# Quantum Mechanics - Two spin 1/2 particles

#### Tangent87

How do we show that the state $$|\chi>=\frac{1}{\sqrt{2}}(|\uparrow>|\downarrow>-|\downarrow>|\uparrow>)$$ has total spin zero? Does it involve acting some combination of the spin operators on it?

I know that the total spin operator $$\underline{S^2}=S_x^2+S_y^2+S_z^2=\frac{3\hbar^2}{4}I$$ where I is the 2x2 identity matrix but I don't see how that helps.

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For two particles you should use

$$S_{12}=S_1\otimes S_2$$

For each spin component. So, for $S_z$ it will be easy. For $S^2_x+S^2_y$ you play with $S_x+iS_y[/tex] and its conjugate, their commutations with [itex]S_z$, or you can play with the explicit form of states and matrices.

#### Dickfore

Act on it with the operator:

$$\hat{S}^{2} = \hat{S}^{2}_{1} + \hat{S}^{2}_{2} + 2 (\hat{\mathbf{S}}_{1} \cdot \hat{\mathbf{S}}_{2})$$

and express the dot product with the ladder operators:

$$(\hat{\mathbf{S}}_{1} \cdot \hat{\mathbf{S}}_{2}) = \hat{S}_{1 x} \hat{S}_{2 x} + \hat{S}_{1 y} \hat{S}_{2 y} + \hat{S}_{1 z} \hat{S}_{2 z}$$
$$= \frac{1}{4} \left[\left(\hat{S}_{1 +} + \hat{S}_{1 -}\right) \left(\hat{S}_{2 +} + \hat{S}_{2 -}\right) - \left(\hat{S}_{1 +} - \hat{S}_{1 -}\right) \left(\hat{S}_{2 +} - \hat{S}_{2 -}\right)\right] + \hat{S}_{1 z} \hat{S}_{2 z}$$
$$= \frac{1}{2} \left(\hat{S}_{1 +} \hat{S}_{2 -} + \hat{S}_{1 -} \hat{S}_{2 +}\right) + \hat{S}_{1 z} \hat{S}_{2 z}$$

You should get:

$$\hat{S}^{2} |\chi\rangle = S (S + 1) |\chi\rangle$$

What value for $S$ do you get?

#### Tangent87

Thank you for your help, I get $$\hat{S}^{2} |\chi\rangle = 0$$, I take it that's correct as it's supposed to be a state of total spin 0? Is there no easier way to do it though as it took quite a bit of work and I see the result stated in so many text books as if it's obvious?

#### Dickfore

There is, through the use of Young tableaux.

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