Quantum Mechanics - Spherical Harmonics

In summary, the wave function of a system takes the form of the spherical harmonics for l=1, and from that we can say that L^2=2. However, I'm unsure whether this is correct, and I can't find the probabilities for L_z.
  • #1
TheBaker
19
0

Homework Statement



The angular part of a system’s wavefunction is

[tex]<\theta, \phi | \psi>\propto (\sqrt{2}\cos\theta + \sin{\theta}e^{−i\psi} - \sin{\theta}e^{i\psi} ). [/tex]

What are the possible results of measurement of (a) [tex]L^2[/tex] , and (b) [tex]L_z[/tex] , and their probabilities? What is the expectation value of [tex]L_z[/tex]?


Homework Equations



[tex]L^2|E> = l(l+1)|E>[/tex]

The Attempt at a Solution



I can see that the wavefunction takes the form of the spherical harmonics for l = 1, and from that I think I can say that [tex]L^2[/tex] = 2. However, I'm unsure whether this is correct, or how to find the probabilities.

I haven't had much luck at all with [tex]L_z[/tex].
 
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  • #2
eigenvalues of [itex]L^2[/itex] are [itex]l(l+1) \hbar^2[/itex]
and eigenvalues of [itex]L_z[/itex] are [itex]m \hbar[/itex] where m ranges from l to -l
 
  • #3
Thanks, so the eigenvalue of [tex]L^2[/tex] is [tex]2 \hbar^2[/tex], which must presumably have a probability of 1 as it's the only eigenvalue.

The eigenvalues of [tex]L_z[/tex] are [tex]-\hbar, 0, \hbar[/tex], but how do I figure out their probabilities?

Sorry if this is a really stupid/easy question - as you can probably tell, QM isn't my forte by a long shot.
 
  • #4
why is l=1? l runs from 0 to n-1
 
  • #5
Because the functions that form the systems wave function are the three spherical harmonics of l=1.
 
  • #6
Try writing the wave function in the form
[tex]\psi(\theta,\phi)=\sum_{m={-1},0,{+1}}c_{1,m}Y_{1,m}(\theta,\phi)[/tex]
and find the [itex]c_{1,m}[/itex] coefficients. Can you say what the probabilities are in terms of these?
 
  • #7
Through a combination of the textbook and lecture notes, I get the constants to be:

[tex]c_1^0 = \sqrt{\frac{6}{8 \pi}}[/tex]
and
[tex]c_1^{\pm 1} = \pm \sqrt{\frac{3}{8 \pi}}[/tex]

However, the sum of the square of these doesn't equal 1 - am I supposed to include the Y function when squaring to find the probability?
 
  • #8
TheBaker said:
Through a combination of the textbook and lecture notes, I get the constants to be:

[tex]c_1^0 = \sqrt{\frac{6}{8 \pi}}[/tex]
and
[tex]c_1^{\pm 1} = \pm \sqrt{\frac{3}{8 \pi}}[/tex]

However, the sum of the square of these doesn't equal 1 - am I supposed to include the Y function when squaring to find the probability?

I don't think the sum of the squares needs to be one in this case since they tell you proportional to and not equals. (Essentially you will need to find the correct constant that normalizes the wave function)
 
  • #9
Right, I think I've got it. So since we know that the coefficient of m=0 is a factor of [tex]\sqrt{2}[/tex] larger than m=±1, we find that...

P(m=0) = 0.5
P(m=1) = 0.25
P(m=-1) = 0.25

Hopefully, that's right. Thanks for everyone's help, I think I understand it all slightly better now (cue everyone telling me that I'm wrong).
 
  • #10
seeing as your coefficients were root(6/8pi) and root(3/8pi) you can see that one is double the other two and you know they have to sum to be 1 so your answer above is correct. i reckon you just hadn't properly normalised them earlier and that's why they weren't summing to 1.
 

Related to Quantum Mechanics - Spherical Harmonics

1. What are spherical harmonics in quantum mechanics?

Spherical harmonics are mathematical functions that describe the quantum states of an electron in an atom. They are solutions to the Schrödinger equation for a spherically symmetric potential. They represent the probability distribution of finding an electron in a particular energy state and angular momentum within an atom.

2. How are spherical harmonics related to the shape of an electron orbital?

The shape of an electron orbital is determined by the values of the spherical harmonics. Each orbital has a unique combination of spherical harmonics that determines its shape and orientation in space. As the quantum numbers for energy and angular momentum change, the shape of the orbital also changes.

3. Why are spherical harmonics important in quantum mechanics?

Spherical harmonics are important in quantum mechanics because they accurately describe the behavior of electrons in atoms. They are essential in understanding the structure and properties of atoms, which is crucial in fields such as chemistry and materials science. Spherical harmonics also play a significant role in explaining the phenomenon of quantum entanglement.

4. How are spherical harmonics used in solving the Schrödinger equation?

The Schrödinger equation is a differential equation that describes the behavior of quantum systems, including atoms. Spherical harmonics are used as solutions to this equation for spherically symmetric systems, such as atoms. By plugging in the values of the quantum numbers for energy and angular momentum, the Schrödinger equation can be solved to determine the wave function and energy levels of the system.

5. Can spherical harmonics be visualized?

Yes, spherical harmonics can be visualized using three-dimensional plots. These plots show the probability distribution of finding an electron in a particular energy state and angular momentum within an atom. The shape of the plot reflects the shape of the orbital, and the color represents the probability of finding the electron in a specific location within the orbital. However, it is important to note that these visualizations are representations and do not accurately depict the actual shape and behavior of electrons in atoms.

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