Probability of measuring an eigenstate of the operator L ^ 2

[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.

 

[stevendaryl]

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.

 

[TSny]

\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 $\int d\varphi \int \sin \theta \, d \theta$

 

[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.

 

[Marioweee]

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.

 

[PeroK]

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.

 

[Marioweee]

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.

 

[PeroK]

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 $N^2 = \frac{15}{32\pi}$?

 

[Marioweee]

Are you sure that's not $N^2 = \frac{15}{32\pi}$?

Yes, you are right, that's $N^2$.

 

[TSny]

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.

 

[Marioweee]

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