Invariance of Schrödinger's equation

[tommy01]

I thought i had a basic to intermediate understanding of quantum physics and group theory, but when reading hamermesh's "group theory and it's application to physical problems" there's something in the introduction i don't understand.

first of all, i know the parity (or space inversion) operator and it's eigenfunctions. so from this point of view the example in the introduction is quit easy but i don't get hamermesh's argumentation.

he start's with the Schrödinger equation in one dimension:
"$$u''+[\lambda - V(x)]u = 0$$
where $$\lambda$$ is the eigenvalue of u.
one dimension => necessarily not degenerate." why?

"We assume that the potential is an even function of x. ($$V(x)=V(-x)$$)"

"replacing x by -x, we see, that if u(x) is a solution, so is u(-x)."
why?

what does he mean? is this a variable substitution $$x \rightarrow -x$$
$$u''(-x)+[\lambda - V(-x)]u(-x) = 0 = u''(-x)+[\lambda - V(x)]u(-x)$$ ?
is such a substitution allowed?

sorry if the answer to my question is obvious but i don't get it at the moment.

thanks and greetings.

 

[tiny-tim]

Greetings tommy01!

(have a lambda: λ )

Don't use x -> -x, it's too confusing.

Use x -> y, where y = -x.

Then d²u/dx² = d²u/dy²,

so d²u/dx² + (λ - V(x))u

= d²u/dy² + (λ - V(-y))u

 

[facenian]

For understanding why the states ara not degenerate I suggest Volume III of L.D. Landau and Lifgarbagez chapter III search for "general properties of motion in one dimmension"

 

[tommy01]

hi all.

thanks for your quick answer.
i'm going to consult landau and lifgarbagez.