Question

Hi all,

I have having difficulty with the attached Schrodinger equation integral problem.

Would anyone be able to assist/point me in the right direction?

http://imgur.com/duOe9

Answer 1

hmmm

Answer 2

thats pretty much all I have got as well :)

Answer 3

what are \(R(r), \Phi(\phi), \Theta(\theta)\) ?

Answer 4

not totally sure in this context. The only other information I have is attached.

Thanks for your reply by the way.

Answer 5

i think you have to normalize

Answer 6

say wat!?

Answer 7

watchu talkin bout Rhaukus?

Answer 8

you 're mostly right @Algebraic! .... just note that:
\[\large\frac{ 1 }{ 4\sqrt{2\pi {(a_0)}^{5}} } \dots \text{ should also be squared}\]

Answer 9

it's already normalized
integrate:
sin^3 (theta) from 0..pi,
sin^2 (phi) from 0..2p

(I left out a sin(theta) also)

Answer 10

@UnkleRhaukus @z61850
\(R(r)\) just means that part of \(\large \Psi\) that has \(r\) as the variable

\[\Large R(r)={1 \over 4\sqrt{2 \pi (a_0)^5}}r \cdot e^{-r/2a_0}\]\[\Large \Phi(\phi)=\sin \left( \phi \right)\]\[\Large \Theta(\theta)=\sin \left( \theta\right)\]
\[\large \Psi=R \times \Theta \times \Phi\]

Answer 11

First link to such integrals
http://quantummechanics.ucsd.edu/ph130a/130_notes/node233.html

Answer 12

deleted prior fail

integrate:
sin^3 (theta) from 0..pi,
sin^2 (phi) from 0..2pi
and
\[\huge \frac{ 1 }{ 32\pi a _{0}^{5}} \int\limits_{0}^{\infty} r ^{2} (re ^{\frac{ -r }{2a _{0} }})^{2}dr\]

multiply results
(I'll probably get this right eventually...)

Answer 13

sorry...
\[\Large R(r)=r \cdot e^{-r/2a_0}\]

\[\large {1 \over 4\sqrt{2 \pi (a_0)^5}} \dots \text{ must be the normalization factor }\Large =N\]

Answer 14

See here (attached)
Appendix B
and Pages 24 -28

Answer 15

First of your Integrals is the expectation value ofn r^2 It is described in the file I've attached

Answer 16

it's gotta be squared, you were right the first time.

Answer 17

@z61850 ?

Answer 18

it works now

Answer 19

See integrals here in "Orthogonality of the Laguerre Polynomials"
http://www.morehouse.edu/facstaff/cmoore/Laguerre%20Polynomials.htm

Answer 20

http://www.physics.drexel.edu/~tim/open/hydrofin/node5.html

Answer 21

Sorry, internet outage. Digesting now...thanks so much guys, really appreciate the input.

Answer 22

Thanks to all who provided input, sadly I still cannot arrive at \[\int\limits_{V}^{}\psi ^{2} dV = 1\].

I arrive at \[\frac{ 1 }{ 4a }\]

Answer 23

\[(i)\large {\int\limits_0^\infty r^2 \cdot R^2(r)dr \\=\int\limits\limits_{0}^{\infty} r ^{2} (re ^{\frac{ -r }{2a _{0} }})^{2}dr \\=\int\limits_0^\infty r^4e^\left( -r/a_0 \right)d}r\\ \Large ={4! \over (1/a_0)^5}=\color{green}{24(a_0)^5}\]

\[(ii) \large {\int\limits_0^{2\pi} \Phi^2(\phi)d\phi=\int\limits_0^{2\pi} \sin^2\left( \phi \right)d \phi \\=2\int\limits_0^{\pi} \sin^2\left( \phi \right)d \phi\\ =2\left({2-1 \over2}\int\limits_0^\pi 1d \theta\right)} = \Large \color{green} \pi \]

\[(iii) \large {\int\limits\limits\limits_0^\pi \Theta^2(\theta)\sin(\theta)d \theta=\int\limits\limits\limits_0^\pi \sin^3\left( \theta\right)d \theta \\ ={3-1 \over 3}\int\limits_0^\pi \sin(\theta) d \theta \\ =\frac 23 \left[ -\cos \left( \theta \right) \right]_0^\pi\\ =\frac23 \left[1-(-1)\right]=\Large \color{green}{\frac 43}}\]

Answer 24

and from: \[\large N={1 \over 4\sqrt{2 \pi (a_0)^5}}\]

we get: \[\huge N^2= \frac{ 1 }{ 32\pi a _{0}^{5}}\]

multiply all that up and you get 1.

Answer 25

Thank you kindly :)