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18

Helium nuclei behave like bosons only in phenomena where their integrity is preserved and they can be assumed as point-like particles. When you compress them a lot this is not the case any more. Single protons and neutrons will start to interact with each other and they are fermions. You have no chance to condensate them in a black hole, indeed what you get ...

8

Great question that exposes some really confusing terminology. This is a rather long answer, and the punchline is basically in the second-to-last paragraph, but I think (hope) it's worthwhile to read the whole answer because I tried to give a somewhat systematic description of fermionic states using a specific, simple example along the way. Firstly, let's ...

8

OK, here you go: No. Consider the one-particle sector, with states $|n\rangle$ having occupation number $n$. Wlog we can suppose that the states are normalized. Then $$a|n\rangle =\sqrt{n}|n−1\rangle,$$ so $$\|a|n\rangle\| = \sqrt n.$$ This means that $a$ is not bounded. The same goes for $a^\dagger$.

8

The simplest Feynman diagram for an interaction between two particles looks like a letter "H". The cross-bar is a force-carrier being exchanged. At each vertex, you have a particle either emitting or absorbing a force-carrier. If the force-carrier has a half-integer spin, then you can't emit or absorb it without violating conservation of angular momentum. ...

8

That's an interesting question, even though it might be biased by the definition of forces, and on what particles they apply. For instance, if you want to describe the force that exists between photon (even though direct photon-photon scattering has not been observed yet), it is mainly due to electron loops, so in that case the `force' is fermionic. On a ...

7

If it is a planetary size mass, then thermal pressure from the kinetic energy of the atoms would prevent collapse. For larger masses, helium fusion would create radiation pressure to stop the collapse. Also, helium plasma would contain free electrons which would cause electron degenarcy pressure. Red giants of less than 2 solar masses initially have ...

6

There simply doesn't exist any container with $\mu\gt \epsilon_0$; that's what the quoted sentence says. What you could try is to try to increase the chemical potential. But the Bose-Einstein distribution says $$\langle n_i\rangle \sim\frac{g_i}{e^{(\epsilon_i-\mu)/kT}-1}$$ and if you chose values $\mu\gt \epsilon_i$, then the exponent in the denominator ...

6

I think what you are talking about is the stimulated emission of radiation. This is part of the process that occurs in a LASER and in fact, gave the LASER it's name - Light Amplification through the Simulated Emission of Radiation. When an atom is in an excited state it can spontaneously decay to a lower energy state with the emission of a photon of a ...

6

Unlike photons, gluons carry the "charge" of the strong force (confusingly, it's called "color"). This means that, unlike photons, gluons interact with each other. The effect is that, rather than spreading out in all directions (as photons do), gluons tend to stick together and form strings. For example, two quarks (which have color) are not connected by ...

5

Our current best experimentally verified theory, quantum field theory, isn't based on matter being particles or waves - all matter consists of excitations in quantum fields. The interactions of the quantum fields may appear particle like or wave like, so the wave-particle duality is a duality in the way the fields interact not a duality in the matter itself. ...

5

If by unification, one means that the bosons' and fermions' properties are linked to each other by a principle, the answer according to everything we know is Yes because the only principle able to link properties is a symmetry and a symmetry mapping bosons to fermions and vice versa is clearly a Grassmann-odd generator which has to carry a half-integral spin ...

5

I think the answer is it depends on distance (relative to the size of your system). Another well known example of a boson which is comprised of fermionic components is the helium-4 atom, which has integer spin (both the nucleus and the neutral atom itself). Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared ...

4

The definition of the partition function is $$Z = \sum_\mathbf{q} e^{-\beta E_\Sigma(\mathbf{q})} \qquad (1)$$ where $\mathbf{q}$ is the set of quantum numbers describing the microscopical state of the system, $E_\Sigma(\mathbf{q})$ is the energy of the system when it is in that microscopical state, $\beta = 1/(k_B T)$ In your case $\mathbf{q}$ is the ...

4

A composite particle consisting of an even (odd) number of fermions behaves like a boson (fermion). Swapping $\require{mhchem}\ce{^3He}$ atoms involves swapping an odd number of fermions (3 nucleons and 2 electrons): an odd number of sign changes of the wave function corresponds to no sign change or bosonic symmetry. Swapping $\ce{^4He}$ atoms involves ...

4

It can. This is exactly what happens when Helium-3 becomes superfluid. It's also what happens in superconductivity, which you mention in your question, when electrons combine into Cooper pairs. Well, it's not exactly what you ask since neither liquid Helium-3 nor electrons are a gas. It's very unlikely a gas of fermions could pair up to form a gas of bosons ...

3

In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms. A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a ...

3

Thats the whole point of dividing the world of particles in bosons and fermions. Because of the fact that they are indistinguishable we could switch two particles and we would NEVER know. So we could switch two particles, let's define the permutation operator $\hat{P}$ that has this effect. Upon applying this operator to a wavefunction that contains two ...

3

Consider the homogeneous ideal Bose gas, where we have: $$\langle N\rangle = \sum_{\vec{k}}\frac{1}{\exp{\beta(\epsilon_{\vec{k}}-\mu)-1}}.$$ The notation $\langle N_{\vec{k}}\rangle$ means the average number of particles in state $|\vec{k}\rangle$, which obviously has to be larger or equal to zero. In the case that $\vec{k} = 0$ we must have $$... 3 I) OP is asking for the composition formula for so-called single mode squeezing operators, see eq. (8) below. We will not here prove the composition formula (8), but only give partial hints and references. The key is to realize that one can identify$$\tag{1} \sigma_{+}~:=~\frac{1}{2} a^{\dagger}a^{\dagger}, \qquad \sigma_{-}~:=~\frac{1}{2} aa, ...

3

As you already wrote, the $(3/2)\partial^2 c$ term is needed for the current to be a one-form i.e. $(1,0)$ tensor field; see also page 131 of Polchinski's String Theory, volume 1. This means that if you compute the OPE $$T(z) j^{BRST}(0)\sim \dots,$$ you want to get $$\dots \sim \frac{1}{z^2} j^{BRST}(0)+\frac{1}{z}\partial j^{BRST}(0),$$ see e.g. ...

2

The way you would manually find the wave functions is more complicated than this. At first, you would need to describe better the one-particle scenario. I don't really understand your concept of the spin-state function, so I'll use mine, I hope it won't make my answer unusable. One particle If the particle has no spin, the wave functions are completely ...

2

My intuition tells me that the ground state energy of bosons should always be lower than the ground state of fermions -- no matter what kind of interactions or other external properties we've chosen. I think your intuition is usually correct; but it's possible to define systems where the ground state fermions would have less energy than the ground ...

2

Did you here that in the context of lasers or BEC? Photons are bosonic particles, this is generally what people refer to when they say 'they want to be together'.

2

I believe, that the derivation is wrong... If you assume a translationally invariant state, such that $G^1(r, r') = G^1(r - r')$ then you can get the result. Rewrite the exponential as $p r - p' r' = p( r- r') + r'(p - p')$. Since, in this case, the left-hand-side of Eq. (2.27) can only depend on $r - r'$ it must be such that $p = p'$ from the second term. ...

2

Let $|\mathbf{n}\rangle = |n_0, n_1, n_2, \dots\rangle$ denote a state with $n_i$ particles in the $i^\mathrm{th}$ energy state of the single-particle hamiltonian. Let $\epsilon_i$ denote the energy of the state with label $i$. The $n_i$ are "occupation numbers" in the standard terminology. Note that my notation is such that $i=0,1,2,\dots$ labels ...

2

The (orbital) wave function for $l=1$ doesn't just "have to be" antisymmetric. It demonstrably "is" antisymmetric. The relevant part of this wave function is completely determined, it's a particular function, so we may see whether it's symmetric or antisymmetric and indeed, it's the latter. Two particles – in this case two pions – orbiting each other are ...

2

Photons are gauge bosons, they do not have spins or magnetic moments! For electrons, bose/fermi atoms in a magnetic field, we have the energy $$E({\bf r})=\boldsymbol{\mu}\cdot {\bf B}({\bf r})$$ where $\boldsymbol{\mu}$ is the magnetic moment. Hence we have force due to the gradient of magnetic field, $${\bf F}=-\nabla E({\bf r})=-\mu\nabla{B}({\bf r})$$ ...

2

Any particle in a system (like a photon or an electron) can be in many different energy state. You may be familiar with the energy states of atoms, these are the states occupied by electrons orbiting the nucleus. But electron are fermions and they don't want to be together (in fact they can't). In order to minimize the energy of the atom any additional ...

2

The idea is actually simple. However, most book usually use sloppy terms, or they have not give explicit discussion on this issue, so it usually confused students. The correct phrase should be: Individual fermion in a system cannot have the same single particle wavefunction It is clear that the whole system itself always described by a total ...

2

Some remarks about both statements: "You need stringent experimental condition to observe the particle behaviour of photons for instance." I feel that you don't need to look very far. Counting photons is quite common in experimental physics. And you likely know about the double slit experiment, where you get those interference patterns - when the ...

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