# Tag Info

17

Spinor fields are sections of a spinor bundle, so you have to be careful when you work with them as if they were functions. For a spinor field, the notions of being parallel, Killing, conformal Killing,... make perfect sense globally as equations on sections of the spinor bundle, but you have to specify which bundle. Spinor bundles are associated vector ...

17

The relation with twistors follows by taking a further square root of Urs's answer. If $(M^n,g)$ is an $n$-dimensional spin manifold with spinor bundle $S$, we have a natural conformally-invariant operator $P: \Omega^1(S) \to C^\infty(S)$, where $C^\infty(S)$ are the smooth sections of $S$ (i.e., smooth spinor fields) and $\Omega^1(S)$ are the smooth ...

15

The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the ...

11

That higher rank tensor which you have in mind is called a (conformal) Killing-Yano tensor . These are skew-symmetric tensors (differential forms) that are covariantly constant in a suitable sense and that serve as "square roots" of Killing tensors in direct analogy to how a vielbein serves as a sqare root for a metric (which is the canonical rank-2 ...

11

Here's my two cents worth. Why Lie Algebras? First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine. The Lorentz Lie Algebra It turns out that the ...

10

Let me first remind you of (or perhaps introduce you to) a couple of aspects of quantum mechanics in general as a model for physical systems. It seems to me that many of your questions can be answered with a better understanding of these general aspects followed by an appeal to how spin systems emerge as a special case. General remarks about quantum states ...

9

What you have is a good start. If we make the usual assignments that ${\partial\over{\partial t}} \to -iE$ and $\nabla \to i{\bf p}$ then we get$$(E - e\Phi)\psi = (\alpha \cdot ({\bf p} - e{\bf A}) + m\beta)\psi.$$Now, pick a particular representation$$\beta = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\text{ }\alpha_i = \begin{pmatrix} 0 & ... 9 Recall a Dirac spinor which obeys the Dirac Lagrangian$$\mathcal{L} = \bar{\psi}(i\gamma^{\mu}\partial_\mu -m)\psi.$$The Dirac spinor is a four-component spinor, but may be decomposed into a pair of two-component spinors, i.e. we propose$$\psi = \left( \begin{array}{c} u_+\\ u_-\end{array}\right),$$and the Dirac Lagrangian becomes,$$\mathcal{L} = ...

9

The mistake you are making is in "daggering" the object $\omega_{\mu\nu}$. For each $\mu, \nu = 0,\dots 3$, the symbol $\omega_{\mu\nu}$ is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; $(\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}$. When we say that $\omega_{\mu\nu}$ is an antisymmetric real matrix, we ...

8

The answer to your question depends on the context, but the basic unifying theme distinguishing different kinds of fields (like vector fields, scalar fields, etc.) is how these fields transform when they are acted on by Lie Groups (and or Lie Algebras) which falls under the mathematical subject of representation theory of Lie groups and Lie algebras. Here ...

8

At the risk of telling you how to "suck eggs" (your level in these things is not altogether clear), here goes. Ingredients: The essential ingredients to this explanation are: A physical "system" which evolves in and whose "events" happen in some space $\mathcal{U}$ (ordinary Euclidean 3-space or Minkowsky spacetime, for example); in physics this space is ...

8

Yes, your confusion is wholly caused by you thinking classically ;) In a hand-wavy way, particles are certain localized excitations of the quantized fields. The QFT picture contains the particle picture in the perturbative approach known as Feynman diagrams (and, relatedly, the LSZ formalism). There, we are given the action of our theory dependent on some ...

7

The question puts the cart before the horse. It is not that you derive that particles described by the Dirac equation have spin $\frac 1 2$. Rather, the Dirac equation is found as the equation for spin $\frac 1 2$ particles. A Dirac spinor $\psi$ is an element of the representation $(0,\frac 1 2) \oplus (\frac 1 2, 0)$ of the Lorentz group.1 In both ...

6

For massless particles, helicity coincides with chirality thus you ask to find the basis such that $$\psi_{\pm}=\left( \psi_{\mp}\right) ^{\star},\quad\gamma_{5}\psi_{\pm}% =\pm\psi_{\pm}.$$ Using the decomposition of hermitian operator: $$\left( \gamma_{5}\right) _{ij}=\left( \psi_{+}\right) _{i}\left( \psi _{+}^{\star}\right) _{j}-\left( ... 6 From the relativistic covariance of the Dirac equation (see Section 2.1.3 in the QFT book of Itzykson and Zuber for a derivation. I also more or less follow their notation.), you know how a Dirac spinor transforms. One has$$\psi'(x')=S(\Lambda)\ \psi(x)$$under the Lorentz transformation$$x'^\mu= {\Lambda^\mu}_\nu\ x^\nu= {\exp(\lambda)^\mu}_\nu\ ...

6

I think its really important to differentiate between helicity and chirality. Helicity is the spin angular momentum of a particle projected onto its direction of motion. For a massive particle this quantity is frame dependent. Furthermore, since angular momentum is conserved, as a particle propagates helicity is conserved. On the other hand, chirality is an ...

6

A gauge field for a particular group $G$ can be thought of as a connection, or a $G$ Lie algebra valued differential form. If we recall the Riemann curvature, $$R(u,v)w = \left( \nabla_u \nabla_v - \nabla_v \nabla_u -\nabla_{[u,v]}\right)w$$ If $[u,v]=0$ the expression simplifies to the usual tensor in general relativity. Similarly, we may think of the ...

6

The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle ...

5

Firstly, it doesn't matter whether the indices are up or down here when you ask what representation an object lives in, because the raised and lowered versions are equivalent. Both $\chi_a$ and $\chi^a$ give objects in the $(\frac{1}{2},0)$ rep, and similarly for dotted. The relation between the two is the linear transformation given by ...

5

There is a definition that $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation is equal to spinor tensor $$\psi_{a_{1}...a_{m}\dot{b}_{1}...\dot{b}_{n}},$$ where $\psi_{\dot{b}}$ transforms as complex conjugation of $\psi_{b}$. Why do we assume that $\left( \frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$ represent spinors? You can think about ...

5

In QFT, the Dirac spinor will also be promoted to a field, whose oscillation mode coefficients are creation and annihilation operators. BUT: For the Dirac spinor it is possible to well-define a probablility density and current: $$\rho^\mu \propto \bar\psi \gamma^\mu \psi$$ This current's zero component is positive definite and using the Dirac equation one ...

5

You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant. Helicity defined $$\hat h = \vec\Sigma \cdot \hat p,$$ commutes with the Hamiltonian, $$[\hat h, H] = 0,$$ but is clearly not Lorentz invariant, because ...

5

Simply think of a Weyl Spinor as a Dirac spinor where the other two components are set to zero. Equivalently, a Weyl spinor (of chirality +1) belongs to the two-dimensional subspace with eigenvalue 1 under the action of the projection operator $P_+=\frac{(1-\gamma^5)}2$. For negative chirality, use the other projection operator, i.e., ...

5

Yes, they're representations of $SO(8)$, more precisely $Spin(8)$ which is an "improvement" of $SO(8)$ that allows the rotation by 360 degrees to be represented by a matrix different from the unit matrix, namely minus unit matrix. ${\bf 8}_v$ transforms normally as $$v\mapsto M v$$ where $MM^T=1$ is the $8\times 8$ real orthogonal $SO(8)$ matrix. The ...

5

There is an interesting way to look at Christoffel connections with spinor fields. The usual Dirac operator is written as $\gamma^\mu\partial_\mu$. It is interesting to change this to $\partial_\mu(\gamma^\mu\psi)$. This then becomes $$\partial_\mu(\gamma^\mu\psi)~=~ \gamma^\mu\partial_\mu~+~(\partial_\mu\gamma^\mu)\psi.$$ The anticommutator ...

5

It's just a choice of basis. Whether you use $$\bigl\{|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle\bigr\}$$ (individual spins) or $$\biggl\{\frac{1}{2}\bigl(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle\bigr),\frac{1}{2}\bigl(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\bigr)\biggr\}$$ (triplet/singlet) they span the same space. ...

5

There are a number of mathematical imprecisions in your question and your answer. Some advice: you will be less confused if you take more care to avoid sloppy language. First, the term spinor either refers to the fundamental representation of $SU(2)$ or one of the several spinor representations of the Lorentz group. This is an abuse of language, but not ...

4

I think confusion arises because the metric tensor is hidden in $SU(2)$, so I'll work in $SU(p,q)$ and then specialize to $SU(2)$. The inequivalent defining reps of the general linear group $GL(m,C)$ carried on the four vector spaces $V_{m},\tilde{V}_{m}, V^{*}_{m}, \tilde{V}^{*}_{m}$ transform vectors as follows. $$v'^{a}=\,[D(g)]^{a}_{\ \ b}v^{b}\\ ... 4 Firstly, I don't think your convention for the derivative is consistent with how you lower the index on \lambda^b. Let us start with$$ \frac{\partial}{\partial \lambda^b} \lambda^a = \delta_b^a $$If you lower the index on \lambda^b you get$$ \lambda_a = \epsilon_{ac} \lambda^c $$Putting this together gives$$ \frac{\partial}{\partial \lambda^b} ...

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