# What generates boosts?

1. Nov 6, 2015

### pellman

Quantum mechanically the generator of time translations is the Hamiltonian H or the energy. That is, $$e^{-iHt'} | \Psi (t) \rangle = | \Psi (t+t') \rangle$$

And the generator of rotations is the angular momentum. The generator of spatial translations is momentum.

What is the generator of boosts (change to another inertial frame)? Is there a classical quantity it corresponds to?

The operator I am asking about is denoted by $C_j$ here https://en.wikipedia.org/wiki/Galilean_transformation#Origin_in_group_contraction

2. Nov 7, 2015

### andrewkirk

The common QM operator that has not yet been used above is $X$, the position operator. Since position is conjugate with momentum, I would expect it to generate momentum boosts, rather than velocity boosts. Maybe what's needed is a multiplication or division by mass then, to get an operator that is conjugate to velocity. As a wild guess, centre of mass sounds like it might be promising, as that is position multiplied by mass.

This is of course all just guess work, as I have not worked through any of the maths. But I reckon looking at an operator of the form $mX$ might be a possible lead.

3. Nov 7, 2015

### vanhees71

The generator for the boosts, written in terms of the usual position and momentum operators is
$$\hat{\vec{K}}=m \hat{\vec{x}}-\hat{\vec{P}} t.$$
This is written for an irreducible representation with fixed mass $m$. Mass in non-relativistic QM is a central charge and thus any irreducible representation is characterized in addition to the spin $s \in \{0,1/2,1,\ldots \}$ by the mass. The true Galilei group in QM is given by the central extensions of the covering group of the classical Galilei group. It's not so simple. For details, see the excellent textbook by Ballentine, Quantum Mechanics: A modern development, Addison-Wesley.

4. Nov 7, 2015

### pellman

Thank you! Lol. Actually I am coming at this from Ballentine. Just thinking it through a little more than he covers. The way he introduces the mass into the commutators of the Galilean Lie algebra is a not very rigorous. I'd like to see another treatment but haven't found one online.

Edit: maybe I don't want to see a more rigorous after all. :) Just had a look at one online. whew!

5. Nov 7, 2015

### vanhees71

I've lecture notes about it, but only in German. To my surprise the students liked it at the time, and I got through the entire Lie-algebra gymnastics. It's not very rigorous. I just took Weinberg's book on the Poincare group and tried to translate it to the Galilei group. Another source was Ballentine and Wigner's papers on the Galilei group too:

http://theory.gsi.de/~vanhees/faq-pdf/quant.pdf

6. Nov 7, 2015

### pellman

I also tried to follow Weinberg's derivation of the Poincare Lie algebra in Quantum Theory of Fields vol 1 and use the same steps to get the Galilei version. But the representation of Poincare group there is non-projective and so following the same steps what you end up with are the commutation relations for the Galilean generators with M=0.

Thanks so much for sharing the notes. I think I am going to be able to follow it.

7. Nov 7, 2015

### dextercioby

8. Nov 14, 2015

### pellman

A translation is given by $$\exp ( i \hat{P}_z a ) \Psi ( x,y,z,t) = \exp ( a \partial / \partial z ) \Psi ( x,y,z,t) = \Psi ( x,y,z+a,t)$$ so why isn't a Galilean boost (x,y,z,t) --> (x,y,z+vt,t) given by $$\exp ( i \hat{P}_z vt ) \Psi ( x,y,z,t) = \exp ( vt \partial / \partial z ) \Psi ( x,y,z,t) = \Psi ( x,y,z+vt,t)$$ ? We see in this the $-\hat{\vec{P}}t$ part of $\hat{\vec{K}}=m \hat{\vec{x}}-\hat{\vec{P}} t$. Where does mx part come in?

9. Nov 14, 2015

### vanhees71

It comes from the fact that you must take a central extension of the classical Galilei group to the quantum Galilei group with the mass $m>0$ of the particle as a central charge, because it turns out that the unitary representations of the classical Galilei group do not lead to any sensible quantum dynamics. Thus the commutation relation with the non-trivial central charge, the mass of the particle, reads
$$[\hat{K}_j,\hat{P}_k]=\mathrm{i} m \delta_{jk} \mathbb{1},$$
and one also has the commutator
$$[\hat{K}_j,\hat{H}]=\mathrm{i} \hat{P}_k.$$
Further, since Galilei boost must be a symmetry of the theory according to Noether's theorem one must have
$$\frac{1}{\mathrm{i}} [\hat{K}_j,\hat{H}]+\frac{\partial \hat{K}}{\partial t}=0.$$
With the commutation relation this implies that $\hat{K}$ must be explicitly time dependent and
$$\frac{\partial \hat{K}_j}{\partial t}=-\hat{P}_j.$$
Since further the Galilei algebra shows that $[\hat{P}_j,\hat{H}]=0$ and since also $\hat{P}_j$ is conserved, because translations must be a symmetry too, $\hat{P}_j$ is not explicitly time dependent, and thus the equation solves to
$$\hat{K}_j=m \hat{X}_j-\hat{P}_j t,$$
where $\hat{X}_j$ is not explicitly time dependent. Now again the (quantum!) Galilei algebra tells us that the $\hat{P}_j$ commute among each other but that
$$[\hat{K}_j,\hat{P}_k]=m [\hat{X}_j,P_k]=\mathrm{i} m \delta_{jk} \mathbb{1}.$$
This shows that with $m \neq 0$ you get the usual Heisenberg algebra, and we have been able to construct the usual position operator for non-relativistic massive particles. As stressed above, for $m=0$ we don't get a sensible quantum dynamics at all, as was shown by Enönü and Wigner in

E. Inönü and E. P. Wigner. Representations of the Galilei group. Il Nuovo Cimento, 9(8):705–718, 1952.

The relativistic case (i.e., the Poincare group) is different, because it does not have any nontrival central charges, i.e., all unitary ray representations are equivalent to unitary representations, and the quantum Poincare group is the same as the classical. Furthermore there are causal local QFTs and thus physically well interpretible models for both massive and massless particles. Mass is not a central charge but a Casimir operator of the Poincare group and thus one of the parameters determining the irreducible representations of it and thus the properties of elementary particles, which are by definition those objects which are describable as irreducible representations of the Poincare group.

10. Nov 14, 2015

### pellman

Thanks for all of that. It was very helpful to my understanding. One question. What does a Galilean boosted function look like?

$\exp(ia \hat{P}) \Psi(x,t) = \Psi(x+a,t)$
$\exp (-i \tau \hat{H} ) \Psi(x,t) = \Psi(x,t+ \tau)$
$\exp (i \hat{K}v) \Psi(x,t) = \exp [i(m \hat{X} - \hat{P}t)v] \Psi (x,t) = ?$

11. Nov 15, 2015

### vanhees71

This is a somewhat lengthy calculation. You also have to be careful with the time evolution, because $\hat{H}$ and $\hat{\vec{K}}$ don't commute. You have
$$\Psi'(\vec{x},t)=\exp(-\mathrm{i} t \hat{H}) \exp(\mathrm{i} \hat{\vec{K}} \cdot \vec{v} ) \Psi(\vec{x},0).$$
You finally get
$$\exp(\mathrm{i} \hat{\vec{K}} \cdot \vec{v}) \Psi(\vec{x},t)=\exp \left[\mathrm{i} m \vec{x} \cdot \vec{v}-\frac{\mathrm{i} m \vec{v}^2}{2} \right ] \psi(\vec{x}-\vec{v} t,t).$$
You find the calculation in my German manuscript for QM 2 (Eq. 6.10.19):

http://theory.gsi.de/~vanhees/faq/quant/node78.html

Note that there I consider a boost with the other sign in the exponential, but that doesn't change much (you just have to set $\vec{w} =-\vec{w}$).

12. Nov 15, 2015

### haushofer

You can easily derive the transformation of psi under a boost if you demand that it keeps the schrodinger eqn. invariant. Psi is not a scalar under boosts, as you can check via the Schrodinger eqn. But it doesnsn't need to be; the probability should be a scalar! So psi could acquire a phase factor under a boost, and invariance of the schrodinger eqn. under boosts gives you the form of this phase factor. See e.g. Ballentine.

13. Nov 15, 2015

### vanhees71

Yes, and what I forgot to mention in my posting is that you also see an interesting effect about the not so trivial structure of the Galilei group which introduces mass as a central extension of this group to extend it to its "quantum version": There cannot be superpositions of states with different mass, because otherwise the different components of such a superposition would get different phase factors, leading to observable interference effects, which means that if you allow such superpositions, QT wouldn't be Galilei invariant. In other words you have to forbid such superpositions in the initial state. Then, due to the Galilei invariance for the allowed states, also time evolution cannot lead to the forbidden superpositions. That's a nice example for a superselection rule due to a symmetry.

14. Nov 15, 2015

### pellman

Your equation didn't display but I looked at your online manuscript. Is that calculation only for a free particle? It matches it what I got when I played with it some more.

15. Nov 15, 2015

### vanhees71

Yes, if you have a single particle and want full Galilei invariance it must be free. If you have, e.g., an external potential (e.g. electrostatic field and a charged particle) it necessarily breaks Galilei invariance. Only closed systems of two or more interacting particles have a Galilei invariant Hamiltonian.

16. Nov 15, 2015

### pellman

17. Jan 6, 2016

### pellman

This continued to bother me. It shouldn't matter what system we are talking about, we should always be able to ask, "What does it look like viewed from another inertial frame?" I worked on it some more and it turns out that $e^{ i/\hbar (Kv + t mv^2 /2) }$ when applied to a wavefunction will always give the corresponding wavefunction in the moving frame, for any potential. I can send you my notes if interested.

18. Jan 7, 2016

### vanhees71

Sure, send it!

19. Jan 7, 2016

### haushofer

If you consider a point particle in a Newton potential, the Galilean symmetries are not broken but extended to so-called Galilei-accelerated ones. See e.g. http://arxiv.org/abs/1206.5176 . Effectively, the boosts become accelerations with arbitrary t-dependency. However, in the context of sigma-models these symmetries are called "pseudo-symmetries", see e.g. http://arxiv.org/pdf/hep-th/9503022.pdf

20. Jan 7, 2016

### pellman

21. Jan 8, 2016

### vanhees71

Yes, and I replied to it via e-mail too.

Anyway, the math is correct, but still the Galilei boost is not a symmetry if there is a non-trivial external potential present, because your math tells you that the boosted wave function solves a problem to a different Hamiltonian, namely for that one where, the external potentail $V(x,t)$ is substituted by
$$V'(x',t')=V(x'+v t',t').$$

22. Jan 8, 2016

### pellman

Right. Thanks!