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- Thread starter lonewolf219
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We want the angular momentum operators to satisfy the commutation relation

##[\hat{J}_i, \hat{J}_j] = i \epsilon_{i j k} \hat{J}_k##

For example, you can verify that the orbital angular momentum operators ##\hat{\vec{x}} \times \hat{\vec{p}}## satisfy this commutation relation. We want the spin angular momentum operators to satisfy the same commutation relation. This forces a certain normalization on the 2x2 matrices we use to represent the angular momentum operators for spin-1/2: the Pauli matrices only satisfy this commutation relation after we multiply them by 1/2. The Pauli matrices themselves satisfy a commutation relation with an extra factor of 2:

##[\sigma_i, \sigma_j] = 2 i \epsilon_{i j k} \sigma_k##.

We remove the extra factor of two by defining the spin angular momentum operators as ##\hat{S}_i = \sigma_i / 2##. Then the ##\hat{S}_i## satisfy the correct commutation relations for angular momentum operators.

##[\hat{J}_i, \hat{J}_j] = i \epsilon_{i j k} \hat{J}_k##

For example, you can verify that the orbital angular momentum operators ##\hat{\vec{x}} \times \hat{\vec{p}}## satisfy this commutation relation. We want the spin angular momentum operators to satisfy the same commutation relation. This forces a certain normalization on the 2x2 matrices we use to represent the angular momentum operators for spin-1/2: the Pauli matrices only satisfy this commutation relation after we multiply them by 1/2. The Pauli matrices themselves satisfy a commutation relation with an extra factor of 2:

##[\sigma_i, \sigma_j] = 2 i \epsilon_{i j k} \sigma_k##.

We remove the extra factor of two by defining the spin angular momentum operators as ##\hat{S}_i = \sigma_i / 2##. Then the ##\hat{S}_i## satisfy the correct commutation relations for angular momentum operators.

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Thanks Duck... so then the 1/2 is not related to fermions having spin of 1/2 ?

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The reason we factor out the [itex]\frac{\hbar}{2}[/itex] is convience. It makes the eigenvalue equation of the [itex]\hat{S} = \frac{\hbar}{2}σ[/itex] operators more clear

[itex]\hat{S}_{n}\left|n\right\rangle = \frac{\hbar}{2}\left|n\right\rangle[/itex]

and it also makes it easier to evaluate rotation operators:

[itex]\hat{R}(\phi\hat{n}) = e^{i\hat{S}_{n}\phi/\hbar}[/itex]

which need to be taylor expanded to give meaning, and will involve powers of the [itex]\hat{S}[/itex] operator.

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- #6

Nugatory

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Google for "bra-ket notation".

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Sure. I highly recommend Townsend's "A Modern Approach to Quantum Mechanics" If you want to really get a hold of spin and bra-ket notation. It's the first chapter and he presents it in a very intuitive way.

[itex]\left|ψ\right\rangle[/itex] (called a "ket") is an expression for a state of a system. When talking only about the intrinsic spin degree of freedom, we express the state as as [itex]\left|±n\right\rangle[/itex] (maybe I should have used a ± in my last post) where [itex]n[/itex] is the axis we are measuring the spin along (ie, x, y, z or anything in between).

So for example a spin 1/2 particle that is "spin up" along the z-axis could be denoted [itex]\left|+z\right\rangle[/itex] and the eigenvalue equation of it with the [itex]S_{z}[/itex] operator would be (I dropped the hats because I don't like the way the look):

[itex]S_{z}\left|+z\right\rangle = \frac{\hbar}{2}\left|+z\right\rangle[/itex]

and similarly for a "spin down",

[itex]S_{z}\left|-z\right\rangle = \frac{-\hbar}{2}\left|-z\right\rangle[/itex]

since [itex]\left|+z\right\rangle[/itex] and [itex]\left|-z\right\rangle[/itex] are eigenstates of the [itex]S_{z}[/itex] operator.

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Aaaah... a light at the end of the tunnel!!! Thank you very much!!!!

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