# Tag Info

4

The double-slit experiment is a one-body experiment, meaning that one is only looking at interferences of one particle with itself. Thus the Bose or Fermi statistics does not play a role in that case. What the OP has in mind in the Hong-Ou-Mandel effect, which for bosons implies that there is an increased probability that two identical bosons will be ...

2

The boost matrix can be chosen in block form as (in $c=1$ units): $$L(\mathbf{p}) = \begin{bmatrix} \frac{E}{M}& \frac{\mathbf{p}^t}{M}\\ \frac{\mathbf{p}}{M}& 1_{(3\times3)}+\frac{\mathbf{p}\mathbf{p}^t}{M(E+M)} \end{bmatrix}$$ where $\mathbf{p}$ is the 3-momentum and $E = \sqrt{\mathbf{p}^2+M^2}$. It can be easily checked that this matrix ...

2

I went through unanswered questions, and stumbled over this... Did you find the original books? The mistake should be in your formula for the $\mu$ of a hollow sphere; the value with $1/5$ you gave is that of a solid sphere... The problem gets more simple I think, if you compare the two things directly: You get both, the angular momentum and $\mu$, from ...

2

Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ...

2

In the case that photons are coherent (having the same wavelength and oscillating in phase), they still separable into two states. This has to do with the two field components of the photon, the electric dipole moment and the magnetic dipole moment. Dipole moment means that at some time there is an electric field with a plus-minus direction and there is at ...

2

The spin $s$ of a particle characterizes how the rotation generators act on it. In $D$ dimensions, you represent the little group $SO(D-1)$ for massive particles and $SO(D-2)$ for massless ones. In fact, you really need to consider its universal cover $\textrm{Spin}(n)$ which happen to be just its double cover. Now, you can define the spin to be the largest ...

2

This is the prototypical example of a superselection rule. The operator $U(2\pi)$ commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation ...

2

Courtesy of its spin the electron has a magnetic dipole moment. That means if we place it in a magnetic field the two states aligned with and against the magnetic field have different energies. The magnitude of the energy difference depends on the strength of the field and the size of the magnetic dipole moment, which in turn depends on the spin. So by ...

2

For force carriers the interacting field theory determines the spin. A scalar field yields spin 0; the Higgs is the only example; a vector field yields spin 1, the photon, W, and Z are examples; a tensor field yields spin 2. Since gravitational field theory requires a tensor field for General Relativity, quantized gravity, in the weak-field, linearized ...

2

Let's say we could perform the experiment with $W^\pm$ bosons. These particles are similar to electrons, but the possible spin states are $-1,0,+1$, that is, three different possibilities. The magnetic moment of these bosons is, therefore, $$\mu_z=\begin{cases} -\mu_W\\\phantom{+}0\\+\mu_W\end{cases}$$ where $\mu_W=6\ 10^{-6}\ \mu_B$ is the $W$ magneton. ...

2

Within the context of first quantization, spin itself is not characterized by the wave function: it exists as a separate (Hilbert) space. The total state of the particle is then a composite of its wave function (often projected onto the configuration or momentum basis) and spin state.

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I assume you are familiar with Wigner's classification in d=4, as you might be implying. The m→0 limit is best appreciated on the Poincaré sphere, but let us skip that here to count particle states. So, reviewing Wigner, for a massive state, we can Lorentz-transform the momentum to the rest frame, (m,0,0,0) so the little group is SO(3) and its vector rep ...

1

Peter is right. The tensor nature requires it to be a spin 2 field, and the graviton is its presumed quanta. But there have been and are theories of gravity that include a spin 0 field. Brans-Dicke theory was one (I think mostly or fully disporved), and some theories for dark energy are spin 0 - quintessence is one, it assumes the cosmological constant is ...

1

The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads $$i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2$$ Now remove $c_2$ using again the first eq., $$c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1$$ and ${\dot c}_2$ using the second eq., ${\dot c_2} = ... 1 First, let's clarify the expression of$|\phi\rangle$. The kets$|+\rangle_z$and$|-\rangle_z$are eigenvectors of$\hat{S}_z$such that $$\hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z$$ This means that in the$\{|+\rangle_z = ...

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The spin quantization was first detected by the Stern-Gerlach experiment. This one ingenious experiment proved that the spin is an observable. Refer to the experiment. I prefer the book: Modern Quantum Mechanics by J.J.Sakurai. chapter 1

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As Praan confirmed, you are fine. Your result is the n=4 case of the general expression for composing spin 1/2 doublets, (so here denoted by their dimensionality, 2), $${\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,$$ where \$\lfloor n/2 ...

1

The company explanation is wrong, really except for the first sentence. The correct hand waving explanation would be much longer. Materials are composed of atoms which contain electrons, and electrons have intrinsic magnetic moments and specific orbitals around atoms that are not affected by temperature. The molecular bonds are affected at high ...

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