- #1

Marioweee

- 18

- 5

- 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.

\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.