# Tag Info

## Hot answers tagged bosons

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 ...

13

The deuterium nucleus is a boson, with spin $\hbar$ (and positive parity). Unlike other stable nuclei, deuterium doesn't have any bound excited states; however if it did they would also have integer spin. The deuterium atom is a fermion, which may have spin $\frac12\hbar$ or $\frac32\hbar$, to be combined with the orbital angular momentum (which is zero in ...

11

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 ...

9

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

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

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 ...

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

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. ...

8

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 ...

7

The chemical potential can be thought of as how accepting the system is of new particles -- how much work you have to do to stick a new particle in the system. Since you can stick as many bosons in a given state as you want, the system is always accepting of new particles. At worst, you have to do zero work to add a boson (corresponding to $\mu=0$), and ...

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

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 ...

6

The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle ...

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

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. ...

5

Comments to the question (v3): In contrast to QED with fermionic matter, in QED with bosonic matter, the full Noether current ${\cal J}^{\mu}$ (for global gauge transformations) tends to depend explicitly on the gauge potential $A^{\mu}$, see e.g. Refs. 1-2 and this Phys.SE post. The reason for this difference is because the QED Lagrangian for fermionic ...

5

It does not mediate flavor changing because the couplings in the gauge basis are of the form $c_L Z_\mu \bar{\psi}^i_L \gamma^\mu \psi^i_L$ and $c_R Z_\mu \bar{\psi}^i_R \gamma^\mu \psi^i_R$ with $i$ flavor index and $c_{L,R}$ flavor independent by gauge invariance (that is, $c_L$ and $c_R$ are proportional to the identity in flavor space). In this basis ...

5

It is not possible to have a state with four indistinguishable particles such that $P_{12} \psi = -\psi$ and $P_{34} \psi =\psi$, for an algebraic reason. Namely, the exchange operators have to form a representation of the permutation group $S_4$. It is rather well known that there are exactly two representations of $S_n$: the trivial representation where ...

5

The correct statement is that the hamiltonian does not affect the total number of particles in a given state. This means that you do cannot equate the state $|n_1,n_2⟩$ with $|n_1\pm1,n_2\mp1⟩$, because they are different states, but they are all eigenstates of the total particle number operator, $$\hat N=a_1^\dagger a_1+a_2^\dagger a_2.$$ Your calculation ...

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 ...

5

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 ...

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

The terminology of a mode of a free quantum field $\phi(x)$ comes from writing it as a Fourier transform, often also called mode expansion: $$\phi(\vec x) = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left(a(\vec p)\mathrm{e}^{\mathrm{i}\vec x\cdot\vec p} + b(\vec p)^\dagger\mathrm{e}^{-\mathrm{i}\vec x\cdot\vec p}\right)$$ where for a ...

4

It's true that "the perturbative series is valid only when the perturbed state is qualitatively similar to the unperturbed state". Generally perturbation theory is acceptable when the coupling is weak, in which case the coupling can be treated as a small perturbation of the free field theory at all energies (for example Yukawa theory and $\phi^{4}$ theory. ...

4

It's pretty darn difficult to describe quantum field theory without technical details but here goes nothing. What we want is some field, $\phi$, which satisfies the equations associated with a scalar (spin 0) fields such as the Higgs. One such equation is the Klein-Gordon equation, which is related to the relationship between energy, momentum and mass found ...

4

This is confusing enough to make me want to scribble down a few equations. This is a creation operator for deuterium: $$D_{\rho}^{\dagger}\left(k\right)=\sum_{\alpha\beta\gamma}\int\mathrm{d}^{3}l\mathrm{d}^{3}p\mathrm{d}^{3}q\delta^{\left(3\right)}\left(l+p+q-k\right)\ ... 4 Essentially you're deriving varients on the ideal gas law:$$ PV = nkT $$The ideal gas law falls out when your system of particles obeys a Boltzmann distribution. From this you can derive the Maxwell-Boltzmann distribution of velocities and from that the pressure. If your system contains bosons then it will obey Bose-Einstein statistics and compared to ... 4 The definition of the partition function is$$ 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 ...

3

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 ...

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