# Ballentine Problem 7.1 Orbital Angular Momentum

• Irishdoug
In summary, the probability distributions of the orbital angular momentum variables ##L^{2}## and ##L_{z}## for the given orbital state functions ##\Psi(x) = f(r) sin(\theta) cos(\phi)## and ##\Psi(x) = f(r) cos^{2}(\theta)## are found by taking the square of the amplitude of the corresponding eigenstates, which are given by the spherical harmonics. The eigenvalues for ##L^{2}## and ##L_{z}## can be determined by measuring the states with the respective quantum numbers. There is an equal probability of 0.5 for obtaining the eigenvalue ##L^{2} = 2\hbar^{2
Irishdoug
Homework Statement
Find the probability distributions of the orbital angular momentum variables ##L^{2}## and ##L_{z}## for the following orbital state functions:
Relevant Equations
##\Psi(x) = f(r) sin(\theta) cos(\theta)##

##\Psi(x) = f(r) cos^{2}(\theta)##
Find the probability distributions of the orbital angular momentum variables ##L^{2}## and ##L_{z}## for the following orbital state functions:
##\Psi(x) = f(r) sin(\theta) cos(\theta)##

##\Psi(x) = f(r) cos^{2}(\theta)##I am aware that the prob. distribution of an observable is ##|<a_{n} | \Psi >|^{2}## were ##a_{n}## are the eigenstates.

I'm at a loss of how to even start the question however. I'm unsure as to how to find l and m to, for example, use the spherical harmonics. Can someone point me in the right direction? Thankyou.

Irishdoug said:
I'm at a loss of how to even start the question however. I'm unsure as to how to find l and m to, for example, use the spherical harmonics. Can someone point me in the right direction? Thankyou.
What do you know about the spherical harmonics? That is the right direction.

I figured I had to work backwards somewhat to write ##´|\Psi> = \frac{1}{2} (\frac{3}{8\pi})^{\frac{1}{2}} (Y_{1}^{-1} - Y_{1}^{1})##

Now, when I do ##|<l,m|\Psi>|^{2} = |<1,1|\Psi>|^{2} and |<1,-1|\Psi>|^{2} ## I get 0 for both. I am doing the integral
\begin{gather*}
\iint (1) sin(\theta) cos(\theta) \,d\theta\,d\phi

\end{gather*}

with 0 - 2 ##\pi## and 0 - ##\pi## as the arguments. It being 0 does not make sense. I am unsure as to what I have done wrong. Any help appreciated.

Irishdoug said:
I figured I had to work backwards somewhat to write ##´|\Psi> = \frac{1}{2} (\frac{3}{8\pi})^{\frac{1}{2}} (Y_{1}^{-1} - Y_{1}^{1})##
If that is correct, then that is all you need, surely. I don't know what the next calculation is supposed to be.

It is the probability distribution.

Irishdoug said:
It is the probability distribution.
It's the wavefunction expressed as a linear combination of the relevant eigenfunctions. You should know how to get the probabilities from that.

Irishdoug
Irishdoug said:

I figured I had to work backwards somewhat to write ##´|\Psi> = \frac{1}{2} (\frac{3}{8\pi})^{\frac{1}{2}} (Y_{1}^{-1} - Y_{1}^{1})##

Irishdoug
Hi, I went back and read Griffiths . I realized I left out the the ##r^{2}dr## and we can normalise seperately, and that I wasn't seeing them as a liner combination like you stated. I've normalised properly now too.

So I get ##\frac{1}{2} (\frac{8\pi}{3})^{\frac{1}{2}}(2) + \frac{1}{2} (\frac{8\pi}{3})^{\frac{1}{2}}(2)## = 0.69. Squared this give a half which I think should be correct.

Irishdoug said:
Hi, I went back and read Griffiths . I realized I left out the the ##r^{2}dr## and we can normalise seperately, and that I wasn't seeing them as a liner combination like you stated. I've normalised properly now too.

So I get ##\frac{1}{2} (\frac{8\pi}{3})^{\frac{1}{2}}(2) + \frac{1}{2} (\frac{8\pi}{3})^{\frac{1}{2}}(2)## = 0.69. Squared this give a half which I think should be correct.
What do ##0.69## and ##0.5## mean in the context of answering the question?

0.69 is the probability amplitude for the ##L^{2}## and ##L_{z}## operators and 0.5 is the probability that the eigenvalues of ##L_{z}## is = ## m\hbar = \hbar ## and ## L^{2} = 2\hbar^{2} ## when measured as m = 1 and l =1

But I am not certain this is the correct answer.

I also assume when I take l=1 and m=-1 I will get half as well and as such I obtain an overall probability of 1 as required.

Last edited:
Irishdoug said:
0.69 is the probability amplitude for the ##L^{2}## and ##L_{z}## operators and 0.5 is the probability that the eigenvalues m and l are 1 and 1 respectively when measured.

But I am not certain this is the correct answer.
Operators don't have probability amplitudes. States or functions have probability amplitudes.

I think you need to be able to express precisely what the question is asking for, and what you are calculating.

Apologies, see updated answer. I didn't express what I meant at all correctly. It is also not formatting when I post, but looks correct when I preview.

So 0.69 is the probability amplitude of the state ##Y_{1}^{1}## and ##Y_{1}^{-1}##

Irishdoug said:

I figured I had to work backwards somewhat to write ##´|\Psi> = \frac{1}{2} (\frac{3}{8\pi})^{\frac{1}{2}} (Y_{1}^{-1} - Y_{1}^{1})##
How did you get that expression for the wave-function?

from the spherical harmonics. So the state I need is ##|\Psi(x)> = f(r)sin(\theta)cos(\phi)##

the spherical harmonic for l=1 and m=1 ; l=1 and m=-1 is ##\frac{+}{-}(\frac{3}{8\pi})^{0.5}sin(\theta)exp(\frac{+}{-}i\phi)##

So doing ##Y_{1}^{-1} - Y_{1}^{1}\frac{1}{2}## gives ##sin(\theta)cos(\phi)##

Last edited:
Irishdoug said:
from the spherical harmonics. So the state I need is ##|\Psi(x)> = f(r)sin(\theta)cos(\theta)##

the spherical harmonic for l=1 and m=1 ; l=1 and m=-1 is ##\frac{+}{-}(\frac{3}{8\pi})^{0.5}sin(\theta)exp(\frac{+}{-}i\phi)##

So doing ##Y_{1}^{-1} - Y_{1}^{1}\frac{1}{2}## gives ##sin(\theta)cos(\theta)##
You might want to double-check that!

##(Y_{1}^{-1} - Y_{1}^{1})\frac{1}{2}## gives ##sin(\theta)cos(\theta)##

??

Irishdoug said:
##(Y_{1}^{-1} - Y_{1}^{1})\frac{1}{2}## gives ##sin(\theta)cos(\theta)##

??
Definitely not! ##sin \theta \cos \phi## maybe.

PeroK said:
Definitely not! ##sin \theta \cos \phi## maybe.
Yes yes sorry! the cos should be ##\phi## not ##\theta##. I have written it wrong here throughout!

Was I correct when I said this:

So 0.69 is the probability amplitude of the state ##Y_{1}^{1}## and ##Y_{1}^{-1}##and 0.5 is the probability that the eigenvalues of ##L_{z}## is = ## m\hbar = \hbar ## and ## L^{2} = 2\hbar^{2} ## when measured as m = 1 and l =1

I feel it is, but I would like confirmation if I am or am not.

Irishdoug said:
Yes yes sorry! the cos should be ##\phi## not ##\theta##. I have written it wrong here throughout!
That explains it!

Back to the question. You have equal weight of the ##l = 1, m = 1## and ##l=1, m = -1## states. I would say that means you will definitely get ##l = 1##, i.e. ##L^2 = 2\hbar^2## and get ##m = \pm 1##, i.e. ##L_z = \pm \hbar## with equal probability of ##0.5##.

You don't have to do any calculations.

But, you shouldn't have the factor of ##\frac 3 {8\pi}##, as that factor is part of the normalised spherical harmonic. You should have simply: $$\Psi = \frac 1 {\sqrt 2} (Y_1^{-1} - Y_1^1)$$

Irishdoug
haha yes apologies for that!

Ok great, it is quite clear to me now. Thankyou for your help and patience! It is must appreciated.

PeroK
Remember that spherical harmonics are already normalised individually, so to normalise a sum of, say, 2 spherical harmonics you divide by sqrt2, etc. So if you see any really weird factors in front something is probably wrong. These exercises get very easy once you get the hang of them!

Irishdoug

## 1. What is Ballentine Problem 7.1 Orbital Angular Momentum?

Ballentine Problem 7.1 Orbital Angular Momentum is a physics problem that deals with the angular momentum of an object in an orbital motion. It is commonly used in quantum mechanics and is used to calculate the angular momentum of an electron in an atom.

## 2. How do you calculate orbital angular momentum?

The orbital angular momentum is calculated by multiplying the orbital radius by the linear momentum of the object. The formula for orbital angular momentum is L = r x p, where L is the angular momentum, r is the orbital radius, and p is the linear momentum.

## 3. What is the significance of orbital angular momentum?

Orbital angular momentum is significant because it is a conserved quantity in a system. This means that it remains constant even if other factors in the system change. It is also used to describe the rotation of an object around a central point, such as an electron orbiting a nucleus.

## 4. How is orbital angular momentum related to quantum mechanics?

In quantum mechanics, orbital angular momentum is used to describe the quantized energy levels of an electron in an atom. The different energy levels are determined by the angular momentum of the electron in its orbital motion around the nucleus.

## 5. Can orbital angular momentum be changed?

Yes, orbital angular momentum can be changed by changing the orbital radius or the linear momentum of the object. It can also be changed by applying a torque to the object, which will cause it to rotate at a different speed and change its angular momentum.

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