Normalising a wavefunction

In summary, the conversation is about a student trying to normalize a system's wavefunction, but encountering difficulties with the integral of psi*psi. They also discuss finding the possible results of measurements of L_z and L^2 and calculating their probabilities. A mistake in the coefficient for cos(theta) is identified and the correct result for L_z is calculated. The conversation ends with the student questioning a discrepancy in their calculations.
  • #1

bon

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Homework Statement



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

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

Now I am 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
 
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  • #2
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
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
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
Well I am 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
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
aha thank you so much!
 
  • #8
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
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
Ok thanks!
Back to my first question - I've 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
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
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.
 

1. What does it mean to "normalise" a wavefunction?

Normalising a wavefunction means to adjust its values so that the total probability of finding a particle in any location is equal to 1. This is an important step in quantum mechanics, as it ensures that the wavefunction accurately describes the behavior of a particle.

2. How is a wavefunction normalised?

A wavefunction is normalised by taking the integral of the absolute value squared of the wavefunction over all space, and then dividing the wavefunction by the square root of that integral. This ensures that the total probability is equal to 1.

3. Why is normalising a wavefunction important?

Normalising a wavefunction is important because it ensures that the wavefunction accurately represents the probability of finding a particle in any location. Without normalisation, the wavefunction may not accurately describe the behavior of a particle and could lead to incorrect predictions.

4. Can a wavefunction be normalised to a value other than 1?

No, a wavefunction must be normalised to a value of 1. This is because the total probability of finding a particle in any location must be equal to 1, as there is a 100% chance that the particle exists somewhere in space.

5. What are the consequences of not normalising a wavefunction?

If a wavefunction is not normalised, it may not accurately represent the behavior of a particle. This could lead to incorrect predictions about the particle's behavior and could potentially violate fundamental principles of quantum mechanics, such as the conservation of probability.

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