# Atomic Physics - Orbital Angular Momentum Probability

• Isaac Pepper
In summary, orbital angular momentum probability is a measure of the likelihood of finding an electron in a specific region of an atom's electron cloud. It is calculated by solving the Schrödinger equation and is influenced by the electron's quantum number, the presence of other electrons, and external forces. The arrangement and distribution of electrons, including their orbital angular momentum probabilities, play a key role in determining the chemical properties of elements. While orbital angular momentum probability cannot be directly observed, experiments can indirectly measure its effects and validate atomic theory predictions.
Isaac Pepper

## Homework Statement

Consider an electron in a state described by angular wavefunction $$\psi(\theta,\phi)=\sqrt{\frac{3}{4 \pi}}\sin \theta \cos \phi$$ Here θ and φ are the polar and azimuthal angles, respectively, in the spherical coordinate system.

i. Calculate the probability that a simultaneous measurement of the electron orbital angular momentum squared (L2) and the Z component of the electron orbital angular momentum (Lz) will give 2ħ2 and ħ respectively.

ii. What is the probability of measuring L2 = 2ħ2 ?
iii. What is the probability of measuring Lz = 0 ?

## Homework Equations

$$\psi(\theta,\phi) = \sum_{l=0}^{l=\infty} \sum_{m=-l}^{m=+l}c_{l,m} Y_{l,m} (\theta, \phi)$$

$$c_{l,m} = \int Y^{*}_{lm}(\theta, \phi) \psi (\theta, \phi) d\Omega$$

## The Attempt at a Solution

[/B]
In my notes I am told that $$\vert c_{l,m} \vert ^2$$ is the probability that a simultaneous measurement of L2 and Lz on a particle described by the wavefunction ψ gives l(l+1)ħ2 AND mħ.
The probability that a measurement of L2 will give l(l+1)2 is simply the sum of the probabilities for each possible m state: $$P(l) = \sum_{m=-l}^l \vert c_{l,m} \vert ^2$$

Looking up the wavefunction in a table, it seems to be a spherical harmonic with l=1 and m=0
which means m can range between -1 and 1.

Now for question i. the answer would simply be $$\vert c_{1,1} \vert ^2$$ ...but how can I calculate what c is?

I guess my problem is similar for part ii. as the answer would be $$P(l) = \sum_{m=-l}^l \vert c_{l,m} \vert ^2$$ where m can be -1, 0 and 1...but I'm stumped on how to calculate cl,m

Any help would be greatly appreciated !
Thank you

Last edited:
Isaac Pepper said:
i. Calculate the probability that a simultaneous measurement of the electron orbital angular momentum squared (L2) and the Z component of the electron orbital angular momentum (Lz) will give ħ2 and ħ respectively.

ii. What is the probability of measuring L2 = ħ2 ?
iii. What is the probability of measuring Lz = 0 ?
Are you sure about the parts I indicated in red?

Isaac Pepper said:
Looking up the wavefunction in a table, it seems to be a spherical harmonic with l=1 and m=0
Why m = 0?

Isaac Pepper said:
Now for question i. the answer would simply be $$\vert c_{1,1} \vert ^2$$ ...but how can I calculate what c is?
Isaac Pepper said:
$$c_{l,m} = \int Y^{*}_{lm}(\theta, \phi) \psi (\theta, \phi) d\Omega$$

Sorry! Edited my post, it was meant to be 2ħ2.

I am having trouble understanding the integral though... What are Yl,m... and dΩ :(
I have a table with values of Y0,0, Y1,0, Y1,1 etc...

So for example $$Y_{0,0} = \sqrt{\frac{1}{4 \pi}}$$

$$c_{0,0} = \int \sqrt{\frac{1}{4 \pi}} \sqrt{\frac{3}{4 \pi}} \sin \theta \cos \phi d\Omega$$
where $$\int d\Omega = \int_0^{2 \pi} d\phi \int_0^{\pi} \sin \theta d\theta$$

$$c_{0,0} = \frac{\sqrt{3}}{4\pi} \int_0^{2 \pi} \cos \phi d\phi \int_0^{\pi} \sin^2 \theta d\theta$$

Leading to a sin φ evaluated between 2π and 0... Which is 0, so c0,0=0 ? Am I on the right track?

If I then apply this with the other values, I could work out each value for c, and put them into the equations to get the probabilities?

So i. would be c1,12
ii. would be c1,-12+c1,02+c1,12

Correct?

Last edited:
Working through the numbers, I have got an answer of $$c_{1,0}=0$$ $$c_{1,1}=-\frac{1}{\sqrt{2}}$$ and $$c_{1,-1}=\frac{1}{\sqrt{2}}$$

Which means the answer to i. is 1/2, ii. is 1/2+1/2 = 1
But how do I calculate the probability of measuring Lz=0?

The probability of m=1 AND l =1 was 1/2 (from part i.)
and the probability of l=1 is 1 from part ii.
This suggests that the probability of m=1 is 1/2...?
Which I'm assuming is the same for the probability of m=-1
As such the probability of m=0 is 0, correct?

Last edited:
Isaac Pepper said:
The probability of m=1 AND l =1 was 1/2 (from part i.)
and the probability of l=1 is 1 from part ii.
This suggests that the probability of m=1 is 1/2...?
Which I'm assuming is the same for the probability of m=-1
As such the probability of m=0 is 0, correct?
That's one way to do it. You can also get it directly from the fact that all coefficients ##c_{l,0}## are zero, therefore the probability of measuring ##m=0## is zero.

Isaac Pepper and PeroK

## 1. What is orbital angular momentum probability?

Orbital angular momentum probability is a measure of the likelihood that an electron will be found in a specific region of an atom's electron cloud. It is based on the concept of an electron's orbital angular momentum, which describes the motion and position of the electron within the atom.

## 2. How is orbital angular momentum probability calculated?

Orbital angular momentum probability is calculated by solving the Schrödinger equation, a mathematical equation that describes the behavior of quantum particles such as electrons. It takes into account the energy, mass, and potential energy of the electron to determine its probability distribution within the atom.

## 3. What factors influence orbital angular momentum probability?

The primary factor that influences orbital angular momentum probability is the quantum number associated with the electron's orbital. This number determines the shape and orientation of the orbital, which in turn affects the probability of finding the electron in a particular region of the atom's electron cloud.

Other factors that can influence orbital angular momentum probability include the presence of other electrons in the atom and the external forces acting on the atom, such as electric or magnetic fields.

## 4. How does orbital angular momentum probability relate to the chemical properties of elements?

The arrangement and distribution of electrons within an atom, including their orbital angular momentum probabilities, directly affect the chemical properties of an element. This is because the number and arrangement of electrons determine the element's reactivity and its ability to bond with other elements to form compounds.

For example, elements with similar orbital angular momentum probabilities tend to exhibit similar chemical properties. This is why elements are organized into groups in the periodic table based on their electron configurations.

## 5. Can orbital angular momentum probability be observed experimentally?

No, orbital angular momentum probability cannot be directly observed experimentally. This is because it is a mathematical concept that describes the likelihood of finding an electron in a certain region of space, rather than a physical property that can be measured.

However, experiments can indirectly measure the effects of orbital angular momentum probability, such as through spectroscopy or electron diffraction techniques. These experiments provide insights into the behavior of electrons within atoms and help validate the predictions of atomic theory.

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