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Jun15-09, 03:49 PM
P: 166
Nick89 -
Quote Quote by Nick89 View Post

We might have a different print of Griffiths, but I cannot find the second conclusion in bold in my book, nor does it make much sense to me.
The first quote (as you know) is from Griffiths)

Sorry - the second was found on a professor's web notes - University of Texas I think - where I was searching for further help on this topic.

Nick89, Cyosis, Bob_for_short, muppet and others -

follow-up question -

I agree that if you "know" that [tex] |\psi|^2[/tex] is a probability density (or behaves like a probability density - I know wording can be sticky) then the area under the curve has to be 1.0.

- If you didn't know how [tex][|\psi|^2] [/tex]behaves, could you make the statements:

“But [tex] |\psi|[/tex] must go to zero as x goes to infinity." - Griffiths

“The above equation is satisfied provided goes to zero as goes to zero." - some professor's notes

by seeing this alone: [tex] \frac {d}{dt} \int_ {-\infty}^{\infty}|\psi|^2 dx= \frac{ih}{2m}(\psi*\frac{d\psi}{dx} - \frac{\psi*}{dx}\psi ) [/tex]

Meaning by analysis alone - does the math tell me that “But [tex] |\psi|[/tex] must go to zero as x goes to infinity." (without knowing the probability density behavior)?

Does my question make sense? - I don't see how the statement can be made "by inspection of the resulting "derivation".