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How does spin justify Pauli's principle?

  1. Jun 11, 2007 #1
    Hi to all

    I don't know if someone has asked this before, but here is my question:

    I have read a little bit of QM (mostly non mathematical). But I don't understand three things. First, if you have a proton and an electron quite far away from each other, what are the necessary conditions in which a hydrogen atom will be formed? If electron and proton are travelling slowly towards each other (on a crash course), will they collide or is there still a probability that they will form an H atom?
    Secondly, how does spin quantum number decide that the two electrons in an orbital will never collide? That is how does spin apply Pauli's exclusion principle.
    Thirdly, why can't an electron in the K shell, radiate energy and fall into the nucleus?
    Please answer in non mathematical terms if you can.

    Love to you all
    Mr V
  2. jcsd
  3. Jun 11, 2007 #2
    if the proton and electron are moving at sufficiently high speeds they will not form Hydrogen. This is analogous to a rocket leaving earth. It must have an escape velocity, otherwise it will remain bound to earth's orbit. An electron with a velocity greater than the escape velocity will likewise escape the proton unbound.

    particle spin determines how particles interact. Spin-half particles (or multiples thereof) cannot occupy the same space (electrons, protons, ... all known forms of matter, are in this catagory). Integer-spin particles on the contrary can occupy the same space (photons, gluons, etc.). If you are interested, look up *Fermi-gas* and contrast this with *Bose-Enstein Condensates*.

    I am too lazy to look up my notes to answer the third question so I will ad-lib. I believe that the electron cannot spiral into the nucleas because of Heisenberg's uncertainty principle. It instead makes a cloud centered around the nucleus and has a finite, but small, probability of being found within the nucleus itself. Electrons are not orbiting balls spinning around the nucleus as Bohr thought. The cloud comparison is the best answer modern physics can currently provide.
  4. Jun 11, 2007 #3
    In 1924, Pauli realized some experimental results like doublets of alkali metals, anomalous Zeeman effect, periodic table etc. demand a new quantum number for the electrons with two possible values. At this time it wasn't clear what that new number represented but it was clear it was not connected to the spatial degrees of freedom of the electrons. So each electron in a multi-electron atom can be described by four quantum numbers (n, l, ml, ms). The Pauli exclusion principle states that no two electrons can have the same four numbers. One year later it was realized that the new quantum number represents the internal spin of the electrons.

    For a system of indistinguishable particles, the wave function either doesn't change when you exchange the quantum numbers of two of them or it is multiplied by -1. In the first case, we call the particles 'bosons', in the second - 'fermions'. The spin-statistics theorem of QED claims that particles with half-integer spin like electrons are fermions.

    Now take the wave function Psi of a multielectron atom. It will depend on the quantum numbers of the electrons, a quartet of numbers for each electron. Denote the quartet of i'th electron by qi:

    Psi (q1, q2, ....)

    Because electrons are fermions we must have

    Psi (q2, q1, ....) = - Psi (q1, q2, ....)

    But if the two electrons have the same quantum numbers q1=q2=q we obtain

    Psi (q, q, .....) = - Psi (q, q, ....)

    From which we conclude that

    Psi (q, q, ....) = 0

    i. e. that state of the system has zero probability of being realized.

    To summarize the connection between spin and Pauli exclusion goes like that:

    electrons have spin = 1/2 ->

    spin-statistics theorem says they are 'fermions' i. e. the wave function of several electrons must changes sign when you exchange the q. numbers of two electrons - >

    wave function must be zero if two electrons have the same q. numbers i. e. the probability to find two electrons with same q. numbers is zero (Pauli Exclusion Principle)

    About your last question, electrons in atoms occupy only states allowed by quantum mechanics. You should jimagine a quantum state of an electron like a 'probability density cloud'. Many of those clouds are non-zero at the center of the atom where the nucleus is so electron can be found very close to the nucleus, it is not restricted 'on an orbit' as the wrong classical picture suggests. The lowest energy state of the electron is called ground state and the electron can't get lower energy by 'falling into the nucleus'.
    Last edited: Jun 11, 2007
  5. Jun 11, 2007 #4
    Thanks for your replies

    So electrons' spin proves them to be fermions. Clears my confusion. Also, an electron's magnetic moment due to spin is not the actual reason that two of them in an orbital never collide due to magnetic repulsion. Am I right?

    Could you also tell me, in which form electron stores its energy when it jumps to a higher orbital, and whether its velocity increases or decreases after gaining this energy. Is all the gained energy converted to kinetic energy?

    Also I could do with a little more theoretical detail as to why it is impossible for the electron of a hydrogen atom to lose its kinetic energy and fall into the nucleus.

    Warm regards
    Mr V
  6. Jun 11, 2007 #5
    Saying that electrons in atoms may 'collide' or 'fall on nucleus' indicates you use classical picture of hard spinning balls following trajectories around the nucleus. Quantum mechanics was created in 1925 out of frustration from the inability of that classical picture to explain the experimental results. Your questions can be answered within QM but to understand the answer you will have to learn QM first.
  7. Jun 12, 2007 #6
    The electron in the lowest shell generally won't fall into the nucleus because an electron bound in an atom is in a negative potential well. A free electron has more energy than one bound in an atom, which requires a specific amount of energy to ionize. If the electron were to continue to radiate and fall in, it would lose even more energy, and the collective energy of the proton and the electron would be much less than that of an neutron, so the two wouldn't even have enough energy to form a neutron.

    So even though bound electrons have overall positive energy, they always have less than free electrons, and the difference is greater the closer they are to the nucleus. When we develop the hypothesis of quantitization, we find that the distribution of energy levels isn't continuous, but discrete, and the energy levels are all easily calculated from the negative potential of the ground state. If hydrogen has an ionization potential of A, then its electron has A less energy in the ground state, then it would have if it suddenly become insensitive to the proton's electric field, and it can only orbit the nucleus where its energy due to the charge potential is E=-A/n^2, n being an integer.

    This might still seem pretty abstract. I found that an easy way to think of it at first, is that if atoms were unstable this way, then free neutrons would be stable. But even a free electron taken with a free proton generally have less energy than a neutron (we're short a neutrino).
    Last edited: Jun 12, 2007
  8. Jun 12, 2007 #7
    So, do quantum rules prohibit electrons from jumping into/hitting the nucleus because 'it doesn't even have energy to form a neutron'. I mean is it established that a low energy electron can only form a neutron with a proton, and if it doesn't have even this much energy then, as it cannot form anyhting else, nature prevents it from falling into the nucleus at all costs?

    alphanzo (see above) says that there is a small probability of finding an electron inside the nucleus. How can there be, if (as you said) the electron can't lose further energy after a certain limit? Also, if this can really happen, then from where does this electron gain energy to come out of the nucleus, and how does it interact with its surroundings while it is there inside the nucleus?

    Mr V
    Last edited: Jun 12, 2007
  9. Jun 12, 2007 #8
    When an electron is in a stable orbital around a nucleus, it's momentum is exactly great enough to counteract the charge force. Compared to what's required to hold the electron inside the nucleus, the charge force is actually tiny.

    Let me digress: a free electron in a vacuum doesn't act like a particle. With the passing of time, it acts like a wave spreading out in every direction with the momentum of the electron. Holding it in place, so it acts like a pointlike particle, can be accomplished by applying a force over the front of the spherical electron wave. So holding an electron in a certain definite position, is basically tantamount to applying pressure to it.

    The charge force between electron and proton is (generally) insufficient to squeeze the electron into the nucleus. An electron wave can spread into the nucleus, therefore there is a chance of an electron being there, but you'll almost never find a particle-like electron in the nucleus.

    This is all implicitly assuming that the electron can orbit at any distance from the proton. But because we know from quantum mechanics that energy transfers are quantized, we also know that the electron has to make discrete, discontinuous jumps between orbits, that it can't have a stable orbit at places inbetween them. If we measure the ionization energy "A" of an atom like hydrogen, which is the amount of energy its electrons need to absorb to break free from orbit around the nucleus, we find that an electron particle can only orbit at positions where the charge energy E=-A/n^2, n being the number of the excitation state (starting with 1).

    The electron wave function, however, spreads continuously over space, and there is a small chance of being able to find an electron particle at any point in the electron wave (but the chances of finding an electron particle in a wave that's spread from it is always exactly 100%, never > or <). So there may be a chance of finding an electron in the nucleus, for a short period of time--but if you look for an electron with the same energy as that one in other atoms of the same substance, you'll generally find them orbiting in the area of the 1st excitation state.

    By the way, in heavy atoms, it's possible that the charge force is great enough to capture an electron and create a neutron with one of the protons in the nucleus. But this doesn't happen with atoms like hydrogen; if it did, we could have more free neutrons floating around than hydrogen atoms. A free neutron in a vacuum will break down into an electron, proton, and an anti-neutrino in a matter of minutes. The hydrogen atom produced takes less energy to stabilize than the neutron.

    [EDIT]: I changed a few words in order to be as specific and precise as possible. This is something I thought a lot about when I first started studying QT (not too long ago), and I wish there were clearer explanations of it on the net. David Bohm's Quantum Theory is a good place to start. He specifically addresses atomic stability early on in the book, but not exhaustively.
    Last edited: Jun 12, 2007
  10. Jun 13, 2007 #9
    Reading this over, I think I wasn't clear enough about at least one point.

    Because the force required to squeeze the electron's wave function in the nucleus is great, you generally won't observe an electron in the nucleus unless it has very high energy. This could be accomplished by colliding a proton and electron with great momentum. We bounce some high energy photons off the two at the approx. point of impact and see them very close together.

    Let's say they don't form a neutron. From here, the electron, which has very high energy, can emit a photon and enter any of hydrogen's electron excitation states, or even still have enough energy to escape the proton entirely. Once again, the charge force is tiny compared to the force necessary to squeeze the electron down into a tight area like this. (In fact, even if they form a neutron, it will soon break down into a hydrogen atom, which is very similar to what happens above.)

    Now we have a stable hydrogen atom, and we bombard it with photons. The electron will pass all the allowed energy levels, getting further away from the proton as it goes, and eventually ionize, break free. At the point of ionization, the electron will still have waaay less energy then it did when it first collided with the proton. So before the electron gains enough energy to possibly remain in the nucleus, it will have travelled out and broken free from the proton.

    The electron doesn't continue to lose energy and fall in, because their comes a point where the force required to compress the wavefunction into a smaller, "denser" orbit is greater than the charge force between electron and proton. It would actually take more energy for the electron to remain inside the nucleus, and the electron would sooner escape orbit than fall in. It turns out that the ground state orbit really is the point of lowest potential energy.
    Last edited: Jun 13, 2007
  11. Jun 14, 2007 #10
    Thanks for your replies :smile:

    As far as I know (and I don't know much :redface:), an electron always behaves as a particle. The wave you are talking about may be the probability wave of finding an electron in vacuum. The probability of finding the electron anywhere in vacuum is 100 %. As you said, holding it in place means defining its position more accurately, which results in increase in its momentum (according to heisenberg's formula). As its momentum increases, its energy also increases, which exerts a kind of opposing force.

    So is the real reason this: that if we are forcing an electron to follow a definite path to a proton, we are applying a force on it, which increases its energy. The attraction of a proton should be greater than this energy for the electron to be present into the nucleus. But it isn't. So an electron in a nucleus has huge energy and the proton does not possess enough attractive energy to keep the electron in that state. The electron inside the nucleus is in a highly unstable state, as now its location can be predicted with high accuracy, which increases its momentum/energy. Therefore it tries to break free of the nucleus. The ground state of the electron of a hydrogen atom is actually that state in which the energy of the electron and the attraction of the hydrogen nucleus balance each other. Am I right?

    Can I also say that even if we don't apply any force on the free electron, and only the attraction of proton is making the electron go towards it, then there comes a point where loss in energy of electron due to attraction becomes equal to the gain in energy due to Heisenberg's principle, and the electron stops? Is this the ground state of electron? At this point if we supply energy to electron to move closer to proton, then though it may reach the nucleus, but it is very unstable. As soon as the applied force is removed, the electron jumps out of the nucleus.

    When you said,
    " The charge force between electron and proton is (generally) insufficient to squeeze the electron into the nucleus."
    I was wondering as to why you were saying that an electron has to be "squeezed" into the proton. But I think now I understand why it is so.

    But after the first shell, why are the other shells also quantized? I mean why can't the electron in its ground state gain a teensy amount of energy and reach the next consecutive shell (not the quantized one) ?
    What type of imbalance prevents electrons from staying in space between quantized shells?
  12. Jun 14, 2007 #11
    This is a tough question to answer here. In classical theory (pre-quantum theory), we assumed that energy transfers occured continuously; we could transfer any amount of energy in an interaction, and an exact amount. If we knew from experiment that we could transfer a specific amount of energy by colliding two particles, we assumed that we could tranfer any fraction of that energy, which is how energy transfers appear to work in macroscopic physics. But major theoretical problems arose with any theory that allowed continuous transfer of energy--like the predicted "ultraviolet catastrophe" that was never observed in experiment. And even when continuous models demonstrated some success in experimentation, other phenomena remained unexplainable, like the photoelectric effect, and what was observed in the dual-slit experiment. I'd love to explain these to you, but I could write a book about them before I got to my actual point. These are the problems and experiments that lead to quantum theory in the first place, and any good text book will explain them. I suggest David Bohm's Quantum Theory, this is all in the first 2 chapters. But to put it plainly, particle physics works only if we assume that a particle has to transfer energy in discrete, discontinuous "jumps".

    Electron wave functions can and do take a value between orbitals, but there are only a few specific possible energy values the electron can have while it orbits the nucleus

    Unfortunately, this isn't the case, and this is where the first part of your post gets off track. Actually, it seems like you have a very good grasp of an atomic model that assumes that the electron is always a particle like a billiard ball, and it's good that you can feel out this model. But we know from experiment that electrons act as waves; they can exhibit an interference pattern (this goes back to the dual slit experiment).

    At first, it seems natural enough to think of an electron wave as nothing but the probability function of finding the electron particle at a certain time. It even spreads out with the momentum of the particle. But if we try to develop this hypothesis, that the wave function is just the spreading probability of finding the ball or point-like particle, it doesn't work out. Again, I'll need to refer you to read up on the dual-slit experiment (especially the variant in which it's performed with a single electron at a time, many times).

    Eh, pretty close. The electron breaks free from the nucleus, because we can only find it in the nucleus when it has a ton more energy than the energy of the charge force--so it's already travelling away from the nucleus pretty fast (or will be shortly). The charge force alone won't bind an electron to the nucleus, we have to apply an external pressure.

    An electron wave naturally "wants" to spread out unless a force is applied to the wave front. If the electron were to lose all momentum, it's wave function would be compressed to an exact location in space, indefinitely. But an electron with no momentum wouldn't have enough energy to hold itself as a point-like particle in a definite location for long, so we find that, paradoxically, having an electron with less momentum requires more energy than an electron with a minimum of momentum. The problem is in thinking of the electron as the billiard ball, which would always have positive potential energy proportional to its distance from the nucleus.

    Also, the charge force and the "centrifugal" force due to the orbit of the electron are balanced at every excitation state, not just the ground state. I guess we can say that the ground state is special, because if the electron were to fall any further into the nucleus from here, there wouldn't be enough charge force to hold the electron in so small a space for long. The more you compress the electron wave, the "denser" it gets, so it becomes increasingly resistant to squeezing.
    Last edited: Jun 14, 2007
  13. Jun 15, 2007 #12
    Thanks a lot Iamu for your explanation.
    It was quite difficult for me to intuitively understand wave-particle duality when it was taught to us the previous year (in the eleventh standard). I found it easier to take electron's wave nature to be its probability wave function.
    Though it is proved experimentally, is there any explanation of how a particle having mass can behave as a wave? I mean how is electron's mass distributed in this wave?
    Then again, a light wave, which we believed to be a wave, is said to consist of quanta. Now if a wave consists of massless energy packets, how do you define an electron wave? Is an electron wave also quantized into energy packets having mass, and when pressure is applied, do these quanta fuse together or become denser to give the appearance of an electron?

    Finally, is this paradoxical nature of electron due to inability of physics to clearly define particle, mass and wave, or what?

    Mr V
    Last edited: Jun 15, 2007
  14. Jun 15, 2007 #13
    Why wouldn't a particle with mass be able to behave like a wave?

    Seriously, we need to analyze our simplest assumptions that may or may not be true. I ask you, how do you know the electron is a ball-like particle when you aren't even looking at it?

    The smallest unit in this case is the electron (or photon) itself, so there are no quanta smaller than this comprising the wave function. There is no apparent distribution of mass. If we have a far-spread electron wave, and we want to find out where the mass in it is, we can bombard it with photons to determine the position of the electron particle--which collapses the wave function to around one apparently random point, where we observe something like a point-like particle, which then begins to spread again.

    But the wave function is not just a guessing game based on the information we have. There may be a more exact explanation underlying QT, but the electron's wave nature is as fundamental as its particle nature. Any particle colliding with the electron we're talking about above constitutes an observation of the electron and collapses it's wave function, even if we don't know the specifics of the interaction. If we know that an electron is constantly interacting with particles as it travels, but we don't know when, where, and how these interactions will occur, we still know that they'll prevent the wave packet from spreading (each collision counts as an "observation").

    It's not us looking inside the box that collapses the wave function. It's the electron bumping up against something else that does it, and then we may or may not observe the collision--but it happens whether we observe it or not. The "observer" becomes very important in quantum theory, but it's not a matter of whether or not there's a guy watching with an electron microscope and making calculations when the electron hits--any particle(/wave) that ineracts with an electron constitutes an "observer".

    It doesn't seem to be a problem of definitions. It seems that not even the electron itself "knows" exactly where it is and where it's going until a passing photon "tells" it.
  15. Jun 15, 2007 #14
    How can we visualize a wave spreading. I mean a wave travells in a straight line (like light waves), so how does it "spread"?

    How does this wavefunction collapse actually happen?

    If a wave spreads uniformly in the whole of the universe, then how is the position of the electron determined by the photon. I mean a photon should interact with this wave as soon as it is emitted ( since the wave is everywhere), and we should always see the electron somewhere quite close to where the photon was emitted. Then why is it that sometimes we see the electron quite close and sometimes quite far off? Or in other words, why wavefunction collapse happens sometimes nearby and sometimes at a distance.

    Mr V
  16. Jun 17, 2007 #15
    Thanks for your message. As I am still at school, it would be a bit difficult for me to seriously study QM maths. Besides, I have chosen computer engineering as my career. But I am sure they teach QM in all engineering colleges.

    Mr V
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