# Probability of measuring an eigenstate of the operator L ^ 2

• Marioweee
In summary, the wavefunction of a particle with an angular momentum of $L ^2$ is normalized if and only if the sum of the squares of the coefficients of the radial function is equal to 1.f

#### Marioweee

Homework Statement
psb
Relevant Equations
$L^2Y^{m}_l=\hbar^2l(l+1)Y^{m}_l$
Calculate, with a relevant digit, the probability that the measure of the angular momentum $L ^2$ of a particle whose normalized wave function is
\begin{equation}
\Psi(r,\theta,\varphi)=sin^2(\theta)e^{-i\varphi}f(r)
\end{equation}
is strictly greater than ##12(\hbar)^2##.
------------------------------------------------
My solution:
First of all, if the measure is sitrictly greater than ##12(\hbar)^2## then l has to be greater than 3 because:
\begin{equation}
L^2Y^{m}_l=\hbar^2l(l+1)Y^{m}_l
\end{equation}
Then, to apply the operator ##L^2## I have tried to express the wave function as linear combination of spherical harmonics.
\begin{equation}
sin^2(\theta)e^{-i\varphi}=\sum_{l=0}^{\inf}\sum_{m=-l}^{l}f^{m}_{l}Y^{m}_{l}
\end{equation}
with the coefficients ##f^{m}_{l}##:
\begin{equation}
f^{m}_{l}=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\theta sin^2(\theta)e^{-i\varphi}(Y^{m}_{l})^{*}
\end{equation}

Realizing that the only nonzero integrals are those with the coefficients l=1,3,5,7... and m=-1.

Therefore, it follows that:
\begin{equation}
sin^2(\theta)e^{-i\phi}=\sum_{l=1,3,5,...}^{\inf}f^{-1}_{l}Y^{-1}_{l}
\end{equation}
Next I have calculated with a software the coefficients ##f^{-1}_{l}## and the probability of measuring a value lower ##12\hbar^2## should be:
\begin{equation}
P(L^2<12\hbar^2)=1-|f^{-1}|^2_{1}-|f^{-1}_{3}|^2
\end{equation}
The problem is that the sum of the infinite coefficients f is not 1 (at least that seems to be when adding quite a few terms) so they cannot indicate the probability. How can I normalize the infinite linear combination so that the sum of the terms is 1 and indicates the probability of measuring an eigenstate of ##L^2##. Would there be another method to solve the exercise?
Thank you very much for your attention and help.

Here's an idea for making the calculation more manageable. If you let ##P(l,m)## be the probability of finding ##L = l## and ##L_z = m##, then you have:

##\sum_{l=0}^\infty \sum_{m=-l}^{+l}P(l,m) = 1##

Now, we can break it up into two parts:

##\sum_{l=0}^3 \sum_{m=-l}^{+l}P(l,m) + \sum_{l=4}^\infty \sum_{m=-l}^{+l} P(l,m) = 1##

The second part is what you're being asked to calculate, but you can rewrite it as:

##\sum_{l=4}^\infty \sum_{m=-l}^{+l} P(l,m) = 1 - \sum_{l=0}^3 \sum_{m=-l}^{+l}P(l,m)##

Now, the right-hand side only involves 4 terms.

• Marioweee
\begin{equation}
f^{m}_{l}=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\theta sin^2(\theta)e^{-i\varphi}(Y^{m}_{l})^{*}
\end{equation}
If you want ##|f^{m}_{l}|^2## to represent a probability, then you should first normalize the function ##\sin^2\theta e^{-i\varphi}##.

When integrating over ##\varphi## and ##\theta##, ##\int d\varphi \int d \theta## should be ##\int d\varphi \int \sin \theta \, d \theta##

Last edited:
• Marioweee
Here's an idea for making the calculation more manageable. If you let ##P(l,m)## be the probability of finding ##L = l## and ##L_z = m##, then you have:

##\sum_{l=0}^\infty \sum_{m=-l}^{+l}P(l,m) = 1##

Now, we can break it up into two parts:

##\sum_{l=0}^3 \sum_{m=-l}^{+l}P(l,m) + \sum_{l=4}^\infty \sum_{m=-l}^{+l} P(l,m) = 1##

The second part is what you're being asked to calculate, but you can rewrite it as:

##\sum_{l=4}^\infty \sum_{m=-l}^{+l} P(l,m) = 1 - \sum_{l=0}^3 \sum_{m=-l}^{+l}P(l,m)##

Now, the right-hand side only involves 4 terms.
This is what I have tried to express with equation 6. Anyways, thank you very much for your attention and for your help.

If you want ##|f^{m}_{l}|^2## to represent a probability, then you should first normalize the function ##\sin^2\theta e^{-i\varphi}##.

When integrating over ##\varphi## and ##\theta##, ##\int d\varphi \int d \theta## should be ##\int d\varphi \int \sin \theta \, d \theta##
I had not even thought about it since the statement said that the function was already normalized but this must be the solution. Thank you very much, in a while I will try and if I have any questions I will comment. Again, thank you very much.

I had not even thought about it since the statement said that the function was already normalized but this must be the solution. Thank you very much, in a while I will try and if I have any questions I will comment. Again, thank you very much.
The wavefunction involves the radial function ##f(r)##. The normalization, therefore, puts a condition on ##f(r)## to balance the angular function you are given.

For example, we can rewrite the normalized wavefunction as:
\begin{equation}
\Psi(r,\theta,\varphi)=N sin^2(\theta)e^{-i\varphi} (\frac 1 N f(r))
\end{equation}

Where ##N## is a normalization constant needed for the spherical harmonic decomposition.

I've calculated N which is equal to ##\dfrac{15}{32\pi}##. Therefore, the probability of measuring ##L^2## greater than ##12h\hbar^2## would be:
\begin{equation}
P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)
\end{equation}
Sorry for so many obvious questions but I am new to quantum mechanics and I still miss the basics but I want to make sure I do things correctly.

I've calculated N which is equal to ##\dfrac{15}{32\pi}##. Therefore, the probability of measuring ##L^2## greater than ##12h\hbar^2## would be:
\begin{equation}
P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)
\end{equation}
Sorry for so many obvious questions but I am new to quantum mechanics and I still miss the basics but I want to make sure I do things correctly.
Are you sure that's not ##N^2 = \frac{15}{32\pi}##?

Are you sure that's not ##N^2 = \frac{15}{32\pi}##?
Yes, you are right, that's ##N^2##.

Therefore, the probability of measuring ##L^2## greater than ##12h\hbar^2## would be:
\begin{equation}
P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)
\end{equation}

That looks right.

Okey, I finally got the answer. Thanks for everyone's help, for my part I conclude this post.

• TSny and PeroK