Ballentine Problem 7.1 Orbital Angular Momentum

[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.

 

[PeroK]

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.

 

[Irishdoug]

Thankyou for your response.

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.

 

[PeroK]

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.

 

[Irishdoug]

It is the probability distribution.

 

[PeroK]

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.

 

[PeroK]

Thankyou for your response.

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

Although you also have to be careful about your normalisation.

 

[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.

 

[PeroK]

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?

 

[Irishdoug]

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.

 

[PeroK]

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.

 

[Irishdoug]

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 $Y_{1}^{-1}$

 

[PeroK]

Thankyou for your response.

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?

 

[Irishdoug]

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 $sin(\theta)cos(\phi)$

 

[PeroK]

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!

 

[Irishdoug]

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

??

 

[PeroK]

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

??

Definitely not! $sin \theta \cos \phi$ maybe.

 

[Irishdoug]

Definitely not! $sin \theta \cos \phi$ maybe.

Yes yes sorry! the cos should be $\phi$ not $\theta$. I have written it wrong here throughout!

 

[Irishdoug]

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.

 

[PeroK]

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.

 

[AndreasC]

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!