- #1

Zarathustra0

- 23

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Zarathustra0
- Start date

In summary: I don't remember the word, but I think it has to do with how the angular momentum of an atom is described. If you have l=1 and s=3/2, then the angular momentum will be described by l+s=4/2. If you have l=1 and s=1, then the angular momentum will be described by l=2 and s=1.

- #1

Zarathustra0

- 23

- 0

Physics news on Phys.org

- #2

Matterwave

Science Advisor

Gold Member

- 3,971

- 329

IIRC it has to do with the fact that the angular momentum quantum numbers correspond to the numbering of the spherical harmonics used to describe the atom (the angular part of the wave-function). Spherical harmonics only come with integer quantum numbers. Spin is not described by spherical harmonics.

Last edited:

- #3

- 24,488

- 15,021

[tex]\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}.[/tex]

If you apply [itex]\hat{L}_z[/itex] to, e.g., the position eigenvector [itex]|\vec{x}=(0,0,z) \rangle[/itex], you'll get 0 since a rotation around the z axis doesn't change this state. You can also easily verify this by direct application of the [itex]\hat{L}_z=\hat{x} \hat{p}_y-\hat{y} \hat{p}_x[/itex] to this state.

Now, you can as well expand this state in orbital-angular-momentum eigenstates, and then the application of [itex]\hat{L}_z[/itex] leads to

[tex]|\vec{x}=(0,0,z) \rangle=\sum_l \sum_{m=-l}^{l} C_{lm} |r=1,l,m \rangle \; \Rightarrow \; \hat{L}_z |\vec{x}=(0,0,z) \rangle = \sum_l \sum_{m=-l}^{l} m C_{lm} |r=|z|,l,m \rangle.[/tex]

Thus, you must have [itex]C_{lm}=0[/itex] for all [itex]m \neq 0[/itex]. Thus only eigen vectors with [itex]m=0[/itex] contribute to the decomposition of [itex]|\vec{x}=(0,0,z) \rangle[/itex], but such eigen states can exist only for integer l.

Now, you can make any position-eigen vector by applying a rotation to [itex]|\vec{x}=(0,0,z) \rangle[/itex]. Thus also all these vectors consist only of superpositions of integer-l orbital-angular-momentum eigen vectors. Since this is a complete set of eigen vectors, l can only take integer values.

- #4

Bill_K

Science Advisor

- 4,158

- 206

- #5

SpectraCat

Science Advisor

- 1,402

- 3

In general, a given angular momentum quantum number can take value EITHER from the set of integers OR from the set of half-integers. As far as I know there is no way to convert between the two sets, at least not in non-relativistic QM.

- #6

Matterwave

Science Advisor

Gold Member

- 3,971

- 329

SpectraCat said:

In general, a given angular momentum quantum number can take value EITHER from the set of integers OR from the set of half-integers. As far as I know there is no way to convert between the two sets, at least not in non-relativistic QM.

I don't think the spin of particles change in general. Electrons always have 1/2 spin, never 3/2 even.

For a composite system it is true, the spin angular momentum addition requires the total spin to be separated by integers. So 2 1/2 spin particles added together may have spin 1, 0, or -1.

This is not quite the same as Angular Momentum where you can increase a single particle's l value simply by imparting to it some angular momentum.

- #7

SpectraCat

Science Advisor

- 1,402

- 3

Matterwave said:I don't think the spin of particles change in general. Electrons always have 1/2 spin, never 3/2 even.

Of course.

For a composite system it is true, the spin angular momentum addition requires the total spin to be separated by integers. So 2 1/2 spin particles added together may have spin 1, 0, or -1.

This is not quite the same as Angular Momentum where you can increase a single particle's l value simply by imparting to it some angular momentum.

You are right that the spins of individual quantum particles are fixed. However, at least for non-relativistic cases, the same algebra is used to describe both spin and angular momentum in quantum systems. Also, the multi-electron states are as much eigenfunctions of the total spin operator as the single electron state.

Anyway, I was more referring to something like a total angular momentum quantum number for an atom:

J=L + S.

L will always be an integer, but S can be either an integer or half-integer value, so J can either be an integer OR a half-integer in the range |L-S| <= J <= L+S. For example, if L=1 and S=3/2, then J can take any value in the set {1/2,3/2,5/2}. If L=1 and S=1, then J can take any value from the set {0,1,2}. But you never get a J that can take values from a set like {0,1/2,1,3/2,2,5/2}.

A similar condition holds for the z-component of angular momentum ... it can only ever increase or decrease in integer multiples of hbar (c.f. angular momentum raising and lower operators). For an electron, s

The angular-momentum quantum number is a quantum number that describes the shape of an electron's orbital in an atom. It is represented by the symbol l and has integer values ranging from 0 to n-1, where n is the principal quantum number. It is also known as the orbital quantum number.

The value of the angular-momentum quantum number determines the type of orbital an electron occupies. For example, when l=0, the orbital is spherically symmetric and is known as an s-orbital. When l=1, the orbital is more elongated and is known as a p-orbital. The shape of the orbital becomes more complex as the value of l increases.

The quantum mechanical model of the atom is based on the principle that electrons exist in specific energy levels and orbitals around the nucleus. The angular-momentum quantum number helps to define these orbitals and describes the distribution of electron density around the nucleus. It also determines the energy of the electron in a particular orbital.

No, the angular-momentum quantum number can only have positive integer values from 0 to n-1. This is because the quantum mechanical model of the atom only allows for specific, discrete values of energy and angular momentum.

As an electron moves to a higher energy level, the value of the angular-momentum quantum number also increases. This is because higher energy levels have more orbitals with different shapes and orientations, requiring a larger range of values for l. For example, the first energy level (n=1) has only one orbital with l=0 (s-orbital), while the second energy level (n=2) has two orbitals with l=0 (s-orbital) and l=1 (p-orbital).

- Replies
- 15

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 2

- Views
- 417

- Replies
- 9

- Views
- 1K

- Replies
- 2

- Views
- 1K

- Replies
- 19

- Views
- 896

- Replies
- 2

- Views
- 2K

- Replies
- 1

- Views
- 405

- Replies
- 2

- Views
- 1K

- Replies
- 3

- Views
- 2K

Share: