What does a spin 1/2 wave function actually look like

  • Thread starter jelathome
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  • #1
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I have only ever seen the wavefunction for a spin 1/2 particle written in the basis set |α> |β>. I was interested in how a wavefunction |ψ> = a|α> + b|β> might be rewritten in a continous basis and hence would need to know what the actual functions of |α> & |β> were.

Thanks
 

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  • #2
vanhees71
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This is a state vector, not a wave function. For a particle with spin, the wave function is given by
[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
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Ok thanks but I still want to represent the state vector |ψ>=a|α> + b|β> in a continuous basis. Is this possible or am I fundamentally misunderstanding something?

Thanks
 
  • #4
kith
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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.

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
vanhees71
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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.
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!
 
  • #6
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OK thanks I think I must be misunderstanding the problem I am trying to solve.
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
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OK thanks I think I must be misunderstanding the problem I am trying to solve.
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
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##.
 

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