New answers tagged

2

Juts to expand on @Chiral Anomaly's answer. Let's say you want to calculate $\sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu)$, and Y can be expressed as (for $D=4$, and implicit index summation): $$ Y = AI + B^\mu\gamma_\mu + C^{\mu\nu}\gamma_\mu\gamma_\nu + D^\mu\gamma_\mu \gamma_5 + E \gamma_5, $$ then $$ Y' := \sum_\mu\gamma^\mu Y\gamma_\mu \\ =4AI -2 B^\...


3

To derive an expression for $\sum_\mu\gamma^\mu_{ab}\gamma_{\mu\ cd}$, note that this is equivalent to requesting an expression for $$ \sum_\mu\text{trace}(X\gamma^\mu Y\gamma_\mu) $$ for aribtrary matrices $X$ and $Y$. To derive such an expression, express $X$ and $Y$ as sums of products of Dirac matrices (with arbitrary coefficients), and use the ...


0

First, my expression for the Laplacian above was off, and the definition of $\nabla^*$ and $\nabla^*\nabla$ is a bit more subtle than I thought. $\nabla^*$ is the adjoint of $\nabla$. In particular, some vector $X$, $\nabla^*_X$ is the adjoint of $\nabla_X = X^\mu \nabla_\mu$. This can be computed from trying to integrate $\langle w, \nabla_X v \rangle = \...


1

Each component of $\Psi$ satisfies the Klein-Gordon equation, and so we can write (cf. this PSE post) $$ \Psi_\alpha(x)=\int\frac{d^3p}{\sqrt{(2\pi)^32E_p}}\Big(a_\alpha(p)e^{-ip\cdot x}+b^\dagger_\alpha(p)e^{+ip\cdot x}\Big) $$ for some operators $a_\alpha,b_\alpha$. If we now require $\Psi$ to satisfy the Dirac equation, we get the algebraic conditions $$ (...


0

Modes decomposition comes from the solution of motion equation. You should start from Dirac equation, $$[i(\gamma\cdot\partial)-m]\psi=0$$ and consider the following ansatz for $\psi$, $$\psi=\sum_s\int_{\bf p}\frac{1}{\sqrt{2E_{\bf p}}}\left(b_su_s(p)e^{-ip\cdot x}+d_s^{\dagger}v_s(p)e^{+ip\cdot x}\right).$$ Then you can rewrite your equation in terms of 2$\...


0

Consider the Lorentz transformation of the spinor parametrizing by the matrix $S$: $$ u(\mathbf p) \to S u(\mathbf{0}) $$ Here $S$ corresponds to the Lorentz transformation $\Lambda_{\mu}^{\nu}p_{\nu,\text{rest}} = p_{\mu}$, where $p_{\nu,\text{rest}} = (m,\mathbf{0})$ and $p_{\mu} = (E,\mathbf{p})$. $\gamma$ matrices have a property $$ S^{-1}\gamma_{\mu}S = ...


2

This is probably easiest to answer in the solid state case, rather than the particle physics case. In particle physics this is a prediction of antimatter, but in a crystalline solid, electrons are forced by Pauli exclusion to occupy higher and higher energy levels, until they more or less occupy a ball in momentum-space whose surface corresponds to an energy ...


0

Let's start with the Dirac equation as you've written (but I choose to use lower indexes only, since they denote 3-vectors here, not 4-vectors): $$\begin{align} \sigma_i(p_i+eA_i)u_B & = (E-m+eA_0)u_A \\ \sigma_i(p_i+eA_i)u_A & = (E+m+eA_0)u_B \end{align} \tag{1}$$ As @Sunyam suggested in his comment you can solve the second equation of (1) for $u_B$...


5

Dirca's equation has is the following form: $$i\hbar\gamma^{\mu}\partial_{\mu}\psi - mc\psi = 0$$ where $\mu = 0,1,2,3$ and $\gamma^{\mu}$ are the $4\times4$ matrices(in Dirac's representation): \begin{align} \gamma^0 = \begin{bmatrix} \mathbb{1} & 0 \\ 0 & -\mathbb{1} \end{bmatrix} \gamma^k = \begin{bmatrix} 0 & \sigma_k \\ -\sigma_k & ...


0

This represents the canonical way to quantize a quantum field theory. You start from a free theory (Klein-Gordon, Dirac,...) and in this way, using plane waves as fundamental solutions, you are able to introduce ladder operators. Then, you can do perturbation theory to evaluate the effects of small perturbations on the free theory. Anyway, the most powerful ...


1

The reason we usually complexify the Clifford algebra is mostly convenience: The representation theory of complex algebras is simpler in general, and if we want to restrict to real representations for some reason later on we can always do that. In particular, Dirac spinors at least exist in all dimensions, while the "real" Majorana spinors are dependent on ...


0

I think the only problem lies in the renormalization, but it's not really a conceptual problem, rather a mathematical one (oh well..). Let me try to explain. Consider a simple tight binding Hamiltonian in one dimensions. The dispersion (one-particle energy) is $-t\cos(k)$. For $-\pi/2 \le k\le \pi/2$ (and zero chemical potential) the dispersion is negative. ...


5

You are using $\mu$ as an index too many times. When you square $\gamma^\mu \partial_\mu$, you have to use a different index for the two factors. Here is a correct derivation: Starting with $(i \gamma^\mu \partial_\mu - m) \psi = 0$, apply $-(i \gamma^\nu \partial_\nu + m)$ from the left: $$ 0 = -(i \gamma^\nu \partial_\nu + m)(i \gamma^\mu \partial_\mu - m)...


-4

The Dirac sea at the time may have seemed an idea worth entertaining. In modern days one has to immediately conclude that it is fundamentally flawed and indefensible. Unfortunately some courses still present it as sensible. The Dirac sea implies uniform infinite charge density throughout the universe. Fluctuations would give large effects. The repulsion ...


-1

The Dirac equation (according to my understanding), is an alternate form of the Klein-Gordon equation. The Dirac equation can be derived from the following equation. $$ Z_{μ\ }Z^{μ}-m^{2}=0 $$ Were,$$ Z_{μ\ }= \left(E\ \ \ P\right)$$ and, $$Z^{μ}= \begin{pmatrix} E\\-P \end{pmatrix} $$ Where "E" is the operator of energy, and "P" is the operator ...


4

In natural units, the Hamiltonian your text tells you was introduced by Dirac is $$ H=\vec{\alpha}\cdot \nabla /i+ \beta m , \implies i\partial_t \psi = H \psi ,\\ \beta =\gamma_0 , \qquad \vec {\alpha}= \gamma_0 \vec{\gamma} . $$ You then have a bona-fide continuity equation $$ \partial_t \rho + \nabla \cdot \vec j =0, \\ \rho =\psi^\dagger \psi > 0, \...


1

The Lie algebra of the Lorentz group can be understood as two copies of the $SU(2)$ Lie algebra. Representations of $SU(2)$ can be labelled by half integers $(0,1/2,1,\ldots)$. (This should be familiar from discussions of angular momentum in quantum mechanics.) This implies that we can label representations of the Lorentz group by using two labels $(i,j)$, ...


6

Here are interpretations for at least two gamma matrices: $\gamma_0$ is the spinor metric. It's role is analogous to the role of the Minkowski metric for four-vectors. We need the Minkowski metric to write down the scalar product of two four-vectors. Analogously, we need $\gamma_0$ to write down the scalar product of Dirac spinors. $\gamma_5 \equiv i \...


7

They are not just mathematical definitions; there is some physics in them. Actually, they are intrinsically connected with the spin structure of the fields. You can see this from the fact that the spin 1/2 representations of the Lorentz group, namely (1/2,0) and (0,1/2) in $SU(2)\times SU(2)$ classification, naturally introduce the definition of the gamma ...


7

Whether there is an $i$ $$ (i\gamma^{\mu}D_{\mu}-m)\psi=0 $$ or no $i$ $$ (\gamma^{\mu}D_{\mu}-m)\psi=0 $$ in the Dirac equation is determined by the metric: For $(\gamma^0)^2 = I$, there is $i$ For $(\gamma^0)^2 = -I$, there is no $i$ In Witten's paper, the metric is (-, +, +), therefore no $i$. Added note: Since the Hermitian of $\gamma^\mu$ is ...


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