If a Bloch electron in a solid is scattered by absorbing / emitting a phonon, does the final state need to conserve the original electron spin?

Similarly, if two Bloch electrons collide due to electron-electron interaction are the final Bloch states to which they scatter restricted to assign the same spin to each electron?

The books I usually consult on solid-state physics and electron scattering do not address this (or at least not anywhere I can find) and I need to know to write proper collision integrals for the relaxation of a perturbation to the electronic distribution.

EDIT: I have yet found nothing in the published literature. The only argument I can think of is in favor of both spin-eigenstates being available after scattering. The argument is that scattering is caused by perturbations which break angular symmetry and thus one cannot impose conservation of angular momentum.

I welcome any new ideas.

  • $\begingroup$ Yes, it does conserve spin. This is easiest to see when looking at, say Fe, where the spin up and spun down electrons have different Bloch functions and band structures. $\endgroup$
    – Jon Custer
    Aug 22, 2017 at 20:44
  • $\begingroup$ Thanks. So I guess this suggests that both scattering mechanisms (electron-phonon and electron-electron) conserve spin, right? Does this also imply that electron-electron scattering may only occur between electrons with the same spin? My guess is not because electron-electron interaction occurs through a screened Coulomb potential, which is unrelated to the electron spin of the electrons exerting or subject to it. Also, could you recommend a source where I could read more about this spin non-degeneracy in Fe? $\endgroup$
    – lcortesh
    Aug 22, 2017 at 20:56
  • $\begingroup$ Phonons have no mechanism to interact with an electron's spin. I might word your second point differently - if spin up and down electrons occupy different bands, it is hard for them to interact... $\endgroup$
    – Jon Custer
    Aug 22, 2017 at 20:59
  • $\begingroup$ Although phonons do not possess spin could they not break angular symmetry and thus angular momentum (spin) conservation? I would argue that if, as is generally the case, an energy band has spin-degeneracy, and the matrix-scattering imposes no condition on spin, both spin eigenstates should be available for an electron to scatter to that state. $\endgroup$
    – lcortesh
    Aug 25, 2017 at 5:27

1 Answer 1


I think I found a satisfactory answer. Carrier scattering in solids is conventionally described using time-dependent pertubation theory. In this formalism, the rate at which an electron in a Bloch state $\mathbf{k}_0$ and definite spin $\sigma_0$ scatters to a state with quantum numbers $\mathbf{k}_1$, $\sigma_1$, $S(\mathbf{k}_0,\sigma_0;\mathbf{k}_1,\sigma_1)$ is given by

$$ S(\mathbf{k}_0,\sigma_0;\mathbf{k}_1,\sigma_1) = \frac{2\pi}{\hbar}|<\mathbf{k}_1,\sigma_1|H_{\operatorname{per}}|\mathbf{k}_0,\sigma_0>|^2\delta(E_1 - E_0 + \Delta E). $$ Here $E_i$ is the energy of the state with wavenumbers $\mathbf{k}_i$, $\sigma_i$. $H_{\operatorname{per}}$ is the Hamiltonian inducing the perturbation. $\Delta E$ is a constant given by the time-dependence of $H_{\operatorname{per}}$ (i.e. $\Delta E = \hbar\omega$ for phonon emission and $\Delta E = 0$ for a constant perturbation).

If $H_{\operatorname{per}}$ depends only on the spatial coordinates of the electron, as is the case for common scattering mechanisms such as phonon scattering and non-magnetic impurity scattering, then the matrix element $<\mathbf{k}_1,\sigma_1|H_{\operatorname{per}}|\mathbf{k}_0,\sigma_0>$ is zero if $\sigma_0 \neq \sigma_1$. To see this, we note that the wave function $\psi_{\mathbf{k}_0,\sigma_0} \equiv <\mathbf{x},\sigma|\mathbf{k}_0,\sigma_0>$ must be of the form

$$ \psi_{\mathbf{k}_0,\sigma_0} = \psi_{\mathbf{k}_0}(\mathbf{x})\delta_{\sigma,\sigma_0} $$

because the electron has a definite spin $\sigma_0$ and the spatial coordinates and spin are compatible, independent observables. Then

$$ <\mathbf{k}_1,\sigma_1|H_{\operatorname{per}}|\mathbf{k}_0,\sigma_0> = \sum_{\sigma,\sigma'} \int\int d^3xd^3x'<\mathbf{k}_1,\sigma_1|\mathbf{x}',\sigma'>\times\\ \times<\mathbf{x}',\sigma'|H_{\operatorname{per}}|\mathbf{x},\sigma><\mathbf{x},\sigma|\mathbf{k}_0,\sigma_0> \\ = \sum_{\sigma,\sigma'} \int\int d^3xd^3x' \left\{\psi^*_{\mathbf{k}_1}(\mathbf{x})\delta_{\sigma',\sigma_1}\right\}\left\{H_{\operatorname{per}}(\mathbf{x})\delta(\mathbf{x - x'})\delta_{\sigma',\sigma}\right\} \left\{\psi^*_{\mathbf{k}_0}(\mathbf{x})\delta_{\sigma,\sigma_0}\right\}\ \\ = \delta_{\sigma_0,\sigma_1}\int d^3x \psi^*_{\mathbf{k_1}}(\mathbf{x}) H_{\operatorname{per}}(\mathbf{x})\psi^*_{\mathbf{k_0}}(\mathbf{x}). $$

And hence, $\sigma_1 \neq \sigma_0$ leads to a vanishing matrix element and thus to a vanishing scattering rate $S(\mathbf{k}_0,\sigma_0;\mathbf{k}_1,\sigma_1)$. As a consequence, scattering due to a spin-independent potential conserves spin. This is the case of phonon scattering and non-magnetic impurity scattering.

Although this explanation relied on Fermi's Golden Rule, which is accurate to first order on the perturbation Hamiltonian, the same argument can be made to arbitrary order on the perturbation strength by substituting $H_{\operatorname{per}}$ in the Golden Rule for the transition operator $T_{\operatorname{per}}$ given by

$$ T_{\operatorname{per}} = H_{\operatorname{per}} + H_{\operatorname{per}}\frac{1}{E - H_0}H_{\operatorname{per}} + H_{\operatorname{per}}\frac{1}{E - H_0}H_{\operatorname{per}}\frac{1}{E - H_0}H_{\operatorname{per}} + \cdots, $$ with $H_0$ being the unperturbed Hamiltonian. If both $H_0$ and $H_{\operatorname{per}}$ are spin-independent, $T_{\operatorname{per}}$ is spin-independent too and the previous argument holds.

The case of electron-electron scattering needs to be treated separately to introduces permutation symmetry considerations, however, but I suspect I should arrive at a similar result.


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