1
$\begingroup$

In the discussion for the addition of angular momentum for composite systems, my lecturer uses the following notation in his notes when referring to a composite system of two spin-half particles:

$ \frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$

There is no further explanations in the notes but I am just curious what this means in the context of composite systems.

$\endgroup$
  • $\begingroup$ For context, do you know what a representation is? $\endgroup$ – knzhou May 26 at 15:22
  • $\begingroup$ Only very vaguely (only had two lectures on representations as part of a very compact course) $\endgroup$ – alfred May 26 at 15:27
  • 1
    $\begingroup$ By the way, in terms of the dimensions of the representations rather than the spin quantum numbers, you may see this written $2 \otimes 2 = 3 \oplus 1$ which looks a lot more intuitive (at least to me)! $\endgroup$ – G. Smith May 26 at 19:45
2
$\begingroup$

This answer is less mathematical and more physical. If you want to understand more about this decomposition you should consult texts on representation theory. Nevertheless, you can understand this without the full mathematical knowledge.

You started with two spins, or two Hilbert spaces $\mathcal{H}_{1,2}={\rm span}\left\{\left|\uparrow\right>,\left|\downarrow\right>\right\}$ each describing one of your particles. In these bases the operators $\hat{S}^{2}_{1,2}$ and $\hat{S}^{z}_{1,2}$ were diagonal. Your next step was to build a new Hilbert space that describes the two-particle system. This is known as the tensor product space

$$\begin{matrix}\mathcal{H}=\mathcal{H}_{1}\otimes\mathcal{H}_{2}&=&{\rm span}\left\{\left|\uparrow\right>\otimes\left|\uparrow\right>,\left|\downarrow\right>\otimes\left|\uparrow\right>,\left|\uparrow\right>\otimes\left|\downarrow\right>,\left|\downarrow\right>\otimes\left|\downarrow\right>\right\}=\\\\&=&{\rm span}\left\{\left|\uparrow\uparrow\right>,\left|\downarrow\uparrow\right>,\left|\uparrow\downarrow\right>,\left|\downarrow\downarrow\right>\right\}\end{matrix}$$

This is what you write as $\dfrac{1}{2}\otimes\dfrac{1}{2}$. You know from addition of angular momentum that it is possible to define a new basis

$$\left\{\begin{matrix}\left|0,0\right>&=&\dfrac{\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>}{\sqrt{2}}\\\left|1,-1\right>&=&\left|\downarrow\downarrow\right>\\\left|1,0\right>&=&\dfrac{\left|\uparrow\downarrow\right>+\left|\downarrow\uparrow\right>}{\sqrt{2}}\\\left|1,1\right>&=&\left|\uparrow\uparrow\right>\end{matrix}\right.$$

This basis diagonalizes $\hat{S}^{2}=\hat{S}_{1}^{2}+\hat{S}_{2}^{2}$ and $\hat{S}^{z}=\hat{S}^{z}_{1}+\hat{S}^{z}_{2}$, with $\hat{S}^{2}$ taking the form

$$\hat{S}^{2}=\hbar^{2}\cdot 1 \cdot\left(1+1\right)\left(\begin{matrix}0&&&\\&1&&\\&&1&\\&&&1\end{matrix}\right)$$

This is interesting. For a single particle of a specific spin $S$ the operator $\hat{S}^{2}$ is just $\hbar^{2}S\left(S+1\right)\hat{I}$ where $\hat{I}$ is the identity matrix. It means that you can interpret the above matrix as a system of spin $S=0$ plus a system of spin $S=1$, which mathematically can be written as $0\oplus1$.

$\endgroup$
  • $\begingroup$ Thank you very much! A physical interpretation is exactly what I am hoping to look for and your answer is really helpful! $\endgroup$ – alfred May 26 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.