- #1

binbagsss

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- TL;DR Summary
- non-commutative properties in contrast to the Newtonian mechanics case

I am reading some literature which is considering translations and boosts in field theory. The reference is Construction of Lagrangians continuum theories, Markus Scholle, 2004, The Royal Society. 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.

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##

(where this, the w.f transformation comes about by ensuring the Schrodinger equation is invariant w.r.t Galilean boosts

[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 applying the boost to 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...( that is, in contrast to Newtonian mechanics and the boost and translation commuting, in field theory they do not

So, this is what the article has done. But, 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 done 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].

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##

(where this, the w.f transformation comes about by ensuring the Schrodinger equation is invariant w.r.t Galilean boosts

[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 applying the boost to 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...( that is, in contrast to Newtonian mechanics and the boost and translation commuting, in field theory they do not

So, this is what the article has done. But, 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 done 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].