- #1
Glenn Rowe
Gold Member
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I've got a question about the identification of SU(2) with O(3) in Ryder's QFT book (2nd edition) pages 34 - 35.
The other posts on this topic I could find don't seem to address this question, so here goes.
He derives the matrix in eqn 2.47:
$$H=
\left[\begin{array}{cc}
-\xi_{1}\xi_{2} & \xi_{1}^{2}\\
-\xi_{2}^{2} & \xi_{1}\xi_{2}
\end{array}\right]$$
He then constructs another matrix ##h## from the position vector and the Pauli spin matrices (eqn 2.49):
$$
h=\boldsymbol{\sigma}\cdot\mathbf{r}=\left[\begin{array}{cc}
z & x-iy\\
x+iy & -z
\end{array}\right]$$
We can identify ##h## with ##H## if we take
$$
\begin{eqnarray}
x & = & \frac{1}{2}\left(\xi_{2}^{2}-\xi_{1}^{2}\right)\\
y & = & \frac{1}{2i}\left(\xi_{1}^{2}+\xi_{2}^{2}\right)\\
z & = & \xi_{1}\xi_{2}
\end{eqnarray}
$$
He says that both ##H## and ##h## transform according to ##H^{\prime}=UHU^{\dagger}## and ##h^{\prime}=UhU^{\dagger}## which seems OK so far. However, he then says that if ##U## belongs to SU(2) and therefore has determinant 1, we can take the determinant of ##h^{\prime}=UhU^{\dagger}## to say that ##x^{\prime2}+y^{\prime2}+z^{\prime2}=x^{2}+y^{2}+z^{2}##.
This seems reasonable, except that when you actually evaluate ##x^{2}+y^{2}+z^{2}## using equations (1) to (3) you get ##x^{2}+y^{2}+z^{2}=0## identically. This also follows by taking the determinant of ##H##.
I guess my question is: how can you get around the fact that this equivalence seems to apply only to a single point at the origin to say anything useful about the relation of SU(2) to rotations?
Thanks for any comments.
The other posts on this topic I could find don't seem to address this question, so here goes.
He derives the matrix in eqn 2.47:
$$H=
\left[\begin{array}{cc}
-\xi_{1}\xi_{2} & \xi_{1}^{2}\\
-\xi_{2}^{2} & \xi_{1}\xi_{2}
\end{array}\right]$$
He then constructs another matrix ##h## from the position vector and the Pauli spin matrices (eqn 2.49):
$$
h=\boldsymbol{\sigma}\cdot\mathbf{r}=\left[\begin{array}{cc}
z & x-iy\\
x+iy & -z
\end{array}\right]$$
We can identify ##h## with ##H## if we take
$$
\begin{eqnarray}
x & = & \frac{1}{2}\left(\xi_{2}^{2}-\xi_{1}^{2}\right)\\
y & = & \frac{1}{2i}\left(\xi_{1}^{2}+\xi_{2}^{2}\right)\\
z & = & \xi_{1}\xi_{2}
\end{eqnarray}
$$
He says that both ##H## and ##h## transform according to ##H^{\prime}=UHU^{\dagger}## and ##h^{\prime}=UhU^{\dagger}## which seems OK so far. However, he then says that if ##U## belongs to SU(2) and therefore has determinant 1, we can take the determinant of ##h^{\prime}=UhU^{\dagger}## to say that ##x^{\prime2}+y^{\prime2}+z^{\prime2}=x^{2}+y^{2}+z^{2}##.
This seems reasonable, except that when you actually evaluate ##x^{2}+y^{2}+z^{2}## using equations (1) to (3) you get ##x^{2}+y^{2}+z^{2}=0## identically. This also follows by taking the determinant of ##H##.
I guess my question is: how can you get around the fact that this equivalence seems to apply only to a single point at the origin to say anything useful about the relation of SU(2) to rotations?
Thanks for any comments.