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Wave-function solution under Galilean transformations

  1. Jan 5, 2016 #1
    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.

    Screen Shot 2016-01-06 at 3.18.03 am.png
    Screen Shot 2016-01-06 at 3.18.17 am.png
     
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  3. Jan 5, 2016 #2

    vanhees71

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    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
    $$\vec{r}=\vec{r}'+\vec{v} t', \quad \vec{p}=\vec{p}'+m \vec{v}, \quad t=t'.$$
     
  4. Jan 5, 2016 #3
    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).
     
  5. Jan 6, 2016 #4

    samalkhaiat

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    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
    [tex]\bar{x} = f(x), \ \ \ \bar{t} = g(t) .[/tex]
    Since the transformations are non-singular, you can solve for [itex]x[/itex] and [itex]t[/itex]:
    [tex]x = f^{-1}(\bar{x}) , \ \ \ t = g^{-1}(\bar{t}) .[/tex]
    Now, substitute the solutions in the original potential function
    [tex]V(x,t) = V(f^{-1}(\bar{x}) , g^{-1}(\bar{t}) ) ,[/tex]
    and call the new function on the right-hand-side [itex]\bar{V}(\bar{x},\bar{t})[/itex].
     
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