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I am only asking this question so that I can write an answer myself with the content found here:

and here:

I learned about these pages in a comment to this answer by Ron Maimon. I think Ron Maimon is the original writer of this content.

This is creative commons, so it's ok to copy it here. It is not on any textbook on non-relativistic quantum mechanics that I know of, and I thought it would be more accessible (if to no one else, at least for myself) and safe here. I hope this type of question is not in disagreement with site policy.

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Really? Not in any textbooks? This is bad! I remember from my undergrad years my third course on quantum mechanics included a broad treatment from this perspective, a personal favourite. It also dispelled myths I was told before, like the spin being an exclusively relativistic property. I understand now why the professor is writing a book with this perspective. – G. Bergeron Aug 15 '15 at 11:16

Free Schrödinger equation

Galilean boosts are transformations which look at the system from the point of view of an observer moving with a steady velocity $-v$. A boost must change the physical properties of a wavepacket in the same way as in classical mechanics:

$$ p' = p + mv $$

$$ x' = x + vt $$

so that the phase factor of a free Schrödinger plane wave:

$$ px - Et = (p' - mv)(x' - vt) - \frac{(p' - mv)^2}{2m} t = p'x' + E't + mvx - \frac{mv^2}{2} t $$

is only different in the boosted coordinates by a phase which depends on $x$ and $t$, but not on $p$.

An arbitrary superposition of plane wave solutions with different values of $p$ is the same superposition of boosted plane waves, up to an overall $x,t$ dependent phase factor. So any solution to the free Schrödinger equation $\psi_t(x)$ can be boosted into other solutions:

$$ \psi'_t(x) = \psi_t(x + vt) e^{imvx - i \frac{mv^2}{2}t} $$

Boosting a constant wavefunction produces a plane-wave. More generally, boosting a plane-wave:

$$ \psi_t(x) = e^{ipx - i \frac{p^2}{2m}t} $$

produces a boosted wave:

$$ \psi'_t(x) = e^{ip(x+vt) - i\frac{p^2}{2m}t + imvx - i\frac{mv^2}{2}t} = e^{i(p+mv)x + i\frac{(p+mv)^2}{2m}t} $$

Bossting the spreading Gaussian wavepacket:

$$ \psi_t(x) = \frac{1}{\sqrt{a + it/m}} e^{-\frac{x^2}{2a}} $$

produces the moving Gaussian:

$$ \psi'_t(x) = \frac{1}{\sqrt{a + it/m}} e^{-\frac{(x+vt)^2}{2a} + imvxx - i\frac{mv^2}{2}t} $$

which spreads in the same way.

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up vote 7 down vote accepted

Operator formalism

Galilean symmetry requires that $H(p)$ is quadratic in $p$ in both the classical and quantum Hamiltonian formalism. In order for Galilean boosts to produce a $p$-independent phase factor, $px - Ht$ must have a very special form -translations in $p$ need to be compensated by a shift in $H$. This is only true when $H$ is quadratic.

The infinetesimal generator of boosts in both the classical and quantum case is

$$ B = \sum_i m_i x_i(t) - t \sum_i p_i $$

where the sum is over the different particles, and $B$, $x$, $p$ are vectors.

The poisson bracket / commutator of $B\cdot v$ with $x$ and $p$ generate infinitesimal boosts, with $v$ the infitesimal boost velocity vector:

$$ [B \cdot v, x_i] = vt $$

$$ [B \cdot v, p_i] = v m_i $$

Iterating these relations is simple, since they add a constant amount at each step. By iterating, the $dv$s incrementally sum up to the finite quantity $V$:

$$ x \rightarrow x_i + Vt $$ $$ p \rightarrow p_i + m_i V $$

$B$ divided by the total mass is the current center of mass position minus the time times the center of mass velocity:

$$ B = M X_\text{cm} - t P_\text{cm} $$

In other words, $B/M$ is the current guess for the position that the center of mass had at time zero.

The statement that $B$ doesn't change with time is the center of mass theorem. For a Galilean invariant system, the center of mass moves with a constant velocity, and the total kinetic energy is the sum of the center of mass kinetic energy and the kinetic energy measured relative to the center of mass.

Since $B$ is explicitly time dependent, $H$ does not commute with $B$, rather:

$$ \frac{dB}{dt} = [H,B] + \frac{\partial B}{\partial t} = 0 $$

This gives the transformation law for $H$ under infinitesimal boosts:

$$ [B \cdot v, H] = - P_\text{cm} v $$

The interpretation of this formula is that the change in $H$ under an infitesimal boost is entirely given by the change of the center of mass kinetic energy, which is the dot product of the total momentum with the infitesimal boost velocity.

The two quantities $(H,P)$ form a representation of the Galilean group with central charge $M$, where only $H$ and $P$ are classical functions on phase-space or quantum mechanical operators, while $M$ is a parameter. The transformation law for infinitesimal $v$:

$$ P' = P + Mv $$ $$ H' = H - P\dot{v} $$

can be iterated as before -$P$ goes from $P$ to $P + MV$ in infinitesimal increments of $v$, while $H$ changes at each step by an amount proportional to $P$, which changes linearly. The final value of $H$ is then changed by the value of $P$ halfway between the starting value and the ending value:

$$ H' = H - (P + \frac{MV}{2}) \cdot V = H - P \cdot V - \frac{MV^2}{2} $$

The factors proportional to the central charge $M$ are the extra wavefunction phases.

Boosts give too much information in the single-particle case, since Galilean symmetry completely determines the motion of a single particle. Given a multiparticle time dependent solution:

$$ \psi_t(x_1, x_2, ..., x_n) $$

with a potential that depends only on the relative positions of the particles, it can be used to generate the boosted solution:

$$ \psi'_t = \psi_t(x_1 + vt,...,x_n + vt) e^{i P_\text{cm} \cdot X_\text{cm} - \frac{M v_\text{cm}^2}{2}t} $$

For the standing wave problem, the motion of the center of mass just adds an overall phase. When solving for the energy levels of multiparticle systems, Galilean invariance allows the center of mass motion to be ignored.

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