Various questions on Galilean group (boosts and spatial translations)

[binbagsss]

1) I am reading some literature which is considering translations and boosts in field theory. It proves something by relying on postulating that Galilei boosts satisfy the Albelian group properties of additivity and identity (it just 'reasons' in analogy with translations in Newton mechanics). Does the additive property seem reasonable to postulate? Could someone help me understand why with a brief explanation?

2) (More of a general question). For the field #\psi# it considers the translation and boost as :

translation:

$\vec{x′}→=\vec{x}+\vec{s}$

$t′=t+τ$

$\psi_′=T_i(\psi_j,\vec{x},t,s,\vec{\tau})$


boost:

$\vec{x′}→=\vec{x}-\vec{u}t$

$t′=t$

$\psi_′=G_i(\psi_j,\vec{x},t,\vec{u})$



and then explains these are local transformations as they do not depend on any derivatives (i.e. it does not depend on #\partial \psi_i / \partial t #, or #\nabla \psi_i# or any higher derivatives). Can someone please explain this to me and how this correlates to the terms local and global transformations?

3) As far as I am aware, the composition of boosts and translations commute. However, in this literature a lemma is used which involves using different expressions for boosts followed by translation and translation followed by boost. This commuting property can't be any different when considering fields instead of Newtonian mechanics can it?

Many thanks

 

[haushofer]

First: We don't have crystal balls at PF, so "some literature", "it considers" and "in this literature" are not specific enough. Please give the correct sources ;)

1) Just apply two infinitesimal boost transformations on the spatial coordinate:

$[\delta_{v_1}, \delta_{v_2}] x^i = \delta_{v_1}(v_2^i t) - \delta_{v_2}(v_1^i t) = 0-0=0$

The same applies trivially on the time coordinate $t$.

2) I'm not sure, but I guess the idea is that if the transformed field contains derivatives, you can develop this term in an infinite Taylor series of terms without derivatives, which makes it non-local.

3) I don't understand this question, and without source I can't check it either. Either be more specific, or give a screenshot.

 

[haushofer]

Another tip: if you want people to help you, it's better to respond.

 

[binbagsss]

1) Just apply two infinitesimal boost transformations on the spatial coordinate:

$[\delta_{v_1}, \delta_{v_2}] x^i = \delta_{v_1}(v_2^i t) - \delta_{v_2}(v_1^i t) = 0-0=0$

The same applies trivially on the time coordinate $t$.

2) I'm not sure, but I guess the idea is that if the transformed field contains derivatives, you can develop this term in an infinite Taylor series of terms without derivatives, which makes it non-local.

1.what is the notation you're using in 1, are the deltas boosts?

2. why wouldn't you be able to write it as a Taylor series if it does not contain derivatives? why does not having derivatives means it's global?

3. The reference is Construction of Lagrangiansin continuum theories, Markus Scholle, 2004, The Royal Society

 

[haushofer]

1.what is the notation you're using in 1, are the deltas boosts?

2. why wouldn't you be able to write it as a Taylor series if it does not contain derivatives? why does not having derivatives means it's global?

3. The reference is Construction of Lagrangiansin continuum theories, Markus Scholle, 2004, The Royal Society

1. Yes. In field theory this is standard notation to write down infinitesimal transformations. Since you were asking about the boosts, I thought that would be clear.

2. Let me state it otherwise: without derivatives, $\psi'$ is determined at one point and hence is local; that's the definition of locality But derivatives compare to different (space/time) points. If $\psi'$ contains derivatives, it depends on multiple space/time points. Hence it wouldn't be local.

3. I strongly urge you to be more specific. I found the reference, but I still don't know which lemma you mean. If you want people to help you, you help them by providing exact page- or lemmanumbers. If my gearbox is broken, I don't take my car to the garage and say "Something is broken, good luck finding it".

But if I have to guess your issue: the Schrodinger equation $i \hbar \frac{\partial \psi}{\partial t} = [- \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V]\psi$ must be covariant w.r.t. boosts $x' = x - vt$ and $t'=t$. However, this is not trivial, because the first order time derivative transforms as follows:

$\frac{\partial}{\partial t'} = \frac{\partial}{\partial t} - v^i \frac{\partial}{\partial x^i}$

while the spatial derivatives are invariant. You can easily show that putting $\psi' (t',x') = \psi(t,x)$ will not work, i.e. the wave function cannot be a simple scalar under boosts (you get an extram term on the LHS of the Schrodinger eqn.). But from Borns rule you know that the wave function is defined up to a complex phase. So there is a way out: try $\psi'(t',x') = e^{i \phi(t,x)}\psi(t,x)$ for some function $\phi$. Plug everything in, perform the tedious algebra, and you will find that the Schrodinger equation is covariant w.r.t. boosts if $\phi(t,x) = mvx+\frac{1}{2}mv^2 t + const.$. On the level of group theory, this means that the wave function is a projective representation of the Galilei algebra, and that this implies a central extension between translations and boosts. The result is the so-called Bargmann-algebra. See e.g. Zee's Group theory in a Nutshell, chapter VIII.1 appendix 1 (page 510 onward).

At the classical level this extension can also be motivated: because the Lagrangian of a non-rel. point particle is not invariant under boosts, the corresponding Noether charge is adjusted. The Poisson brackets of the Noether charges of translations and boosts therefore becomes non-zero. Because these brackets are isomorphic to the underlying Lie algebra, this suggests a central extension between boosts and translations, which can be verified by the Jacobi identities. See

https://pure.rug.nl/ws/portalfiles/portal/34926446/Complete_thesis.pdf

page 48 onwards.

 

[binbagsss]

1. Yes. In field theory this is standard notation to write down infinitesimal transformations. Since you were asking about the boosts, I thought that would be clear.

2. Let me state it otherwise: without derivatives, $\psi'$ is determined at one point and hence is local; that's the definition of locality But derivatives compare to different (space/time) points. If $\psi'$ contains derivatives, it depends on multiple space/time points. Hence it wouldn't be local.

3. I strongly urge you to be more specific. I found the reference, but I still don't know which lemma you mean. If you want people to help you, you help them by providing exact page- or lemmanumbers. If my gearbox is broken, I don't take my car to the garage and say "Something is broken, good luck finding it".

But if I have to guess your issue: the Schrodinger equation $i \hbar \frac{\partial \psi}{\partial t} = [- \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V]\psi$ must be covariant w.r.t. boosts $x' = x - vt$ and $t'=t$. However, this is not trivial, because the first order time derivative transforms as follows:

$\frac{\partial}{\partial t'} = \frac{\partial}{\partial t} - v^i \frac{\partial}{\partial x^i}$

while the spatial derivatives are invariant. You can easily show that putting $\psi' (t',x') = \psi(t,x)$ will not work, i.e. the wave function cannot be a simple scalar under boosts (you get an extram term on the LHS of the Schrodinger eqn.). But from Borns rule you know that the wave function is defined up to a complex phase. So there is a way out: try $\psi'(t',x') = e^{i \phi(t,x)}\psi(t,x)$ for some function $\phi$. Plug everything in, perform the tedious algebra, and you will find that the Schrodinger equation is covariant w.r.t. boosts if $\phi(t,x) = mvx+\frac{1}{2}mv^2 t + const.$. On the level of group theory, this means that the wave function is a projective representation of the Galilei algebra, and that this implies a central extension between translations and boosts. The result is the so-called Bargmann-algebra. See e.g. Zee's Group theory in a Nutshell, chapter VIII.1 appendix 1 (page 510 onward).

At the classical level this extension can also be motivated: because the Lagrangian of a non-rel. point particle is not invariant under boosts, the corresponding Noether charge is adjusted. The Poisson brackets of the Noether charges of translations and boosts therefore becomes non-zero. Because these brackets are isomorphic to the underlying Lie algebra, this suggests a central extension between boosts and translations, which can be verified by the Jacobi identities. See

https://pure.rug.nl/ws/portalfiles/portal/34926446/Complete_thesis.pdf

page 48 onwards.

i meant to write the page, apologies.

 

[binbagsss]

1. Yes. In field theory this is standard notation to write down infinitesimal transformations. Since you were asking about the boosts, I thought that would be clear.

2. Let me state it otherwise: without derivatives, $\psi'$ is determined at one point and hence is local; that's the definition of locality But derivatives compare to different (space/time) points. If $\psi'$ contains derivatives, it depends on multiple space/time points. Hence it wouldn't be local.

3. I strongly urge you to be more specific. I found the reference, but I still don't know which lemma you mean. If you want people to help you, you help them by providing exact page- or lemmanumbers. If my gearbox is broken, I don't take my car to the garage and say "Something is broken, good luck finding it".

But if I have to guess your issue: the Schrodinger equation $i \hbar \frac{\partial \psi}{\partial t} = [- \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V]\psi$ must be covariant w.r.t. boosts $x' = x - vt$ and $t'=t$. However, this is not trivial, because the first order time derivative transforms as follows:

$\frac{\partial}{\partial t'} = \frac{\partial}{\partial t} - v^i \frac{\partial}{\partial x^i}$

while the spatial derivatives are invariant. You can easily show that putting $\psi' (t',x') = \psi(t,x)$ will not work, i.e. the wave function cannot be a simple scalar under boosts (you get an extram term on the LHS of the Schrodinger eqn.). But from Borns rule you know that the wave function is defined up to a complex phase. So there is a way out: try $\psi'(t',x') = e^{i \phi(t,x)}\psi(t,x)$ for some function $\phi$. Plug everything in, perform the tedious algebra, and you will find that the Schrodinger equation is covariant w.r.t. boosts if $\phi(t,x) = mvx+\frac{1}{2}mv^2 t + const.$. On the level of group theory, this means that the wave function is a projective representation of the Galilei algebra, and that this implies a central extension between translations and boosts. The result is the so-called Bargmann-algebra. See e.g. Zee's Group theory in a Nutshell, chapter VIII.1 appendix 1 (page 510 onward).

At the classical level this extension can also be motivated: because the Lagrangian of a non-rel. point particle is not invariant under boosts, the corresponding Noether charge is adjusted. The Poisson brackets of the Noether charges of translations and boosts therefore becomes non-zero. Because these brackets are isomorphic to the underlying Lie algebra, this suggests a central extension between boosts and translations, which can be verified by the Jacobi identities. See

https://pure.rug.nl/ws/portalfiles/portal/34926446/Complete_thesis.pdf

page 48 onwards.

ok ; so to summarise in Galilean group the translations and boosts do not commute, but with this central extension the properties are ofc different?

 

[haushofer]

ok ; so to summarise in Galilean group the translations and boosts do not commute, but with this central extension the properties are ofc different?

No: the translations and boosts DO commute in the Galilei group. You can easily check this by applying infinitesimal transformations (P=translation, B=boost) on the coordinates:

$[\delta_P, \delta_B]x^i = \delta_P (v^i t) -\delta_B(s^i) = 0 - 0 = 0$

and similarly for $t$. But because the wavefunction transforms non-trivially under boosts, this does not hold anymore for the wavefunction, hence the central extension between boosts and translations.

 

[binbagsss]

No: the translations and boosts DO commute in the Galilei group. You can easily check this by applying infinitesimal transformations (P=translation, B=boost) on the coordinates:

$[\delta_P, \delta_B]x^i = \delta_P (v^i t) -\delta_B(s^i) = 0 - 0 = 0$

and similarly for $t$. But because the wavefunction transforms non-trivially under boosts, this does not hold anymore for the wavefunction, hence the central extension between boosts and translations.

apolgies ofc i meant to say they do commute in the Galilei group.

 

[binbagsss]

First: We don't have crystal balls at PF, so "some literature", "it considers" and "in this literature" are not specific enough. Please give the correct sources ;)

1) Just apply two infinitesimal boost transformations on the spatial coordinate:

$[\delta_{v_1}, \delta_{v_2}] x^i = \delta_{v_1}(v_2^i t) - \delta_{v_2}(v_1^i t) = 0-0=0$

The same applies trivially on the time coordinate $t$.

2) I'm not sure, but I guess the idea is that if the transformed field contains derivatives, you can develop this term in an infinite Taylor series of terms without derivatives, which makes it non-local.

3) I don't understand this question, and without source I can't check it either. Either be more specific, or give a screenshot.

So 1 here would need to be re-done , applied to wave-functions to see if it seems reasonable or not?

...the translations and boosts DO commute in the Galilei group....But because the wavefunction transforms non-trivially under boosts, this does not hold anymore for the wavefunction....

Should a group definition include what its acting on or not? i.e. what is meant by galillean group translations and boosts commute exactly , s.t. this statement is not including 'wave-functions'.

Then, what here is meant by a wave-function; one that satisfies this Schrodinger equation right?

Thanks.

 

[haushofer]

So 1 here would need to be re-done , applied to wave-functions to see if it seems reasonable or not?




Should a group definition include what its acting on or not? i.e. what is meant by galillean group translations and boosts commute exactly , s.t. this statement is not including 'wave-functions'.

Then, what here is meant by a wave-function; one that satisfies this Schrodinger equation right?

Thanks.

1) Yes. You should do the calculation again for the wavefunction. To be honest, I'd have to check it for myself since it's been a while.

2) No, this action does not define the group, but its representation. The same group has different actions on different representations. Compare with the Lorentz group: infinitesimally its corresponding Lie-algebra is defined by the commutators, but these infinitesimal transformations act differently depending on the representation.

Here, the wavefunction is indeed defined as a function satisfying the Schrodinger equation. But it's not a simple scalar representation of the Galilei group, which one could naively think. If that would be the case, one would have that $\psi'(t',x')=\psi(t,x)$ under Galilei boosts, which doesn't keep the Schrödinger equation covariant as we saw.

IIRC Ballentine's book on quantum mechanics has a good overview of the symmetries of the Schrödinger equation.

 

[binbagsss]

1) Yes. You should do the calculation again for the wavefunction. To be honest, I'd have to check it for myself since it's been a while.

2) No, this action does not define the group, but its representation. The same group has different actions on different representations. Compare with the Lorentz group: infinitesimally its corresponding Lie-algebra is defined by the commutators, but these infinitesimal transformations act differently depending on the representation.

Here, the wavefunction is indeed defined as a function satisfying the Schrodinger equation. But it's not a simple scalar representation of the Galilei group, which one could naively think. If that would be the case, one would have that $\psi'(t',x')=\psi(t,x)$ under Galilei boosts, which doesn't keep the Schrödinger equation covariant as we saw.

IIRC Ballentine's book on quantum mechanics has a good overview of the symmetries of the Schrödinger equation.

OK thank you. But so when one speaks of Galilean boosts and translations commuting, this is referring to when this acts on what? i.e. when one defines a group action the objects that it is operating on must surely be defined as well?

 

[haushofer]

OK thank you. But so when one speaks of Galilean boosts and translations commuting, this is referring to when this acts on what?

On the coordinates.

 

[binbagsss]

Hi, sorry just re-looking at this, I am wondering if anyone is able to help me understand applying the transformation. I can see what is done, but I do not understand why.

So the translation is defined by:

$\vec{x} \to \vec{x} + \vec{s}$
$t \to t$
$\psi \to \psi$

the boost by:

$\vec{x} \to \vec{x} - \vec{u_0}t$
$t \to t$
$\psi \to \psi- \vec{u_0}.\vec{x}+\frac{1}{2}\vec{u_0^2}t$

[1]the article then will now consider the translation followed by the Galilean and this is done by:

First I get :

1)$\vec{x} \to \vec{x} + \vec{s}, \psi \to \psi$

Then I now apply the boost and get:

2)$\vec{x} \to \vec{x} + \vec{s} - \vec{u_0}t, \psi \to \psi- \vec{u_0}.\vec{x}+\frac{1}{2}\vec{u_0^2}t$

[2]And now the Galilean boost then the translation:

Boost:
$\vec{x} \to \vec{x} - \vec{u_0}t$
$\psi \to \psi- \vec{u_0}.\vec{x}+\frac{1}{2}\vec{u_0^2}t$

And the translation then gives:

$\vec{x} \to \vec{x} - \vec{u_0}t-\vec{s}$
$\psi \to \psi- \vec{u_0}.(\vec{x}+\vec{s})+\frac{1}{2}\vec{u_0^2}t$

so the argument is that while $\vec{x}$ is the same $\psi$ differs...

So, this is what the article has done. But, to be honest, I do not really understand this. To me, the $x$ from 1) needs to be plugged into the $\psi$, not treating them separately. I thought it has to be a simultaneously transformation since $\psi$ is a function of $x$ and do not understand how they can be considered separately like this? so that is, i would be getting the same as the result of [2].

 

[binbagsss]

anyone thanks?