QFT: Invariant Measures & Rotational Invariance Explained

[whynothis]

I am reading through sidney colemans lectures on QFT and I am stuck on what seem to be a silly question: He talks about the fact that the measure used in a calculation should be invariant in order to prove unitarity and later on that operators transform properly. He uses the example of rotational invariance.

$$U^{-1}(R)\int d^{3}k|k><k| U(R)$$
$$\int d^{3}k |R^{-1}k><R^{-1}k|$$

Change of variable: Rk' = k
Now this is the part I don't get (I must be confused)
$$d^{3}k' = d^{3}k$$
It seems to me like there should be a factor of R or something. However, the strange thing is that R is a matrix (isn't it?) so I don't really get it. Can someone explain what is going on here?

 

[reilly]

When in doubt, use brute force. Grind it out. That is, here, compute the appropriate Jacobean, and thus prove the equality. (Done in countless texts.)
Regards,
Reilly Atkinson

 

[strangerep]

[Coleman] talks about the fact that the measure used in a calculation should be invariant [...]

$$U^{-1}(R)\int d^{3}k|k><k| U(R)$$
$$\int d^{3}k |R^{-1}k><R^{-1}k|$$

Change of variable: Rk' = k
Now this is the part I don't get (I must be confused)
$$d^{3}k' = d^{3}k$$

In general, when you perform a change of variable in an integral, you
must include the Jacobian determinant of the transformation. In this case
the Jacobian is $| \partial k'_i/\partial k_j |$.
(Consult Wiki for more detail.)

In the current case, $d^3k$ is an infinitesimal volume element
in 3-momentum space, and are preserved by rotation transformations,
so the Jacobian turns out to be 1. But that's not necessarily so in more
general transformations.

 

[ismaili]

I am reading through sidney colemans lectures on QFT and I am stuck on what seem to be a silly question: He talks about the fact that the measure used in a calculation should be invariant in order to prove unitarity and later on that operators transform properly. He uses the example of rotational invariance.

$$U^{-1}(R)\int d^{3}k|k><k| U(R)$$
$$\int d^{3}k |R^{-1}k><R^{-1}k|$$

Change of variable: Rk' = k
Now this is the part I don't get (I must be confused)
$$d^{3}k' = d^{3}k$$
It seems to me like there should be a factor of R or something. However, the strange thing is that R is a matrix (isn't it?) so I don't really get it. Can someone explain what is going on here?

From the point of view of group theory, what he used is the so-called rearrange lemma. It's most easily to be understood in finite dimensional.
For example, you have a cyclic group $$G = \{e,a,a^2\}$$ with $$a^3 = e$$. Let $$g\in G$$, for example, say $$g = a$$, then $$g\{e,a,a^2\} = \{a,a^2,e\}$$, meaning, $$g$$ operates on the all group element would reproduce all group elements. Hence, if we consider a summation over the group elements, say $$\sum_{g\in G}f(g'g)$$ where $$f$$ is a function of the group element. By the rearrangement lemma, we may safely rewrite the summation as $$\sum_g f(g)$$

In the continuous group case, the rearrangement lemma is somewhat more involved. Since we have infinitely many ways to parametrize a Lie group, so we have to be more careful. We define that, a parametrisation $$g(\xi)$$, together with a weight function $$\rho_g(\xi)$$ such that the following equation holds
$$\int dg f(g) = \int dg f(s^{-1}g)$$
where $$s\in G$$ and $$dg = \rho_g(\xi)d\xi$$ is called to provide an invariant measure. And one can prove that the weight function for SO(3) group is just 1. (It happens to be the Jacobian factor).

 

[lbrits]

We define that, a parametrisation $$g(\xi)$$, together with a weight function $$\rho_g(\xi)$$ such that the following equation holds
$$\int dg f(g) = \int dg f(s^{-1}g)$$
where $$s\in G$$ and $$dg = \rho_g(\xi)d\xi$$ is called to provide an invariant measure.

Is this the Haar measure?

 

[mrandersdk]

looks like it, if you nomalize it to 1 when you integrate 1 over the whole manifold (group).

 

[whynothis]

Thanks everyone for the help. I was forgetting about the fact that the determinant of the rotation matrix was 1... oops. Thanks for the further insight ismaili that is very interesting and I will have to look further into your comment.