If two or more identical fermions cannot occupy the same quantum state simultaneously, then shouldn't the entire universe have no two identical fermions in the same quantum state at any given time?  Intuition tells me this isn't true, so how do you define the boundaries for a quantum system in which the Pauli exclusion applies?
122861
Can two electrons be in the same state, when they belong to two different atoms, which are "far enough" (whatever that means) apart from each other?  With "same state" I mean that (as far as specifiable) the states are really identical, except for the position of the electrons.  More specifically: I am still not clear of how the separation of particles is taken into account for the Pauli exclusion principle. E.g. in a crystal the electrons seem close enough for the exclusion principle to become meaningful, for particles "a universe apart" from each other, it seems that it is kind of redundant (is there specific maths to that?), but what about everything in between?   (This is closely related to the talk Brian Cox gave, see this question: physics.stackexchange.com/q/18527/16689 )
27056
I simply know that a single free quark does not exist. What is the reason that we can not get a free quark?  If we can't get a free quark then what is single-top-quark?
103485
In practice, when entangling two particles (say two electrons), is creating an entangled state purely a matter of bringing them close enough together?  What I mean is, to make their wave functions correlated, is it just a matter of bringing them close enough together (regardless of what is required to accomplish this), or are there other processes necessary to cause the entanglement?
65161
In a normal atom, there is a limit of 2 electrons per obital due to the Pauli-Exclusion principle. I have seen people talking about replacing an electron with a muon, but since muons and electrons are distinguishable, would this then allow 2 electrons and 2 muons to occupy the same orbital, or would only 2 leptons be allowed to have the same set of quantum numbers?  For example, would a Beryllium atom with 2 muons and 2 electrons use only the 1S orbital, or would the 2S be used as well? Spectroscopic methods should be able to distinguish between the two.
62569
In the past few days I've become increasingly intrigued by the QHE, mainly thanks to very interesting questions and answers that have appeared here. Unfortunately, I am as of yet very confused by all the (seemingly disparate) stuff I learned.  First, here are some random points that I've been able to gather   I(nteger)QHE occurs due to the presence of Landau levels IQHE is an embodiment of topological order and the states are characterized by the Chern number that tells us about topologically inequivalent Hamiltonians defined on the Brillouin zone IQHE requires negligible electron-electron interations and so is dependent on the presence of impurities that shield from Coulomb force F(ractional)QHE occurs because of formation of anyons. In this case Coulomb interaction can't be neglected but it turns out an effective non-interacting description emerges with particles obeying parastatistics and having fractional charge FQHE has again something to do with topology, TQFT, Chern-Simons theory, braiding groups and lots of other stuff FQHE has something to do with hierarchy states   So, here are the questions        Most importantly, do these points make sense? Please correct any mistakes I made and/or fill in other important observations   How do explanations 1. and 2. of IQHE come together? Landau quantization only talks about electron states while topological picture doesn't mention them at all (they should be replaced by global topological states that are stable w.r.t. perturbations)   How do explanations 4., 5. and 6. relate together   Is there any accessible introductory literature into these matters?   Do IQHE and FQHE have anything (besides last three letters) in common so that e.g. IQHE can be treated as a special case? My understanding (based on 3.) is that this is not the case but several points hint into opposite direction. That's also why I ask about both QHE in a single question
104175
I am a bit confused about the terminology concerning identical fermions. In quantum mechanics, identical fermions need to obey certain anticommutation relations i.e. have an antisymmetric total wavefunction (product of spatial and spin parts) under the exchange of two particles. But when exactly are two fermions considered to be 'identical'? As far as I understand, two electrons are identical in the sense that they for example have the same electric charge, mass and spin quantum number $s=\frac{1}{2}$. (These are shared by all electrons.) How about the secondary spin (or spin projection) quantum number $m_s=\pm \frac{1}{2}$? If in an atom we have two electrons with the same principal quantum number, orbital quantum number and magnetic quantum number, but different $m_s$, are these two electrons 'identical' or not? The electrons are definitely identical if they additionally have the same $m_s$, but the Pauli exclusion principle forbids that. How does the consideration of the quantum numbers fit with the electrons having the same electric charge, mass, and spin $s$, which are independent of atomic orbitals? Furthermore, are 'identical' and 'indistinguishable' fermions different things or not? If 'identical' particles cannot be told apart by an experiment, how about the Stern-Gerlach experiment? This links back to the above question whether $m_s=\pm \frac{1}{2}$ electrons are identical. My partial answer is that if we have electrons labelled 1 and 2, we cannot say which of the electrons has $m_s=-\frac{1}{2}$ and which has $m_s=\frac{1}{2}$. So in an experiment, you cannot say that 'The electron with $m_s=-\frac{1}{2}$ is electron 1' even if you were able to separate the $m_s=\pm \frac{1}{2}$ states. Or something along those lines. In other words, electrons are identical regardless of the quantum numbers. Further confusion arises when we go beyond electrons. I have learned that atoms with an odd number of neutrons in an 'optical lattice' can be used to mimic electrons in a metal. Then the role of $m_s=\pm \frac{1}{2}$ is played by different hyperfine states of the atom as far as I can tell. If I now have atoms of the same mass (same species) in two different hyperfine states corresponding to spin-up and spin-down electrons, how are these atoms classified? Are they 'identical', 'indistinguishable', 'distinguishable', or what? Do these atoms obey the fermionic anticommutation relations? I thought I understood what's going on i.e. two atoms of the same species are identical but can be in different internal states representing the spin projections and thus the atoms obey the usual fermionic anticommutation relations of spin-$\frac{1}{2}$ electrons. But then I found this paper () which says that atoms in the $| F=\frac{1}{2}, m_F=\pm \frac{1}{2} \rangle$ states of $^6$Li are distinguishable. How can distinguishable atoms play the role of electrons? How can they obey the anticommutation relations of spin-$\frac{1}{2}$ electrons? If I have understood correctly, there are no symmetry requirements for distinguishable particles in quantum mechanics. Or are the authors just possibly using sloppy language by calling two identical atoms in different internal states distinguishable? I would call two atoms of different species, e.g. $^6$Li and $^{40}$K, distinguishable. Sorry for the long text, but hopefully I made my confusion sufficiently clear. Any clarifications are appreciated.
78045
It seems to be related to , but I can't penetrate the Wikipedia article. What has the Pauli exclusion principle to do with indistinguishability?
33514
I am trying to get an intuitive idea of how the  affects Quantum computation. My understanding is that given a qubit $Q$ in superposition $Q_0 \left| 0 \right&gt; + Q_1 \left| 1 \right&gt;$, NCT states another Qubit $S$ cannot be designed such that $S$ is equivalent to the state of $Q$.   Now the catch is, what does Equivalent mean? It could mean either that:   $S = S_0 \left| 0 \right&gt; + S_1 \left| 1 \right&gt;$ such that $S_0 = Q_0, S_1 = Q_1$. Or it could mean that $S = Q$, meaning that if $S$ is observed to be some value ( for example 0) then $Q$ MUST be that same value, and vice versa.    So it seems that point 2, occurs anyways in entangled systems (particularly cat-states), so I can eliminate that option and conclude that that No Cloning states, given a qubit $Q$, it's impossible to make another qubit $S$ such that:  $S = S_0 \left| 0 \right&gt; + S_1 \left| 1 \right&gt;$ such that $S_0 = Q_0, S_1 = Q_1$.  Is this correct?
4258
Suppose we have an atom. It is commonly said that because of the PEP, two electrons can't be in the ground state unless they have opposite spins, because no two electrons can have the same wavefunction.  What bugs me is that spin up and spin down aren't the only possible spin states. There's a whole continuum of linear combinations of them, and as far as I can tell the PEP wouldn't exclude the possibility of having lots of electrons, all sharing the same spatial wavefunction but with different combinations of $\mid\uparrow \rangle$ and $\mid\downarrow\rangle$. Why doesn't this happen?
106849
(I was requested to edit the question to explain why it is different that a proposed duplicate question. This seems counterproductive to do here, inside the question it self, but that is what I have been asked by the site and moderators. There is no way for me to vote against their votes. So, here I go: Please stop voting this as a duplicate so quickly, which will eventually lead to this question being closed off. Yes, the other question linked to asks the same math, but any newcomer to the problem who was exposed to it via physics, as I was, will prefer this question instead of the one that is purely mathematically. I beg the moderators to not be pedantic on this one. This question spills into physics, which is why I did the cross post to the physics forum as well.)  How does the sum of the series “1 + 2 + 3 + 4 + 5 + 6…” to infinity = “-1/12”, in the context of physics?  I heard Lawrence Krauss say this once during a debate with Hamza Tzortzis (). I found a transcript of another debate between Krauss and William Lane Craig which has the same sum. Here is the paragraph in full:     Let’s go to some of the things Dr. Craig talked about. In fact, the   existence of infinity, which he talked about which is   self-contradictory, is not self-contradictory at all. Mathematicians   know precisely how to deal with infinity; so do physicists. We rely on   infinities. In fact, there’s a field of mathematics called “Complex   Variables” which is the basis of much of modern physics, from   electro-magnetism to quantum mechanics and beyond, where in fact we   learn to deal with infinity; without the infinities we couldn’t do the   physics. We know how to sum infinite series because we can do complex   analysis. Mathematicians have taught us how. It’s strange and very   unappetizing, and in fact you can sum things that look ridiculous. For   example, if you sum the series, “1 + 2 + 3 + 4 + 5 + 6…” to infinity,   what’s the answer? “-1/12.” You don’t like it? Too bad! The   mathematics is consistent if we assign that. The world is the way it   is whether we like it or not.   -- Lawrence Krauss, debating William Lane Craig, March 30, 2011  Source:   CROSS POST: I'm not sure if I should post this in mathematics or physics, so I posted it in both. Cross post:   EDIT: I did not mean to begin a debate on why Krauss said this. I only wished to understand this interesting math. He was likely trying to showcase Craig's lack of understanding of mathematics or logic or physics or something. Whatever his purpose can be determined from the context of the full script that I linked to above. Anyone who is interested, please do. Please do not judge him out of context. Since I have watched one of these debates, I understand the context and do not hold the lack of a full breakdown as being ignorant. Keep in mind the debate I heard this in was different from the debate above.
18988
I apologize if this question is posted on the wrong portion of the website, but I knew no better place to ask.  I've come to see that some effects, such as super-conduction may happen at either extremely low temperatures, or at extreme temperatures (the recent discovery of superconducting iron in the bowels of the earth, for one), and this is fascinating.  My question is: can (does?) the same principle be applied to quantum effects? Would it be applicable to, for example, say that although quantum effects happen at nano scales and below they may happen on a grand scale of say, the entire Universe (as a solitary unit)?  Can said quantum effects happen at moments of say, picoseconds after the Big Bang due to the turbulence of forces involves in such an event? Or even within the core of a dying star?  Thank you for any answers or clarification of my misinformation.
106838
The standard interpretation of $|\psi|^2$ is taken as the probability density of the wave-function collapsing in the given infinitesimal small region. The way this probability is interpreted (at least in the text-book by Griffiths) is that if I prepare a large number of identical states and then perform a measurement on each of them then the probability associated with $|\psi|^2$ actually represents the statistical results of the measurements made (individually) on all these wave-functions of the entire ensemble. This seems like a good enough argument to accept $|\psi|^2$ as the probability density of the wave-function collapsing in a given infinitesimal small region even if I have only one wave-function.  But the  suggests that it is fundamentally impossible to make two states that are completely identical and thus, to my understanding, it makes absolutely no sense to talk about those identical wave-functions that were used to project $|\psi|^2$ as the probability density.  Put in other words, if I can't replicate a wave-function at all then how do I confirm in my laboratory that the probabilities obtained by $|\psi|^2$ actually represents the likely-hood of the collapse happening in a given region?   Edit Is it the catch that the No-Clone Theorem suggests only that a given state can't be evolved to a state identical to another state and it allows the two states being identical if they are so from eternity?
116896
I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question.
35982
In Moore's introductory physics textbook Six Ideas that Shaped Physics, he describes a set of qualitative rules that first-year physics students can use to sketch energy eigenfunctions in a 1D quantum-mechanical potential.  Many of these are basically the WKB formalism in disguise;  for example, he introduces a notion of "local wavelength", and justifies the change in amplitude in terms of the classical particle spending more time there.  He also notes that the wavefunction must be "wave-like" in the classically allowed region, and "exponential-like" in the classically forbidden region.  However, there is one rule that he uses which seems to work for many (but not all) quantum potentials:     The $n$th excited state $\psi_n(x)$ of a particle in a 1D potential has $n$ extrema.   This is true for the particle in a box (either infinite or finite), the simple harmonic oscillator, the , and presumably a large number of other 1D quantum potentials.  It is not true, however, for a particle in a double well of finite depth;  the ground state, which has a symmetric wavefunction, has two maxima (one in each potential well) and one minimum (at the midpoint between the wells).  The following questions then arise:   Are there conditions can we place on $V(x)$ that guarantee the above quoted statement is true?  For example, is the statement true if $V(x)$ has only one minimum?  Is the statement true if the classically allowed region for any energy is a connected portion of $\mathbb{R}$?  (The second statement is slightly weaker than the first.) Can we generalize this statement so that it holds for any potential $V(x)$?  Perhaps there is a condition on the number of maxima and minima of $V(x)$ and $\psi_n(x)$ combined?   I suspect that if a statement along these lines can be made, it will come out of the orthogonality of the wavefunctions with respect to some inner product determined by the properties of the potential $V(x)$.  But I'm not well-enough versed in operator theory to come up with an easy argument about this.  I would also be interested in any interesting counterexamples to this claim that people can come up with.
64431
I have read that all electrons are identical and I assume the same for protons, etc. That is, all electrons are of the exact same mass and charge.  But do we really know this? Could there be slightly heavier electrons with slightly higher charge or is there a reason why such particles could not exist or would be unstable?
105091
Are  and the  really the same thing?  Not too long ago, Susskind and Bousso uploaded the article "" with the thesis that the many-worlds interpretation and the multiverse of eternal inflation are one and the same thing. The parallel worlds of one are exactly the same thing as the parallel worlds of the other.   First, they claim decoherence can't happen over a complete description of the future light-cone of the measurement. Then, they apply that principle to eternal inflation. Without decoherence, superpositions of nucleating bubbles and metastable vacua can't decohere. According to the anthropic principle, most bubbles have no conscious observers, but an exponentially small minority do. Apply black hole complementarity to causal horizons.  Then, somehow, in a way I can't follow, they combine causal diamond worlds into a global multiverse.  Then they claim decoherence is reversible.  My head is spinning. What are your opinions on this paper?
90983
The many-worlds interpretation of quantum mechanics has always been explained to me at a high level using examples of binary events (e.g. atom either did or did not decay at any given moment in time), which leads to a conceptually clean idea of "branching" into two distinct universes. But how does branching work if you have, say, a 70% probability of something happening (e.g. measuring the spin of an electron having gone through an SG apparatus oriented at an arbitrary angle)?  Do you say that 7 universes got to spin up and 3 that got spin down (to account for the 70% probability)? Does that mean those 7 universes are in-every-way identical copies of each other? But then what if you had something with a 71.87154819...% chance of happening? You would need an uncountably infinite number of branches to be able to represent arbitrarily precise probability ratios, and then subsets of those branches would contain uncountably infinite universes that are 100% degenerate and identical to each other. Is this what the standard many worlds interpretation assumes?  If not this, then what? You can't say that there's a 70% chance of universe A happening and a 30% chance of universe B happening if you're saying both happen. How does the many-worlds interpretation put a "weighting function" on different branches or outcomes?