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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 ...


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

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 ...


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

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 ...


7

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 ...


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

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 ...


5

Nah, they're the same. Even Planck's constant comes from fields; he was looking at the electromagnetic radiation (which is all field, all the time) getting kicked off by a warm black body. Also, while most people use $h \nu$ for the energy of a photon, in grad school (physics) we often used $\hbar \omega$. Also, usually I've seen the Dirac constant defined ...


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

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\ ...


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

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 ...


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

The QFT is strongly based on the group theory formalism. Often when people say about some QFT theory they primarily say about the symmetries of the theory - invariance of lagrangian of theory (or about covariance of equations of motion) under sets of transformations. The group theory formalize these statements and help to construct theories which corresponds ...


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

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 ...


5

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 $$\psi = (u_{+},u_{-})^{T},$$ and the Dirac Lagrangian becomes, $$\mathcal{L} = ...


4

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 ...


4

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 ...


4

Fundamental fermions like quarks and leptons are described by the spinor field, while gauge bosons like photons are described by the vector field. They together with the Higgs bosons are currently what we have in the Standard Model for elementary particles.


4

Literally speaking the answer is negative. The charge of the state has to be always defined in view of the charge superselection rule. Thus for a particle described by Dirac equation, there are no things like coherent superpositions of electron states and positron states. A Dirac particle always stays in a quantum state which is proper of an electron or a ...


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

When you write the Dirac equation in a curved spacetime, in the context of General Relativity (which allows curvature, but not torsion) , you have a spin connection : $$\nabla_\mu\psi=\left(\partial_{\mu}-\frac i4\omega_{\mu}^{IJ}\sigma_{IJ}\right)\psi$$ Now, the Einstein-Cartan theory is not General Relativity, because it allows curvature, but also ...


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} ...


4

In a coordinate-free description, we get two 4-component spinors $u$ and $v$ in projections to orthogonal (and in particular different) eigenspaces of dimension 2. But the above chooses suitable bases and then considers the coefficient vectors $\xi$ and $\eta$, which are vectors in the same space $C^2$. Their physical meaning is not determined by this ...


4

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 ...


3

It can be instructive to see the applications of Clifford algebra to areas outside of quantum mechanics to get a more geometric understanding of what spinors really are. I submit to you I can rotate a vector $a = a^1 \sigma_1 + a^2 \sigma_2 + a^3 \sigma_3$ in the xy plane using an expression of the following form: $$a' = \psi a \psi^{-1}$$ where $\psi = ...



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