Question

A second order homogeneous linear differential equation has odd-even parity. Prove that if one of its solutions is an even function, the other can be constructed as an odd function.

Answer 1

Parity are the behavior of a physical system,

or it is one of mathematical functions that describe such a system, under reflection.

There are two "kinds" of parity:

Even and odd

• If f(x)=f(−x) means even parity
• If f(x)=−f(−x) means odd parity

for many functions, none of these onditions are true, and in that case function f having indefiniteparity.

−ℏ/2m d2/dx2ψ(x)+V(x)ψ(x)=Eψ(x)

x→−infinty

−ℏ/2m d2/dx2ψ(−x)+V(−x)ψ(−x)=Eψ(−x)

If we have symmetric (even) potential, V(x)=V(−x)

this is exactly the same as the original equation except that we've transformed ψ(x)→ψ(−x)

ψ(x)&ψ(−x) satifies the same eqn, we will have the same solutions for them,

except for an overall multiplicative constant

ψ(x)=mψ(−x)

Normalizing ψ needs |a|=1 which leaves 2 cases

m=+1(even parity) &m= −1(odd parity)