I was going through Le Bellac's Quantum Physics book.In the "symmetry" chapter 1st page(Classical physics), he makes the following comments a part of which look a bit weired to me...Each statement starts with "Invariance of the potential energy".Do you think this is meaningful? *Invariance of the potential energy under time-translations implies conservation of mechanical energy E = K + V , the sum of the kinetic energy K and the potential energy V . *Invariance of the potential energy under spatial translations parallel to a vector n implies conservation of the momentum component p_n . *Invariance of the potential energy under rotations about an axis n implies conservation of the component j_n of the angular momentum.
Meaningful yes, but also simplified. I presume this is a quantum mechanics book aimed at undergraduate students taking a first course in QM? If that's the case you might not have come across Lagrangian mechanics yet, which is a very useful formulation when you talk about quantum field theory, and symmetry is crucial to QFT. A more accurate way of saying your three sentences would be: Noether's theorem implies that the invariance of the Lagrangian (basically the kinetic energy minus the potential energy) under ... translations implies conservation of ...
In the most likely scenario in classical mechanics, you're dealing with two-body interaction potentials of the form \(V(\bar{r}_{i},\, \bar{r}_{j})\) where \(\bar{r}_{i}\) and \(\bar{r}_{j}\) are the position vectors of your two interating particles. If the potential only depends on the position difference between the two particles: \(V(\bar{r}_{i},\, \bar{r}_{j}) \,=\, V(\bar{r}_{i} \,-\, \bar{r}_{j})\)then it'll remain invariant under a given translation \(\bar{r}_{i} \,\rightarrow\, \bar{r}_{i} \,+\, \bar{a}\) of all particle positions. If it only depends on the norm of this position difference: \(V(\bar{r}_{i},\, \bar{r}_{j}) \,=\, V\bigl(|\bar{r}_{i} \,-\, \bar{r}_{j}|\bigr)\)then it'll also be invariant under rotations of all the position vectors. And of course as long as the potential term doesn't depend explicitly on \(t\) it is invariant under a translation of the time variable. You can characterise an entire theory by specifying it's kinetic and potential energy terms. The reason they're only discussing the symmetries of the potential is that the classical kinetic energy terms \(T \,=\, \frac{1}{2} m v_{i}^{2} \,=\, \frac{p_{i}^{2}}{2m}\)already possess all these symmetries. As prometheus said, the author is stating special case results of Noether's theorem, which associates a conserved current with every global symmetry in a theory. It applies as long as the theory can be derived from an action principle, which is pretty much the standard way of formulating modern theories nowadays (so in practice it's not a significant restriction).
Yea...the thing is other way around,actually.There is no need to invoke Noether's theorem and its conserved current.The author puts like this: conservation of the Lagrangian [\(\delta\ L=0\)]under space translation,time translation and space rotation gives rise to conservation of linear momentum,energy and the angular momentum. However,whether \(\delta\ L=0\) or not is decided by the potential function V(r1-r2)...Hence,if the potential energy is conserved,we must have \(\delta\ L=0\) w.r.t. appropriate variables and the corresponding quantities are conserved.
Yes, but this can be interpreted as proving a special case of Noether's theorem by a simpler method without going through the general argument. Actually if you look at Hamilton's equations the momentum conjugate to a cyclic variable is conserved. A cyclic variable is a variable that the Lagrangian (so the Hamiltonian) does not explicitly depend on.
