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I originally asked this question on math SE but I'm asking it again here due to the lack of responses. I should note that I come from a mathematical background and not a physics one so I am not comfortable defining things in terms of how they transform and I have a hard time understanding spinors as introduced in many physics books. Here is the original post:

I am trying to understand spinors from a mathematical view. I've seen similar questions on this website but I'm still unclear on what they are exactly. On Wikipedia they state:

Although spinors can be defined purely as elements of a representation space of the spin group (or its Lie algebra of infinitesimal rotations), they are typically defined as elements of a vector space that carries a linear representation of the Clifford algebra.

What confuses me is say we are working on space time so that $\mathbb{R}^4$ is our vector space. Going by the above passage, a spinor would be an element of $\mathbb{R}^4$ that carries a representation of the spin group, let us denote this pair as $(x, \rho)$ where $\rho$ is the representation. However $x$ is itself also a vector since $x \in \mathbb{R}^4$. So why are vectors and spinors referred to as two different objects? In other words, how does the representation play any role in describing the spinor itself (which as I understand is simply a vector in the underlying vector space)?

My current guess is that spinors are always to be taken as a pair consisting of a vector and a representation, for example $(x, \rho)$ above. Thus, spinors are actually also vectors, but the difference is that when you talk about rotations (i.e. the action of O$(3)$) then spinors "rotate" differently than an ordinary vector in $\mathbb{R}^4$. However I am not sure if this is right since most textbooks do not describe anything along these lines. For example, no physics textbook describes a spinor as a pair $(x, \rho)$. Is this implicitly assumed?

In the example above involving $\mathbb{R}^4$, since $4 = 2n + 1 \implies n = 3/2$, this spinor is said to be spin 3/2. Thus using common physics terminology, a Dirac spinor is the unique 2-dimensional vector space (up to isomorphism) representing the action of $SO(3)$. Is this right?

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  • $\begingroup$ Going by the above passage, a spinor would be an element of $\mathbb{R}^4$… That’s not what the passage says. How well do you understand what a “representation space” of a Lie group or Lie algebra is? Often they are complex vector spaces. $\endgroup$
    – Ghoster
    Nov 26, 2023 at 23:09
  • $\begingroup$ @Ghoster A representation space is the underlying vector space $V$. If we take $V = \mathbb{R}^4$, then what is wrong with seeing spinors as an element of $\mathbb{R}^4$? $\endgroup$
    – CBBAM
    Nov 26, 2023 at 23:28
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    $\begingroup$ That’s a poor understanding of a representation space. In general, the four components of Dirac spinors are complex numbers. And Weyl spinors live in a two-dimensional complex vector space. $\endgroup$
    – Ghoster
    Nov 26, 2023 at 23:32
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    $\begingroup$ @Ghoster Thank you. So let $V$ be a representation space for $SU(2)$. Would it be correct to say that a spinor is an element of $V$ where the dimension of $V$ is such that the SU$(2)$ representation on $V$ does not reduce to an irreducible representation of SO$(3)$? $\endgroup$
    – CBBAM
    Nov 26, 2023 at 23:50
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    $\begingroup$ I’d say “where $V$ can carry a representation of SU(2) but not one of SO(3)”. $\endgroup$
    – Ghoster
    Nov 26, 2023 at 23:56

4 Answers 4

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The spin group has a multiple irreducible representations of dimension 4. Two of them are the left- and right-handed spin-3/2 representations. The other one is called the vector representation. "Ordinary" 4-vectors belong to (copies of) the latter representation, which is not considered one of the "spinor representations". Elements of these representations do all "rotate differently" from each other under elements of the spin group. In general, a representation of the spin group is labelled $(p,q)$ for $p,q$ nonnegative half-integers, with $n=p+q$ being the spin and the dimensionality being $(2p+1)(2q+1).$

In math, "vector" just means "element of a vector space". Spinors are by definition always "vectors" in the math sense. But in physics, "vector" usually means more than it does in math. Intuitively, a physical vector is "a magnitude with a direction" in space(time). An element of $\mathbb R^{100}$ is not going to be a physical vector. Even an element of $\mathbb R^4$ should only be considered a physical vector if you've fixed a coordinate system etc. to interpret it as a magnitude and a direction in spacetime. In fact, one structure you'd like to see on any space of physical vectors is a vector representation of the spin group on it, which tells you how to rotate the elements. (Again, the vector representation is the $\left(\frac12,\frac12\right)$ one, not the $\left(\frac32,0\right)$ or $\left(0,\frac32\right)$ spinor ones.) Hopefully this clarifies why in physics we distinguish "vectors" from "spinors".

Note that treating a spinor $v$ as a pair $(v,\rho)$ where $\rho$ is the representation is much like treating a (math) vector $v$ as a pair $(v, V)$ where $V$ is a vector space. It's somewhat wrong-headed. The word "vector" in math means "member of a vector space" and says nothing about what the object actually is. If you want to talk about vectors, you first fix your vector space and then you talk about its elements. The same is true for spinors. If you want to talk about spinors, you pick some spinor representation, perhaps postulated by some physical theory, and then you start talking about its elements. You don't consider the representation to be part of the spinor; it's the other way around.

The Dirac equation is a differential equation for a field $\psi:M^{3,1}\to V,$ where $V$ is the space of "Dirac spinors". Specifically, we write $V=\left(\frac12,0\right)\oplus\left(0,\frac12\right)$ to denote that $V$ is a four-dimensional space carrying a reducible representation of the spin group, where two of the dimensions transform under the (irreducible) $\left(\frac12,0\right)$ representation (left-handed spin-1/2) and the other two independently transform under the $\left(0,\frac12\right)$ irrep (right-handed spin-1/2), such that $V$ is a direct sum of two smaller spaces with the specified representations.

  • Note that $V$ carries yet another 4-dimensional representation of the spin group that is distinct from the 4-vector representation. Even though the Dirac spinor has four components, you can't apply a Lorentz transform with the usual $\mathbf\Lambda$ matrices we use for 4-vectors, but must come up with a new ones. The Dirac spinor rotates differently to a 4-vector.
  • Note that there is no such thing as the unique 2-dimensional representation of the spin group. There are two. The members of these 2-D representations are called Weyl spinors.
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    $\begingroup$ Thank you, this is exactly the kind of information I was looking for! $\endgroup$
    – CBBAM
    Nov 27, 2023 at 6:23
  • $\begingroup$ Which spin group is referred to in the first sentence? Spin(3)? Spin(3, 1)? Something else? $\endgroup$
    – Jagerber48
    Jan 1 at 6:43
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Here's a purely mathematical answer. The Lie algebra $\mathfrak{so}(1,3)$ of the Lorentz group is a simple 6-dimensional real Lie algebra which is isomorphic to $\mathfrak{sl}_2(\Bbb{C})$ viewed as a real Lie algebra.

The representations studied are complex representations so we take the complexification which is $\Bbb{C} \otimes_\Bbb{R} \mathfrak{so}(1,3)\cong \mathfrak{sl}_2(\Bbb{C}) \oplus \mathfrak{sl}_2(\Bbb{C})$ (isomorphism of complex Lie algebras.)

From here it's shown that the irreducible representations of $\mathfrak{so}(1,3)$ are a tensor product of an irreducible representation of the left factor by an irreducible representation of the right factor. Now, irreducible (complex) representations of $\mathfrak{sl}_2(\Bbb{C})$ (which are the same as those of $\mathfrak{su}(2)$) are characterized by their 'spin' which in physics is taken as a non-negative half-integer. The dimension of the representation for spin $l$ is $2l+1$, and the Casimir operator acts as a scalar $-l(l+1)$. (In physics this is multiplied by $(i\hbar)^2$ to give $l(l+1)\hbar^2$).

Here's a concrete view via a basis. $\mathfrak{so}(1,3)$ has a basis $J_1, J_2, J_3, K_1, K_2, K_3$ with commutation relations $$[J_i, J_j] = \varepsilon_{ijk}J_k\quad [J_i, K_j] = \varepsilon_{ijk}K_k\quad [K_i, K_j] = -\varepsilon_{ijk}J_k$$

Then over the complexification you define $$N^+_i=\frac{1}{2}(J_i + i K_i)\quad N^-_i=\frac{1}{2}(J_i - i K_i)$$ and then $$[N^+_i, N^+_j] = \varepsilon_{ijk} N^+_k\quad [N^-_i, N^-_j] = \varepsilon_{ijk} N^-_k\quad [N^+_\star, N^-_\star] = 0$$ which shows the split into a direct sum. The Casimirs are $$(N^+)^2 = \sum_{i=1}^3 (N^+_i)^2\quad (N^-)^2 = \sum_{i=1}^3 (N^-_i)^2$$ and then for an irreducible representation, $(N^+)^2$ acts as a scalar $-p(p+1)$ and $(N^-)^2$ acts as a scalar $-q(q+1)$ which is where the two 'spins' $(p, q)$ come from.

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  • $\begingroup$ Thank you, as someone who comes from a math background this makes everything very clear! $\endgroup$
    – CBBAM
    Nov 27, 2023 at 22:14
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...$x$ is itself also a vector since $x \in \mathbb{R}^4$. So why are vectors and spinors referred to as two different objects?

My current guess is that spinors are always to be taken as a pair consisting of a vector and a representation, for example $(x, \rho)$ above. Thus, spinors are actually also vectors, but

No, they are functions of a spacetime vector...

Just like a scalar function is a function of spacetime vector $x$ that produces a scalar value $\phi(x)$, which transforms trivially (i.e., via a $1\times 1$ representation where all the elements are $1$). So too, there is a spinor function that takes a space-time vector $x$ as input and produces a spinor $\psi(x)$, which transforms according to a $2\times 2$ representation.

So, under some transformation $x\to \Lambda x$, a scalar field ("field" in the physics sense just means "function") transforms like: $$ \phi(x) \to \tilde\phi(\tilde x) = \phi(\Lambda^{-1}\tilde x) $$ and a spin-1/2 spinor field transforms like: $$ \psi_a(x) \to \tilde\psi_a(\tilde x) = \sum_{b=1}^2D^{(1/2)}_{a,b}(\Lambda)\psi_b(\Lambda^{-1}\tilde x) $$ and a spacetime vector field transforms like: $$ V^\mu(x) \to \tilde V^{\mu}(\tilde x) = \sum_{\nu=0}^3\Lambda^\mu_\nu V^\nu(\Lambda^{-1}\tilde x) $$

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    $\begingroup$ Thank you for your answer. Wouldn't viewing spinors as functions make them spinor fields? If we remove the field aspect and view everything on a fixed vector space, is a spinor not an element of the representation space (i.e. the underlying vector space)? $\endgroup$
    – CBBAM
    Nov 26, 2023 at 23:32
  • $\begingroup$ For example, Wikipedia says: "Given a representation specified by the data $(V, Spin(p,q) \rho)$ where $V$ is a vector space and $\rho$ is a homomorphism $\rho: Spin(p,q) \rightarrow GL(V)$, a spinor is an element of the vector space $V$. $\endgroup$
    – CBBAM
    Nov 26, 2023 at 23:33
  • $\begingroup$ @CBBAM I'm not sure. I think someone else will have to help me out here... $\endgroup$
    – hft
    Nov 27, 2023 at 0:00
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    $\begingroup$ @hft it is common in physics to refer to (what a mathematician would call) tensor fields, spinor fields etc. simply as tensors or spinors $\endgroup$
    – Callum
    Nov 28, 2023 at 23:15
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The understanding of spinors in general is not easy, it requires quite a lot of knowledge on representation theory. In particular the treatment of the (restricted) Lorentz group in math lectures and books is very swift, although the construction of the group representations is quite complex.

The whole "hassle" with different objects transforming under different group representations is related with the fact that modern physics laws have to be covariant under Lorentz transformations (or at least rotations when only classical physics is considered), i.e. they should not change their form if they are studied in another reference frame. This is a fundamental requirement. This is the reason why a vector $(x,\rho_1)$ is so fundamentally different from another $(y,\rho_2)$ one. Imagine one would add both vectors $x + y$ (of course we assume that $x$ and $y$ of same dimension/length), upon a change of the reference system one would expect something like $x' + y'$, but due to the different transformation properties of $x$ and $y$ the result would be a complete mess. Such an expression is not covariant.

In particular in a vector space with representation $\rho_1$ a physicist always assumes that there is a scalar product, a scalar product which has the very important property that it is invariant under transformations of $\rho_1$. However, a vector transforming under a different representation would require another scalar product, a product that's invariant under $\rho_2$. So spaces with vectors of different representations are not compatible (well except if the other vector were a result of a product of the $\rho$-representation ($\rho \times \rho$), i.e. a tensor, a scalar product for the product representation can be constructed which can live aside of the simple $\rho$-representation on the same space). In particular if it is said that a vector lives in $\mathbf{R}^4$, it would suggest that the scalar product of the vector with itself is the Euclidean norm. And it might very well be that the transformation of the representation of the vector (x,$\rho$) does not keep the Euclidean scalar product invariant.

The vector space the spinors live in has nothing to do with the "well-known" spacetime, it is a space apart (a state space where typically 2 states for spin projections on the z-axis exist for +1/2 and -1/2; higher representations also with more states exist, see below), and the concept vector space only tells us about the basic calculation rules which govern in that space.

Actually the example of spin 3/2 is actually misleading, because in the actual standard model of elementary particle physics there are no 3/2 spin particles, but in hypothetical theories (Supergravity) there are. I think you used a formula for spinors being representations of SO(3), whereas the 4-dim. spinors you certainly refer to are representations of the (restricted) Lorentz-group where there is another formula.

When one constructs spinor representation one has first fix the base manifold over which the representations are constructed. In non-relativistic physics this is 3-dimensional Euclidean space, and the group would be SO(3). Then spinors --- actually one can imagine these as fibers of a bundle --- appear as projective representations of SO(3). They also appear as fundamental representations of the group SU(2) which is the universal cover of SO(3). In that case the spinors have representations of dimension 2s+1. For spin s=1/2 (fundamental representation) one gets a 2-dimensional state vector.

In 4-dimensional Minkowski space-time this is already different. It has hyperbolic geometry. Spinors are representations of the (restricted) Lorentz-group and actually 2 fundamental representations are possible, "standard" and "complex conjugated of standard" (this naming is not standard BTW). But both are again 2-dimensional.

However, one can also construct a reducible representation of a combination of a "standard" and a "complex conjugated of standard" spinor which then has 2+2=4 components --- this is certainly the spinor you refer to --- The construction looks complicated, but it turned out that these objects show up in the Dirac equation which describe the relativistic behaviour of electrons (and their anti-particles) in particle physics. For simplicity they are also called spinors --- simply because these are those the physicists have to deal with most frequently --- or sometimes Dirac-spinors but actually they are bispinors. Actually they were not constructed on purpose, instead the solutions of the Dirac-equation turned out to be bispinors. They describe 2 different spin states and electrons and their anti-particles. This is actually a consequence of being a representation of the Lorentz-group and relativity, in 3D classical Newton physics they would not exist.

The relationship with Clifford-algebra is actually not very helpful for the understanding. The group representations for the spinors can be constructed via the corresponding Lie-algebra, and the generators of the corresponding Lie-algebra are also members of an (appropiate) Clifford-algebra. Physicists only start talking about Clifford-algebra when dealing with the (bi)spinors as solutions of the Dirac-equation. Members of the $Cl_{1,3}$ appear directly in the Dirac-equation and (other) members appear in the Lie-algebra representation which serves for the construction of the (bi)spinor representation.

Actually this happens also for 2-D spin representations of SO(3)/SU(2): The Pauli-matrices are the generators of the fundamental Lie-algebra spin representation. And together with other elements the Pauli-matrices form also a Clifford-algebra. This fact, as already said, is not very important in physics (may be for some Fierz-identities). What really matters are the Lie-algebras. Inspite the complicated topology of the (restricted) Lorentz-group the group representations can be constructed from the Lie-algebra.

The math description of a spinor as $(x,\rho)$ is indeed not used in physics. The understanding of symbols is often implicit and contextual, the physicist knows from the context what he deals with. But the use of indices is still wide-spread. This means that if indices are used different index types are used for spinors and [vectors in 3D or 4D space].

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  • $\begingroup$ Thank you for this wonderful post! Yes you are right, to to obtain the spin 3/2 I used a formula for spinors representing SO(3). The mistakes I made was not not realizing QFT spinors are different than those in QM (the base manifolds differ and the symmetry groups, i.e. Lorentz vs SO(3), differ). Do you have any (preferably mathematical) resources you can recommend to learn more about spinors in this level of generality? $\endgroup$
    – CBBAM
    Nov 27, 2023 at 22:48
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    $\begingroup$ @CBBAM Well, my knowledge is based on many sources. One is "Relativity, groups & particles" from Sexl&Urbantke, but it is written for physicists. It covers most of what I said in the post, but it has also its weaknesses. The construction of the Lorentz group representations is not so clear. Actually, wikipedia is quite good: Representations of the Lorentz group is an excellent article, but difficult. Also the other articles on Lie groups and their representations are quite good. I had a look into some math books, at max they touch upon complex Lorentz group, but physics needs the real group. $\endgroup$ Nov 27, 2023 at 23:37
  • $\begingroup$ Thank you for the resources. $\endgroup$
    – CBBAM
    Nov 28, 2023 at 1:57

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