# 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.
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

\Psi(r,\theta,\varphi)=sin^2(\theta)e^{-i\varphi}f(r)

is strictly greater than ##12(\hbar)^2##.
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My solution:
First of all, if the measure is sitrictly greater than ##12(\hbar)^2## then l has to be greater than 3 because:

L^2Y^{m}_l=\hbar^2l(l+1)Y^{m}_l

Then, to apply the operator ##L^2## I have tried to express the wave function as linear combination of spherical harmonics.

sin^2(\theta)e^{-i\varphi}=\sum_{l=0}^{\inf}\sum_{m=-l}^{l}f^{m}_{l}Y^{m}_{l}

with the coefficients ##f^{m}_{l}##:

f^{m}_{l}=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\theta sin^2(\theta)e^{-i\varphi}(Y^{m}_{l})^{*}

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

Therefore, it follows that:

sin^2(\theta)e^{-i\phi}=\sum_{l=1,3,5,...}^{\inf}f^{-1}_{l}Y^{-1}_{l}

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:

P(L^2<12\hbar^2)=1-|f^{-1}|^2_{1}-|f^{-1}_{3}|^2

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
Marioweee said:

f^{m}_{l}=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\theta sin^2(\theta)e^{-i\varphi}(Y^{m}_{l})^{*}
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
stevendaryl said:
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.

TSny said:
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.

Marioweee said:
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:

\Psi(r,\theta,\varphi)=N sin^2(\theta)e^{-i\varphi} (\frac 1 N f(r))

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:

P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)

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.

Marioweee said:
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:

P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)

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}##?

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

Marioweee said:
Therefore, the probability of measuring ##L^2## greater than ##12h\hbar^2## would be:

P(L^2>12\hbar^2)=1-\dfrac{15}{32\pi}(|f_{1}^{-1}|^2+||f_{3}^{-1}|^2)

That looks right.

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

TSny and PeroK

## 1. What is the significance of measuring an eigenstate of the operator L ^ 2?

The operator L ^ 2 represents the total angular momentum of a quantum system. Measuring an eigenstate of this operator allows us to determine the magnitude of the angular momentum of the system, which is a fundamental property in quantum mechanics.

## 2. How is the probability of measuring an eigenstate of L ^ 2 calculated?

The probability of measuring an eigenstate of L ^ 2 is calculated by taking the square of the absolute value of the inner product between the state and the corresponding eigenstate. This is known as the Born rule in quantum mechanics.

## 3. Can the probability of measuring an eigenstate of L ^ 2 be greater than 1?

No, the probability of measuring an eigenstate of L ^ 2 cannot be greater than 1. This is because the total probability of all possible outcomes must add up to 1, as per the laws of probability.

## 4. How does the probability of measuring an eigenstate of L ^ 2 change with different quantum systems?

The probability of measuring an eigenstate of L ^ 2 is dependent on the specific quantum system being measured. Different systems have different energy levels and eigenstates, which affect the probability of measuring a particular eigenstate of L ^ 2.

## 5. What is the relationship between the probability of measuring an eigenstate of L ^ 2 and the uncertainty principle?

The uncertainty principle states that certain pairs of physical properties, such as position and momentum, cannot be known simultaneously with absolute precision. In the case of measuring an eigenstate of L ^ 2, the probability of measuring a particular eigenstate is related to the uncertainty in the measurement of the angular momentum of the system.

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