Let us consider operators $c_{\uparrow}$ and $c_{\downarrow}$ which destroy a fermion with spin up and a fermion with spin down, respectively. These operators can be found, for example, in the Hubbard Hamiltonian:

$$ H = -\sum_{\langle ij \rangle\sigma} (c_{i\sigma}^{\dagger}c_{j\sigma} + H.c.) + U\sum_i n_{i\uparrow}n_{i\downarrow} $$ where $$ n_{i\uparrow}=c_{i\uparrow}^\dagger c_{i\uparrow}, $$ $$ n_{i\downarrow}=c_{i\downarrow}^\dagger c_{i\downarrow}. $$

It is clear that these fermion operators satisfy anticommutation relations (I drop, for simplicity, the site index $i$): $$ \{c_\uparrow,\, c_\uparrow^\dagger \}=1, $$ $$ \{c_\downarrow,\, c_\downarrow^\dagger \}=1. $$ One can then write the spinor $$ \Psi= \begin{align} \begin{pmatrix} c_{\uparrow} \\ c_{\downarrow} \end{pmatrix} \end{align}. $$ One can then apply an unitary transformation, i.e. enact a change of basis of the type: $$ \Psi^\prime=\mathbb{M}\Psi $$ where $\mathbb{M}$ is a $2\times 2$ unitary matrix, i.e. a matrix belonging to the group $U(2)$. For example, matrix $\mathbb{M}$ can be chosen as $$ \mathbb{M}= \left( {\begin{array}{cc} e^{i\varphi_1} & 0 \\ 0 & e^{i\varphi_2} \\ \end{array} } \right). $$ Choosing a diagonal matrix simply allows one not to mix $c_\uparrow$ and $c_\downarrow$. So it seems that one is free to "rotate" the operator $c_\uparrow$ of angle $\varphi_1$ and "rotate" the operator $c_{\downarrow}$ of angle $\varphi_2$. The two "rotations", $\varphi_1$ and $\varphi_2$ look completely independent.

At this point, one has to take into account that the symmetry group of the Hubbard Hamiltonian and, more in general, the symmetry group of rotations in spin space, is $SU(2)$ and not $U(2)$. This circumstance immeditaely introduces a constraint on the possible values of $\varphi_1$ and $\varphi_2$. More specifically, matrix $\mathbb{M}$ is an element of the group $SU(2)$ provided that $$ \varphi_1=-\varphi_2. $$ In conclusion, due to the fact that spin rotations are known to have $SU(2)$ symmetry (and not the $U(2)$ symmetry!) one cannot "rotate" the two components of the spinor, $c_\uparrow$ and $c_\downarrow$, in an independent way.

My questions are:

  • Why do spin rotations have $SU(2)$ symmetry and not $U(2)$ symmetry?
  • Which is the physical meaning/origin of the constraint $\varphi_2=-\varphi_1$, which reduces the symmetry from $U(2)$ to $SU(2)$?
  • Why do textbooks usually say that the Hubbard Hamiltonian has $SU(2)$ symmetry and not $U(2)$ symmetry? It seems that authors usually factor out a global $U(1)$ phase factor.
  • 2
    $\begingroup$ why $U(2)$ and not $GL(2)$? the symmetry group depends on the Hamiltonian -- care to write it down? $\endgroup$ Mar 5, 2020 at 1:13
  • $\begingroup$ I have written down the Hubbard Hamiltonian. I hope that now the post is more clear. To anser your question, I think that $GL(2)$ would not be ok, as it does not "preserve the norm". I think matrix $\mathbb{M}$ should be regarded as a change of basis. $\endgroup$
    – AndreaPaco
    Mar 5, 2020 at 1:29
  • $\begingroup$ Thanks. $GL(n)$ is a perfectly good change of basis (of vector spaces; not necessarily normed vector spaces, but that is another question). Anyway, it is not obvious to me that your $H$ is invariant under independent rephasing. It looks like $c^\dagger c$ is only invariant if both operators transform with the same phase -- otherwise you pick up a phase $e^{\varphi_1-\varphi_2}$, right? what am I missing? $\endgroup$ Mar 5, 2020 at 1:47
  • $\begingroup$ Well, all operators with spin UP must undergo the same rotation, say $\varphi_1$. Similarly, all operators with spin DOWN must undergo the same rotation, say $\varphi_2$. Recall that, in the Hamiltonian, $i$ and $j$ are site indices while $\sigma$ is the spin index. Therefore, i understand that operators at different sites must undergo the same rephasing. But the spin-up and the spin-down families look like two independent channels. Therefore I would be tempted to rephase them in an INDEPENDENT way. But, it seems that I cannot, because of the SU(2) symmetry which introduces the constraint. $\endgroup$
    – AndreaPaco
    Mar 5, 2020 at 2:02

2 Answers 2


I think the model you are talking about does have a $$ U(2) = U(1)\times SU(2)$$ symmetry. The $U(1)$ part is just a global phase and responsible for particle number conservation (essentially you only have $c^\dagger c + c c^\dagger$ type terms and no $cc + c^\dagger c^\dagger$). The $SU(2)$ symmetry is spin-rotation symmetry and implies that your energy eigenstates form representations of $SU(2)$ (and can therefore have degeneracies).

So I think that you have the full $U(2)$ symmetry, it's just more meaningful to talk about the $U(1)$ and $SU(2)$ parts separately as they have different physical meaning.

In the matrix above you can reparametrize the phases as $\phi_1=\phi+\psi$ and $\phi_2=\phi-\psi$ to decouple the global $U(1)$ rotation related to charge and the $SU(2)$ spin-rotation component: $$ M = e^{i \phi}\begin{pmatrix} e^{i \psi} & 0\\ 0 & e^{-i \psi}\\ \end{pmatrix} $$

So you do have both, it's just physically more meaningful to separate these.

If your question is more deeper about why $SU(2)$ is associated to spin (in general, not just in this model) and not $U(2)$. Then you need to study the induced representation theory of the Poincare group (see here). For massive particles there is a so-called 'little group' from which representations of the Poincare group can be constructed from, this group is $SO(3)$. Since we are interested in projective representations in quantum physics (Wigner's theorem), we need to use the universal cover which is $SU(2)$ which is the relativistic origins of spin. But this is a whole story by itself.

  • 2
    $\begingroup$ s-wave pairing is probably the smoking gun here: $c_\uparrow c_\downarrow + H.c.$. Breaks the $U(1)$ part but keeps the $SU(2)$. $\endgroup$ Mar 5, 2020 at 2:05
  • 2
    $\begingroup$ FWIW: $U(2)$ is not equal to $U(1)\times SU(2)$. There is a central quotient (which is usually rather important; especially here, since it does not really allow you to think of $U(1)$ and $SU(2)$ as independent symmetries). $\endgroup$ Mar 5, 2020 at 2:12
  • $\begingroup$ Thanks a lot for your very detailed answer. Ok, so the Hubbard Hamiltonian has U(2) symmetry. Let us leave aside the U(1) part and let us focus on the SU(2) part. Can you tell me an intuitive reason why the two rotations must be opposite? Does it have something to do with the concept of “not breaking the quantization axis”, i.e. not breaking the property that spin-up and spin-down operators must be “collinear”? $\endgroup$
    – AndreaPaco
    Mar 5, 2020 at 2:19
  • $\begingroup$ Can you also please address the last question which I inserted in the post. Thanks in advance. $\endgroup$
    – AndreaPaco
    Mar 5, 2020 at 14:13

Spinors, unlike vectors, need to pick up a negative sign when the coordinate system (continuously) undergoes a full $2\pi$ rotation. They are meant to transform, roughly, like the square-root of a vector. They became important to physicists after Dirac was looking for (what’s now called) the formal square root of a Laplacian.

The unitary group clearly doesn’t accomplish this, but yes there is a $U(1)$ global phase factor on states in, for fermions, the alternating algebra of a (separable) Hilbert space.

Spin$(n)$ is the group of transformations that captures the two topologically distinct classes (homotopy classes) of paths on the space that spinors “notice” and vectors don’t. As an example, Spin$(3)$ is isomorphic to $SU(2)$; from these, there exists a surjective 2:1 homomorphism to $SO(3)$. The 2:1 cover is essential and while $U(n)$, with its norm-preserving goodness, is important, it doesn’t capture how spinors transform under rotation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.