Translation operator have the form \begin{equation} T_R = e^{i\frac{\hat{p}}{\hbar}\cdot R} \end{equation} and this can be easily proved, with $\hat{p}$ being the total momentum operator; but in mamy textbooks is stated that the generator for lattice translation is the Crystal momentum operator and that is not related in any way to the total momentum $\hat{p}$ (Ashcroft - Appendix M). How can the lattice translation operator have two form different form?

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    $\begingroup$ The translation symmetry of a crystal is a discrete symmetry. Momentum is the generator of continuous translation symmetry. Where did you get the above form for $T_R$? $\endgroup$
    – SRS
    Jul 13, 2020 at 15:44
  • $\begingroup$ If you apply $T_R=e^{i\frac{\textbf{p}}{\hbar}R}$ to a generic wave function you would obtain $\psi(r+R)$ but if you define $T_R=e^{i\frac{\textbf{K}}{\hbar}R}$, with $\textbf{K} \psi = \hbar k \psi$, and apply it to a generic wave function you also obtain $\psi(r+R)$ due to the Bloch Theorem; so i am a little confuse, which one Is the translation operator? My source physics.stackexchange.com/questions/536039/… and physics.stackexchange.com/questions/538516/… $\endgroup$ Jul 13, 2020 at 17:49

2 Answers 2


Strictly speaking, the crystal momentum operator is not a well-defined operator. This is because crystal momentum is only defined modulo reciprocal lattice vectors $\vec{G}$. It's better to think of the crystal momentum as a label for the eigenstates of the translation operator, which is indeed well-defined.

Usually, when people say a unitary transformation is "generated" by a Hermitian operator, what they have in mind is a continuous unitary transformation. But the lattice translations are a discrete group of unitary transformations, so it's not really appropriate to say they're "generated" by either momentum or crystal momentum.

Nevertheless, it's certainly true that the discrete translations are a subgroup of the continuous translations, and you can write them as $$ T_{\vec{R}} = e^{i \vec{p} \cdot \vec{R} / \hbar} $$ as you have done. Now you may ask: if you have an eigenstate of $T_{\vec{R}}$, is it an eigenstate of definite momentum, or of definite crystal momentum? Well, since momentum eigenstates with eigenvalues $\vec{p}$ and $\vec{p} + \hbar \vec{G}$ share the same eigenvalue for $T_{\vec{R}}$, you can superpose them to obtain a new eigenstate of $T_{\vec{R}}$. This is why people say that crystal momentum is not the same as real momentum: you can mix many different states with different momentum, so long as they all differ by reciprocal lattice vectors $\vec{G}$, and obtain a state with well-defined crystal momentum.

The final point I want to make: you say that crystal momentum and real momentum are in no way related. From the above discussion, this is clearly not true. You can think of the crystal momentum as an equivalence class of real momentum, where two momenta are related if they differ by a reciprocal lattice vector. The representatives of each equivalence class are those unique momenta belonging to the first Brillouin zone.

  • $\begingroup$ can you give any references to this answer ? $\endgroup$ Sep 18, 2023 at 7:21
  • $\begingroup$ To which part in particular? $\endgroup$
    – Zack
    Sep 18, 2023 at 14:19
  • $\begingroup$ To your explanation. $\endgroup$ Sep 21, 2023 at 13:39
  • $\begingroup$ Again, to which part of my explanation in particular? If you just want to know about crystal momentum in general, you can look at (for example) Ashcroft and Mermin. $\endgroup$
    – Zack
    Sep 22, 2023 at 20:49

Translation operator commutes with the energy, and with common eigenbasis, its eigenvalue is pure phase and doesn't incorporate the periodicity of the lattice.

To incorporate periodicity of lattice, we use Bloch wavefunction which has extra factor $u(x)$ multiplied to the translation operator and is periodic. But this wavefunction is not momentum eigenstate and therefore its eigenvalue is not the momentum, however it tells how the state transforms under translations.

"To see the mathematics regarding this, please have a look at Robert Eisberg book on Quantum Mechanics"

  • $\begingroup$ My question was about how it is possibile that $T_R=e^{i\frac{p}{\hbar}R}$ and at the same time $T_R=e^{i\frac{K}{\hbar}R}$ where K Is the Crystal momentum operator $\endgroup$ Jul 13, 2020 at 7:15

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