Pauli spin matrices and hbar/2

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SUMMARY

The Pauli spin matrices are represented as σi/2 to ensure they satisfy the angular momentum commutation relations for spin-1/2 particles. The normalization by 1/2 is necessary because the Pauli matrices alone yield an extra factor of 2 in their commutation relations. This adjustment allows the spin angular momentum operators, defined as &hat;Si = σi/2, to correctly represent the intrinsic spin values of ±ℏ/2. The use of bra-ket notation is also discussed, with recommendations for Townsend's "A Modern Approach to Quantum Mechanics" for further understanding.

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lonewolf219
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Why are the Pauli matrices multiplied by 1/2 ?? Why are they represented as σ1/2 σ2/2 and σ3/2 and not just σ1 σ2 σ3 ?
 
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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.
 
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Thanks Duck... so then the 1/2 is not related to fermions having spin of 1/2 ?
 
It's because the Pauli matrices are used for spin 1/2 particles, which have intrinsic spins of either \frac{\hbar}{2} or \frac{-\hbar}{2}.

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

\hat{S}_{n}\left|n\right\rangle = \frac{\hbar}{2}\left|n\right\rangle

and it also makes it easier to evaluate rotation operators:

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

which need to be taylor expanded to give meaning, and will involve powers of the \hat{S} operator.
 
Wotanub, thank you... if you don't mind, maybe you could explain what the bracket notation means? I would GREATLY appreciate it! All I can see from it is that S is an operator, and maybe n is the principal quantum number? What exactly does | signify... "over"?
 
lonewolf219 said:
Wotanub, thank you... if you don't mind, maybe you could explain what the bracket notation means? I would GREATLY appreciate it! All I can see from it is that S is an operator, and maybe n is the principal quantum number? What exactly does | signify... "over"?

Google for "bra-ket notation".
 
lonewolf219 said:
Wotanub, thank you... if you don't mind, maybe you could explain what the bracket notation means? I would GREATLY appreciate it! All I can see from it is that S is an operator, and maybe n is the principal quantum number? What exactly does | signify... "over"?

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.

\left|ψ\right\rangle (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 \left|±n\right\rangle (maybe I should have used a ± in my last post) where n 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 \left|+z\right\rangle and the eigenvalue equation of it with the S_{z} operator would be (I dropped the hats because I don't like the way the look):

S_{z}\left|+z\right\rangle = \frac{\hbar}{2}\left|+z\right\rangle

and similarly for a "spin down",

S_{z}\left|-z\right\rangle = \frac{-\hbar}{2}\left|-z\right\rangle

since \left|+z\right\rangle and \left|-z\right\rangle are eigenstates of the S_{z} operator.
 
:biggrin: Aaaah... a light at the end of the tunnel! Thank you very much!
 

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