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In my quantum physics book (Quantum Mechanincs Second edition B.H. Bransden & C.J. Joachain) in a chapter of Galilean transformation and Schrödinger equation there is a couple of weird equations.

Let $\vec r, \vec u , \vec v \, \in \mathbb R^3$ and $t\in \mathbb R$ and $\Psi:\mathbb R^4 \to \mathbb R$. Which are related by the following relations: $$\vec r=\vec u+\vec vt \tag{1}$$ $$\frac{\partial}{\partial t}\Psi(\vec r,t)=\frac{\partial}{\partial t}\Psi(\vec u+\vec vt,t)-\vec v \cdot \nabla_{\vec u}\Psi(\vec u+\vec vt,t) \tag{2}$$ $$\nabla_{\vec r}\Psi(\vec r,t) = \nabla_{\vec u}\Psi(\vec u+\vec vt,t) \tag{3}$$

What I find very peculiar is the second term on the right hand side of (2). If I were to substitute $\vec r$ into the right hand side of (2) then for (2) to maintain its equality the following should be true $$\vec v\cdot \nabla_{\vec u}\Psi(\vec u + \vec v t,t)=\vec v\cdot \nabla_{\vec u}\Psi(\vec r,t)=\vec v \cdot \nabla_{\vec r}\Psi(\vec r,t)=0$$

The above could very well be true but then what is the purpose of this redundant-looking term. And if this should be on physics.stack instead please migrate.

Thanks in advance!

Quantum Mechanincs Second edition B.H. Bransden & C.J. Joachain page 255

ludz
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  • That last term is not zero. Bear in mind substituting an explicitly $t$-dependent formula for $\vec{r}$ into $\partial_t\Psi(\vec{r},,t)$ gives $\Psi(\vec{r},,t)$ an extra reason for having a partial time derivative. – J.G. May 19 '21 at 18:17

1 Answers1

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To expand on my comment, (2) is easier to explain if rearranged as$$\partial_t\Psi(\color{red}{u_j+v_jt},\,\color{blue}{t})=\color{red}{\sum_{j=1}^3v_j\partial_{u_j}\Psi(u_j+v_jt,\,t)}+\color{blue}{\partial_4\Psi(r_j,\,t)},$$where $j$ runs over the spatial dimensions. This is just a multivariable chain rule. The red parts are differentiation with respect to the first three arguments of $\Psi$; the blue parts are differentiation with respect to its final argument. In particular, I've called the blue derivative on the right-hand side $\color{blue}{\partial_4}$ to emphasize which argument of $\Psi$ provides it, but the black $\partial_t$ tallies up all sources of explicit time-dependence. On the right-hand side, the red $\color{red}{\partial_{u_j}}$ term tracks the source of time-dependence introduced by taking $\vec{r}=\vec{u}+\vec{v}t$, which is why $\partial_t\ne\color{blue}{\partial_4}$.

J.G.
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