# Galilei Algebra and QM

1. Jul 10, 2015

### andresB

I have a problem with the derivation of the the form of QM starting from the Lie algebra of the Galilei Group like the one given in Ballentine's cap 3.

My Issue is that the procedure is shown almost as unavoidable, And my feeling is that there have to be more postulates that I'm not seeing because they are not stated explicitly.

For example, the operator of translation is identified with the Momentum, and in that way they show the quantum commutation relations. I basically understand the Math of the generators of the Galilei group, But that procedure can't be unique because there is another theory in which the momentum operator is not identified with the translation operator and commute with the position operator, namely classical mechanics

http://arxiv.org/abs/1105.4014

So, what is really happening?

2. Jul 10, 2015

### DrDu

Momentom is always the generator of translations due to Noether's theorem.

3. Jul 10, 2015

### andresB

Not in the same sense. In Hamiltonian mechanics the momentum is the generator of translation via the Poisson brackets and canonical transformation, but when you see that bracket as an operator the real generator of translation is d/dx as in QM.

The issue is that in the operational approach to CM d/dx is not the momentum operator.

4. Jul 10, 2015

### Avodyne

I don't have Ballentine, but you cannot "derive" quantum mechanics from anything. Quantum mechanics is a postulated set of physical laws that (so far) agrees with experiment. Classical mechanics is a postulated set of physical laws that agrees with experiment in everyday, macroscopic settings, but fails to account for the observed behavior of very small things like atoms. That is why quantum mechanics had to be invented in the first place.

5. Jul 10, 2015

### samalkhaiat

Well, you have answered your own question. The Lie bracket in QM is realized by commutator, while in CM the Lie bracket is realized by Poisson bracket which, as one can show, is the classical limit of the commutator.

6. Jul 10, 2015

### andresB

That is somewhat my point, QM should not be derivable from the structure of the Galilei Lie algebra, there have to be some other assumptions in the process, and I'm failing to understand them.

7. Jul 10, 2015

### andresB

In hamiltonian Mechanics, not in the Hilbert space formulation of CM, see for example equation (8) and the equations just before (11) in the article I quoted (I think is clear there that the classical momentum Operator is different from the translation operator).

8. Jul 10, 2015

### aleazk

I'm not sure what you are trying to say here. I will sketch the standard argument (which can be found in, e.g., Varadarajan's "Geometry of Quantum Theory") so you can point out which are the points you find problematic.

In classical mechanics, linear momentum is indeed the generator of translations. Let $M$ be a Riemannian manifold representing space. Consider now the cotangent bundle of $M$, $T^{*}[M]$, as phase space. Let $\varphi_{t}$ be a one-parameter group of diffeomorphisms in $M$ and $X^{\varphi}$ its tangent field. We can associate an observable to it (called "momentum associated to this particular one-parameter group"), $h_{\varphi}:T^{*}[M]\longrightarrow\mathbb{R}$, via the following definition: $h_{\varphi}(P)\doteq v_{x}(X^{\varphi}\mid_{x})$, for all $P\in T^{*}[M]$ and remembering that points of phase space are actually cotangent vectors in $M$ and so we use that $P\equiv v_{x}$ with $v_{x}\in T_{x}^{*}$ for some $x\in M$.

In this sense, given the action of a symmetry, we define an observable associated to it and call it momentum. In the particular case of the action in $M$ of the full translation group, we can form the corresponding one-parameter groups. The associated observables are called "linear momentum in the $j$-direction". If we take the action in $M$ of the rotation group $SO(3)$, we can form the corresponding one-parameter groups and the associated observables are called "angular momentum with respect to a rotation in the $i-j$ plane". And so on. The thing that determines the type of momentum (e.g., linear, angular) is the underlying group. Doing the calculation on some canonical chart, one obtains the familiar formulas for these observables.

If we take the Poisson bracket defined by the natural symplectic form in $T^{*}[M]$, i.e., $\Omega\doteq\mathrm{d}\tau$ where $\tau$ is the tautological 1-form, we get: $h_{aX+bY}=ah_{X}+bh_{Y}$ and $h_{\left[X,Y\right]}=\left\{ h_{X},h_{Y}\right\}$. Thus, we obtain a realization of the Lie algebra of the Lie group $G$ acting in $M$ (since the tangent field to the one-parameter groups is a representative of an element of this algebra via $(Xf)(x)\doteq\left[\frac{\mathrm{d}}{\mathbf{d}t}f\left(\left(exp\, tX\right)\cdot x\right)\right]\mid_{t=0}$) in terms of the previously defined momentum observables associated to the action.

This basic structure is completely analogous in QM. The action of the symmetry group is implemented in terms of, e.g., Wigner symmetries. If $\pi$ is a unitary representation of the Lie group $G$, we take the corresponding one-parameter unitary groups and (using Stone's theorem) define the associated quantum momentum operators via $\mathrm{d}\pi(X)(\varphi)=i\frac{\mathrm{d}}{\mathrm{d}t}\left[\pi\left(exp\left(tX\right)\right)\left(\varphi\right)\right]\mid_{t=0}$. Again, we get a Lie algebra representation: $i\mathrm{d}\pi(\left[X,Y\right])=\left[\mathrm{d}\pi(X),\mathrm{d}\pi(Y)\right]$. So, again, the thing that determines the type of momentum is the underlying group

If the Galilei group acts as symmetries in the quantum system, we obtain unitary projective representations $U_g$ of this group. Using group extension techniques, one can get true unitary representations of the extension and therefore to use some theorems to classify all the possible representations. Done this, one finds that all representations are equivalent to certain special representations which have the familiar form $\left[U_{h,\overrightarrow{u}}f\right](\overrightarrow{x})=s(h)f\left(\delta^{-1}(h)\left(\overrightarrow{x}-\overrightarrow{u}\right)\right)$ in $L^{2}(\mathbb{C}^{s};\mathbb{R}^{3})$
for the part corresponding to the Euclidean subgroup. The linear momentum is as usual defined as the infinitesimal generator of the translation group part. So, we get the usual $P_{i}\propto\frac{\partial}{ \partial x_{i} }$ in this realization.

Last edited: Jul 10, 2015
9. Jul 10, 2015

### Staff: Mentor

But you can use different starting points.

The starting point in Ballentine is that the probabilities from Born's rule is invariant to Galilean transformations from which one obtains exactly the same form for the equations as classical mechanics ie the energy operator has the classical form of the Hamiltonian.

The key physical assumption is, because it has exactly the same form, that when one has a classical Hamiltonian it translates over. That is if you take the expectation values you get the classical equation so it is a reasonable thing to do - so reasonable and natural unless you actually think about it you possibly don't even realise its an assumption. But there is a well known ambiguity in doing that, that even more advanced treatments like Aleazk mentioned doesn't fix:

Its to do with operators not commuting so you sometimes have an order ambiguity in deciding on the correct quantum Hamiltonian.

The treatment in Ballentine is mathematically very beautiful and much clearer than other treatments - but that very beauty can obscure assumptions are made. Its a bit like Gleason's Theorem which is also very beautiful and you can forget it depends crucially on non-contextuality.

Just to be clear what's going on I will restate it. From the invariance of probabilities one proves the usual quantum operators of energy and momentum have exactly the same form as classically eg for a free particle the momentum operator is mV where V is the velocity operator defined in Ballentine . The physical assumption is when given a classical system the equations translate over. For momentum since the definition of the velocity operator Ballentine gives is very reasonable you naturally accept its the proper definition of momentum. Its so intuitive you likely don't even realise its an assumption - but an assumption it certainly is.

Thanks
Bill

Last edited: Jul 10, 2015
10. Jul 11, 2015

### DrDu

The point where qm differs from classical mechanics in terms of galilei group representation is in that in qm, mass enters as a central extension of the algebra.
You may want to read the original articles by Levy Leblond.