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My question is whether Peierls substitution really holds true for time-dependent electromagnetic (EM) potentials and, if yes, why.

To implement an electromagnetic field in a condensed matter system described by a Bloch Hamiltonian, I have often seen people make the following substitution of the hopping integrals

$t_{12}\rightarrow t_{12}' = e^{i\frac{q}{\hbar}\int_{\vec{R}_1}^{\vec{R}_1}\vec{A}(\vec{r})d\vec{r}} t_{12} \tag{1}\label{subst}$

with the EM vector potential $\vec{A}$ and connected lattice sites $\vec{R}_i$. This is correct as a result of the Peierls substitution (as long as closed integrals over $\vec{A}$ can be assumed $0$). The proof on the Wikipedia page is very nice, actually.

My problem is that I have also seen substitution $\ref{subst}$ for time-dependent EM potentials $\vec{A}(\vec{r},\tau)$ with time $\tau$. If one checks the proof on Wikipedia (every other proof I have seen is conceptually equivalent), one can easily see that for a time-independent EM potential, the Schrödinger equation is still satisfied: Assume

$ H(\vec{r})=\frac{(\vec{p})^2}{2m}+U(\vec{r}) \qquad\text{and}\qquad t_{12}=-\int\phi_{\vec{R}_1}(\vec{r})^\dagger H\phi_{\vec{R}_2}(\vec{r}) \tag{2}\label{tDef} $

with Wannier functions $\phi_{\vec{R}_i}$. If we now have eigenfunctions

$ \Psi_k(\vec{r})=\frac{1}{\sqrt{N}}\sum_{\vec{R}_i} e^{i\vec{k}\vec{R}_i}\phi_{\vec{R}_i}(\vec{r}) \qquad\text{with}\qquad H\Psi_k(\vec{r})=E(\vec{k})\Psi_k(\vec{r}), \tag{3}\label{energy} $

we can use them to obtain

$ i\hbar \frac{d}{d\tau} \left( e^{-i\frac{E(\vec{k})}{\hbar}\tau} \Psi_k(\vec{r}) \right) = E(\vec{k}) \left( e^{-i\frac{E(\vec{k})}{\hbar}\tau} \Psi_k(\vec{r}) \right) = H \left( e^{-i\frac{E(\vec{k})}{\hbar}\tau} \Psi_k(\vec{r}) \right) \tag{4}\label{satisfies} $ and to thus satisfy the time-dependent Schrödinger equation.

Now we introduce an EM potential, such that $ \tilde{H}=\frac{(\vec{p}-q\vec{A}(\vec{r}))^2}{2m}+U(\vec{r}). \tag{5}\label{newHami} $

Upon Peierls substitution $ \phi_{\vec{R}_i}(\vec{r})\rightarrow \tilde{\phi}_{\vec{R}_i}(\vec{r})= e^{i\frac{q}{\hbar}\int\vec{A}(\vec{r})d\vec{r}} \phi_{\vec{R}_i}(\vec{r}), \tag{6}\label{subst2} $ equation $\ref{satisfies}$ is still satisfied as

$ i\hbar \frac{d}{d\tau} \left( e^{-i\frac{E(\vec{k})}{\hbar}\tau} \tilde{\Psi}_k(\vec{r}) \right) = \tilde{H} \left( e^{-i\frac{E(\vec{k})}{\hbar}\tau} \tilde{\Psi}_k(\vec{r}) \right) \tag{7}\label{satisfies2} $ with $\tilde{\Psi}(\vec{r})$ like in definition $\ref{energy}$, but with all $\phi_{\vec{R}_i(\vec{r})}$ replaced with $\tilde{\phi}_{\vec{R}_i}(\vec{r})$ and with the original $E(\vec{k})$ (easy to check).

If one now wants to express this in the basis of the $\tilde{\phi}_{\vec{R}_i}(\vec{r})$ (like e.g. for a numerical evaluation where the hopping integrals are known), one indeed makes the substitution $\ref{subst}$. However, equation $\ref{satisfies2}$ remains true only if

$ \frac{d}{d\tau}\int\vec{A}(\vec{r},\tau)d\vec{r}=0 \tag{8}\label{question} $ for all the integrals inside the $\tilde{\phi}_{\vec{R}_i}(\vec{r})$. Otherwise, on the left hand side of equation $\ref{satisfies2}$, countless obstructive prefactors arise, but they do not appear on the right hand side.

I personally want to use Floquet formalism (not immediately relevant for this discussion) to desribe a time-dependent system. However, I wonder whether I am truly allowed to use Peierls substitution. So far, I am using another gauge instead, such that $\vec{A}(\vec{r},\tau)$ vanishes and I am left with a scalar potential $\Phi(\vec{r},\tau)$. This, however, leads to terms which are not diagonal in $\vec{k}$ which is not nice either.

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  • $\begingroup$ Yes it obviously holds you do not need all elementary calculations. pierls substiution is a result of gauging u1 symmetry in lattice. Just forget about the wierd formalism you use. 1 write continumm hamiltonian in second qnt picture. 2 enforce local u1 symmetry 3. Do it same in lattice forget about wsnnier orbitals just discretize the space naively. $\endgroup$ – physshyp Oct 28 at 11:50
  • $\begingroup$ Would you mind elaborating it a little? Because to me, and apparantly to some other people as well, it is not obvious at all. If it was an actual symmetry, the formalism should not matter, as long as no errors are made during its execution. If I understand you correctly, then you only want to gauge the Hamiltonian itself and not the wave functions. Fair enough. However, I do not see any way to perform this without getting into trouble along the way. For the time-independent case, the situation is easy and obvious. For the time-dependent case, however, things do not seem as simple to me. $\endgroup$ – Fred Oct 28 at 12:11
  • $\begingroup$ just forget about wavefunctions just work in second quantized picture. you get exactly same info with 1st quantized picture, but much more straightforwardly, and clearly. $\endgroup$ – physshyp Oct 28 at 13:19
  • $\begingroup$ Thank you very much for your answers. Right now, I'm calculating things through. I've already unchecked my old (bad) response as an answer and will respond as soon as I am confident about everything. $\endgroup$ – Fred Oct 28 at 13:36
  • $\begingroup$ becarfeul though it holds in action! just go through my answer its very clear. $\endgroup$ – physshyp Oct 28 at 13:57
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Ok since I have to give a full, valid answer to the question, that can stand on its own:

The Peierls substitution does NOT work for time-dependent Vector potentials. The proof at Wikipedia is wrong, as it stands.

Here is my proof:

We start from the field operator representation of the Hamiltonian, for simplicity we assume that the Vector potential is spatial homogeneous (long wave length approximation) \begin{align} \label{Eq.ONE} H(t) = \int \mathrm{d} r \Psi^\dagger(r) \left[ \frac{(-i \hbar \nabla - e A (t) )^2}{2m} + V(r) \right] \Psi(r) \end{align}

We express the Field operators in time-independent atomic basis which are site diagonal (i.e. Wannier orbitals) \begin{align} \Psi(r) = \sum\limits_l \phi(r-R_l) c_l \end{align} Substituting yields \begin{align} H(t) = \sum\limits_{l l'} J_{l l'} (t) c_l^\dagger c_{l'} \end{align}

with \begin{align} \label{Eq.Hop} J_{l l'} (t) = \int \mathrm{d} r \phi^*(r-R_l) \left[ \frac{(-i \hbar \nabla - e A (t) )^2}{2m} + V(r) \right] \phi(r-R_{l'}) \end{align}

Note that the operators $c_l$ are time independent, the time dependence is only in the hopping integral.

We introduce a new set of Bloch basis states, \begin{align} \tilde{\psi}_k = \frac{1}{\sqrt{N}} \sum\limits_{R_l} e^{i\int_{R_l}^{r} e A \mathrm{d}r} e^{ikR_l} \phi_{R_l}(r) \end{align} and the corresponding inverse transformation \begin{align} \phi_{R_l}(r) = \frac{1}{\sqrt{N}} \sum\limits_k e^{-ik{R_l}} e^{-i\int_{R_l}^r e A \mathrm{d}r} \tilde{\psi}_k(r) \end{align} where $\phi_{R_l}(r) = \phi(r-{R_l})$. Note that $\tilde{\psi}_k$ are instantaneous Eigenstates of the operator \begin{align} H(t) = \left[ \frac{(-i \hbar \nabla - e A (t) )^2}{2m} + V(r) \right], \end{align} they are time-dependent, however they only enter in the calculation of the hopping matrix elements. They are not used to solve the physical time-dependent Schroedinger equation. The eigenequation reads \begin{align} H(t) \tilde{\psi}_k = \tilde{E}_k \tilde{\psi}_k \end{align} We use the representation of the Wanier orbitals in the diagonal Bloch states to rewrite \begin{align} J_{l l'} &= \int \mathrm{d}r \sum\limits_{k,k'} \frac{1}{N} e^{ikR_l} e^{i\int_{R_l}^r eA \mathrm{d}r'} \tilde{\psi}^*_k(r) H(t) \tilde{\psi}_k(r) e^{-ik'R_{l'}} e^{-i \int_{R_l'}^r eA \mathrm{d}r'} \\ &= e^{ i \int_{R_l}^{R_{l'}} eA \mathrm{d}r'} \sum\limits_{k,k'} e^{-ik'R_{l'}} e^{ikR_{l}} \int \mathrm{d}r\tilde{\psi}_k^*(r) H(t) \tilde{\psi}_{k'}(r) \\ &= e^{ i \int_{R_l}^{R_{l'}} eA \mathrm{d}r'} \sum\limits_{k} e^{ik(R_l-R_{l'})} \tilde{E}_k \\ \end{align}

Here comes the main problem. Only if $\tilde{E}_k = E_k$ holds we can deduce that the Peierls substitution works (i.e. its just a phase factor). But that is in general not true.

If $A$ is not time dependent, we can introduce a new set of Wannier Orbitals and indeed proof that the peierls substition works, as you do in your derivation. But if $A$ is time dependent, we should not introduce time depentent Wannier Orbitals, as that changes the form of the time dependent Schrödinger equation.

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  • $\begingroup$ Thank you for your comprehensive answer. This is indeed exactly what I tried to convey, but written in a much better way. Two notes, however: 1. $A$'s spacial dependence must behave nicely enought to plug the integrals in the exponents together. If there were e.g. singularities, one would need an extra term. (That's irrelevant for the argument, but since you wrote your proof in such detail, I wanted to mention it.) $\endgroup$ – Fred Oct 28 at 9:25
  • $\begingroup$ 2. What colleagues have complained to me about, was the notion of $\tilde{\Psi}$ in the first place. They claimed that everything would be fine if I just added the Peierls phase to the phases inside the Hamiltonian $H(\vec{r})$ from my eq. (2) (and not inside the wave function). According to them, I could then solve my new time-dependent Hamiltonian by other means and get a solution to my original problem. Since I don't know where to start to check this, I don't know how to disprove it. However, I think you are absolutely right. $\endgroup$ – Fred Oct 28 at 9:31
  • $\begingroup$ One final note, reassuring that it probably does not work, is the following paper with its references [4-7]: journals.aps.org/prb/pdf/10.1103/PhysRevB.14.2239 The paper seems to be the first time this method has beeen called Peierls substitution. It states "Perhaps the most difficult step to justify on physical grounds is the following one, which I shall refer to as the "Peierls substitution"". This certainly sounds more doubtful than all the modern sources just citing the original Peierls paper. Also, [4] argues that it is only a good approximation for slow time variation. $\endgroup$ – Fred Oct 28 at 9:39
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It holds but the Wikipedia, wannier orbitals and hamiltonian formalism unnecessarily blurs this very obvious fact.

First assume that we have an action of Dirac fermions(this can also be non relativistic fermions as in your case it won't effect the argument) in $2+1$ flat space-time.

$$S_F=\int d^3x\bar{\psi}\gamma^{\mu}i(\partial_\mu)\psi+m_0\bar{\psi}\psi$$ now this is invariant under global $U(1)$ symmetry that is if I let $$\psi(x)\to\psi(x)e^{i\alpha}$$ for some alpha and substitute it to above action, the action would not change. However if $\alpha$ depends on $x$ than the above action would change. So we write a new action which would not change under such transformations with $x$ dependent $\alpha$ that is

$$S_F=\int d^3x\bar{\psi}\gamma^{\mu}i(\partial_\mu+iA_\mu)\psi+m_0\bar{\psi}\psi$$ such that we defined a new quantity $A$ which transforms as $$A_\mu(x)\to A_\mu(x)+\partial_\mu\alpha(x)$$ so this is how we actually add an electromagnetic coupling to any system, just make the system invariant under $x$ dependent gauge transformation, it automatically adds the gauge coupling.

Ok, now the next step is just doing this in the lattice. First we wick rotate, than we enforce the local gauge invariance and then, we discretize, only the spatial coordinates, and leave the temporal coordinate continuous then the action is now, \begin{equation} S_{WF}[A]=\int d\tau \bigg[\sum_{\vec{r}}\bar{\psi}(r)(\gamma^{0}_E(\partial_0+iA^{E}_0)\psi(r)+(m_0+2R)\sum_{\vec{r}}\bar{\psi}(r)\psi(r)-\frac{1}{2}\sum_{\vec{r}i}[\bar{\psi}(r)(R-\gamma^i_E)\psi(r+\hat{i})U_{ir}+\bar{\psi}(r+\hat{i})(R+\gamma^i_E)\psi(r)U^{\dagger}_{ir} ]\bigg] \end{equation} where $r=(\tau,\vec{r})$, $R$ is the wilson parameter which is dropped in continuum limit and as long as it is finite, the particular value is not important and $U_{ir}=\exp(iA_{ir})$

Here the $U$ is actually what you called peierls substitution, but the actual name is Schwinger integral. Somehow they usually teach this very fundamental stuff in a very obscure weird and unecesarrly useless peirls point of view.

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  • $\begingroup$ I had to read up a lot about action integrals again, since I haven't woked with them for two years now. So maybe I still have some misunderstandings. However, if I do understand it correctly, we are now at the point where we can obtain a time-dependent Hamiltonian from our action via Legendre transformation. This Hamiltonian then looks pretty much like the unperturbed (real-space) Hamiltonian, but with the Peierls phase on some part of it. I need to check whether I can still properly apply Bloch thrm to obtain a k-dependent Hamiltonian. I may take some days to answer, since I am quite busy rn. $\endgroup$ – Fred Oct 28 at 15:10
  • $\begingroup$ about legendre tranformation i am not sure just use transfer matrix method to be safe. $\endgroup$ – physshyp Oct 28 at 18:06
  • $\begingroup$ you just take momentum rep of fields and substitute them. $\endgroup$ – physshyp Oct 28 at 18:08
  • $\begingroup$ Just adding a comment so it doesn't seem like I had forgotten. I have a lot to do this month, so I did not yet find time to thoroughly check everything. As soon as I find the time, I will delete this comment and ammend a proper answer. Thank you, already, for your answers provided thus far. $\endgroup$ – Fred Nov 9 at 9:32
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It seems the derivation I presented is flawed in the sense that usually one would derive the Peierls phase inside the hopping terms differently (see e.g. the other derivation on Wikipedia). The proof I linked seems to just be an example of another derivation agreeing with the original one in the case of a time-independent EM potential. In the time-dependent case, however, there is no reason for the Schrödinger equation 7 to still hold true in the first place, not only because of the additional terms on the left hand side (because of the time-dependent basis functions), but also because of the right hand side now being time-dependent as well.

Edit: J. M. Luttinger, Phys. Rev. 84, 814 (1951) shows that Peierls Substitution is a good approximation for slowly varying fields. For fast variation, however, further precautions seem necessary.

Edit 2: After reading through the paper and its reference $1$ to know whether "not too rapidly varying" means a variation in time or space (it means space in that case), the second equation above eq. $(2')$ seems to actually confirm my original complaint with Peierls substitution. It includes an additional term on the time-derivative side of the Schrödinger equation before another approximation is made. I will keep this edit as an addition to my original answer, because the proof in the paper is rather close to the Wikipedia proof I originally used. If I find another proof, I will further edit this answer.

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