netflix Modeling non-quantum objects (in finance, sociology etc) using fermionic fields

3down votefavorite 2	I want to know if deuterium is a fermion or boson. Please give me a descriptive answer. I tried the formula that is the combination of protons and electrons which gives odd number but the answer is boson. particle-physics fermions bosons share|cite|improve this question edited Dec 17 '14 at 9:22 abc 1 asked Dec 17 '14 at 9:07 Rocky mishra 162
	1   possible duplicate of Atoms: boson or fermion? – Rob Jeffries Dec 17 '14 at 9:51 add a comment |

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The deuterium nucleus is a boson, with spin ℏ $\hslash$ (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 12 ℏ $\frac{1}{2}\hslash$ or 32 ℏ $\frac{3}{2}\hslash$ , to be combined with the orbital angular momentum (which is zero in the ground state). However, the total atomic spin isn't a good quantum number in a nonzero magnetic field: since the electron and the nucleus have different magnetic moments, you can use the magnetic energy to tell which is polarized which way and can't generate a proper symmetric or antisymmetric mixture where the spin of each is superposed. The deuterium molecule, like all diatomic molecules, is a boson. The electrons combine into an antisymmetric spin singlet. The two nuclei must combine symmetrically, because they are identical bosons. That requirement introduces a coupling between the nuclear spin and the molecular angular momentum quantum number L $L$ , because a molecule with orbital angular momentum L $L$ has a sign change (−1) L  $(-1{)}^L$ under parity. This means the symmetric spin states, called "orthodeuterium" with spin 0 or spin 2ℏ $2\hslash$ , may only have even L $L$ , while the antisymmetric state "paradeuterium" with spin ℏ $\hslash$ must have odd L $L$ . This means that when deuterium is cooled near absolute zero it's impossible to remove all the nuclear angular momentum; you generate pure orthodeuterium, but it has some spin-2 fraction that can't be removed. This behavior is very different from ordinary dihydrogen molecules, where the nuclei are fermions. In that case the ground state is parahydrogen, with spin zero and even L $L$ , and cooling liquid or solid hydrogen eventually produces a material without any angular momentum. This makes a big difference in the heat capacities for cold hydrogen and cold deuterium, and also for their interaction with low-energy spin systems, most notably cold neutrons. If you were to call up your gas supplier and order a cylinder of deuterium, you'd get D 2  ${\mathrm{D}}_2$ gas, which is bosons. And the reason that cold deuterium acts differently from cold hydrogen is that the deuterium nucleus obeys Bose-Einstein statistics, while the hydrogen nucleus obeys Fermi-Dirac statistics — the two molecules have the same electronic configuration. If you held a gun to my head and forced me to check "boson or fermion" without any other context, I'd say boson get shot in the middle of this little lecture. Modeling non-quantum objects (in finance, sociology etc) using fermionic fields? up vote10down votefavorite 5 Please provide (if any) applications of fermionic field theory in non-physics macro contexts (finance, sociology etc). I see only bosonic fields being used mostly. The only (minor) application of fermionic fields that I have come across is in section 7.9.1 in Baaquie's Quantum Finance book where nonlinear forward interest rates are modeled using fermionic fields. Is there an intuitive reason for such asymmetry between bosonic and fermionic field usage at such scales? Edit (6 Jan 2015): After posting this question, I came to know that bosons and fermions (along with bosonic and fermionic exclusion principles) are just two special cases of a more general statistic called Haldane's exclusion statistics. While this may sound like the familiar anyon statistics in condensed matter physics, they are not exactly the same (for example, while anyons exist only in 2+1 dimensions, Haldane's statistics can be formulated for any dimension). For more on Haldane statistics: G. S. Canright, M. D. Johnson, "Fractional statistics: alpha to beta" M. V. N. Murthy, R. Shankar, "Exclusion Statistics: From Pauli to Haldane" So, the question now is, does Haldane's generalized statistics be used to describe non-quantum objects (in finance, sociology etc)? fermions statistics models spin-statistics anyons share|cite|improve this question edited Jan 6 '15 at 14:53 asked Feb 14 '13 at 1:55 crackjack 411212 add a comment |  2 Answers 2 active oldest votes up vote11down vote Actually a paper recently came out, and highlighted in Popular Science, discussing using fermionic field concepts to model crowd avoidance at Netflix. You can imagine that the same concept could be used to consider in any situation where there are large numbers of people competing for limited preferred items. Update Now that we have a few minutes, rather than give an expose of linked paper, I find it more appropriate to discuss some theory. First off, fermionic behavior is actually very relevant and widely used outside physics when it comes to economic problems. To illustrate this, first consider the graph below.  $$ People should recognize this as the standard textbook supply and demand diagram. There are different variations available, but I have made a few simplifications to aid in the explanation. Here both price and quantity are discrete (e.g. all quantities and prices are rational numbers), and are always positive. To review some features: The demand curve is downward sloping and represents how price responds to the decreasing utility of increasing quantities. This means that as an individual acquires more of some product, the product has less and less value to the individual. This is reflected in the price a person is willing to pay for an additional item of the same product after they already have several. The supply curve represents the marginal cost a supplier incurs for each additional unit of a product they have. This can be a complicated concept to convey, because one usually thinks on the demand side of the equation (e.g. price goes down as quantity goes up...e.g. learning curves etc). One way to think about the supply curve is to realize that as a seller increases their quantity of goods, it becomes increasingly difficult to sell the good at a profitable price, so overhead and carrying costs begin to dominate the sellers cost, and so carrying large quantities of goods increases the sellers marginal cost. The point where the supply curve and the demand curve meet is the equilibrium point for the particular product. The equilibrium point has an associated equilibrium price and quantity. It is the point of where maximum efficiency and profit is achieved for a particular product. For our example, I am assuming that our seller is a retailer, and therefore does not produce its own products. So here the retailer buys a product on the wholesale market for a fixed price, and then adds some minimum markup to each item in order to reflect their basic fixed operating expenses (and in some cases their minimum profit margin). The dynamics of the wholesale market are not directly represented here, but we include a max quantity line where one would have to adjust the wholesale price if that quantity were exceeded. For our retailer, this represents an upper bound on total quantity of any one product that the retailer could possibly sell (this is one type of "fermionic" attribute...if it were "bosonic", we would assume that the wholesale price would be completely independent of quantity, and the quantity could be infinite). Now the seller/retailer (or producer in our graph) wants the price to be as high as possible, and the individual buyer wants the price to be as low as possible. If the price is the equilibrium price, and the supply and demand curves hold true, then seller and buyer will maximize their benefits at the equilibrium price. When the price is lower, the buyer thinks they are "getting a bargain", and when the price is higher then the seller will make more profit off an individual sale. However, if the curves hold true, then the changes in price change the quantities that are sold, and a higher price will result in fewer sales, and a lower price will begin to reduce profits, and the utility of the product to the buyer. The equilibrium price has an equilibrium quantity associated with it. This is another "fermionic" attribute, it is a natural limit to the quantities of a product that is being sold. It does not benefit the seller/retailer to carry more than the equilibrium quantity of any good within the context of that good. The statement "within the context of that good" is important when one starts considering the case of multiple products, where the function to be minimized is now a function of many goods. Getting to some equations: For an individual product, if we say S(Q) $S(Q)$ is the supply price as a function of quantity, and D(Q) $D(Q)$ is the demand price as a function of quantity, and Q $Q$ is quantity sold, then we could say the equilibrium is reached when: (S(Q)−D(Q))Q=0 $$(S(Q)-D(Q))Q=0$$ This just tells us that when the difference of the price functions is zero for a particular quantity, then you are at the equilibrium quantity, and the measured inefficiency is zero. For our particular graph we could write: (K S Q−K D 1  Q+K D 2  )Q=K S Q 2 −K D 1  Q 2 +K D 2  Q=[(K S −K D 1  )Q 2 +K D 2  Q]=0 $$(K_{S}Q-K_{D_1}Q+K_{D_2})Q=K_S{Q}^2-K_{D_1}{Q}^2+K_{D_2}Q=[(K_S-K_{D_1})Q^2+K_{D_2}Q]=0$$ Where the K $K$ 's are constants controlling the slopes of the curves and the demand curve has two constant values, one controlling the downward slope, the other controlling the individual's "ground state" price, or rather the price they would pay when the possess none of the commodity (it should be pointed out that this is example specific). We can expand the minimizing function to account for individual buyers i $i$ and individual commodities j $j$ as follows: ∑ i,j [(S(Q i )−D i (Q i ))Q i ] j =0 $$\sum _{i,j}{\left[(S(Q_i)-D_i(Q_i))Q_i\right]}_j=0$$ ∑ i,j Q ij =Q total  $$\sum _{i,j}{Q}_{ij}=Q_{total}$$ it is assumed that the supply curve S(Q) $S(Q)$ for a commodity j $j$ is not dependent on the buyer i $i$ (e.g. the seller's curve is not tailored to an individual buyer...which is not always the case) but is determined by the total quantity Q j  $Q_j$ for some commodity j $j$ , and that Q total  $Q_{total}$ is the total number of items sold across the entire set of commodities j $j$ . In our specific example we could write this equation as: ∑ i,j [(K S −K D 1  i  )Q 2 i +K D 2  i  Q i ] j =0 $$\sum _{i,j}{[(K_S-K_{{D_1}_i})Q_i^2+K_{{D_2}_i}{Q}_i]}_j=0$$ and defining: (K S −K D 1  i  )=M i  $$(K_S-K_{{D_1}_i})=M_i$$ we can write: ∑ i,j [M i Q 2 i +K D 2  i  Q i ] j =0 $$\sum _{i,j}{[M_i{Q}_i^2+K_{{D_2}_i}{Q}_i]}_j=0$$ ∑ i,j Q ij =Q total  $$\sum _{i,j}{Q}_{ij}=Q_{total}$$ It should be recognized that in our simple example, the equation to be minimized ∑ i,j [M i Q 2 i +K D 2  i  Q i ] j =0 $$\sum _{i,j}{[M_i{Q}_i^2+K_{{D_2}_i}{Q}_i]}_j=0$$ can be thought of as a series of hypersurfaces where inefficiency is present when the sum ≠0 $\ne 0$ . In real systems, it is not likely that sum will =0 $=0$ . However, although there are squared quantities, and are not linear individually, the quantities are linear in our summation making it possible to consider linear optimization as a problem solving technique. In addition, the first derivatives of the individual equations should also sum to zero, taking away any particular concern about the squared quantities. ∑ i,j ∂ Q i  [M i Q 2 i +K D 2  i  Q i ] j =∑ i,j [2M i Q i +K D 2  i  ] j =0 $$\sum _{i,j}{\mathrm{\partial }}_{Q_i}{[M_i{Q}_i^2+K_{{D_2}_i}{Q}_i]}_j=\sum _{i,j}{[2M_i{Q}_i+K_{{D_2}_i}]}_j=0$$ Since the K D 1  i   $K_{{D_1}_i}$ 's are constants they can be summed and moved to the right hand side of the equation: ∑ i,j [K D 2  i  ] j =K total  $$\sum _{i,j}{\left[K_{{D_2}_i}\right]}_j=K_{total}$$ ∑ i,j [2M i Q i ] j =−K total  $$\sum _{i,j}{\left[2M_i{Q}_i\right]}_j=-K_{total}$$ Discussion It should be pointed out that there is only one optimal equilibrium point, however, if in real systems it can be shown that the optimal point is unobtainable and is instead replaced by some other minimum value, then there may be multiple solutions to a given problem. This means there may be multiple allocation schemes that are deemed optimal. It should also be pointed out that individuals who understand this simple theory are able to manipulate a given situation by temporarily moving a system from its equilibrium point. This is often a focus of wall street trading, but can make itself apparent in other areas of life. One example that comes to mind is the current use of qualifications in order to limit job opportunities to a select set of individuals. Although job qualifications can be good by maintaining a minimum standard for employment, they can also be manipulated to limit job allocation to a smaller population. The inefficiency appears when people who are otherwise capable of performing a job are excluded due to some arbitrary set of constraints unrelated to job performance. Recent evidence indicates that some of the current economic malaise is directly associated with a market inefficiency trend in the form of qualification standards. This and many other examples can be shown to represent "fermionic" behavior in financial and sociological settings. To get to the specific point as to whether "fermionic" models are better than "bosonic" models, the simple answer is that they already have been used and are regularly used, in particular, even the linked paper above discusses how many of these problems simplify to standard type linear resource allocation problems. UPDATE 2 As discussed in the comments below and in the linked paper, occupancy constraints by restricting quantities or by reducing utility with increased quantities reflect "crowd avoidance" and thus represent "fermionic" behavior. Edit Adjusted so that the supply and demand curves were combined in a more intuitive system based on signs of their slope. Further corrected to reflect the need for an additional constant to represent the downward sloping demand curve. share|cite|improve this answer edited Feb 17 '13 at 19:15 answered Feb 14 '13 at 2:17 user11547 3,580626 I'd +1 if you could summarize the essential point. Links do go bad. In particular, what's the advantage of fermionic fields over a bosonic field that avoids itself due to charge? Say, a box full of π +  ${\pi }^{+}$ particles. – DarenW Feb 14 '13 at 4:22 @DarenW incentives are always good, I will have to come back to this later today or tomorrow – user11547 Feb 14 '13 at 11:36 Hal, I've explained fermions as particles that need "personal space", and that get very upset (require more high-cost energy) if they don't get enough of it. For real particles this "space" has addresses with three parts (location, momentum, and spin), but many types of address spaces could be used without changing the idea. So, I found your answer fascinating... and alas, utterly incomprehensible! So, um... could you maybe point out to me (and likely others) exactly where the "fermionic" part of all that was? Again, I'm interested but baffled. – Terry Bollinger Feb 17 '13 at 5:44 1   @HalSwyers, we're in sync then. From my first very quick scan, I just didn't pick up that the critical anti-crowding component, which is indeed what makes a fermion a fermion. We see it everyday -- two objects don't like to be in the same space, because ordinary matter is fermionic. In the past I've had some brief but interesting email exchanges with Ginestra Bianconi, whose work was pointed out to me by Albert-László Barabási, on both bosonic and fermionic networking models. While my conclusion was that one must be careful in defining the entities involved, both concepts can be useful. – Terry Bollinger Feb 18 '13 at 18:52 1   @TerryBollinger Concur with your statements. What is interesting though is that I think we are coming full circle since economics has, as you point out, been fermionic in nature since its beginning. What would be interesting is how economic insights might translate into research into objects like Marjorana fermions – user11547 Feb 18 '13 at 19:24  |  show 8 more comments up vote0down vote According to this book (p. 61) "[this] paper discusses in detail how a Fermion and Boson field can be used to express the state of the human psyche." I cannot expand. share|cite|improve this answer edited Feb 22 '13 at 13:32 Terry Bollinger 11.9k3584 answered Feb 17 '13 at 13:15 Keep these mind 4,64511641 Apologies, I guess this doesn't qualify as a "macro context". – Keep these mind Feb 17 '13 at 13:18 @TerryBollinger Thanks? I'm not quite sure what you did to my answer. The line in the book is "The paper discusses in detail how a Fermion and Boson field can be used to express the state of the human psyche." I replaced 'The' with '[this]', as per convention. You removed '"[' and ']'. What did you "fix"? – Keep these mind Feb 22 '13 at 8:49 My bad; I just rolled back to your original correct version. I saw the square brackets and must have interpreted it as a slightly mangled link attempt. – Terry Bollinger Feb 22 '13 at 13:34 @TerryBollinger Thanks again. :) – Keep these mind Feb 22 '13 at 13:40 Fundamental forces behind covalent bonding up vote16down votefavorite 7 I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other? Also, am I right to think that "repulsion occurs when atoms are too close together" comes from electrostatic interaction? bond share|improve this question edited Jun 29 '12 at 12:15 ManishEarth♦ 8,17843788 asked Jun 29 '12 at 10:44 wolfik 11115 add a comment |  3 Answers 3 active oldest votes up vote17down vote I understand that covalent bonding is an equilibrium state between attractive and repulsive forces, but which one of fundamental forces actually causes atoms to attract each other? The role of Pauli Exclusion in bonding It is an unfortunate accident of history that because chemistry has a very convenient and predictive set of approximations for understanding bonding, some of the details of why those bonds exist can become a bit hard to discern. It's not that they aren't there -- they most emphatically are! -- but you often have to dig a bit deeper to find them. They are found in physics, in particular in the concept of Pauli exclusion. Chemistry as avoiding black holes Let's take your attraction question first. What causes that? Well, in one sense that question is easy: it's electrostatic attraction, the interplay of pulls between positively charged nuclei and negatively charged electrons. But even in saying that, something is wrong. Here's the question that points that out: If nothing else was involved except electrostatic attraction, what would be the most stable configuration of two or more atoms with a mix of positive and negative charges? The answer to that is a bit surprising. If the charges are balanced, the only stable, non-decaying answer for conventional (classical) particles is always the same: "a very, very small black hole." Of course you could modify that a bit by assuming that the strong force is for some reason stable, in which case the answer becomes "a bigger atomic nucleus," one with no electrons around it. Or maybe atoms as Get Fuzzy? At this point some of you reading this should be thinking loudly "Now wait a minute! Electrons don't behave like point particles in atoms, because quantum uncertainty makes them 'fuzz out' as they get close to the nucleus." And that is exactly correct -- I'm fond of quoting that point myself in other contexts! However, the issue here is a bit different, since even "fuzzed out" electrons provide a poor barrier for keeping other electrons away by electrostatic repulsion alone, precisely because their charge is so diffuse. The case of electrons that lack Pauli exclusion is nicely captured by Richard Feynman in his Lectures on Physics, in Volume III, Chapter 4, page 4-13, Figure 4-11 at the top of the page. The outcome Feynman describes is pretty boring, since atoms would remain simple, smoothly spherical, and about the same size as more and more protons and electrons get added in. While Feynman does not get into atoms how such atoms would interact, there's a problem there too. Because the electron charges would be so diffuse in comparison to the nuclei, the atoms would pose no real barrier to each other until the nuclei themselves begin to repel each other. The result would be a very dense material that would have more in common with [neutronium[(http://en.wikipedia.org/wiki/Neutronium) than with conventional matter. For now I'll just forge ahead with a more classical description, and capture the idea of the electron cloud simply by asserting that each electron is selfish and likes to capture as much "address space" (see below) as possible. Charge-only is boring! So, while you can finagle with funny configurations of charges that might prevent the inevitable for a while by pitting positive against positive and negative against negative, positively charged nuclei and negatively charged electrons with nothing much else in play will always wind up in the same bad spot: either as very puny black holes, or as tiny boring atoms that lack anything resembling chemistry. A universe full of nothing but various sizes of black holes or simple homogenous neutronium is not very interesting! Preventing the collapse So, to understand atomic electrostatic attraction properly, you must start with the inverse issue: What in the world is keeping these things from simply collapsing down to zero size -- that is, where is the repulsion coming from? And that is your next question: Also, am I right to think that "repulsion occurs when atoms are too close together" comes from electrostatic interaction? No; that is simply wrong. In the absence of "something else," the charges will wiggle about and radiate until any temporary barrier posed by identical charges simply becomes irrelevant... meaning that once again you will wind up with those puny black holes. What keeps atoms, bonds, and molecules stable is always something else entirely, a "force" that is not traditionally thought of as being a force at all, even thought it is unbelievably powerful and can prevent even two nearby opposite electrical charges from merging. The electrostatic force is enormously powerful at the tiny separation distances within atoms, so anything that can stop charged particles from merging is impressive! The "repulsive force that is not a force" is the Pauli exclusion I mentioned earlier. A simple way to think of Pauli exclusion is that identical material particles (electrons, protons, and neutrons in particular) all insist on having completely unique "addresses" to tell them apart from other particles of the same type. For an electron this address includes: where the electron is located in space, how fast and in what direction it is moving (momentum), and one last item called spin, which can only have on of two values that are usually called "up" or "down." You can force such material particles (called fermions) into nearby addresses, but with the exception of that up-down spin part of the address, doing so always increases the energy of at least one of the electrons. That required increase in energy is a nutshell is why material objects push back when you try to squeeze them. Squeezing them requires minutely reducing the available space of many of the electrons in the object, and those electrons respond by capturing the energy of the squeeze and using it to push right back at you. Now, take that thought and bring it back to the question about where repulsion comes from when to atoms bond at a certain distance, but no closer. They are the same mechanism! That is, two atoms can "touch" (move so close, but no closer) only because they both have a lot of electrons that require separate space, velocity, and spin addresses. Push them together and they start hissing like cats from two households who have suddenly been forced to share the same house. (If you own multiple cats, you'll know exactly what I mean by that.) So, what happens is that the overall set of plus-and-minus forces of the two atoms is trying really hard to crush all of the charges down into a single very tiny black hole -- not into some stable state! It is only the hissing and spitting of the overcrowded and very unhappy electrons that keeps this event from happening. Orbitals as juggling acts But just how does that work? It's sort of a juggling act, frankly. Electrons are allowed to "sort of" occupy many different spots, speeds, and spins (mnemonic s 3  $s^3$ , and no, that is not standard, I'm just using it for convenience in this answer only) at the same time, due to quantum uncertainty. However, it's not necessary to get into that here beyond recognizing that every electron tries to occupy as much of its local s 3  $s^3$ address space as possible. Juggling between spots and speeds requires energy. So, since only so much energy is available, this is the part of the juggling act that gives atoms size and shapes. When all the jockeying around wraps up, the lowest energy situations keep the electrons stationed in various ways around the nucleus, not quite touching each other. We call those special solutions to the crowding problem orbitals, and they are very convenient for understanding and estimating how atoms and molecules will combine. Orbitals as specialized solutions However, it's still a good idea to keep in mind that orbitals are not exactly fundamental concepts, but rather outcomes of the much deeper interplay of Pauli exclusion with the unique masses, charges, and configurations of nuclei and electrons. So, if you toss in some weird electron-like particle such as a muon or positron, standard orbital models have to be modified significantly and applied only with great care. Standard orbitals can also get pretty weird just from having unusual geometries of fully conventional atomic nuclei, with the unusual dual hydrogen bonding found in boron hydrides such as diborane probably being the best example. Such bonding is odd if viewed in terms of conventional hydrogen bonds, but less so if viewed simply as the best possible "electron juggle" for these compact cases. "Jake! The bond!" Now on to the part that I find delightful, something that underlies the whole concept of chemical bonding. Recall that it takes energy to squeeze electrons together in terms of the main two parts of their "addresses," the spots (locations) and speeds (momenta)? I also mentioned that spin is different in this way: the only energy cost for adding two electrons with different spin addresses is that of conventional electrostatic repulsion. That is, there is no "forcing them closer" Pauli exclusion cost like you get for locations and velocities. Now you might think "but electrostatic repulsion is huge!", and you would be exactly correct. However, compared to the Pauli exclusion "non-force force" cost, the energy cost of this electrostatic repulsion is actually quite small -- so small that it can usually be ignored for small atoms. So when I say that Pauli exclusion is powerful, I mean it, since it even makes the enormous repulsion of two electrons stuck inside the same tiny sector of a single atom look so insignificant that you can usually ignore its impact! But that's secondary, because the real point is this: When two atoms approach each other closely, the electrons start fighting fierce energy-escalation battles that keep both atoms from collapsing all the way down into a black hole. But there is one exception to that energetic infighting: spin! For spin and spin alone, it become possible to get significantly closer to that final point-like collapse that all the charges want to do. Spin thus becomes a major "hole" -- the only such major hole -- in the ferocious armor of repulsion produced by Pauli exclusion. If you interpret atomic repulsion due to Pauli exclusion as the norm, then spin-pairing two electrons becomes another example of a "force that is not a force," or a pseudo force. In this case, however, the result is a net attraction. That is, spin-pairing allows two atoms (or an atom and an electron) to approach each other more closely that Pauli exclusion would otherwise permit. The result is a significant release of electrostatic attraction energy. That release of energy in turn creates a stable bond, since it cannot be broken unless that same energy is returned. Sharing (and stealing) is cheaper So, if two atoms (e.g. two hydrogen atoms) each have an outer orbital that contains only one electron, those two electrons can sort of look each other over and say, "you know, if you spin downwards and I spin upwards, we could both share this space for almost no energy cost at all!" And so they do, with a net release of energy, producing a covalent bond if the resulting spin-pair cancels out positive nuclear charges equally on both atoms. However, in some cases the "attractive force" of spin-pairing is so overwhelming greater for one of the two atoms that it can pretty much fully overcome (!) the powerful electrostatic attraction of the other atom for its own electron. When that happens, the electron is simply ripped away from the other atom. We call that an ionic bond, and we act as it if it's no big deal. But it is truly an amazing thing, one that is possible only because the pseudo force of spin-pairing. Bottom line: Pseudo forces are important! My apologies for having given such a long answer, but you happened to ask a question that cannot be answered correctly without adding in some version of Pauli "repulsion" and spin-pair "attraction." For that matter, the size of an atom, the shape of its orbitals, and its ability to form bonds similarly all depend on pseudo forces. share|improve this answer edited Oct 4 '12 at 17:12 answered Jul 1 '12 at 2:24 Terry Bollinger 1,994517 1   I would have expected mention of Slater's treatment using the quantum virial theorem, and the important result that it is electrostatics that keeps atoms together. Also, your answer seems to ignore two important observations 1) nuclear fusion is possible and 2) the Pauli principle leads to the vanishing of the wave function when electrons of the same quantum number share the same region of space, again cf. Slater determinants for antisymmetrized wave functions. (I do not deny the importance of the Pauli principle in specific solutions to HΨ=EΨ $H\mathrm{\Psi }=E\mathrm{\Psi }$ , but that is not the issue here.) – Eric Brown Jun 2 '14 at 6:07 add a comment |  up vote4down vote Thinking about this like a physicist, there are four fundamental forces: the strong nuclear force, the weak nuclear force, the electromagnetic force, and gravity. The strong nuclear force holds the protons and neutrons in the nucleus together. The weak nuclear force causes beta decay. Those two might be considered chemistry. I don't, but some people do. Gravity is much too weak to have any effect on chemistry. So that leaves the electromagnetic force to control nearly all of chemistry. On a simple conceptual level, that's all there is. The nuclei are both positively charged, so they repel each other. The electrons are negatively charged, so they are attracted to their respective nuclei. When the electron clouds get close enough to interact with both nuclei, then they begin to pull the nuclei together. The deeper explanation requires quantum mechanics. When the atoms are separated, you can use the Schrödinger equation with the electric potential from the nucleus. That gives you the electron orbitals for an atom all by itself. When the two atoms get close together, you use the electric potential for both nuclei in the Schrödinger equation. The solution is then the molecular orbital rather than the atomic orbitals. Because the Schrödinger equation is impossible to solve exactly for a molecule, chemists need an approximation. The usual approximation is to build the molecular orbital out of the atomic orbitals by adding and subtracting the atomic orbitals. This is where the ideas of sp $sp$ -, sp 2  $s{p}^2$ -, and sp 3  $s{p}^3$ -hybridizaton, and π $\pi$ - and σ $\sigma$ -bonding come from. For further information, most introductory college-level general chemistry texts should discus this. As an example, I pulled most of the above explanation from Zumdahl's Chemistry. In the 5th edition, this is in chapters 8 and 9 (the current edition appears to be the 7th). This is a much more important idea in organic chemistry, so those textbooks usually review it in the first one or two chapters. The organic chemistry book I have in front of me at the moment is McMurray's Organic Chemistry. This is discussed in chapter 1 of the 3rd edition of that book (current edition is the 8th). share|improve this answer answered Jun 29 '12 at 21:02 Colin McFaul 1,1902721 add a comment |  up vote2down vote Both the attraction and repulsion are the result of the electromagnetic interaction. At long distances, two atoms attract each other because of dipole-dipole interactions. When they get close enough together, the electrostatic repulsion of the nuclei takes over (as well as the exchange interaction acting on the non-valence electrons of the atoms and forcing them into higher energy states). This makes the atoms repel each other at short distances