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.

  • $\begingroup$ 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, 2020 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, 2020 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, 2020 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, 2020 at 13:36
  • $\begingroup$ becarfeul though it holds in action! just go through my answer its very clear. $\endgroup$
    – physshyp
    Oct 28, 2020 at 13:57

3 Answers 3


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.

Edit To be more clear:

Let us assume we have a three dimensional Hamiltonian in lattice that is, $$H =a^2\bigg[(m_0a+3R)\sum_n\bar{\psi}_n\psi_n -\frac{1}{2}\sum_{n\mu}[\bar{\psi}_n(R-\gamma_\mu)\psi_{n+\hat{\mu}} +\bar{\psi}_{n+\hat{\mu}}(R+\gamma_\mu)\psi_n]\bigg], $$ where $a$ is the lattice spacing, $\gamma^\mu$ are the usual pauli matrices $\mu$ runs from $1$ to $3$ which represents coordinates of 3 dimensional spatial space $x,y,z$. Now we want to make this locally gauge invariant under transformations, $$\psi_n\to G_n\psi_n$$ where $G\in U(1)$ thus notice that if $\psi$ defined to transform as above we would also have the following transformation (just substitute above to LHS of below and you get the RHS of below), $$\bar{\psi}_n\psi_{n+\hat{\mu}} +\bar{\psi}_{n+\hat{\mu}}\psi_n\to \bar{\psi}_nG^{-1}_{n}G_{n+\hat{\mu}}\psi_{n+\hat{\mu}} +\bar{\psi}_{n+\hat{\mu}}G^{-1}_{n+\hat{\mu}}G_{n}\psi_n$$ so if we define the locally gauge invariant Hamiltonian as, $$H=a^2\bigg[(m_0a+3R)\sum_n\bar{\psi}_n\psi_n -\frac{1}{2}\sum_{n\mu}[\bar{\psi}_n(R-\gamma_\mu)U_{n,n+\hat{\mu}}\psi_{n+\hat{\mu}} +\bar{\psi}_{n+\hat{\mu}}(R+\gamma_\mu)U_{n,n+\hat{\mu}}^\dagger\psi_n]\bigg]$$

and define the transformation rule on $U$ as, $$U_{n,n+\hat{\mu}}\to G(n)U_{n,n+\hat{\mu}}G^{-1}(n+\hat{\mu})$$

we see that the above Hamiltonian is indeed locally gauge invariant (Just apply the transformation rules we have just shown on $\psi$ and $U$ in that hamiltonian, and you see that it is indeed invariant and stay same). So we have derived the locally invariant hamiltonian. Finally, we can define another field $A$ to express $U$ more conveniently, $$U=e^{iaA_\mu(n)}$$

  • $\begingroup$ 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, 2020 at 15:10
  • $\begingroup$ about legendre tranformation i am not sure just use transfer matrix method to be safe. $\endgroup$
    – physshyp
    Oct 28, 2020 at 18:06
  • $\begingroup$ you just take momentum rep of fields and substitute them. $\endgroup$
    – physshyp
    Oct 28, 2020 at 18:08
  • $\begingroup$ I'm sorry for taking so long to answer again. Please pardon if the following is a misundewrstanding, but I lack experience with this formalism. In the last line of your definition of $S_{\text{WF}}$, you have $i$ and $\hat{i}$. My interpretation would be that these are different, and $i$ indices to your gamma matrices while the $\hat{i}$ are lattice vectors which also need to be summed over. Is this assumption correct? In that case, I think to understand and agree with you. However, can you recommend a good source to read up all methodology used here? I would like to fill my knowledge gaps. $\endgroup$
    – Fred
    Dec 21, 2020 at 14:00
  • $\begingroup$ i is direction since the system has 3 dimension its summed over that $\endgroup$
    – physshyp
    Dec 25, 2020 at 12:15

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.

  • $\begingroup$ 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, 2020 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, 2020 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, 2020 at 9:39

I have been informed in the comments that I made a mistake here. However, since I consider it interesting, I want to keep my answer and only add an additional comment: In the following, one needs to be able to cancel the vector potential in the Hamiltonian with that from the phase. For that to be possible, one needs an indefinite integral of the vector potential in the phase. The existence of an indefinite integral, however, means that the vector potential is rotation-free and thus does not give rise to a magnetic field.

I interpret Peierls substitution as a means to handle an electric vector potential $\vec{A}$ in a Schrödinger equation (I explicitly do not write Hamiltonian, since further down, the time differential will be changed in the process, too, and one should not overlook that part). Assume the simplest SDE for this case, i.e.

$i\frac{\partial}{\partial t}\psi(\vec{r},t) = \left( \frac{\left(\vec{p}-\vec{A}(\vec{r},t)\right)^2}{2m}+U(\vec{r},t) \right)\psi(\vec{r},t). $

To simplify it, one modifies the wave function via

$\psi(\vec{r},t) \rightarrow \psi(\vec{r},t)'= \psi(\vec{r},t)\cdot \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right) $

with some $\vec{r}_0$ that does not matter for now. Upon this substitution, one obtains

$H(\vec{r},t)\psi(\vec{r},t)'\\ = \left( \frac{\left(\vec{p}-\vec{A}(\vec{r},t)\right)^2}{2m}+U(\vec{r},t) \right) \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right) \psi(\vec{r},t)\\ = \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right) \left( \frac{\left(\vec{p}+\vec{A}(\vec{r},t)-\vec{A}(\vec{r},t)\right)^2}{2m}+U(\vec{r},t) \right) \psi(\vec{r},t)\\ = \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right) \left( \frac{\vec{p}^2}{2m}+U(\vec{r},t) \right) \psi(\vec{r},t) $


$i\frac{\partial}{\partial t}\psi(\vec{r},t)'= \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right) \left(-q\Phi(\vec{r},t)\right)~ i\hbar\frac{\partial}{\partial t}\psi(\vec{r},t). $

with the new scalar potential

$ \Phi(\vec{r},t)= \int_{\vec{r}_0}^{\vec{r}} \frac{\partial\vec{A}(\vec{r}',t)}{\partial t} d\vec{r}'. $

Plugging together the equations above, one obtains the new SDE

$ \quad\exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right)\cdot i\hbar\frac{\partial}{\partial t}\psi(\vec{r},t)\\ = \exp\left( i\frac{q}{\hbar} \int_{\vec{r}_0}^{\vec{r}} \vec{A}(\vec{r}',t) d\vec{r}' \right)\cdot \left( \frac{\vec{p}^2}{2m} +\Phi(\vec{r},t) +U(\vec{r},t) \right) \psi(\vec{r},t), $

where the new scalar potential is ultimately a gauge transformation originating from the vector potential. For a time-independent potential $\vec{A}$, the new $\Phi$ is zero. For a time-dependent one, however, this transformation just changes an SDE involving a vector potential into one involving a scalar potential. And yes, this scalar potential needs to be taken into account when working with this substitution.

To apply this to a condensed matter system, one should take a look at the Wikipedia proof mentioned in the initial statement. Basically, it adds a sum over lattice sites and gives the lower integration limit a meaning. This will then lead to the modified hopping terms.

As a final note, 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 are necessary. So people have been aware of this issue back then already.

  • $\begingroup$ I forgot to write $q\vec{A}$ instead of just $\vec{A}$ in many places. I will edit it later. Please just ignore those mistakes for now. $\endgroup$
    – Fred
    Feb 11, 2021 at 23:25
  • $\begingroup$ this is completly wrong. You can not gauge away electro magnetic vector potential. $\endgroup$
    – physshyp
    Feb 12, 2021 at 2:17
  • $\begingroup$ What you do here is only correct if there no magnetic field in the system, but it is wrong for general system. but if there is no magnetic field. then what you do is just changing the gauge. $A=(\phi,\vec{A})$ thus, $E=-\nabla\phi-\partial_t A$ thus, you don't change the physics $E$ is still same you can change $phi$ and $\vec{A}$ without changing $E$ $\endgroup$
    – physshyp
    Feb 12, 2021 at 2:28
  • $\begingroup$ You're right. I forgot to take into account that for all this to rigorously work, we ultimately need to be able to find an indefinite integral of $\vec{A}$ which means that there really cannot be a magnetic field. I overlooked it, since I only considered it as a line integral. Thank you for noting that. And yes, of course $E$ does not change. To change the fields was never my intention. $\endgroup$
    – Fred
    Feb 12, 2021 at 11:26

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