# 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.

• I don't think the Schrodinger equations as you've written it makes sense unless $\Psi$ is a two component spinor. One the left side you just have a scalar, but on the right side you have a 2x2 matrix. If you write the equation correctly with $\Psi$ as a spinor, then you end up with equations for each spin flavor. So the answer to "where has the spin degree of freedom gone" is that you've taken it out by hand, in the process making your equation mathematically inconsistent.
– d_b
Commented Feb 20, 2014 at 17:28
• @user37496 You're perfectly right, but it doesn't change the conclusion: I can define an energy level for each spinor component. Written as it is $\Psi$ is a spinor, I should call $\Psi_{0}$ a constant spinor instead of a constant. Then better to say $\Psi_{0}^{2}=\Psi_{0}^{\dagger}\Psi_{0}=1$. Nevertheless, the energy can be defined the way I did, and only a polariser + analyser experiment will measure a spin. But that's true my energy is more a statistical averaged quantity than a quantum averaged one. At least that's what I think, but maybe people have better ideas. Commented Feb 20, 2014 at 18:01

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.

• Thanks for this beautiful answer. I agree with everything, except I believe what we measure in the Stern-Gerlach experiment is precisely the entanglement (I would call it a correlation since there is only one system to my mind, but it makes no difference here and this of purely rhetorical interest) between the space and spin degree of freedom, not the spin. Moreover, we really measure a position, not a spin. We infer the spin from the correlation you point out. The funny thing is that my gauge transform which absorbs the Zeeman term does not work for the Stern-Gerlach experiment, ... Commented Feb 25, 2014 at 10:51
• ... due to the space-dependency of the Zeeman term. So somehow, the Stern-Gerlach experiment is a interference effect between gauge degrees of freedom, the thing I suspected when I asked my question. Thanks for your answer once again. Commented Feb 25, 2014 at 10:53

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:

[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)

• +1 to you. I indeed forgot the paramagnetic contribution to the current, see Landau vol. III on quantum mechanics. The orbital effect you mentioned in your answer is of no help to my concern, but you're entirely right: the current should get a gauge term $\mu\cdot\sigma$, with $\mu$ the gyromagnetic ratio. How do we measure this term, though ? I'm not aware of any measurement of it. Do you have an idea, or a reference ? Thanks in advance for further explanations, and thanks again for your answer. Commented Feb 25, 2014 at 10:56
• Sorry, I just check the reference you gave in your answer. I believe the term there mentioned is of much higher order than the gyromagnetic term I mentioned in my previous comment. As your orbital effect, it explains nothing for me, because I discussed an uncharged spin particle. Whether or not this particle exists is an other story. My question in my previous comment concerns uncharged yet spinful particle, for which a gyromagnetic term exists, see Landau III. Commented Feb 25, 2014 at 11:04
• By the way, I'm not aware of the reference mentioned in the Wikipedia article [Quantum mechanics, E. Zaarur, Y. Peleg, R. Pnini, Schaum’s Easy Oulines Crash Course, Mc Graw Hill (USA), 2006] (I suppose it should read outlines or online...), and I would not trust it too much. The term on Wikipedia scales really strangely as $\sim s^{-1}$, when $s$ is the spin number ! To my mind, it should be larger and larger for larger spin numbers, and it must disappears when there is no spin ! At least as written on Wikipedia, I would say this term is barely wrong... it should a spin-orbit term I guess Commented Feb 25, 2014 at 11:08
• @Oaoa I've added some further remarks. Commented Feb 25, 2014 at 21:06
• Thanks a lot for the update, and the references. The term on wikipedia is clearly a spin-orbit effect, the term in Landau is clearly a gyromagnetic effect. The spin-orbit effect is of higher order than the gyro one. Wikipedia should be corrected about this point. One of the resulting effect of the spin-orbit term is the so-called Aharonov-Casher effect. Thanks again for your update. Commented Feb 27, 2014 at 10:59