The Schrödinger Equation in Three Dimensions

The Schrödinger Equation in Three Dimensions

Particle in a Rigid Three-Dimensional Box (Cartesian Coordinates)

To illustrate the solution of the time-independent Schrödinger equation (TISE) in three
dimensions, we start with the simple problem of a particle in a rigid box. This is the three-
dimensional version of the problem of the particle in a one-dimensional, rigid box. In one
dimension, the TISE is written as
 2 d 2ψ ( x)
− + U ( x)ψ ( x) = Eψ ( x). (1)
2m dx 2

In three dimensions, the wave function will in general be a z
function of the three spatial coordinates. In the Cartesian y
coordinate system, these coordinates are x, y, and z. Also,
the potential energy U will in general be a function of all 3 x
 dψ
2 2
coordinates. Now, in the 1-D TISE, the term − Lz Lx
2m dx 2
2 2 2
p  kx
can be identified with the kinetic energy x = of the Ly
2m 2m
 2 d 2ψ
particle because − = [ E − U ]ψ . [Try, for example, the free-particle wave function
2m dx 2
ψ = Aei ( kx −ω t ) . ]

In three dimensions, the KE is ( px2 + py2 + pz2 ) / 2m , so we suspect that additional second
derivative terms will be needed to represent the additional kinetic energy terms. In fact, the TISE
in three dimensions is written as
 2 ⎛ ∂ 2ψ ( x, y, z ) ∂ 2ψ ( x, y, z ) ∂ 2ψ ( x, y, z ) ⎞
− ⎜ + + ⎟ + U ( x, y, z )ψ ( x, y, z ) = Eψ ( x, y, z ). (2)
2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠

∂ψ (x, y, z)
The symbol represents the partial derivative of ψ (x, y, z) with respect to x, which is
∂x
simply the derivative of ψ (x, y, z) with respect to x with y and z held constant. Thus,
∂ψ
if ψ = x 2 yz, then = 2xyz .
∂x

In the particle in a box problem, U(x, y, z) = 0 inside the box and U = ∞ at the walls.1 So, inside
the box, the TISE becomes
 2 ⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞
− ⎜ + + ⎟ = Eψ . (3)
2m ⎝ ∂x 2 ∂y 2 ∂z 2 ⎠


1
Inside the box, 0<x<Lx, 0<y<Ly, and 0<z<Lz. (See figure of box above.)


1

, To solve partial differential equations (the TISE in 3D is an example of these equations), one can
employ the method of separation of variables. We write

ψ (x, y, z) = X(x)Y (y)Z(z), (4)
where X is a function of x only, Y is a function of y only, and Z is a function of z only.
Substituting for ψ in Eq. (3) yields
2 ⎡ d2X d 2Y d 2Z ⎤
− Y ( y ) Z ( z ) + X ( x )Y ( z ) + X ( x )Y ( y ) = EX ( x)Y ( y ) Z ( z ). (5)
2m ⎢⎣ dx 2 dy 2 dz 2 ⎥⎦

Note that the partial derivatives have disappeared, since X, Y, and Z are functions of one variable
2
only. Dividing both sides by − X ( x)Y ( y ) Z ( z ) yields
2m
⎡ 1 d 2 X 1 d 2Y 1 d 2 Z ⎤ 2mE
⎢ X dx 2 + Y dy 2 + Z dz 2 ⎥ = −  2 . (6)
⎣ ⎦

For a given solution, E is a constant. Further, note that the first term in the square brackets is a
function of x only, the second term is a function of y only, and the third term is a function of z
only. For Eq. (6) to be valid for all values of x, y, and z, each of the three terms must be constant.
If this were not true, and, for example, all the terms were variable, then if one held y and z
constant and changed x, the sum on the left hand side of the equation would change, violating the
equation. Eq. (6) therefore becomes three separated ordinary differential equations:
1 d2X
= −k 2x , (7)
X dx 2
1 d 2Y
2
= −k 2y , (8)
Y dy
and
1 d2Z
2
= −k 2z , (9)
Z dz


with
2mE
k x2 + k y2 + k z2 = . (10)
2

The separation constants are written as −k 2x , −k 2y , and −k 2z in analogy with the 1-D particle in a
box problem. Eqs. (7), (8), and (9) are identical to the equation obtained in the 1-D problem and
the boundary conditions are the same. For example, X(x) = 0 at x = 0 and x = Lx since the wave
functions cannot penetrate the wall. The boundary condition at x = 0 leads to X(x) = A1sin kxx.
nπ
The boundary condition at x = Lx leads to k x = x , where nx = 1, 2,… Similar solutions are
Lx
obtained for Y(y) and Z(z). Hence, we find that



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