s/1. The one-dimensional
version of the Laplacian is the
second derivative. It is positive
here because the average of the
two nearby points is greater than
the value at the center. 2. The
Laplacian of the function A in
example 20 is positive because
the average of the four nearby
points along the perpendicular
axes is greater than the function’s
value at the center. 3.∇^2 C= 0.
The average is the same as the
value at the center.
Three dimensions
For simplicity, we’ve been considering the Schr ̈odinger equation
in one dimension, so that Ψ is only a function ofx, and has units of
m−^1 /^2 rather than m−^3 /^2. Since the Schr ̈odinger equation is a state-
ment of conservation of energy, and energy is a scalar, the general-
ization to three dimensions isn’t particularly complicated. The total
energy termE·Ψ and the interaction energy termU·Ψ involve noth-
ing but scalars, and don’t need to be changed at all. In the kinetic
energy term, however, we’re essentially basing our computation of
the kinetic energy on the squared magnitude of the momentum,p^2 x,
and in three dimensions this would clearly have to be generalized to
p^2 x+p^2 y+p^2 z. The obvious way to achieve this is to replace the second
derivative d^2 Ψ/dx^2 with the sum∂^2 Ψ/∂x^2 +∂^2 Ψ/∂y^2 +∂^2 Ψ/∂z^2.
Here the partial derivative symbol∂, introduced on page 220, indi-
cates that when differentiating with respect to a particular variable,
the other variables are to be considered as constants. This operation
on the function Ψ is notated∇^2 Ψ, and the derivative-like operator
∇^2 =∂^2 /∂x^2 +∂^2 /∂y^2 +∂^2 /∂z^2 is called the Laplacian, and was
introduced briefly on p. 655. It occurs elsewhere in physics. For
example, in classical electrostatics, the voltage in a region of vac-
uum must be a solution of the equation∇^2 V = 0. Like the second
derivative, the Laplacian is essentially a measure of curvature. Or,
as shown in figure s, we can think of it as a measure of how much
the value of a function at a certain point differs from the average of
its value on nearby points.Examples of the Laplacian in two dimensions example 20
.Compute the Laplacians of the following functions in two dimen-
sions, and interpret them:A=x^2 +y^2 ,B=−x^2 −y^2 ,C=x^2 −y^2.
.The first derivative of functionAwith respect tox is∂A/∂x =
2 x. Sincey is treated as a constant in the computation of the
partial derivative∂/∂x, the second term goes away. The second
derivative ofAwith respect toxis∂^2 A/∂x^2 = 2. Similarly we have
∂^2 A/∂y^2 = 2, so∇^2 A= 4.
All derivative operators, including∇^2 , have the linear property
that multiplying the input function by a constant just multiplies the
output function by the same constant. SinceB = −A, and we
have∇^2 B=−4.
For functionC, thexterm contributes a second derivative of 2,
but theyterm contributes−2, so∇^2 C= 0.
The interpretation of the positive sign in∇^2 A= 4 is thatA’s graph
is shaped like a trophy cup, and the cup is concave up. ∇^2 B<
0 is becauseBis concave down. FunctionCis shaped like a
saddle. Since its curvature along one axis is concave up, but the
curvature along the other is down and equal in magnitude, the910 Chapter 13 Quantum Physics