How to find the generator of translation?

by kakarukeys
Tags: generator, translation
kakarukeys is offline
Nov14-04, 08:32 AM
P: 190
The Galilei group contains rotations, Galilean transformations, space translation and time translation.

It is easy to work out generators for rotations and Galilean transfromations in matrix form.

And they obey:
[tex][J^i, K^j] = i \epsilon^{ijk}K^k[/tex]

Can one work out the generator for space translation, [tex]P[/tex]? so that one can show explicitly that:

[tex][K^i, P^j] = 0[/tex]

and same for time translation.
[tex][K^i, H] = i P^i[/tex]


there is no matrix form for these two generators?
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pervect is offline
Nov16-04, 01:58 AM
Sci Advisor
P: 7,443
Nobody else has taken a shot at this, so I'll put my $.02 in, though I'm afraid I can't give you as definite an answer as you'd like (well, for that matter, as I'd like).

Usual matrix notation is linear, so you can write down (x,y) -> ax + by in matrix form. However, you want to write a transform of the form x -> x+a. This isn't a linear tranform. It appears to me that you can do this by just defining a "variable" that's equal to a constant. To avoid winding up with non-square matrices, you'll have to add a dummy line, that describes how a constant transforms. Well, a constant is always equal to itself, it doesn't depend on the other variables, so the matrix entry for how a constant transforms will have to say that it's equal to 1 x itself, no other variable affects it.
da_willem is offline
Nov16-04, 03:12 AM
P: 603
Isn't the generator for space translations [tex]e^{i \hat{p}a/\hbar}[/tex]? Or is a generator something different?

kakarukeys is offline
Nov23-04, 07:36 PM
P: 190

How to find the generator of translation?

I do not quite understand what you said.
But I found the answer.
The matrix can be found by expanding the 4D space to 5D, provided that [tex]x_5[/tex] is always 1.

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