# Is momentum conservation for the classical Schrödinger equation due to non-relativistic or due to some more exotic invariance?

I had no problem appliying the Neothers theorem for translations to the non-relativistic Schrödinger equation

$\mathrm i\hbar\frac{\partial}{\partial t}\psi(\mathbf{r},t) \;=\; \left(- \frac{\hbar^2}{2m}\Delta + V(\mathbf{r},t)\right)\psi(\mathbf{r},t)$

$\Longrightarrow\ \mathcal{L}\left(\psi, \mathbf{\nabla}\psi, \dot{\psi}\right) = \mathrm i\hbar\, \frac{1}{2} (\psi^{*}\dot{\psi}-\dot{\psi^{*}}\psi) - \frac{\hbar^2}{2m} \mathbf{\nabla}\psi^{*} \mathbf{\nabla}\psi - V( \mathbf{r},t)\,\psi^{*}\psi$

$\Longrightarrow\ \pi=\frac{\partial \mathcal{L}}{\partial \dot{\psi}} \propto \psi^{*}$

$T[\psi]\propto \mathbf{\nabla} \psi$

$\Longrightarrow\ I_{\ \psi,{\ T_\text{(translation)}}}=\int\text d^3x\ \pi\ T[\psi]\propto \int\text d^3x\ \psi^{*} \mathbf{\nabla} \psi = \langle P \rangle_\psi$

But I actually wonder why that works out, given that the Schrödigner equation is not invariant under Galileian transformations.

It might well be that the Schrödinger group, which I'm not familiar with, is close enough to the Galileian group, that the fourth line $T[\psi]\propto \mathbf{\nabla} \psi$ is just the same and that's the reason. I'd like to know if the evaluation of the infinitesimal transformation is the only point at which one has to know the transformations one is actually dealing with. Is my guess right?

Also, regrding the "trick" to establish Galilei-invariance after the conventional transformation via multiplication of the Schrödinger field by a phase (a phase which, among other things, is mass dependend):

Some authors change $\psi(r,t)$ to $\psi(r',t')=\psi(r-vt,t)$, like here in the paper referenced on wikipedia (there is also a two year old version of it online (google)), but other authors, like the writers of the page in the first link, also transform $p$ to $p+mv$ in $\phi$ (which doesn't change the fact that they still have to add a phase). This is all before the phase multiplication. So what is the "right way" here? If I do this transformations involving a multiplication of the phase, do I only transform the actual arguments of the scalar field $\psi(r,t)$ or do I also transform the objects like $p$, which classically transform too, but are really just parameters (and the Eigenvalues) or the field - and not arguments?

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The Schrodinger equation changes form under Galilean transformations, but it is invariant in a quantum sense under these, since you cancel out the change with a phase factor. I wonder why you are confused, because translations and Galilean transformations are both mathematically and logically independent--- you can make a translation symmetry ignoring galilean symmetry, like in a crystal, where you have discrete translations and no boosts, or in He4, where you have continuous translation symmerty but again no boosts. – Ron Maimon Aug 10 '12 at 19:26
@RonMaimon: You're right, I just did the computation for the translations (because that's easy) and here I was just assuming there is some conserved quantity for boosts as well. Is that not the case? And furthermore, are there interesting conserved quantities via Noether due to the new symmetry group (the Schrödinger symmetries)? – NikolajK Aug 11 '12 at 0:32
Yes, there are further non-obvious conserved quantities, the location of the center of mass. This shows up as phase relations in scattering, and in separation theorems, like the reduced-mass/total-mass decomposition for the two-body problem. The center of mass law is independent of the conservation of momentum, although this is counterintuitive. Is this your question? I will answer this way, but it's not clear from what you ask. – Ron Maimon Aug 11 '12 at 3:57

The conserved quantity corresponding to translation is the generator of translations. This is P, and you can see this because $e^{iPa}$ acting on a state $|x\rangle$ produces $|x+a\rangle$.

By P-X symmetry, the operator $X$ generates translations in $P$, so that $e^{iXa}$ takes $|p\rangle$ to $|p-a\rangle$ (the minus sign is dictated by the orientation of the phase space, but you can also explicitly see it from the usual form of the X,P operators). So the naive generator of boosts is

$$mvX$$

Because this shifts the momentum by $mv$. But this is nonsense, because it doesn't commute with H! So it is not a symmetry. But the reason is because you need a time-dependent phase factor to fix the phase space. Once you do this, the correct conserved quantity B is

$$vB = v(mX - Pt)$$

Which shifts the momentum eigenstates by $mv$ and multiplies by an additional phase. The quantity $mX - Pt$ is the additional conservation law for boost invariance, and it is the location of the center of mass. For several particles, the generator of boosts is:

$${\sum_i m_i X_i - Pt}$$

which shifts each of the momenta by $m_i v$, and corrects by a total phase.

The Hamiltonian

$${p^2\over 2} + {p^4\over 4} + V(x)$$

Is an example of an H that is not Boost invariant but is translation invariant. Motion in this H doesn't conserve center of mass, but conserves momentum. Another example is a crystal, where the p-dependence goes like $1-\cos(p)$, so again, you have translation invariance (discrete translation invariance--- p is periodic), but no boost invariance. In the crystal case, boost invariance is an accidental symmetry at low p.

To see how boosts work in the Lagrangian picture, look here: Galilean invariance of classical lagrangian .

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