A paradox with spin: Is spin a physical degree of freedom? Suppose I want to calculate the state associated with a spin particle under a magnetic field. I suppose the particle interacts via the Zeeman effect, and only through it. Then I want to resolve the Schrödinger equation: 
$$\mathbf{i}\hbar\dfrac{\partial\Psi}{\partial t}+\dfrac{\hbar^{2}}{2m}\dfrac{\partial^{2}\Psi}{\partial x^{2}}=h_{Z}\sigma_{z}\Psi\left(x,t\right)$$ 
with $h_{Z}$ the Zeeman term. All other terms are usual. To resolve, I apply the transformation 
$$\Psi=e^{-\mathbf{i}h_{Z}\sigma_{z}t/\hbar}e^{-\mathbf{i}Et/\hbar}e^{\mathbf{i}kx}\Psi_{0}$$ 
with $\Psi_{0}$ a constant taken to be real for commodity, and the dispersion relation reads
$$E=\dfrac{\left(\hbar k\right)^{2}}{2m}$$ 
as usual for a free particle. There is no mark of the spin splitting we are used to, like the energy doublet $E_{\pm}=E_{0}\pm h_{Z}$ due to the Zeeman term. Of course this is just a redefinition of the energy origin, done in two different ways for spin up and spin down (or something like that). If I calculate observable like the density $\rho\propto\left|\Psi\right|^{2}=\Psi_{0}^{2}$ or the current $j\propto\Im\left\{ \Psi\partial_{x}\Psi^{\ast}\right\} =k\Psi_{0}^{2}$ again I find no measurable signature of the spin degree of freedom... 
So where is the spin degree of freedom gone ? Do we need to discuss only space and/or time dependent spin degree of freedom to make any sense of it ? And then the spin is only a relative degree of freedom (in other word, I know what is a spin mismatch, but not what is a spin, a bit like a phase: only phase difference matters). Or anyone gets a better idea of what a spin is? Thanks in advance.
 A: As the other poster said, the thing for which you use $E$ in your expression is not the eigenvalue of the Hamiltonian, but rather the eigenvalue of the kinetic part of the Hamiltonian.
The Hamiltonian is:
$$ \hat H = \frac{\hat p^2}{2m} \otimes \mathbb{1}  + \mathbb{1} \otimes h_z\sigma_z$$
and the eigenvalues are:
$$E_\pm(p) = \frac{p^2}{2m}\pm h_z$$
However, this doesn't really answer your question of are spin degrees of freedom physical in an absolute sense or merely relative; the answer is that the spin degrees of freedom are physical in an absolute sense, but several initial results suggest that they're not.
Firstly, measurement: the results of a measurement of spin could be anything you like, so there is no absolute meaning to spin in that context. Consider the following operator:
$$ \hat A = a_\uparrow\left|\uparrow \right\rangle\left\langle \uparrow\right| + a_\downarrow\left|\downarrow \right\rangle\left\langle \downarrow\right| = \left( \begin{array}{ccc}
a_\uparrow & 0 \\
0 & a_\downarrow \end{array} \right) $$
This is hermitian and has eigenvalues $a_\uparrow$ and $a_\downarrow$, which can be any real numbers as long as they aren't the same, so the results of a measurement of spin can be anything you like.
This is actually a generic properties of measurement in quantum mechanics, that the results can be relabeled in any way you like, so it doesn't really tell us anything about the physical significance of spin.
Next: time evolution of the spin state in a magnetic field. Let's consider if the values of the spin operator had some constant offset, ie:
$$ \hat S = S_0 + \frac{\hbar}{2}\sigma_z $$
Now the Zeeman terms in the Hamiltonian:
$$ \hat H _Z = a \hat S = a S_0 + a \frac{\hbar}{2}\sigma_z $$
where $a$ is some constant. Now consider the time evolution of a general state:
$$ \hat U(t)\left(\alpha \left|\uparrow \right\rangle + \beta \left|\downarrow \right\rangle \right) =  \alpha \exp\left(\frac{aS_0t}{\hbar}+\frac{at}{2}\right)\left|\uparrow \right\rangle + \beta \exp\left(\frac{aS_0t}{\hbar}-\frac{at}{2}\right)\left|\downarrow \right\rangle $$
$$ =  \exp\left(\frac{aS_0t}{\hbar}\right)\left(\alpha \exp\left(+\frac{at}{2}\right)\left|\uparrow \right\rangle + \beta \exp\left(-\frac{at}{2}\right)\left|\downarrow \right\rangle\right) $$
So the offset in the spin operator only contributed to a total phase offset, which makes no difference to the probabilities of spin measurements. This supports what you suggest---that only t difference in spin is physically relevant.
The absolute relevance of spin degrees of freedom, however, comes when you consider the intrinsic magnetic moment of particles with spin. The magnetic moment operator for a spin half particle is given by:
$$ \hat\mu_z = g\hat S_z=g\frac{\hbar}{2}\hat\sigma_z$$
where $g$ is the gyromagnetic ratio of that particle. The physical significance of this can be seen, for instance, in Stern-Gerlach experiments.
In these experiments you have a state where, in general, the position degrees of freedom are entangled with the spin degrees of freedom, ie:
$$ \left| \Psi \right\rangle = \alpha|\psi_\uparrow\rangle \otimes \left| \uparrow \right\rangle + \beta| \psi_\downarrow \rangle \otimes \left| \downarrow \right\rangle $$
I can't easily go over the derivation of the effect here, but the short story is that the evolution of the spatial state depends, via the entanglement, on the direction of the spin state to which the spatial state is entangled. Then the final position of the particle on the screen depends on the spin, hence the experiment constitutes a measurement of spin.
The final position of a wavepacket does depend on the absolute value of spin, giving absolute physical significance to the spin degrees of freedom.
A: I would argue that while the expression is indeed a solution of the Schrödinger-equation for the given Hamiltonian, the interpretation of $E$ as the eigenenergy of $H$ is not quite correct. The general solution for a time-independant Hamiltonian is of the form
$$ \Psi(t) = \exp\left(-it\hat{H}\right)\Psi(t=0)=\sum_nc_n \exp\left(-itE_n\right)\Psi_n $$
The last equality being the expansion wrt. eigenfunctions of $\hat{H}$.
Given a solution of the form $\Psi(t)=\exp\left(-it\omega\right)\Psi_0$ as in your example, whatever appears in the exponential as conjugate to $t$ is the eigenenergy $(\hbar\equiv 1)$ - in the present case $E\pm h_z$. This splitting manifests itself for example as Zeeman-splitting in spectroscopic experiments.
As far as the current is concerned, the usual expression $$ \mathbf{j}\propto \Im\{\Psi^*\hat{p}\Psi\}$$ needs modification in the presence of electromagnetic interactions. Even for charged spin 0 particles one should consider the minimal coupling $$\hat{p}\rightarrow \hat{p}-q\mathbf{A}$$ which leads to an additional term $$ -2q\mathbf{A}\left|\Psi\right|^2$$ For non-zero spin you'll get an additional contribution due to the magnetic moment spinful particles posses. See e.g wikipedia:Probability current [1]
I believe, at least for spin 1/2 particles, the fate of this last contribution is best understood by tracing it back to the non-relativistic reduction of the Dirac equation coupled to an electromagnetic potential. Details of this calculation should be found in most textbooks on relativistic quantum theory.
Edit: The term under discussion is $$ \frac{\mu_S}{S}\nabla\times (\Psi^\dagger\mathbf{S}\Psi) $$
Unfortunatly, the source [2] mentioned in [1] does not give a derivation. However, I'd say this term is not of the order $S^{-1}$, because the spin-operator is of order $S$.
In contrast to the gauge contribution which couples to the charge density, this term holds even for uncharged particles as long as they posses a spin. The neutron would be an example. It has no charge, but spin due to its quark substructure.
I could not find a text giving a rigorous derivation for arbitrary spin, but in this paper, it is shown for a spin $1/2$ particle by reduction of the Dirac equation. Considering that spin is a fundamentally relativistic property, it seems to make sense this to be the right way to fix the term.
In section §115 of Landau & Lifschitz Vol. 3 an expression for the current is derived under the assumption that a spin couples to the magnetic field like $-\hat{S}\cdot\mathbf{B}$.
I'm not an experimentalist guy myself, but i think gyromagnetic ratios are measured using magnetic resonance techniques.
References:
[1] https://en.wikipedia.org/wiki/Probability_current
[2] Y. Peleg et al.,Schaum's outline of theory and problems of quantum mechanics, McGraw Hill Professional 1998
[3] Marek Nowakowski, The quantum mechanical current of the Pauli equation, Am. J. Phys. 67, 916 (1999)
