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I'm rather confused, and I was hoping if someone could help me figure out this (probably rather elementary) issue. I have two particles with spin 1, whose state I describe by $m_S$ and $m_I$ respectively. Both can take on the values -1, 0, and 1.

Now, what I want to do is compute a tensor product. Now, this is where I might already be going wrong, but I of course have to pick a basis. Would it be possible to say that $\left|m_s = 1\right> = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\left|m_s = 0\right> = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, $\left|m_s = -1\right> = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$, $\left|m_I = 1\right> = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\left|m_I = 0\right> = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$, $\left|m_I = -1\right> = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix}$

I suppose that this might already be where things start going wrong, as maybe I can't pick the same basis vectors for two particles? Anyhow, IF the above is true, I would then for example like to calculate

$\frac{1}{\sqrt{2}}(\left|m_s = 0\right> -i\left|m_s = -1\right>) \otimes \left|m_I = -1\right>$

Now the way I see it, this would simply be a 9x1 vector equal to \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\\frac{-i}{\sqrt{2}} \\ \end{pmatrix}

However, I'm unsure if what I'm writing here is correct. In the next part of my calculations I introduce a hamiltonian term, which causes the spin of the first particle to evolve depending on the spin of the second particle, and working out the terms it seems as if things go horribly wrong. As the hamiltonian is simply a constant times the tensor product of the two z pauli spin matrices for a spin 1 particle, there's not much that could go wrong there, so I figured the error must be in here somewhere.

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Hm, quick comment, I just noticed something I did wrong in the part where not much could go wrong. Maybe I can delete the post. – user129412 Feb 27 '14 at 23:09
Even better: post the solution. – Faq Feb 27 '14 at 23:38
up vote 3 down vote accepted

The two particles $m_s$ and $m_I$ live in different vector spaces, so you are actually not picking the same basis vectors (because the basis vectors of the different particles belong to two separate vector spaces).

Secondly, the tensor product between the basis vectors of the two different vector spaces will form the basis vectors of a new $3 \times 3 = 9$ dimensional vector space. For instance: \begin{equation} |m_s=1\rangle \otimes |m_I=0\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} \end{equation} So the equation you have written down is correct.

I would strongly recommend these short set of notes to get a better feeling of the full machinery.

Edit in response to comment:

We need to use the tensor product to add angular momenta. For simplicity, consider a system with two particles, and the spin of both particles is $1/2$. Then, these particles can be combined in the following way: \begin{equation} \begin{array}{cccc} \uparrow \uparrow \; ,& \uparrow \downarrow \; ,& \downarrow \uparrow \; ,& \downarrow \downarrow \end{array} \end{equation} This means that the two-particle Hilbert space (i.e. the Hilbert space corresponding to the system) is spanned by four basis vectors: \begin{equation} |s_1,m_1; s_2,m_2 \rangle \equiv | s_1,m_1 \rangle \otimes |s_2,m_2\rangle \end{equation} Subsequently, as I have sketched in this answer, we decompose this tensor product to determine what the possible eigenvalues for the magnitude and $z$-component of the total system (i.e. the system formed by the two spin-$1/2$ particles) can be.

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I'm sure I'm missing something but aren't we adding angular momentum eigenstates? So shouldn't the answer be a linear combination of states weighted by their Clebsch Gordan coefficients? – JeffDror Feb 28 '14 at 0:20
@JeffDror I have updated my message. Hope this answers your question (roughly). – Hunter Feb 28 '14 at 0:37
@JeffDror also, the OP is not directly asking to add the angular momentum of the two particles. – Hunter Feb 28 '14 at 0:45
You're right, thanks! Now I understand. – JeffDror Feb 28 '14 at 0:49
Thanks a bunch for the help, and also especially for those notes, very helpful! – user129412 Feb 28 '14 at 9:53

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