# Another one on Lorentz Invariance

1. Feb 26, 2007

### alphaone

I recently read an author making the following argument in QFT:
if <m|A^0(t,0)|n>=B then <m|A^mu(t,0)|n>=(B/p^0)*p^mu by Lorentz invariance. Can anybody tell me under which circumstances this holds and how it comes about? I understand that <m|A^mu(t,0)|n> had to transform as a 4-vector but why should it be the momentum 4-vector?

2. Feb 26, 2007

### dextercioby

Who's $A^{\mu}$ ? What does it stand for ?

3. Feb 26, 2007

### alphaone

That is part of my question: Does the relation hold for arbitrary operators which form a 4-vector? I think not, but maybe I am wrong. The context in which I saw the argument, A^mu was a conserved current but otherwise arbitrary. Does that help?

4. Mar 1, 2007

### alphaone

Ok I think I understand the argument partly now and must admit that the question did not really make sense the way I phrased it earlier. What I should have asked is: Suppose we have <0|A^0(t,0)|p> = B where <0| is the vacuum |p> is an eigenstate of momentum p and A^0(t,0) is the zero component of an operator forming a 4-vector evaluated at x_i=0. Given this does it then follow that <0|A^mu(t,0)|p> = (B/E) * p^mu ? I think it should and I think the argument goes something like this: the LHS only really depends on p^mu and therefore the only 4-vector that can possibly appear on the RHS is the momentum 4-vector p^mu. However this argument is not totally clear to me as A^mu itself appears on the LHS and so why should the RHS not be something proportional to that? Anyway if you can explain this argument to me pleae let me know.