Normalising a wavefunction

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  • #1
bon
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Homework Statement



Ok so im told that the angular part of a system's wavefunction is:

Psi (theta, phi) = root2 cos(theta) -2i sin(theta) sin (phi)

Now im trying to normalise it..

Homework Equations





The Attempt at a Solution



Psi * (theta, phi) = root2 cos(theta) + 2i sin(theta) sin (phi)

Therefore, when we integrate over all space, we need the integral of psi* psi sin(theta) d theta d phi, but I keep getting this to be 0! We get a cos^2theta sintheta which integrates over 2pi to give 0 and a sin^3 theta which also integrates over 2pi to give 0! what's going wrong!?
THanks
 

Answers and Replies

  • #2
LeonhardEuler
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You need to find the integral of
[tex]|\psi|^2=\psi\psi^{*}[/tex]
to normalize, not the integral of [tex]\psi[/itex].
(I've made the same mistake before, so I know your pain)
 
  • #3
bon
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Thanks, but that is what i found:

I found |psi|^2 = 2 cos^2 (theta) + 4 sin^2 (theta) sin^2 (phi)

but then when I integrate this I have the factor of sin theta that comes from the sin(theta) d theta d phi,

so both the theta integrals come out to 0!?!!

thanks!!
 
  • #4
LeonhardEuler
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Can you show the integral because this is a non-negative function that is positive almost everywhere and the integral can not possibly be 0, so there is some other mistake in your integration.
 
  • #5
bon
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Well im just integrating cos^2 theta sin (theta) d theta from 0 to 2pi = 0

and adding this to the integral of sin^3 (theta) d theta from 0 to 2pi = 0

thanks for your help
 
  • #6
LeonhardEuler
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Oh, your mistake is that you take the factor of integration to be sin(theta) over 0,2pi. In fact, depending on the convention, theta is the angle from the z-axis and the integral only goes to pi, or theta is the polar angle in the x-y plane and the integral factor is in fact sin(phi)
 
  • #7
bon
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aha thank you so much!
 
  • #8
bon
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One related question if you don't mind:

Im told a system's wavefunction is proportional to sin^2 theta and I'm asked what the possible results of measurements of L_z and L^2 and asked for probabilities of each...

How do I go about doing this?

Thanks again!
 
  • #9
LeonhardEuler
Gold Member
859
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I would start by representing L_z and L^2 in terms of phi and theta. For instance,
[tex]L_z = \i\hbar\frac{\partial}{\partial \phi}[/tex]
and then putting this into the expectation formula
 
  • #10
bon
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Ok thanks!
Back to my first question - ive now worked out A = 1/root (8pi)
But looking at the wavefunction, i can see by inspection that Lz has to be 1, -1 or 0

now since A = A = 1/root (8pi), the amplitude to find +1 = amplitude to find -1 = A = 1/root (8pi)

and the amplitude to find 0 = root(2) /root (8pi)

but if i square and add these, it doesn't come to 1. what has gone wrong?

thanks again
 
  • #11
LeonhardEuler
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I'm not getting the same A as you, though it's entirely possible I made a mistake.

Edit: Did make a mistake. Read the coefficient of cos(theta) in psi as 1/root2 instead of root2 for some reason. I'll calculate angular momentum now.
 
  • #12
LeonhardEuler
Gold Member
859
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Ok, I'm sure I got the right answer now for L_z and it's not what you got. Make sure you're doing the [itex]\phi[/itex] integral correctly, it comes out very easily.
 

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