What does it mean to satisfy the Schrodinger equation?

warfreak131
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What does it mean to "satisfy" the Schrodinger equation?

Homework Statement



Show that the 2p wave functions of the hydrogen atom satisfy the radial Schrodinger eq.

One of the radial equations for the 2p state is \frac{1}{\sqrt{96 \pi a^{3}}} \frac{r}{a} e^{\frac{-r}{2a}}

Homework Equations


The Attempt at a Solution



[\frac{-\hbar^2}{2m}\frac{1}{r^2}\frac{d}{dr}(r^2\frac{d}{dr})+\frac{l(l+1)\hbar^2}{2mr^2}+V(r)]R=ER

I took the derivative with respect to r, and followed all the subsequent derivatives, and the answer is really messy. What exactly am I looking for when something "satisfies" the equation?
 
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You're looking to see if the operator on the left, when acting upon the wave function (or in this case part of it), returns a constant times that same wave function.

So if I had something like f(x)=e^{-3x/5} and were asked if f(x) satisfies

{{df}\over{dx}} = Af(x).

What you get when you do the derivative of f(x) is the -3/5 times the function again. Thus, f(x) satisfies the equation with A = -3/5. Same idea here
 


Ok, I see. So what should I do since all the derivates and sums don't come out to a nice constant?
 


Check your work :P Or show us your work
 


okay, but quick question first
\frac{d}{dr}(r^2\frac{d}{ dr})

is this line asking you to take the derivative of R, multiply it by r^2, and then take the derivative of the resulting equation?
 


warfreak131 said:
okay, but quick question first
\frac{d}{dr}(r^2\frac{d}{ dr})

is this line asking you to take the derivative of R, multiply it by r^2, and then take the derivative of the resulting equation?

Yup!
 


okay, i used mathematica, i created functions for each term, did all the necessary derivatives and additions, and what i got was

(2 a^2 e^2 m - 8 a h^2 pi epsilon + h^2 pi r epsilon)/(8 a^2 m pi r epsilon)

this answer was close with the exception of r, everything else is a constant
 

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Express the Bohr radius in terms of the other constants. You should find the first two terms in the numerator will cancel.
 


vela said:
Express the Bohr radius in terms of the other constants. You should find the first two terms in the numerator will cancel.

ill try that, thanks
 
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