the right-hand side of the
Schr ̈odinger equation is whatgivesus∂Ψ/∂t. But the Schr ̈odinger
equation has some other properties that match up with those of
Newton’s laws.
A plane wave example 3
Consider a free particle of massmin one dimension, with the
wavefunction
Ψ=ei(k x−ωt),
wherek = 2π/λ=p/~is called the wavenumber. Ifkandωare
both positive, we can tell that the particle is moving to the right,
because the signs inside the exponential are such thatx could
increase astincreases while keeping the phase the same. This
would happen fork∆x−ω∆t= 0, orv=ω/k, which is the phase
velocity (not the same as the group velocity, sec. 13.3.2, p. 894).
Suppose that the particle is in free space, so thatUis constant,
and for convenience takeU= 0. Application of the Schrodinger ̈
equation,i~∂Ψ/∂t=−(~^2 / 2 m)∂^2 Ψ/∂x^2 , gives~ωe(...)=~
(^2) k 2
2 me
(...),
Section 14.4 The underlying structure of quantum mechanics, part 1 967