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Is the Jacobian determinant of a unitary transformation equal to one?

I ask because I get that impression from the appendix of this paper. They have spherical coordinates for two particles, $\{\theta_1,\phi_1,\theta_2,\phi_2\}$, and define the spinor variables

$u_k=\cos(\theta_k/2)e^{i\phi_k/2}$

$v_k=\sin(\theta_k/2)e^{-i\phi_k/2}$.

They are computing an integral over all coordinates and introduce the new variables $u_2'$ and $v_2'$:

$u_2=u_1u_2'-v_1^*v_2'$

$v_2=v_1u_2'+u_1^*v_2'$

From the continued calculation it's clear that they use a Jacobian determinant of 1 for the substitution.

I've tried to compute the determinant for myself (on paper and using Mathematica), and I don't get 1. But it's a bit complicated; maybe I'm doing something wrong.

Another possibility is that they neglect the Jacobian because they are dividing with a similar integral further down, and this has the same Jacobian (assuming the latter doesn't actually depend on the coordinates). But computing the integral numerically before and after the substitution with different parameters does not support this (since the substitution doesn't depend on any parameters I assume the determinant can only be a constant if it is independent of the coordinates).

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Is the (absolute values of the) Jacobian determinant of a unitary transformation equal to one? of course it is.

Notice that if you have a unitary transformation

$$y_i=U_{ij}x_j$$

the Jacobian matrix is necessarily the unitary matrix you have in your coordinate transformation

$$\frac{\partial{}y_i}{\partial{}x_j}=U_{ij}$$

since the determinant of a unitary matrix is either one or minus one, the absolute value of the determinant of the Jacobian matrix is 1.

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    $\begingroup$ Is this an answer to my question? Of course it is. $\endgroup$
    – jorgen
    Apr 14 '15 at 14:53
  • $\begingroup$ Does it mean the volume element in y space and the volume element in x space equal? $\endgroup$ Jun 25 at 2:37

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