- #1
snoopies622
- 846
- 28
I see how one can arrive at the Schrodinger equation by starting with a complex plane wave
[tex]
\psi = \psi _0 e ^{i(k \cdot x - \omega t)}
[/tex]
taking its first partial derivative with respect to time, second partial derivative with respect to space, making the quantum substitutions
[tex]
k=p/ \hbar \hspace{10 mm} \omega = E / \hbar
[/tex]
as well as the classical one
[tex]
E=p^2/2m + V(x,y,z)
[/tex]
and putting it all together.
But why does this work when the [itex] \psi [/itex]'s that one finds in quantum mechanics are typically not plane waves? Luck? I've read that one should accept the Schrodinger equation as an axiom and not worry about "deriving" it, and yet...
[tex]
\psi = \psi _0 e ^{i(k \cdot x - \omega t)}
[/tex]
taking its first partial derivative with respect to time, second partial derivative with respect to space, making the quantum substitutions
[tex]
k=p/ \hbar \hspace{10 mm} \omega = E / \hbar
[/tex]
as well as the classical one
[tex]
E=p^2/2m + V(x,y,z)
[/tex]
and putting it all together.
But why does this work when the [itex] \psi [/itex]'s that one finds in quantum mechanics are typically not plane waves? Luck? I've read that one should accept the Schrodinger equation as an axiom and not worry about "deriving" it, and yet...