- #1
tommy01
- 40
- 0
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:
"[tex]u''+[\lambda - V(x)]u = 0[/tex]
where [tex]\lambda[/tex] is the eigenvalue of u.
one dimension => necessarily not degenerate." why?
"We assume that the potential is an even function of x. ([tex]V(x)=V(-x)[/tex])"
"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 [tex]x \rightarrow -x[/tex]
[tex]u''(-x)+[\lambda - V(-x)]u(-x) = 0 = u''(-x)+[\lambda - V(x)]u(-x)[/tex] ?
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.
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:
"[tex]u''+[\lambda - V(x)]u = 0[/tex]
where [tex]\lambda[/tex] is the eigenvalue of u.
one dimension => necessarily not degenerate." why?
"We assume that the potential is an even function of x. ([tex]V(x)=V(-x)[/tex])"
"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 [tex]x \rightarrow -x[/tex]
[tex]u''(-x)+[\lambda - V(-x)]u(-x) = 0 = u''(-x)+[\lambda - V(x)]u(-x)[/tex] ?
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.