# Total spin of system of two spin-$1/2$ particles

Consider a quantum system of two spin one half particles. Let $\alpha(1)$ be 'spin up' for first system, and $\beta(1)$ 'spin down' for first system, and likewise for second system. We have $$\chi = \frac{1}{\sqrt{2}} \bigg( \alpha(1) \beta(2) - \beta(1) \alpha(2) \bigg).$$ If I want to compute the total spin of the system, do I just write $$S = S_1 + S_2 = (S_{1x} + S_{1y} + S_{1z}) + (S_{2x} + S_{2y} + S_{2z} )$$ and act with this operator on $\chi$? Then the components of $S_1$ would act only on state $1$, and the components of $S_2$ on state $2$. Is that the correct approach to compute the total spin?

• Just like in case of a single spin, the complete quantum state of a system is specified by the total spin and just one component (the default choice being the z-component, or you could say that we choose the z-direction to correspond to whatever component we choose). The 3 different spin components don't commute with each other, $S^2$ and $S_z$ form a complete set of commuting observables. So, what you need to do here is compute $S^2$ and $S_z$ for this state. – Count Iblis Jun 12 '16 at 17:40
• That's not how I did it. I used the relations $S_x \alpha = \frac{\hbar}{2} \beta, S_y \alpha = \frac{i\hbar}{2} \beta$, etc. And computed it like that. Is that correct? – Kamil Jun 12 '16 at 18:07
• The space of states of one $\:s=1/2\:$ particle is a 2-dimensional Hilbert space. The space of states of two $\:s=1/2\:$ particles is a 4-dimensional Hilbert space. You must study how to construct this 4-dimensional Hilbert space from the two identical copies of a 2-dimensional Hilbert space and how to construct operators on the 4-dimensional space from those on the 2-dimensional ones. I think that this is the only way to understand the addition of angular momenta (orbital and/or spin) in quantum mechanics. – Frobenius Jun 12 '16 at 23:43

There are two quantum numbers to consider, the total spin = 1/2 and the azimuthal or projection quantum number on an axis, say z with angular momentum +-hbar/2. This quantum number is also 1/2 for electrons. (The x and y components are undefined by the uncertainty principle as spin quantum number and spin z component are defined) Thus for 2 electrons there are 4 states produced, one with electron spin paired (total spin =0) which a singlet state with and three triplet states with total spin 1. In the absence of any external fields these have the, same energy. The figure below gives more details.

If you look at some text books on atomic spectroscopy you will find this described in detail.
The simplest description is in 'Modern Molecular Photochemistry' by N. Turro, (Chapter 2) more detailed are 'Modern Spectroscopy' by Hollas; (chapter on electronic spectroscopy); 'Molecular Quantum Mechanics' by Atkins & Friedman,(Chapter 4) and 'Molecules & Radiation' by J. Steinfeld. (Chapter 2). Look for Russell Saunders Coupling for an explanation of adding angular momentum in general.

• What a nice figure! Can you edit you answer to include a link back to its source? – rob Jun 15 '16 at 23:01
• I made the figures myself but the 'vector model' of spin and angular momentum can be found in several textbooks. The simplest description is in 'Modern Molecular Photochemistry' by N. Turro, (Chapter 2) more detailed are 'Modern Spectroscopy' by Hollas; (chapter on electronic spectroscopy); 'Molecular Quantum Mechanics' by Atkins & Friedman,(Chapter 4) and 'Molecules & Radiation' by J. Steinfeld. (Chapter 2). Look for Russell Saunders Coupling for an explanation of adding angular momentum in general. Hope this helps. – porphyrin Jun 16 '16 at 12:43

In quantum mechanics you cannot "compute the total spin" of a certain quantum state: all you can compute are the expected values of the spin components.

(I'm going to use Dirac notation because I find it much clearer) To compute expected values in your case it is convenient to use the $\mid s,m_z\rangle$ representation. If $\vec S = \vec S_1 + \vec S_2$ is the total spin of the system, you have

$$S^2 \mid s,m_z\rangle =(\vec S_1 + \vec S_2)^2 \mid s,m_z\rangle = s(s+1) \hbar^2 \mid s,m_z\rangle$$

and

$$S_z \mid s,m_z\rangle = (S_{z,1}+S_{z,2}) \mid s,m_z\rangle = m_z \hbar \mid s,m_z\rangle$$

The state you want to consider is the singlet state, which we will write in the $\mid s,m_z\rangle$ representation (see here for example):

$$\mid \chi \rangle = \frac 1 {\sqrt{2}} (\mid \uparrow,\downarrow \rangle - \mid \downarrow,\uparrow \rangle) = \ \mid s=0,m_z=0\rangle \ = \ \mid 0,0 \rangle$$

So you will have:

$$\langle 0,0\mid S^2 \mid 0,0 \rangle = 0$$

$$\langle 0,0\mid S_z \mid 0,0 \rangle = 0$$

To calculate $\langle 0,0\mid S_x \mid 0,0 \rangle$ and $\langle 0,0\mid S_y \mid 0,0 \rangle$, use the fact that

$$S_x=\frac{S_+ + S_-}{2}$$ $$S_y = \frac{S_+-S_-}{2i}$$

where

$$S_{\pm}\mid s,m_z\rangle = \hbar \sqrt{s(s+1)-m_z(m_z\pm1)} \mid s,m_z \pm 1\rangle$$