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3

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...


3

Indeed, the photon doesn't have spin, what it has is POLARIZATION. For a photon emitted from an atom the polarization is CIRCULAR. That means, if we look in a fixed plane perpendicular on its motion, we see that the electric vector does not have a fixed position, it rotates. If so, if the electric vector rotates counter-clockwise the circular polarization ...


3

Consider a one-particle state of a relativistic quantum field theory, and let this state be an eigenstate of the 4-momentum operator, $\hat P^\mu |p^\mu\rangle = p^\mu|p^\mu\rangle$. Other than the 4-momentum, what other quantum numbers can the state have, and how should they transform? That is, if there are any quantum numbers collectively labeled by $s$, ...


2

Let $\mathcal{H}$ be the Hilbert space for one particle. Then, $S_{x}\in\mathcal{B}(\mathcal{H})$ is a bounded, self-adjoint operator. Now, if you want to have the Hilbert space for two particles, remember that this is the tensor product, i.e. $\mathcal{H}=\mathcal{H}_1\otimes \mathcal{H}_2$ (where $\mathcal{H}_1$ is the Hilbert space of the first and ...


2

Both $h$ and $\tilde{h}$ are usually called weights. Their sum, $\Delta=h+\tilde{h}$ is the (scaling) dimension of the operator, while the difference, $s=h-\tilde{h}$ is called the spin. This is due to their association with scale transformations (dilatations) and rotations, respectively. To see this, note that the dilatation operator is given by ...


2

I guess what you are missing is the following: given a representation $\rho(g)$ of $g\in$SU(2) acting on some vector space $V$. We define the representation $\rho_\otimes$ of SU(2) (not of SU(2)$\times$SU(2)) on $V\otimes V$ as $$\rho_\otimes(g) (v_1 \otimes v_2) = \rho(g) v_1 \otimes \rho(g) v_2.$$ So in fact we are defining the tensor product of two ...


2

Fermion wavefunctions are antisymmetric under the interchange of two particles. Spatial inversion flips the spatial coordinate, but does not interchange particles. In other words, let's say we have a two particle wave function, $\psi(x_1, x_2)$ (where $x_1$ is the position of particle 1, and $x_2$ is the position of particle 2). Being odd under parity ...


2

Spin of an elementary particles is not necessarily the result of a movement of the particle around itself i.e. around some rotation axis that passes through the particle.If there were such an axis, the projection of the spin in the plane perpendicular to that axis were zero. But, this is not the case. So, along whatever axis we would measure the spin, we ...


2

The photon as an elementary particle is special: the quantum mechanical wave equation whose solutions squared will describe the probability of finding the photon at a given phase space point is the Maxwell equation that describes the classical electromagnetic wave, except it is the potential form of it and it acts as an operator on the wave function of ...


1

Step 0: Outline. We are going to define a candidate Hamiltonian. We know that we get the right answer if it has all the right behavior. While checking everything the Hamiltonian might possibly do seems daunting, things are simplified by two important facts: The Hilbert space is small -- only 4-dimensional. In fact, we're going to group three basis ...


1

Unlike in QFT where you can derive spin from more basic principles, in ordinary non-relativistic QM spin is essentially defined into existence as a group of operators $S^i = (\hbar/2) \sigma^i$ that satisfy the algebra $$[\sigma^i, \sigma^j] = i \epsilon^{i j}_{\,\,k} \sigma^k.$$ The dimensions in the Hilbert space on which the Pauli operators act are ...


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In order to appreciate the periodicity of graphene one has to recognize that it consists out of two interpenetrating hexagonal Bravais sublattices, A and B, which together make up the honeycomb lattice. The two sublattices are like two degrees of freedom, and the electron can have an amplitude to be on sublattice A, and an amplitude to be on the sublattice ...


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Generally speaking, the choice of what the $z$-axis (equivalently $x,y$) is is arbitrary. You can choose any direction to be your $z$-axis, as long as you do the calculations consistently with this choice. If the system has a priviliged direction (like that imposed by the magnetic field in the Stern-Gerlach case) that is usually choosen to be the $z$-axis. ...


1

In short: the laws of conservation (angular momentum, charge, mass-energy, etc.) still work during the process of creation of a black hole. So if a star had some angular momentum/charge before it collapsed, the resulting black hole will also have some (assuming the angular momentum/charge was not radiated away during the collapse). Also, the claim that ...



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