- #1

- 6

- 0

Thanks

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 jelathome
- Start date

- #1

- 6

- 0

Thanks

- #2

- 17,961

- 8,930

[tex]\psi_{\sigma}(t,\vec{x})=\langle \vec{x},\sigma|\psi(t) \rangle,[/tex]

where I used the Schrödinger picture of time evolution. [itex]\sigma \in \{-s,-s+1,\ldots,s-1,s \}[/itex] is the eigenvalue of the spin-z component. The wave function of nonrelativistic quantum theory is thus a spinor field with (2s+1) components.

- #3

- 6

- 0

Thanks

- #4

kith

Science Advisor

- 1,395

- 481

The question is, why can we treat the spin part independently from the spatial wavefunction. The answer is that for purposes like the Stern Gerlach experiment, we can approximate the time evolution of the spatial wavefunction by a classical trajectory and use the quantum description only for the spin.

- #5

- 17,961

- 8,930

It's important to keep in mind that this is true only for non-relativistic quantum theory. In relativistic QFT it's even difficult to make a unique split of total angular momentum into orbital and spin angular momentum!Your state is only the spin part of the complete state vector. The corresponding Hilbert space is 2-dimensional, so all bases consist of two vectors and not of (uncountable) infinite many. The full quantum description of a spin-1/2 particle is the direct product of the spin part and the usual (spinless) spatial wavefunction.

- #6

- 6

- 0

If anyone could point me in the right direction that would be great.

The paper is http://arxiv.org/pdf/1407.2139v1.pdf

And my problem is how they go from 17 to 18

Thanks

- #7

DrClaude

Mentor

- 7,607

- 4,009

They consider spin as a classical 3D vector. If #you take the usual convention of the z axis defining ##|\alpha\rangle## and ##|\beta\rangle##, then ##|\alpha\rangle## corresponds to a function in ##(\theta,\phi)## that is zero everywhere except when ##\theta = 0## and ##\phi = 0##, and ##|\beta\rangle## is zero everywhere except when ##\theta = \pi/2## and ##\phi = 0##.

If anyone could point me in the right direction that would be great.

The paper is http://arxiv.org/pdf/1407.2139v1.pdf

And my problem is how they go from 17 to 18

Thanks

Share: