Simple Nature - Light and Matter

Three dimensionsFor simplicity, we’ve been considering the Schrödinger equationin one dimension, so that Ψ is only a function of x, and has units ofm −1/2 rather than m −3/2 . Since the Schrödinger equation is a statementof conservation of energy, and energy is a scalar, the generalizationto three dimensions isn’t particularly complicated. The totalenergy term E ·Ψ and the interaction energy term U ·Ψ involve nothingbut scalars, and don’t need to be changed at all. In the kineticenergy term, however, we’re essentially basing our computation ofthe kinetic energy on the squared magnitude of the momentum, p 2 x,and in three dimensions this would clearly have to be generalized top 2 x +p 2 y +p 2 z. The obvious way to achieve this is to replace the secondderivative 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 216, indicatesthat when differentiating with respect to a particular variable,the other variables are to be considered as constants. This operationon 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. It occurselswehere in physics. For example, in classical electrostatics, thevoltage in a region of vacuum must be a solution of the equation∇ 2 V = 0. Like the second derivative, the Laplacian is essentially ameasure of curvature.Examples of the Laplacian in two dimensions example 20⊲ Compute the Laplacians of the following functions in two dimensions,and interpret them: A = x 2 + y 2 , B = −x 2 − y 2 , C = x 2 − y 2 .⊲ The first derivative of function A with respect to x is ∂A/∂x =2x. Since y is treated as a constant in the computation of thepartial derivative ∂/∂x, the second term goes away. The secondderivative of A with respect to x is ∂ 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 propertythat multiplying the input function by a constant just multiplies theoutput function by the same constant. Since B = −A, and wehave ∇ 2 B = −4.For function C, the x term contributes a second derivative of 2,but the y term contributes −2, so ∇ 2 C = 0.The interpretation of the positive sign in ∇ 2 A = 4 is that A’s graphis shaped like a trophy cup, and the cup is concave up. The negativesign in the result for ∇ 2 B is because B is concave down.Function C is shaped like a saddle. Since its curvature alongone axis is concave up, but the curvature along the other is downand equal in magnitude, the function is considered to have zeroconcavity over all.872 Chapter 13 Quantum Physics