Galiliei transformations explicit proof

[ma18]

Homework Statement

Show explicitly that

e^i*ε*κ_u * e^i*ε*κ_v * e^-i*ε*κ_u* e^-i*ε*κ_v = Identity + ε² [K_v,K_u + O (ε³)

The Attempt at a Solution

K_v,K_u = K_v*K_u - K_u*K_v

I'm not sure exactly how to approach this problem. I know that

U (tau) = ∏ e^i*s_u*K_u

and that for operators O --> O' = U O U^†

I have this information but I don't know how to put it together, any help would be greatly appreciated

 

[Orodruin]

I suggest expanding the exponentials up to order $\epsilon^2$ and then simply checking that the expression reduces to the given one.

 

[ma18]

I suggest expanding the exponentials up to order $\epsilon^2$ and then simply checking that the expression reduces to the given one.

When I expand the exponentials to the second order I get the expression given here

http://www.wolframalpha.com/input/?i=(1+i*e*x-e^2*x^2/2)(1+i*e*v-e^2*v^2/2)(1-i*e*x-e^2*x^2/2)(1-i*e*v-e^2*v^2/2)

Where I used x for K_u and v for K_v
But I don't see how this reduces to the other expression

 

[Orodruin]

Dont let wolfram do it for you, just multiply the terms together and keep only terms up to order two in epsilon.

 

[ma18]

two

Dont let wolfram do it for you, just multiply the terms together and keep only terms up to order two in epsilon.

Alright, If I do that I get

(1+i*e*v-e^2*v^2/2 +i*ex-e^2*x*v-e^2*x^2/2)(1-i*e*v-e^2*v^2/2-i*e*x-e^2*v*x-e^2*x^2/2)

then expanding that leads to many terms

which doesn't lead to the correct answer, perhaps I am making an algebraic mistake

Dont let wolfram do it for you, just multiply the terms together and keep only terms up to order two in epsilon.

 

[Orodruin]

Go order by order. First check that all the linear terms cancel.