0
$\begingroup$

I've cross listed this post on math SE in case it is more appropiate there. That post can be found here: https://math.stackexchange.com/q/4833722/.

I am approaching this from a Clifford algebra point of view.

Let $\mathbb{C}l(4)$ be a Clifford algebra over $\mathbb{C}^4$ with the usual bilinear product. Then there is an isomorphism $$\mathbb{C}l(4) \cong \text{End}(\mathbb{C}^4)$$ and thus we may represent $\mathbb{C}l(4)$ using the following map: $$\rho: \mathbb{C}l(4) \rightarrow \text{End}(\mathbb{C}^4).$$ An element of $\mathbb{C}^4$ is a Dirac spinor.

Now if we restrict $\mathbb{C}l(4)$ to only those products that consist of an even number of vectors, which we denote $\mathbb{C}l^0(4)$, we get the following $$\mathbb{C}l^0(4) \xrightarrow{\cong}\text{End}(\mathbb{C}^2) \oplus \text{End}(\mathbb{C}^2).$$ We call elements of this representation space Weyl spinors, which thus belong to $\mathbb{C}^2$.

Hence if we choose a Dirac spinor that comes from an even number of elements in the Clifford algebra, it can be written as two Weyl spinors. However if the Dirac spinor comes from an element in the Clifford algebra that has an odd number of elements then it is not true that it can be written as two Weyl spinors.

Have I made a mistake somewhere in my understanding? If not, what justifies the claim often seen in physics books that "every Dirac spinor can be written as two Weyl spinors"?

$\endgroup$
9
  • 1
    $\begingroup$ I think you are confusing the number of Lie algebra basis generator elements with the dimensions of the spinors. The former is the one better mapped to the Clifford algebra elements. If anything, it is the Weyl spinors that are closer to Clifford algebra, and you can express the Dirac spinors as a pair of them. In particular, Weyl spinors are your $\mathbb C^2$, usually written as a 2x1 vector matrix that Pauli spin matrices act on, whereas Dirac spinors are your $\mathbb C^4$, usually 4x1 matrix that Dirac gamma matrices act on. $\endgroup$ Dec 26, 2023 at 3:24
  • $\begingroup$ @naturallyInconsistent Yes I think I was making the mistake you mentioned, thanks for pointing that out. So really when we say every Dirac spinor can be written as two Weyl spinors we just mean that every $\mathbb{C}^4$ vector can be written as two $\mathbb{C}^2$ vectors? $\endgroup$
    – CBBAM
    Dec 26, 2023 at 3:27
  • 1
    $\begingroup$ It is not that every $\mathbb C^4$ vector can be written as two $\mathbb C^2$ vector; irreps cannot be decomposed. Instead, it is that Dirac spinors are decomposable into two Weyl spinors; Dirac spinors are an irrep if you take CPT into account, because at least one of them swap between the two Weyl spinors, which are of different types. However, if you just want to express the future-oriented parity-upright continuous Lie subgroup of the Poincaré group, then one single Weyl spinor is enough. $\endgroup$ Dec 26, 2023 at 3:40
  • 1
    $\begingroup$ I think that you would have a far more fruitful time if you invested them in the standard derivations of this correspondence. That is, either you compose two Weyl spinors and derive the Dirac equation, or you work with the solutions of the Dirac equation and massage them into manifestly two Weyl spinors. I suggest the latter: Instead of solving the Dirac equation in conventional mass basis, do it in chirality basis this time, especially considering the massless limit, whereby the two Weyl spinors decouple. $\endgroup$ Dec 26, 2023 at 6:29
  • 2
    $\begingroup$ Following the argument of both Weyl and Wigner in the group representation of spinors (as in QFT), computing them outright, made them make most sense; Clifford algebra also makes a lot of sense, but then we have to think about how to merge those pictures together. $\endgroup$ Dec 26, 2023 at 9:02

1 Answer 1

2
$\begingroup$

Your notion that a "spinor comes from an element in the Clifford algebra" seems misguided. (Dirac) Spinors are elements of $\mathbb C^4,$ as you say. $\rho$ doesn't turn Clifford algebra elements into (Dirac) spinors. It turns Clifford algebra elements into linear operators on Dirac spinors, also as you say ($\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$). There's no meaningful way to get "a" unique spinor out of a Clifford algebra element.

In any case, the most important physical object when discussing spinors is the Lorentz group. We demand that physical quantities carry representations of the Lorentz group (projective representations, if quantum), because we demand that we can ("easily") predict the properties of an object after rotating or boosting it. Representations of the Clifford algebra are just a convenient proxy for (projective) representations of the Lorentz group: a copy of (the universal cover of) the Lorentz group sits inside $\mathrm{Cl}(4).$ This is strictly $$\mathrm{Spin}_\mathbb C(1,3)=\{v_1\ldots v_{2k}\mid v_i\in\mathbb C^4,|v_i|=1\}.$$ (Note: this is smaller than your $\mathrm{Cl}^0(4)$, since in addition to only taking the even elements I also impose a condition of unit norm. As $\rho$ is linear, this doesn't matter too much.) Yes, it may be true that, mathematically, the unrestricted representation $\rho:\mathrm{Cl}(4)\to\mathrm{End}(\mathbb C^4)$ does not split into two smaller representations. But this $\rho$ is not the physically relevant Dirac representation; not every element of $\mathrm{Cl}(4)$ represents a Lorentz transform. The Dirac representation is the restriction $\rho|_{\mathrm{Spin}_\mathbb C(1,3)},$ and this one does split.

That is, if you choose the right basis on $\mathbb C^4,$ you have $\rho|_{\mathrm{Spin}_\mathbb C(1,3)}(\Lambda)=\rho_L(\Lambda)\oplus\rho_R(\Lambda)$ for two distinct representations $\rho_L,\rho_R:\mathrm{Spin}_\mathbb C(1,3)\to\mathrm{End}(\mathbb C^2),$ the Weyl representations. I.e. there is more to the decomposition of a Dirac spinor into two Weyl spinors than just "2 + 2 = 4" in terms of dimensions. Specifically, a Dirac spinor rotates under Lorentz transforms as two independent objects of 2 dimensions each; the components of the two Weyl spinors don't get mixed up. (Compare to a 4-vector, which cannot be decomposed into 4 scalars, because Lorentz transforms mix them all up.)

$\endgroup$
6
  • $\begingroup$ Thank you very much for your answer! I understand that we may decompose the Dirac representation on $\mathbb{C}^4$ into two (Weyl) representations on $\mathbb{C}^2$. This means for any operator $A$ coming from the Dirac representation we can rewrite it as two operators $B_1$ and $B_2$ coming from the Weyl representation, i.e. $A = (B_1, B_2)$. However what I am still confused about is how this decomposition of operators on spinors implies a decomposition of the actual spinors themselves. $\endgroup$
    – CBBAM
    Dec 26, 2023 at 5:21
  • $\begingroup$ That is, how does $A = (B_1, B_2)$ imply that any Dirac spinor $X$ can be written as $X = (Y_1, Y_2)$ where $Y_1, Y_2$ are Weyl spinors. $\endgroup$
    – CBBAM
    Dec 26, 2023 at 5:21
  • $\begingroup$ So because $\mathbb{C}^2 \oplus \mathbb{C}^2 = \mathbb{C}^4$ we can choose any two Weyl spinors $v_1, v_2 \in \mathbb{C}^2$ to create a Dirac spinor. The reason this is still meaningful is if we start with an operator $A \in \text{End}(\mathbb{C}^4$ acting on $v$, then the decomposition $$\text{End}(\mathbb{C}^2) \oplus \text{End}(\mathbb{C}^2)$$ tells us how $A$ affects $v_1$ and $v_2$ by giving us two new operators $A_1, A_2 \in \text{End}(\mathbb{C}^2)$? $\endgroup$
    – CBBAM
    Dec 26, 2023 at 7:08
  • $\begingroup$ @CBBAM Weyl spinors are 2-dimensional vectors. Dirac spinors are 4-dimensional vectors. 2 + 2 = 4, so of course a Dirac spinor may be split into two Weyl spinors. The only trick is finding a meaningful decomposition: the rep. decomposition tells you how. Here, if $\{l_1,l_2,r_1,r_2\}$ is the basis in which $\rho_\mathrm{Dirac}=\rho_L\oplus I+I\oplus\rho_R,$ then (by definition of basis) every Dirac spinor is $v=c_1l_1+c_2l_2+c_3r_1+c_4r_2=(c_1l_1+c_2l_2)+(c_3r_1+c_4r_2)$ so the sum of two Weyl spinors ($\mathrm{span}\{l_1,l_2\}$ carries the Weyl rep $\rho_L\oplus I$, similar for $\rho_R$). $\endgroup$
    – HTNW
    Dec 26, 2023 at 7:09
  • $\begingroup$ @CBBAM You can't decompose any operator in $\mathrm{End}(\mathbb C^4)$ into two operators in $\mathrm{End}(\mathbb C^2).$ You have only a decomposition for $\rho_\mathrm{Dirac}.$ The equation $\rho_\mathrm{Dirac}=\rho_L\oplus I_2+I\oplus\rho_R$ says "to rotate/boost a Dirac spinor, rotate/boost 2 of the components as a left-handed Weyl spinor and the other two components as a right-handed Weyl spinor." It does not (and cannot, count the dimensions!) work for general linear transforms. That's part of the point of the answer: we care about Lorentz transforms more than other operators. $\endgroup$
    – HTNW
    Dec 26, 2023 at 7:16

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.