# Wave-function solution under Galilean transformations

1. Jan 5, 2016

### Happiness

How do we get (5.381)?

The term involving $V$ in (5.378) is $V(r' + vt, t)\ \Psi(r' + vt, t)$. After dividing on both sides of (5.378) by the exponential term $e^{[i(mv.r' + mv^2t/2)/\hbar]}$ [which appears in (5.379)], the term becomes $V(r' + vt, t)\ \Psi(r', t)$. But the term as given in (5.380) is $V'(r', t)\ \Psi(r', t)$. This means that we must have $V'(r', t) = V(r' + vt, t) = V(r, t)$, which contradicts (5.381).

I've worked out the other terms and they are correct.

2. Jan 5, 2016

### vanhees71

I think you are right. Under Galileian transformations one should have
$$V'(\vec{r}',t')=V(\vec{r},t)=V(\vec{r}'+\vec{v} t',t'),$$
because
$$\vec{r}=\vec{r}'+\vec{v} t', \quad \vec{p}=\vec{p}'+m \vec{v}, \quad t=t'.$$

3. Jan 5, 2016

### Happiness

Is there an explanation (intuitive or otherwise) how $$\vec{r}=\vec{r}'+\vec{v} t', \quad \vec{p}=\vec{p}'+m \vec{v}, \quad t=t'$$ implies or is consistent with $$V'(\vec{r}',t')=V(\vec{r},t)$$? I mean apart from doing the calculations after substituting (5.379) into (5.378).

4. Jan 6, 2016

### samalkhaiat

Because the potential is a scalar function of the coordinates and time, i.e., its value stay the same if you transform it. Suppose you have the following non-singular transformations
$$\bar{x} = f(x), \ \ \ \bar{t} = g(t) .$$
Since the transformations are non-singular, you can solve for $x$ and $t$:
$$x = f^{-1}(\bar{x}) , \ \ \ t = g^{-1}(\bar{t}) .$$
Now, substitute the solutions in the original potential function
$$V(x,t) = V(f^{-1}(\bar{x}) , g^{-1}(\bar{t}) ) ,$$
and call the new function on the right-hand-side $\bar{V}(\bar{x},\bar{t})$.