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1

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. ...


0

Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: $$ \tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + ...


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 ...


0

well, fermions' "spatial wave function" can also be antisymmetric. I think it's the whole wave function(spin+spatial) that matters.


0

Just to elaborate on ACuriousMind's answer in case it it not immediately clear what he means. Think of angular momentum fundamentally being defined as the generator of rotation. If we have any system, in this case an isolated quantum system described by a state, how does this change if we rotate it (or if we rotate the frame from which we describe it). Since ...


-1

Any angular momentum in QM is quantized. Spin is not an exclusion. A hydrogen atom may also have spin even though "constructed" with spinless particles. And a free atom is not localized in space, just like a free electron. So, the angular momentum is a property of elementary and non elementary "particles" in QM. Spin is the angular momentum in the system ...


0

To say, in non-relativistic QM, that a state has spin $\frac{1}{2}$ means that it transforms in the representation of $\mathrm{SU}(2)$ with highest weight $\frac{1}{2}$, which is a two-dimensional space. In general, to say that a state has spin $s$ means to say that it transforms in the representation with highest weight $s$.


0

The hexagonal Graphene lattice can be considered as a superposition of two identical sub-lattices set off by one one carbon-carbon bond length. As a result, it has two sets of wavevectors k,that are picked out by the lattice, inequivalent (since the two sublattices really are distinct) but otherwise identical (since it's semantics to say which sublattice is ...


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 ...


1

Objects in orbit tend to lose their spin on their own axis. However they do not completely lose their rotation and end up rotating with a period that is the same as the orbital period, so that they face always the same side towards the other body. The best know example is the Moon that shows always the same side to Earh. The phenomenon is called tidal ...


0

The dot product is between the same unit vectors. You get |n|^2 * cos theta, where theta is 0. That leaves you with that magnitude of n squared. Magnitude is 1. The cross product becomes zero because the cross product of the same vector is zero. This is because n x n is |n|^2 sin theta. Here sin theta evaluates to zero.


3

To show that $$ \left(\sigma\cdot\mathbf{n}\right)^2=\mathbf n\cdot\mathbf n+i\sigma\cdot\left(\mathbf n\times\mathbf n\right)\tag{1} $$ consider writing the above as \begin{align} \left(\sigma\cdot\mathbf a\right)\left(\sigma\cdot\mathbf b\right)&=\sum_j\sigma_ja_j\sum_k\sigma_kb_k\\ ...


0

Use conservation of angular momentum here. Angular momentum is a vector quantity, it points along the axis of rotation, so that when you look "along" the momentum vector, the rotation is clockwise from your point of view. Formation of a black hole preserves angular momentum. It doesn't necessarily make sense to talk about the speed of a black hole's ...


-4

The top of the first black holes will be located on the side furthest away from the point of the singularity. Each of which rotate in the same direction. As it feeds a thread of matter, no bigger than a fine thread, emits from its underside. Undectable from our view. The thread then then reaches a point of limitation while the black hole continues to ...


0

Ok I think I got it: $$\langle \psi' | H| \psi\rangle =\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^+S_{i+1}^-)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle+\sum_{n'=1}^N\frac{1}{\sqrt{N}}\langle n'|e^{-ikn' }(-\frac{J}{2}\sum_{i=1}^NS_i^-S_{i+1}^+)\frac{1}{\sqrt{N}}\sum_{n=1}^Ne^{ikn}|n\rangle ...


0

The pion is a pseudoscalar particle, which behaves like a scalar, except that it changes sign under a parity inversion while a true scalar does not. For details, see this post by @LuboŇ° Motl and links there: What is a Pseudoscalar particle?


0

I think this is a really good question, and one that unfortunately I haven't found addressed when learning about this stuff, and I think that avoiding it by saying that you need Dirac's equation neglects that the question can be interpreted classically. So yes, we need relativity, but we shouldn't need quantum mechanics to understand this effect. To ...


0

Notice that both $B$ and $B^\star$ are unitary matrices and, therefore, $B B^\star$ must be unitary as well, which implies that its determinant must be either $1$ or $-1$. Since you have already determined that $B B^\star = c\,I$, this narrows the possible values of $c$ down to $c = \pm 1$. About the expression $B^\prime B^{\prime\star} = U B B^\star ...


0

An electron in a homogeneous magnetic field $\vec B$ has only 2 possible spin states, typically called 'spin up/down'. These two states do not refer to the total spin being perfectly aligned/anti-aligned with $\vec B$. The total quantized angular momentum is $S=\sqrt{3}/2\hbar$ while the z-component is $S_z=\pm1/2\hbar$. Thus the state of precession with ...


0

The spin of a particle is not created by a rotation of the particle around itself. It is a relativistic effect. Please ask yourself why the electron in a hydrogen atom can be found at certain distances from the nucleus, and NOT at ANY distance. It is because the wave-function interferes with itself and there are forbidden distances because there, the ...


0

Neutrons provide nothing to nuclear spin due to even number, but uncouple proton in 1d5/2.To maximize the Iz (uncouple proton) it should be in 3/2 state because couple protons in 5/2.So I=5/2


1

I seem to have found an answer in the reference On the Variation of Ortho-hydrogen and Para-hydrogen Ratio with Magnetic Field Strength at Low Temperature. At high enough field strength, orthohydrogen rather than parahydrogen is favored.


6

The Poincare group has two Casimir Invariants - namely $p^2$ and $W^2$ where $$ W_\mu = \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} J^{\nu\rho} p^\sigma $$ is the Pauli-Lubanski pseudo-vector. Thus, representations of the Lorentz group are labelled by the eigenvalues of both $p^2$ and $W^2$. When $p^2 = -m^2$, we have the property $W^2 = -m^2 {\bf J}^2$. ...


1

There are two other interpretation of the Pauli matrices that you might find helpful, although only after you understand JoshPhysics's excellent physical description. The following can be taken more as "funky trivia" (at least I find them interesting) about the Pauli matrices rather than a physical interpretation. 1. As a Basis for $\mathfrak{su}(2)$ 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 ...


3

Groups are abstract mathematical structures, defined by their topology (in case of continual (Lie) groups) and the multiplication operation. But it is almost impossible to talk about abstract groups. That is why usually elements of groups are mapped onto linear operators acting on some vector space $V$: $$ g \in G \rightarrow \rho(g) \in \text{End}(V), $$ ...


0

Your own reference makes it very clear that a magnetic field affects the ortho-para equilibrium. A magnetic field acts as a catalyst to facilitate the conversion of orthohydrogen to parahydrogen, as the temperature of the dihydrogen is decreased.


2

As @ACuriousMind pointed out in his comment, the definition of a pure state is one that is a vector of $\mathcal{H}$, in this case $\Bbb{C}^2\otimes\Bbb{C}^2$. If you take for example $\rho=|\varphi^+\rangle\langle\varphi^+|$, the reduced density matrix on the first space is \begin{align}\rho_1&=\operatorname{Tr}_2\rho\\ ...



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