Consider a time-dependent problem. A particle is placed in a periodic potential with time-dependent amplitude

$$H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V_0(t)\cos^2(x)$$

Instantenous eigenstates are labelled by quasi-momentum $k$ nad band index $n$ - Bloch states $\phi_{n,k}(x,t)$. I wonder if there are some consequences of the discrete translational symmetry of lattice on the time-evolution? Does the initial state $\phi_{n,k}(x,t=0)$ evolves into a superposition of different Bloch states $\phi_{m,k}(x,t)$ but with the same quasi-momentum $k$ as the inital state?


I checked some literature and found only one reference with time-dependent potential, but I think it does not answer my question completely.

Consider the translation operator $\hat{T}_{a_j}$ which has the property $$\hat{T}_{a} \psi(x,t) = \psi(x - a,t).$$ This operator commutes with the Hamiltonian due to periodicity of the potential. We have $[H(x,t), \hat{T}_{a}] = 0$. Consider what happens to the initial state of the Bloch form $$\phi_0(x) = e^{i q x} u_0(x)$$ where $u_0(x)$ is a periodic function with the same periodicity as the lattice itself. It can be proved that it evolves to a state of the form $$\psi(x,t) = e^{iqx}u(x,t)$$ where $u(x,t)$ has the same periodicity as $u_0(x)$. This is what the author calls: 'conservation of crystal momentum'. For me it does not mean that if I start with the Bloch eigenstate $\phi_{q}(x) = e^{iqx} u_q(x)$ I will end up with Bloch eigenstate at some other moment of time $\phi_q(x,t) = e^{iqx} u_{q}(x,t)$ (I ommited inter-band scattering). This would indicate a perfect adiabatic evolution no matter how fast the evolution is. I suppose the overlap between different Bloch states $$\int dx \phi^{*}_{q}(x,t)\frac{\partial }{\partial t}\phi_{k}(x,t)$$ is non-zero so perfect adiabatic evolution is not possible.


In the following review(p. 21) the author mentions what happens with the single atom when the potential amplitude is changed in time.

Importantly, the time dependence enters only in a scale factor for the potential, which remains periodic in space with the same period at all times (the lattice translation operator commutes with H(t) at all times). Therefore Bloch’s theorem applies, and the Bloch wave functions are the eigenstates of the Hamiltonian at all times (including t = 0 when there is no lattice). This also means that the quasi-momentum is conserved. The band index, however, can change.

As I wrote in the first update, the only consequence of translational invariance is that at the end of the ramp time we end up with the state of the Bloch's form $$\psi(x,t) = e^{iqx} u(x,t)$$ which is not necessarily the eigenstate of the Hamiltonian (translational invariance alone does not impose that). It can be decomposed in the Bloch states basis $$e^{iqx}u(x,t) = \sum\limits_{n,k}C_{n,k}(t) e^{ikx}u_{n,k}(x,t)$$ where $$C_{n,k}(t) = \int dx\ e^{ix(q-k)}u_{n,k}^{*}(x,t) u(x,t).$$ This analysis does not imply that all $C_{n,k = q}$ are 0. In contrary, this is what the paper says.

  • $\begingroup$ The case where the time-dependence of the Hamiltonian is periodic has been extensive studied - it's called Floquet theory. $\endgroup$
    – tparker
    Apr 11, 2017 at 4:08
  • $\begingroup$ @tparker Thanks for answer, but my Hamiltonian is not periodic in time, so Floquet's theorem does not apply. In fact, for space periodic problems Floquet's theorem is called the Bloch's theorem. $\endgroup$
    – WoofDoggy
    Apr 12, 2017 at 10:26

1 Answer 1


Because the time-dependent Hamiltonian $\hat H(t)$ commutes with $\hat T_a$, we find that $\hat T_a$ commutes with the evolution operator $\hat U(t)$ ($\hat U(t)$ solves $i\partial_t \hat U(t)=\hat H(t) \hat U(t)$).

Now, if the initial state $|\psi(0)\rangle$ is an eigenstate of $\hat T_a$, with eigenvalue $e^{-i q a}$ (i.e., it has quasi-momentum $q$), one readily finds that $|\psi(t)\rangle=\hat U(t)|\psi(0)\rangle$ is also an eigenstate of $\hat T_a$ with the same eigenvalue, that is, quasi-momentum is conserved during time-evolution.

Update : The discussion above only insures that the quasi-momentum stays the same during the evolution. However, nothing prevents the system to have interband transitions, in particular if the driving is fast. In particular, even if the system is prepared in, say, the lowest band of $\hat H(0)$, in general, $$|\psi(t)\rangle=\sum_n c_n(t) |\phi_{q,n}(t)\rangle,$$ where $|\phi_{q,n}(t)\rangle$ are the instantaneous eigenstate of $\hat H(t)$ with quasimomentum $q$.

  • $\begingroup$ I understand that if the translation operator commutes at every moment of time with the Hamiltonian and the spectrum is non-degenerate then they have the same set of eigenstates. If the state $| \psi(t)\rangle$ is an eigenstate of $\hat{T}_a$, then it is also an eigenstate of $H(t)$ (up to a global phase). Am I right? $\endgroup$
    – WoofDoggy
    Apr 6, 2017 at 9:56
  • $\begingroup$ No, not necessarily. You can have interband transition if the driving is fast enough. Only quasi-momentum is conserved for sure. $\endgroup$
    – Adam
    Apr 6, 2017 at 10:02
  • $\begingroup$ If the spectrum of translation operator is degenerate due to band index then maybe in general I can say that if initially the quasi-momentum was $q$ then at the end I will end up in a superposition of Bloch states from all the bands, but with the same quasi-momentum $q$? $\endgroup$
    – WoofDoggy
    Apr 6, 2017 at 10:12
  • $\begingroup$ You do not need to have any degeneracy to have interband transition. See update above. $\endgroup$
    – Adam
    Apr 6, 2017 at 10:30
  • 1
    $\begingroup$ Yes indeed. Remembering that $|\phi_{q,n}(t)\rangle$ are instantaneous eigenstates of $\hat H(t)$, and not truly evolving state,you can show is that $\langle \phi_{k,m}|i\partial_t|\phi_{q,n}(t)\rangle$ is proportional to the matrix element $\langle \phi_{k,m}|(i\partial_t\hat H(t))|\phi_{q,n}(t)\rangle$. Using the fact that $\partial_t\hat H(t))$ commutes with $\hat T_a$, one shows this property. However, this tells us nothing about the case $k=q$ and $m\neq n$. $\endgroup$
    – Adam
    Apr 6, 2017 at 14:48

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