# Show Galilean covariance of Schrödinger eq

I'm trying to show the Galilean covariance of the (time-dependent) Schrödinger equation by transforming as follows:

$$\left\{\begin{eqnarray}\psi(\vec{r},t) &=& \psi(\vec{r}'-\vec{v}t,t),\\ \frac{\partial\psi(\vec{r},t)}{\partial t} & = & \frac{\partial\psi(\vec{r}'-\vec{v}t,t)}{\partial t}-\vec{v}\cdot\vec{\nabla}_{r'}\psi(\vec{r}'-\vec{v}t,t)\\ \vec{\nabla}_{r}\psi(\vec{r},t)&=&\nabla_{r'}\psi(\vec{r}'-\vec{v}t,t)\end{eqnarray}\right.$$ Which gives: $$i\hbar\left[\frac{\partial}{\partial t}-\vec{v}\cdot\vec{\nabla}_{r'}\right]\psi(\vec{r}'-\vec{v}t,t)=\left[-\frac{\hbar^2}{2m}\nabla_{r'}^2+V(\vec{r}'-\vec{v}t,t)\right]\psi(\vec{r}'-\vec{v}t,t)$$

Now, supposedly, using the unitary transformation: $$\psi(\vec{r}'-\vec{v}t,t)\to \psi(\vec{r}',t)\exp\left[\frac{i}{\hbar}m\vec{v}\cdot\vec{r}'+\frac{i}{\hbar}\frac{mv^2}{2}t\right],$$

it should be possible to recover the original form of the Schrödinger equation if there is no potential. When trying this, however, I can't seem to make it work. Is it just algebra or am I missing a crucial step here?

My lecturer also hinted that there was a physical reason that there may be no potential for Galilean invariance to hold. What is this reason and how can it intuitively be seen?

• This approach is known as the Schrödinger group and is a low light of physics. The coordinate transformation depends on mass via a phase term. This overstretches the concept of a coordinate system. – my2cts Aug 22 '19 at 10:22
• A comment related to the potential part. There is no potential invariant to a dynamical subgroup (i.e. containing Lorentzian or Galileian boosts) of the full dynamical group. That would make the potential velocity dependent, hence a dissipative system. Quantum theories are constructed around conservative systems. – DanielC Aug 22 '19 at 10:37
• Possible duplicates: physics.stackexchange.com/q/56024/2451 and links therein. – Qmechanic Aug 22 '19 at 10:39

## 1 Answer

I'm not convinced that your transformations are correct. This is one place where it is important, if $$x=x'-vt$$, to distinguish between $$\left(\frac{\partial \psi}{\partial t}\right)_x$$ and $$\left(\frac{\partial \psi}{\partial t}\right)_{x'}.$$ They are not the same thing. You notation is uclear as to what you are keeping fixed in your partials. I find it it easier to distinguish the derivatives by writing
$$\xi=x-vt,\\ \tau=t.$$ Then the chain rule gives $$\frac{\partial}{\partial t}= \frac{\partial}{\partial \tau}-v \frac{\partial}{\partial \xi}\\ \frac{\partial}{\partial x}= \frac{\partial}{\partial \xi}$$ where the derivative wrt $$t$$ keeps $$x$$ fixed, and the derivative wrt $$\tau$$ keeps $$\xi$$ fixed. Similarly the derivative wrt $$x$$ is at fixed $$t$$ and the derivative wrt $$\xi$$ is at fixed $$\tau$$. The equation in the frame in which the potential moves past you $$i\hbar \frac{\partial \psi}{\partial t}=- \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}+V(x-vt)\psi$$ becomes, in the frame in which it is stationary $$i\hbar\left( \frac{\partial }{\partial \tau}-v \frac{\partial }{\partial \xi}\right)\psi=- \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial \xi^2}+V(\xi)\psi.$$ You can now absorb the extra $$\xi$$ derivative into the phase transformation that you mention to find a wavefunction $$\tilde \psi(\xi,\tau)$$ which obeys $$i\hbar \frac{\partial\tilde \psi }{\partial \tau}\psi=- \frac{\hbar^2}{2m} \frac{\partial^2 \tilde \psi}{\partial \xi^2}+V(\xi)\tilde \psi,$$ in which the only change in the Schroedinger equation is in the time dependence of the potential.

I think I have a minus sign in the convective time derivative compared to your transformed equation.

This transformation is discussed in Baym's quantum mechanic textbook.