In my quantum mechanics courses I have come across this notation many times: $$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$$ but I feel like I've never fully understood what this notation actually means. I know that it represents the fact that you can combine two spin 1/2 as either a spin 1 (triplet) or a spin 0 (singlet). This way they are eigenvectors of the total spin operator $(\vec S_1+\vec S_2)^2.$ I also know what the tensor product (Kronecker product) and direct sum do numerically, but what does this notation actually represent?

Does the 1/2 refer to the states? Or to the subspaces? Subspaces of what exactly (I've also heard subspaces many times but likewise do not fully understand it). Is the equal sign exact or is it up to some transformation?

And finally is there some (iterative) way to write a product of many of these spin 1/2's as a direct sum? $$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\dots=\left(\mathbf{1}\oplus\mathbf 0\right)\otimes\mathbf{\frac 1 2}\dots=\dots$$

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    $\begingroup$ A comment on an alternative way to understand this: Suppose you have two spin-1/2 fields, which we represent by the SU(2) spinors $\psi_{\alpha}$ and $\chi_{\beta}$ . We take their tensor product: $\psi_{\alpha}\chi_{\beta}$. But we would like to decompose this into an irreducible representation, and $\psi_{\alpha}\chi_{\beta}$ is not irreducible since the irreps are realised on totally symmetric spinors. $\endgroup$ Mar 6, 2020 at 14:00
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    $\begingroup$ It's not too hard to show that $\psi_{\alpha}\chi_{\beta}=\psi_{(\alpha}\chi_{\beta)}+\frac{1}{2}\varepsilon_{\alpha\beta}\psi^{\delta}\chi_{\delta}$, where $\varepsilon_{\alpha\beta}$ is the anti-symmetric spinor metric (it is an SU(2) invariant). The first (totally symmetric) term, $\psi_{(\alpha}\chi_{\beta)}$, belongs to the spin $\bf{1}$ irrep, whilst the contraction $\psi^{\delta}\chi_{\delta}$ belongs to the scalar rep $\bf{0}$. This kind of decomposition may be represented schematically by the notation $\bf{\frac{1}{2}}\otimes \bf{\frac{1}{2}}=\bf{1}\oplus \bf{0}$. $\endgroup$ Mar 6, 2020 at 14:09

4 Answers 4


This is a really deep question and I urge you to go ahed and read about it in the literature i'll give at the end. I'll try to give a glimpse of what this actually means.

In physics we can construct our theories based solely upon symmetries of a system. When talking about angular momentum and spin in non relativistic quantum mechanics, we are talking about a specific set of symmetry, namely $SU(2)$ symmetry. $SU(2)$ is a lie group and, being a group, it's an abstract object. To make it something useful we use, what is called as a representation. There are many representation of $SU(2)$ and the one we're interested in is the spinorial representation.

The spinorial is the fundamental representation of $SU(2)$ since all representations can be constructed from tensor product of spinors. In physical terms this means that you can construct composite systems just by using spin $1/2$ particles. What you gave is how to construct a spin $1$ or spin $0$ from two spin $1/2$ systems $$\mathbf{\frac{1}{2}\otimes\frac{1}{2} = 0\oplus 1} $$

what do this numbers indicate? A number written in boldface gives the dimension (which is $2j+1$ where $j$ is the boldface number) of an irriducible representation of that group. What this implies is that you can decompose a composite system of two spin $1/2$ particles into two irriducible representation of a spin $0$ system and a spin $1$ system.

If all of this seems confusing, it's normal, it's a lot of stuff. I would suggest the following reading if you want to get a better undestanding

  • Lie algebras in particle physics, Georgi
  • Group Theory in a Nutshell for Physicists, Zee
  • Group Theory, a physicist's survey, Ramond
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    $\begingroup$ Great answer. I'd add the (usually well known) example of two spin 1/2 particles coupling to a singlet state and a triplet state, where those two exactly correspond to the boldface 0 and 1 in your post. Making this connection helped me understand this topic much better. $\endgroup$
    – ersbygre1
    Mar 7, 2020 at 8:19
  • $\begingroup$ @Stephan I'm happy it helped! $\endgroup$
    – Quiver
    Mar 7, 2020 at 10:12
  • $\begingroup$ Interesting suggestion of books, none of which are particularly well written or clear IMO. Anyways a range of possible textbooks can be found here: physics.stackexchange.com/q/6108/36194 $\endgroup$ Mar 7, 2020 at 17:06

The $\otimes$ sign denotes the tensor product. Given two matrices (let’s say $2\times 2$ although they can be $n\times n$ and $m\times m$) $A$ and $B$, then $A\otimes B$ is the $4\times 4$ matrix \begin{align} A\otimes B =\left( \begin{array}{cc} A_{11}B&A_{12}B\\ A_{21}B&A_{22}B \end{array}\right)= \left(\begin{array}{cccc} A_{11}B_{11}&A_{11}B_{12}&A_{12}B_{11}&A_{12}B_{12}\\ A_{11}B_{21}&A_{11}B_{22}&B_{12}B_{21}&A_{12}B_{22}\\ A_{21}B_{11}&A_{21}B_{12}&A_{22}B_{11}&A_{22}B_{12}\\ A_{21}B_{21}&A_{21}B_{22}&A_{22}B_{21}&A_{22}B_{22} \end{array}\right) \, . \end{align} A basis for this space is spanned by the vectors \begin{align} a_{1}b_{1}&\to \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)\, ,\quad a_1b_2 \to \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}\right)\, ,\quad a_2b_1\to \left(\begin{array}{c} 0 \\ 0 \\ 1 \\0\end{array}\right)\, ,\quad a_2b_2\to \left(\begin{array}{c} 0\\0\\0\\1\end{array}\right) \end{align} In terms of $a_1\to \vert +\rangle_1$, $a_2\to \vert -\rangle_1$ etc we have \begin{align} a_1b_1\to \vert{+}\rangle_1\vert {+}\rangle _2\, ,\quad a_1b_2\to \vert{+}\rangle_1\vert{-}\rangle _2 \, ,\quad a_2 b_1\to \vert{-}\rangle_1\vert {+}\rangle _2 \, ,\quad a_2b_2\to \vert{-}\rangle_1\vert{-}\rangle_2\, . \end{align} In the case of two spin-$1/2$ systems, $\frac{1}{2}\otimes \frac{1}{2}$ implies your are taking $\sigma_x\otimes \sigma_x$, $\sigma_y\otimes \sigma_y$, $\sigma_z\otimes \sigma_z$, since these are operators acting on individual spin-$1/2$ systems. The resulting matrices can be simultaneously block diagonalized by using the basis states $a_1b_1$, $\frac{1}{\sqrt{2}}(a_1b_2\pm a_2b_1)$ and $a_2b_2$. There is a $3\times 3$ block consisting of $a_1b_1, \frac{1}{\sqrt{2}}(a_1b_2+a_2b_1)$ and $a_2b_2$ and a $1\times 1$ block with basis vector $\frac{1}{\sqrt{2}}(a_1b_2-a_2b_1)$.

The $3\times 3$ block never mixes with the $1\times 1$ block when considering the operators $S_x=s_x^{1}+s_x^{2}$ etc. The basis vectors of the $3\times 3$ block transform as states with $S=1$, in the sense that matrix elements of $S_x$, $S_y$ and $S_z$ are precisely those of states with $S=1$; the basis vector of the $1\times 1$ block transforms like a state of $S=0$. Hence one commonly writes \begin{align} \frac{1}{2}\otimes \frac{1}{2} = 1\oplus 0 \end{align} with the $\oplus$ symbol signifying that the total Hilbert space is spanned by those vectors spanning the $S=1$ block plus the vector spanning the $S=0$ part; note that those vectors are product states of the type $a_1b_1$ etc.

  • $\begingroup$ I accepted this answer because it shows most explicitly what is happening. So are you saying that the systems are related by a unitary basis transformation $U$ that is constructed from the basis states? $\endgroup$ Mar 8, 2020 at 22:25
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    $\begingroup$ Yes. The correct combination of basis states is actually given by the Clebsch-Gordan coefficients for the coupling of $A$ and $B$. Although many are familiar with CGs in the context of angular momentum, they are defined quite generally for the decomposition of any product of representations of the type $A\otimes B$. $\endgroup$ Mar 8, 2020 at 22:53

I should concur with the other answers that there is no substitue for reading up in good texts and WP.

You are right that, in a given basis, there is a similarity (equivalence) transformation implied in the equation of your title: it basically means that the tensor product on the l.h.s. is reducible, by an orthogonal basis change to the r.h.s.; that is, in words,

  • The Kronecker product of two 2-vectors (spinors; in general you have (2s+1)-dim vectors!) is a 4-vector. But rotations keep two subspaces in it separate: a 3-vector subspace, and a 1-vector (scalar) subspace. However, this is invisible to the naked eye. There is an orthogonal basis change, the Clebsch matrix, which visibly separates these two subspaces, so rotations act on these visibly separately, by block matrix action. (In your singlet case, by no action at all! the rotation matrices are the identity, 1).

Can you find this 4×4 Clebsch matrix $\cal C$ in Problem 4 here for your exact problem? (Hint: mix up just the 2nd and 3rd components by a rotation of $\pi/4$.) The "right is detail-in-left" convention in the tensor product amounts to $$ \begin{pmatrix} a_1\\a_2\end{pmatrix} \otimes \begin{pmatrix} b_1\\b_2\end{pmatrix} = \begin{pmatrix} a_1 b_1\\a_1 b_2 \\ a_2 b_1\\ a_2 b_2\end{pmatrix} \leadsto \begin{pmatrix} \uparrow \uparrow\\ \uparrow \downarrow \\ \downarrow \uparrow\\ \downarrow \downarrow \end{pmatrix} , $$ in the spherical basis notation. The second and 3rd component, then, mix up into $(\frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2}},\frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} )$, the triplet component and the singlet component.

The upshot is a direct sum of a 3-vector (components 1,2, & 4) and a singlet (component 3): $$ \begin{pmatrix} \uparrow \uparrow\\ \frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2} } \\ \downarrow \downarrow \end{pmatrix} \oplus \frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} . $$

Your title formula, however, never picks a basis.

Finally, there are elaborate formulas for recursive compositions of spins, pioneered by Bethe and elaborated by several authors afterwards. Your case is particularly simple, as WP details. I copy the WP formula, which uses dimensionality, instead of spin indices (2s+1 instead of your s), since you can do instant arithmetic checks by ignoring the circles in × and + !

Combining n doublets (your spin 1/2s) nets you $$ {\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,$$ where $\lfloor n/2 \rfloor$ is the integer floor function; the number preceding the boldface irreducible representation dimensionality (2 s+1) label indicates multiplicity of that representation in the representation reduction. The random walk that takes you there reconstructs the celebrated Catalan's triangle.

For instance, from this formula, addition of three spin 1/2 s yields a spin 3/2 and two spin 1/2s, ${\mathbf 2}\otimes{\mathbf 2}\otimes{\mathbf 2}={\mathbf 4} \oplus{\mathbf 2}\oplus{\mathbf 2} $; four spin 1/2 s yields two singlets, three spin 1 s, and one spin 2, and so forth.

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    $\begingroup$ Just to make sure, the bold numbers in your last paragraph are the dimensionalities of the representations and not the spins, unlike the other answers. So in other notation $2\otimes2\otimes 2 = 4\oplus 2 \oplus 2 $ equals $1/2\otimes 1/2 \otimes 1/2 = 3/2 \oplus 1/2 \oplus 1/2 $, using spin quantum numbers. $\endgroup$
    – Hans Wurst
    Apr 13, 2021 at 22:09
  • $\begingroup$ Yes. Spin notation, by contrast, is only good for SU(2), and it obscures the trivial dimensionality check which is more than a check: it is a conceptual enforcer.... $\endgroup$ Apr 14, 2021 at 0:12

This is actually the decomposition for the tensor product of irreducible representation of SU(2). We can set your $1/2$ as $j$, which means the (2j+1) dimimension irreducible representation of SU(2). Generally, Clebsch–Gordan series gives: $$D^{\left(j_{1}\right)} \otimes D^{\left(j_{2}\right)}=\bigoplus_{J=\left|j_{1}-j_{2}\right|}^{j_{1}+j_{2}} D^{(J)}$$

thus, it can explains the reason behind $\frac{1}{2} \otimes \frac{1}{2}=0 \oplus 1$. Physically, the $1/2,0,1$ here just means the $S=1/2,0,1$. And the dimension of representation, i.e. $j$, means the number of states in this space, e.g. $s_z=-1, 0, 1$ for $S=1$.


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