- #1
gnieddu
- 24
- 1
Homework Statement
Hi,
I'm trying to self-study quantum mechanics, with a special interest for the group-theoretical aspect of it. I found in the internet some lecture notes from Professor Woit that I fouund interesting, so I decided to use them as my guide. Unfortunately I'm now stuck at a point which suggests I'm perhaps not as fluent with a couple of concepts - namely "invariance" and "group action" - as I thought. Hopefully someone will be able to help me and suggest the best way to progress.
It all started with a discussion about the tensor product of vector spaces, and the application of this concept to spinors. In order to give a practical example, the notes consider the tensor product of two spinors ##V^1 \otimes V^1##. Taking ##\begin{pmatrix}1 \\ 0\end{pmatrix}## and ##\begin{pmatrix}0 \\ 1\end{pmatrix}## as a basis for ##V^1##, a basis for the space resulting from the tensor product is easily found by taking all possible tensor products of the two basis vectors. So far, so good.
At this point, the notes make the following statement: "one can show that the vector:
$$\frac 1 {\sqrt 2}(\begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix} - \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}) \in V^1 \otimes V^1$$
is invariant under SU(2), by computing either the action of SU(2) or of its Lie algebra su(2)".
2. The attempt at a solution
Now, in order to understand all this, I first started to explicitly compute the value of the vector which, unless I'm wrong, is:
$$\frac 1 {\sqrt 2}(\begin{pmatrix}1 \\ 0\end{pmatrix} \otimes \begin{pmatrix}0 \\ 1\end{pmatrix} - \begin{pmatrix}0 \\ 1\end{pmatrix} \otimes \begin{pmatrix}1 \\ 0\end{pmatrix}) = \frac 1 {\sqrt 2}\begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}$$
The next step, in my mind, was to take a generic element of SU(2), i.e. the matrix:
$$\begin{pmatrix}\alpha & \beta \\ -\bar \beta & \bar \alpha\end{pmatrix}$$
(with ##\alpha\bar\alpha + \beta\bar\beta = 1##), and verify that, by multiplying it with the vector results in the vector itself, i.e. to verify that:
$$\frac 1 {\sqrt 2}\begin{pmatrix}\alpha & \beta \\ -\bar \beta & \bar \alpha\end{pmatrix} \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix} = \frac 1 {\sqrt 2}\begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}$$
However, by doing the calculations I get:
$$\frac 1 {\sqrt 2}\begin{pmatrix}\alpha & \beta \\ -\bar \beta & \bar \alpha\end{pmatrix} \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix} = \frac 1 {\sqrt 2}\begin{pmatrix}-\beta & \alpha \\-\bar\alpha & -\bar\beta\end{pmatrix}$$
whihch doesn't look like the initial vector. I also tried to see if conjugation would work, but:
$$\frac 1 {\sqrt 2}\begin{pmatrix}\alpha & \beta \\ -\bar \beta & \bar \alpha\end{pmatrix} \begin{pmatrix}0 & 1 \\-1 & 0\end{pmatrix}{\begin{pmatrix}\alpha & \beta \\ -\bar \beta & \bar \alpha\end{pmatrix}}^{-1}$$
doesn't seem to work as well. I'm clearly missing something here, but can't tell what.
Can anyone help me shed some light on this subject?
Thanks
Gianni