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I have two questions regarding this topic.

1.

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I captured the part of the section I'm referring to. If I didn't my question would probably not make sense. My first question is to the second to last sentence above:

"Notice also that the center-of-mass position vector $R$ does not appear in the Hamiltonian at all, which, classically, is a reflection of the fact that the momentum of the center of mass is conserved because there are no external forces."

Could someone elaborate as to why having no external forces allows the center of mass to be conserved?

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"...the center-of-mass wave function contributes only an overall phase to the system wave function and so has no effect on calculating probabilities of relative motion quantities."

What does he mean by an overall phase? How do I see this mathematically, that the center-of-mass wave function is an overall phase? What does it even mean, that it's "an overall phase"?

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Answer to your first question:

Centre of mass (a position vector $\mathbf{R}$) is given by \begin{align} \mathbf{R} \;= \; \frac{1}{M} \sum_{i} m_{i} \mathbf{r}_{i} \end{align} where $m_{i}$ is the mass of the $i$th point particle, and the corresponding position is denoted by $\mathbf{r}_{i}$. $M$ is the total mass of the system.

Now, we can define the momentum of the centre of mass as \begin{align} \mathbf{P} \; = M \dot{\mathbf{R}} \; = \; \sum_{i} m_{i} \dot{\mathbf{r}}_{i} \end{align} where the overdot is time derivative.

Taking another time derivative, we obtain the Newton's Second Law for the centre of mass \begin{align} \dot{\mathbf{P}} \; = \; M\ddot{\mathbf{R}} \; = \; \sum_{i}m_{i}\ddot{\mathbf{r}}_{i} \; = \; \sum_{i} \mathbf{F_{i}} \; = \; \mathbf{F}_{\mathrm{sys}} \end{align} where $\mathbf{F}_{i}$ is the net force acting on the $i$th point particle, while $\mathbf{F}_{\mathrm{sys}}$ is the net force acting on the whole system.

If there is no external force acting on the whole system, $\mathbf{F}_{\mathrm{sys}} = \mathbf{0}$, then we have \begin{align} \dot{\mathbf{P}} \; &= \; \mathbf{0}\\ \implies \mathbf{P} \; &= \; \mathbf{P}_{0} \end{align} where $\mathbf{P}_{0}$ is a time independent constant vector. Thus, the momentum of the centre of mass is conserved.

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  • $\begingroup$ Great. Thank you. $\endgroup$ – mikanim May 28 at 6:30

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