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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

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 ...


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 ...


10

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 ...


8

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 ...


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( ...


5

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

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

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

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 ...


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

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} ...


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

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., ...


4

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} = ...


3

"First quantization" and "second quantization" are widely used names for the same procedure applied to two different classical systems – classical mechanics and classical field theory. In both cases, the operation introduces Planck's constant and it's always the same constant. For example, the angular momentum $J_z$ is a multiple of $\hbar/2=h/4\pi$ both in ...


3

The fact that $h$ appears in first quantization and $\hbar$ is purely a matter of convention. Since $$\hbar = \frac{h}{2 \pi} $$ we may as well write equations like $$E = h \nu $$ as $$E = \hbar \omega $$ and in fact that is a frequent form of that equation. Multiplication and division by a constant doesn't make the two independent of each other. The ...


3

The factor $\frac{1}{2\pi}$ it is a matter of convenience. Dirac constant, or Planck's reduced constant can show up and indeed it does in quantum statistical mechanics. If the inverse of $2\pi$ is going to appear a lot of times,you should use $\hbar$ instead of $h$ but, as far as I know, it has nothing to do with your proposal.


3

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 ...


3

OP's two methods are isomorphic. In general one is only interested in classifying representations modulo isomorphism. The point is that for the Lie group $SU(2)$, the spinor representation ${\bf 2}$ and the complex conjugate spinor representation $\bar{\bf 2}$ are equivalent representations ${\bf 2}\cong \bar{\bf 2}$. The equivalence is precisely given by ...


3

The parallel/antiparallel picture is not quite correct. Mostly because it hides the fact that the electrons are indistinguishable from our intuition. In a (very rare) cases when you may consider electrons distinguishable, this picture is correct. In more natural case when electrons are completely equal triplet/singlet picture is much better. Maybe few more ...


3

The matrices $A^\mu$ are clearly just the inverse matrices that multiply the "bispinor components" of a vector to get the usual vector component. So $A^{\mu\dot\alpha \alpha}$ is the inverse to $\sigma^\mu_{\alpha\dot\alpha}$ – you treat the $\alpha,\dot\alpha$ indices as the rows and columns – and this inverse may also be obtained by a simple raising of the ...


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 = ...


3

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

There is a fairly comprehensive book by Thaller, called ''The Dirac equation'', published 1992, which discusses a spinor particle from many angles. Possibly, the book can be downloaded from one of the links at http://ebookee.org/Thaller-The-Dirac-Equation_141995.html


3

The decomposition $p=\lambda \tilde\lambda$ is only valid for null vectors, indeed. In loop integrals, the loop momentum may generally be off-shell but because of the delta function in these integrals, only the null values of the momentum contribute, so it's enough to deal with the null momenta and they can be factorized to the spinors in this way. The ...



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