In Quantum Mechanics it is said that the Galileo transformation $$\hat{\mathbf{r}}\mapsto \hat{\mathbf{r}}-\mathbf{v}t\quad \text{and}\quad \hat{\mathbf{p}}\mapsto \hat{\mathbf{p}}-m\mathbf{v}\tag{1}$$ is given by the operator $$\hat{G}(\mathbf{v},t)=\exp\left[\dfrac{i}{\hbar}\mathbf{v}\cdot (m\hat{\mathbf{r}}-\hat{\mathbf{p}}t)\right].\tag{2}$$ Now I want to understand how does one show that this is the operator that implements the Galileo transformation.

I just can't understand, because for me, since we want $\hat{\mathbf{r}}\mapsto \hat{\mathbf{r}} -\mathbf{v}t$ it seems that the operator should just be a translation by $\mathbf{v}t$ which would be

$$\hat{G}(\mathbf{v},t)=\exp\left[\dfrac{i}{\hbar}\hat{\mathbf{p}}\cdot \mathbf{v}t\right]$$

but that's not it. There's also the $m\hat{\mathbf{r}}$ part which I don't understand where it comes from.

I've tried two things: first, defining $\tilde{\psi}(\mathbf{r},t)=\psi(\mathbf{r}+\mathbf{v}t,t)$ to be the transformed wavefunction. It also leads me to the translation only.

The second thing was to define


with ${\varepsilon}$ infinitesimal and impose the conditions

$$\hat{G}(\mathbf{v},t)^\dagger \hat{\mathbf{R}}(t)\hat{G}(\mathbf{v},t)=\hat{\mathbf{R}}(t)-\mathbf{v}t$$ $$\hat{G}(\mathbf{v},t)^\dagger \hat{\mathbf{P}}(t)\hat{G}(\mathbf{v},t)=\hat{\mathbf{P}}(t)-m\mathbf{v}$$

in terms of the infinitesimal operator this becomes

$$\dfrac{i}{\hbar}[\hat{\mathbf{R}}(t),\hat{\varepsilon}(\mathbf{v},t)]=\mathbf{v}t, \quad\dfrac{i}{\hbar}[\hat{\mathbf{P}}(t),\hat{\varepsilon}(\mathbf{v},t)]=m\mathbf{v}$$

but this doesn't leads very far.

So what is the reasoning behind the $\hat{G}$ usually presented being the operator that implements Galileo transformations?

  • 4
    $\begingroup$ Reference needed for 'it is said'. $\endgroup$
    – my2cts
    Commented May 10, 2021 at 22:46
  • $\begingroup$ A Galileo transformation is a coordinate transformation and therefore the same for all matter. The operator $\hat{G}(\mathbf{v},t)$ depends explicitly on the mass of the Schrödinger particle at hand. It cannot represent a Galileo transformation. $\endgroup$
    – my2cts
    Commented May 10, 2021 at 22:58
  • $\begingroup$ Related: physics.stackexchange.com/a/747493/226902 and this cool answer: physics.stackexchange.com/a/301058/226902 $\endgroup$
    – Quillo
    Commented Jan 27, 2023 at 23:50

2 Answers 2


Note that first of all, in a Galilean transformation, from $S$ to $S'$, a particle of constant momentum will have its energy be different in $S'$ than $S$ because kinetic energy is proportional to the velocity squared. We know that wavefunctions (in a time-independent potential) oscillate like $$ e^{-i\dfrac E\hbar t} $$ so we have to take this into account when we look in our different frames.

Let's apply these results. What we'll find that it's accounting for this change in momentum gives you the term that is confusing you, and the change in energy gives you the term that you suspect would be the answer.

Consider a particle in $S$ with momentum $\vec p$ and mass $m$. Its wavefunction $\psi$ is then

$$ \psi = e^{i\left(\dfrac{\vec p \cdot \vec x}{\hbar} - \dfrac{p^2}{2m\hbar}t\right)} $$

In $S'$, with velocity $\vec v_0$, it's momentum goes to $\vec p' = \vec p - m \vec v_0$, and likewise it's energy changes. $E' = \frac{(p')^2}{2m}$, so we find that

$$ E' = E - \vec p \cdot \vec v_0 + \frac 12 m v_0^2 $$

Since the wave function must have a similar form as in $S'$,

$$ \psi' = e^{i\left(\dfrac{\vec p' \cdot \vec x}{\hbar} - \dfrac{E'}{\hbar}t\right)} $$

Plugging in these new values of energies and momentum yields this form,

$$ \psi' = e^{i\left(\dfrac{\vec p \cdot \vec x}{\hbar} - \dfrac{m\vec v_0\cdot \vec x}{\hbar}-\dfrac{Et}{\hbar} + \dfrac{\vec p \cdot \vec v_0}{\hbar}\right)} e^{-i\dfrac {m v_0^2t}{2\hbar} } $$

we drop the very last term because it is invariant with respect to momentum. It's just a global overall phase (if there is one and only one type of mass). Rearranging leads us to find that,

$$ \psi' = e^{i\dfrac{\vec p \cdot \vec v_0 t}{\hbar}}e^{-i \dfrac{m\vec v_0\cdot \vec x}{\hbar}}\psi $$

This is exactly the transformation you claim is the Galilean transformation. Replacing $\vec p$ with an operator gives,

$$ \psi' = \hat G \psi $$

where $\hat G = e^{i\dfrac{\vec p \cdot \vec v_0 t}{\hbar}}e^{-i \dfrac{m\vec v_0\cdot \vec x}{\hbar}}$ Since the momentum states are a complete basis, this holds for any superposition of momentum states. So in general, it's true, and we've derived $\hat G$. It's because of energy and momentum change in different frames.

  • 1
    $\begingroup$ Sorry for commenting on an old post, but in your equation for $\psi'$, shouldn't you have $x'$ instead of $x$? That would change everything below too. $\endgroup$
    – GRrocks
    Commented Apr 10, 2019 at 11:09
  • $\begingroup$ I was going to post an answer to your Mathematica question you just posted, but you just deleted it. If you want the answer, you can undelete the question., $\endgroup$
    – Nasser
    Commented Jul 22, 2020 at 21:03

It's straightforward to check that OP's eq. (2) indeed generates Galilean transformations. Rather it seems OP is asking

How to derive formula (2) from first principles?

Sketched derivation of formula (2):

  1. Consider first the classical theory. The Hamiltonian Lagrangian for a free non-relativistic particle is $$\begin{align} L_H~=~&\sum_{k=1}^3p_k\dot{x}^k-H, \cr H~:=~ & \frac{1}{2m}\sum_{k=1}^3p_k p_k.\end{align}\tag{A}$$

  2. Show that an infinitesimal Galilean transformation $$\begin{align}\delta x^k~=~&t ~\delta v^k, \cr \delta p^k~=~&m ~\delta v^k, \cr \delta t~=~&0,\end{align}\tag{B}$$ is a quasi-symmetry $$ \delta L_H ~=~\frac{d}{dt}\sum_{k=1}^3m x_k~\delta v^k \tag{C} $$ for the Hamiltonian Lagrangian (A). [Concerning quasi-symmetry, the reader may also enjoy reading this related Phys.SE post.]

  3. Use Noether's theorem to find the corresponding full Noether charge $$ Q_k~=~ tp_k - m x_k.\tag{D}$$ [The first term $tp_k$ is the bare Noether charge, while the second term $m x_k$ comes from the rhs. of eq. (C).]

  4. As a check, note that the Noether charge (D) generates the infinitesimal Galilean transformation (B), $$ \delta ~=~\sum_{k=1}^3 \{~\cdot~ , Q_k\} ~\delta v^k ,\tag{E}$$ as it should, cf. my Phys.SE answer here.

  5. Use the correspondence principle between classical & quantum physics to deduce that $$ \delta ~=~ \sum_{k=1}^3\frac{1}{i\hbar}[~\cdot~ , \hat{Q}_k] ~\delta v^k, \tag{F}$$ where $$ \hat{Q}_k~=~ t\hat{p}_k - m \hat{x}_k.\tag{G}$$ is the Noether charge operator.

  6. Use standard arguments to integrate the infinitesimal Galilean transformation (F) into a finite Galilean transformation in order to achieve OP's sought-for formula (2). $\Box$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.