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Pauli exclusion principle: a Force or not?

  1. Jun 9, 2010 #1
    A small part of my brain has been bugged by this for a while now, so I figured I'd ask. According to most teachings, as well as the wikipedia entry, there are only 4 fundamental forces. The strong and weak interaction, electromagnetism and gravity, and as far as I know the Pauli exclusion principle (PEP) has nothing to do with either one of those.

    What's been bugging me is that for all practicle purposes the PEP seems to act like a force. For example, on the wiki page on forces it list the four fundamental forces, and then proceeds to list a number of forces that can all be dervied from these, but on several occasions throughout the text, e.g. while explaining the normal force, they say that it comes from the PEP, with no further link back to any of the four fundamental forces.

    Further, in the case of a neutron star the situation is even more telling, namely that the size of the star is given simply by the relation of the gravity pressing matter inwards, and the fermi pressure, casused directly by the PEP, pressing matter outwards. This makes it seem very much like a force to me.


    So, question: Why is the Pauli exclusion principle not the fifth fundamental force?
     
  2. jcsd
  3. Jun 9, 2010 #2

    Borek

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  4. Jun 9, 2010 #3
    Classical concepts are very often deficient for describing the quantum. As I stated in another thread:

    "It is the immaterial principle that makes a thing to be what it is, not little atomic billiard balls. Think of it in terms of the exclusion principle of Pauli and the stability of matter. To really understand this, we have to learn to disregard our imagination."

    Post 159 taken from:

    https://www.physicsforums.com/showthread.php?t=396540&page=10

    At the fundamental level, matter is almost entirely a set of relations, principles and rules.
     
  5. Jun 10, 2010 #4

    tom.stoer

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    Looking at the mathematical implementation of the exclusion principle using creation operators it just says that

    [tex]X = Y \to a^\dagger_X a^\dagger_Y |\ldots> = 0[/tex]

    That means that trying to "create" to identical fermions always gives zero. So there is no quantum mechanical state which can contain two identical fermions, because you are unable to create this state (even mathematically). Here X and Y label everything, all variables by which the two fermions could be distinguished: X = {momentum, spin, isospin, color, ...}

    If you compare this rule for the field operators with expressions describing forces you will see that it's totally different.
     
  6. Jun 10, 2010 #5
    Thanks for the links and comments, howerer, I'm afraid I'm not quite happy yet. As it seems to me that in the example with the neutron star, there is a force (gravity) getting cancelled by something that comes from the PEP. This should imply that the two things are similar enough to have the same units at the very least, no?

    Well, I haven't gotten to QFT yet (am an experimentalist :wink: ) but as I just mentioned above, if they are so different, how come you can cancel a force by something coming from the PEP?
     
  7. Jun 10, 2010 #6
    If something can't be in a space because that space is occupied why is it a force? It's just a constraint on the possible configuration. A force is mediated by particle exchanges, the only doubt is gravity which is hypothesised to be a force mediated by gravitons which have yet to be observed, it may be that gravity is just a "classical" property of spacetime geometry, but this model doesn't seem to work at planck scale.
     
  8. Jun 10, 2010 #7

    Borek

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    When you put marbles in the box, and the box gets filled, why can't you put more marbles in the box? Is there a force involved, or not?

    Honestly - I have no idea why the Pauli exclusion principle works the way it does, the only answer I know is "because that's the way it is". However, I understand where does the thread question comes from.
     
  9. Jun 10, 2010 #8

    DrDu

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    The equilibrium of forces only holds in static equilibrium. However, the particles in a neutron star move with high velocity and it is rather a quantum mechanical extension of Newtons law (i.e. the acceleration ma and the mechanical forces are equal) which is relevant here. Similar things happen even in one-electron systems like, e.g., H_2^+ (Dihydrogen cation) where the Pauli principle is irrelevant. The forces holding the two nuclei together are the result of an equilibrium of nuclear repulsion, proton-electron attraction and the change of the kinetic energy of the electron with the distance of the protons. The latter "forces" are often called Hellmann-Feynman forces in this context.
     
  10. Jun 10, 2010 #9
    Newton's equations define our perception of what a force is. Anything which doesn't look like it is in uniform motion has a force acting on it, our brains tell us. Well trained physicists will know that Newton's equations are only valid in an inertial frame in the classical, non-relativistic limit, where all particles are distinguishable though, and these days the word 'force' tends to mean particular interaction terms in a Lagrangian instead.

    But since our perceptions are still basically informed by Newtonianism, we should answer the question from that perspective: almost any violation of Newton's equations of motion can be 'explained' by introducing arbitrary forces acting on particles. A famous example is centrifugal force.

    In a rotating frame, Newton's equations are invalid, so we should just give up at trying to describe our experiences in that mathematical framework. We persist though, and after a bit of thought, we can conclude that we can still use Newton's equations to describe and understand our experiences (hurrah!), but only if we allow for a new force called the centrifugal force. The force is fictitious in the sense that it was invented purely for the purpose of being able to use Newton's equations. Luckily, the maths shows that motion in a rotating frame is exactly described by Newton's equations, as long as coriolis and centrifugal forces are included, so we don't feel so guiltly about applying Newton's equations outside of their realm of applicability.

    Analogously, Newton's equations are not valid for describing the behaviour of quantum indistinguishable particles, so we should just give up (and use a fully quantum description). After a bit of thought, you can show that the behaviour of quantum particles can be described in Newtonian language, if we allow for a fictitious force called Pauli Exclusion, which is repulsive when electrons get too close. In this case, there is no exact equivalence: there is no modification of Newton's laws with arbitray forces that is the same as QM, but inventing a few forces like Pauli Exclusion gets you closer to an intuition about QM based on modified Newtonianism.

    So if Pauli exclusion a force? No, it's just a constraint on allowed QM states, not a result of a potential energy term. Does it look, feel, and smell like a force? Yes, because it causes a violation of Newton I, and we are trained by experience to call every violation of Newton I a force.
     
  11. Jun 10, 2010 #10
    In that case I would say there is a force, the electromagnetic force.
     
  12. Jun 10, 2010 #11
    Thanks for the long explanation peteratcam, it was helpful, and was infact what I was somewhat guessing. But this basically just means that the classical concept of "force" is somewhat outdated, and that it would have been better to explain violations of newton's equations of motion using a more general class of fundamental "interactions" rather than only the four forces typically done today (like on the mentioned wiki articles). And that these could include not only things strictly defined as coming from the potential energy term, but everything that plays a force-like role.
     
  13. Jun 10, 2010 #12

    Cleonis

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    I agree.

    It is common in the history of physics that when there are new developments the same words are still used but they are given new meaning. In relativistic physics the words 'space' and 'time' are still used, but there is a profound shift in meaning as compared to what the words 'space' and 'time' mean in classical context.

    It would seem that the four fundamental forces are to be seen as members of a larger group. This larger group can be characterized as: factors that play a part in the outcome of physical processes.

    It seems to me our current physics theories involve six such factors:
    - Inertia,
    - Gravitation,
    - Electromagnetism,
    - Pauli exclusion principle
    - Strong nuclear force
    - Weak nuclear force.

    I think that currently the most notable unification is the scheme that unifies Electromagnetism, Strong nuclear force and Weak nuclear force.

    Perhaps a successor to quantum physics will offer a unification of the Pauli exclusion principle with other fundamental factors.


    By the way, I suppose there is another way of casting the difference:
    In the case of the four fundamental forces it is physically possible to have a volume of space that is devoid of any presence of either of the four. (Or more precisely: the physics laws allow that you move arbitrarily far away from any force source.)

    But there is no such thing as 'moving away from inertia', or 'moving away from the exclusion principle'. Inertia and the exclusion principle are always and everywhere.
     
    Last edited: Jun 10, 2010
  14. Jun 10, 2010 #13

    tom.stoer

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    It is misleading to talk about forces as we know them from Newton; we should talk about interactions as they are defined in quantum field theory. I come back to the example I started with.

    Let's assume we have a unique vacuum state from which we construct the space of all possible states, the so-called Fock space. The Fock space is a direct sum of all 0-, 1-, 2-, ... -particle Hilbert spaces. As I said above there must not be any state in tis space which contains two identical fermions. So in principle one can construct all possible states by considering all products of creation operators acting on the vacuum

    [tex]\prod_n a^\dagger_{X_n} |0>[/tex]

    The Pauli principle then tells us that

    [tex]X_m \neq X_n \;\forall\; m \neq n[/tex]

    otherwise the action of the product on the state produces 0 (not the vacuum state, just 0).

    Up to now we have not introduced any interaction, so what I was saying is valid for a free theory in which no interaction takes place. If this is the case then the state I have just constructed remains invariant under time-evolution.

    Example: if I create a state with one electron and one positron it stays exactly this: an electron-positron pair. They do not interact, they do not annihilate, they do not scatter, ... Nevertheless it is forbidden to have to identical electrons. A state with two identical electrons simply does not exist by construction (of course there's a deeper reason for the Pauli principle which we can discuss later).

    Now we switch on the interaction. In quantum field theory that means that there is an operator which is constructed from at least three operators.As an example we could have something like

    [tex]b^\dagger_{X_i} a_{X_m} a_{X_n}[/tex]

    Such an operator can e.g. destroy two fermions and create one photon. When acting on a state as defined above we either get zero (if there are no two electrons with matching properties which can be destroyed) or we get a new state with one photon. So an interaction transforms incoming states (e.g. containing two electrons) into outgoing states (e.g. with one photon).

    Now we come back to the Pauli exclusion principle: It does not transform particles into other particles. It simply tells you on which states interactions can act (and which states are forbidden). It constrains the allowed states and the allowed interactions. An interaction forbidden by the Pauli principle is not an interaction which does nothing (like the identity operation), it is an interaction that is described by the 0-operator. It does not exist. It is not their.

    Consider again a product like

    [tex]a^\dagger_{X_n} a^\dagger_{X_n}[/tex]

    It annihilates all states in the theory. So not only is

    [tex]a^\dagger_{X_n} a^\dagger_{X_n} |0> = 0[/tex]

    but even

    [tex]a^\dagger_{X_n} a^\dagger_{X_n} = 0[/tex]

    The two-particle operator itself does not exist. It is zero.

    So this means that the probability that something in your theory turns into two identical fermions vanishes. And this principle is valid for all theories containing fermions. It does not depend on the specific interaction.

    So the Pauli principle is more fundamental than "forces" or "interactions"!

    Let's make a simple example: consider a mathematical theory containing only real numbers. In this theory you must not calculate the square root of -1. This number simply does not exist. You can not act on it, you cannot multiply it, devide by it, nothing. So there is no mathematical operation acting on the square root of -1. This is not because the operation is not defined, it is because already the square root of -1 is not defined. There is nothing on which an operaton could act.

    If you ask which operation (= "force") prevents imaginary numbers from being created, this is misleading. There is no force doing something with the imaginary numbers turning them into real numbers. There is no operation +, -, *, /, ... which is responsible for avoiding imaginary numbers. There is an underlying principle which from the very beginning exludes the existence of imaginary numbers.
     
    Last edited: Jun 10, 2010
  15. Jun 10, 2010 #14

    Cleonis

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    This raises the question whether the metaphors 'electron degeneracy pressure' and 'neutron degeneracy pressure' should be in use at all. The word 'pressure' is quite evocative, but is it an appropriate word here?

    What happens when a extinguished star collapses moves to the state of Neutron star? Let's say that for a while electron degeneracy pressure was just sufficient to prevent further collapse, but subsequently a large amount of matter accretes and the extuinguished star goes to further collapse.

    In collapsing to a neutron star the matter itself transforms to another form. Normally that transformation is prohibitively unlikely. I suppose that in the case of collapse to neutron star the gravitational potential energy that is released by further contraction is larger than the energy cost of the matter transformation that is required.

    Seen in that way the electron degeneracy pressure arises from the energy cost of the matter transformation that is required for further contraction.
     
  16. Jun 10, 2010 #15

    tom.stoer

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    I think this is not a metaphore but due to a calculation for an thermodynamic quantity "pressure".
     
  17. Jun 11, 2010 #16

    DrDu

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    I want to highlight the problem from another side. Consider a gas of noninteracting fermions. The Pauli principle then forces the particles to occupy excited one-particle states up to the Fermi energy. However, the way these particles excert forces on their surrounding is not fundamentally different from how classical particles do. Consider e.g. electrons confined to a trap consisting of two charges. The electrons will get reflected when approaching the charges but the way they excert force is of purely electromagnetic type.
     
  18. Jun 11, 2010 #17
    This calculation may be the source of the confusion at least for the fermi gas example. Your description a few posts up is a good one concerning the micro-scale of a few fermions, and as you said, the PEP simply expresses that the probability is zero for two identical fermions. However, since I've read several times that the gravitational force in a neutron star really is balanced in some since by something coming for the PEP, I assume this to really be so, but what I'm really missing is how exactly.

    What I mean is, it seems to me that you start with a probability (dimensionless), and then an integration is performed overall fermions in the neutron star, afterwhich you arrive at something that has the unit of Newton (because how can it otherwise cancel the gravity force?). And this intermediate integration is very unclear to me, exactly at what point does the PEP become a force?

    On one place I saw an attempt at this explanation, and it basically stated that since the fermions in a neutron star are so tightly packed, i.e. their position is so well known, that their momentum is very unknown (from the HUP) and that this would lead to large momentum changes that on average would balance the gravity pressure. I don't know how true this explanation is though, but at least it gives a sense of units matching.
     
  19. Jun 11, 2010 #18

    tom.stoer

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    I am no expert in calculations of neutron star properties, but I doubt that you will get something like force [Newton]. What you get is that the star does not collapse and you interpret this as a force.

    Can you show me a reference where they do the math?
     
  20. Jun 11, 2010 #19
    Zarqon,

    In spite of the responses you have been given thus far, I am here to tell you that your intuition about the PEP being a force, is in fact spot on! The de Broglie-Bohm formulation of QM explains electron degeneracy pressure in terms of a 'quantum force' (the gradient of the quantum potential) which, for example, in a collapsing neutron star, counter-balances the gravitational implosion. There was a rather extensive discussion about the quantum force explanation of the PEP in a previous thread. See for example (and make sure to read all the posts by zenith8):

    https://www.physicsforums.com/showthread.php?t=364464

    See also page 34 of this talk by Cambridge physicist Mike Towler:

    http://www.tcm.phy.cam.ac.uk/~mdt26/PWT/towler_pilot_waves.pdf

    Feel free to let me know if you have further questions.
     
  21. Jun 11, 2010 #20
    See the second link in my post above.
     
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