# How to get the expectation value of the spin of a generate spin triplet states? [closed]

For two spin-1/2 electrons, the general spin triplet states is a linear combination of the three basis states: $\left.|\uparrow\uparrow\right>, \left.|\uparrow\downarrow\right>+\left.|\downarrow\uparrow\right>, \left.|\downarrow\downarrow\right>$, which are the simultaneously eigenstates of $S^z=S_1^z+S_2^z$ and $\bf{S}^2=(S_1+S_2)^2$.

When denote these three states in the matrix form respectively(named as direct product of states?): $\chi^{1}=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \chi^{0}=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \chi^{-1}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$, the most generate form of a spin triplet states can be written as a linear combination of them, explicitly: $$\chi= \sum_i \alpha^i\chi^i$$

Question is, how can I get the expectation value of $\bf{S}$ of this generate spin state using its matrix form? I don't know how to express the operator $\bf{S}$ in matrix form, it seems that $\bf{S}=\bf{S_1}\otimes1+1\otimes S_2$ is a $4\times 4$ matrix and I don't know how to apply this on this $2\times 2$ state.

ps: I know the method using the operator form, which requires that $\bf S_1$ only operate on the first spin and $\bf S_2$ only operate on the second spin, etc... I just want to see how the matrix form of spin state can be used to do the calculation...

## closed as unclear what you're asking by Norbert Schuch, user36790, HDE 226868, Gert, Sebastian RieseJan 3 '16 at 18:50

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The expressions you wrote for the $\chi^j$ are not exactly what you want. You'd like to define a composite state in the basis $$\vert\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix},\quad \vert\downarrow\rangle = \begin{pmatrix}0\\1\end{pmatrix}$$ which is $$\vert\uparrow\uparrow\rangle\equiv\vert\uparrow\rangle \otimes \vert\uparrow\rangle = \begin{pmatrix}1\\0\end{pmatrix} \otimes \begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$ by the common definition of the Kronecker product. As you noted correctly, by adopting this convention operators like $S^1\otimes 1$ are represented by $4\times 4$ matrices. Again, their representations in any given basis is by the Kronecker Product, e.g

$$S^3\otimes 1 = \frac{1}{2}\begin{pmatrix} 1 &&& \\ & 1 && \\ && -1 & \\ &&& -1\end{pmatrix}$$

Instead of this more canonical choice you wrote e.g

$$\chi^1 = \begin{pmatrix}1&0\\0&0\end{pmatrix}$$ which is also possible, but uncommmon. One would rather identify this expression with $$\vert\uparrow\rangle \otimes \langle\uparrow\vert \in \mathbb{C}^2\otimes (\mathbb{C}^2)^*$$

Since $\mathbb{C}^4 \simeq \mathbb{C}^{2\times 2} \simeq \mathbb{C}^2\otimes (\mathbb{C}^2)^* \simeq\mathrm{End}(\mathbb{C}^2)$, both choices are equivalent. Work out the isomorphism relating the two representations! From there it should not be hard to derive the action of the composite operators on $\chi^j$. Be guided by what you already know, namely that by definition

$$(A\otimes B)\cdot (u\otimes v) = (Au)\otimes(Bv)$$

It then just comes down to choosing a basis for the tensor product space as either $\mathbb{C}^4$ or $\mathbb{C}^{2\times 2}$.

• Thank you for your answer, many points you made are both important and clear. Regarding to your answer: thses seemingly uncommon expressions for the spin states are widely used in unconventional superconductors. It seems in your answer you claimed the isomorphism between these two representation. However, it is still unclear how to calculate the expectation value directly from the matrix form. The last expression in your answer still not utilize the matrix expression, but decompose it in two vector space. I think people written the spin states as the matrix form must have the reason that cal – an offer can't refuse Jan 4 '16 at 4:20
• …calculation using the matrix form is easier, but from your last expression, you just go backwards, decompose the matrix, then calculate in two sperate spin space and combine them again. – an offer can't refuse Jan 4 '16 at 4:24