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7.2: Exclusion Principle

  • Page ID
    17222
  • Antisymmetry and Exclusion

    An odd thing happens when a two-particle state exhibits negative exchange symmetry and the two particles have the same quantum number(s). Setting \(n_1\) equal to \(n_2\) in Equation 7.1.12 gives a two-particle wave function that is exactly zero! Importantly, it is not just zero for some specific values of \(x\) (like at nodes), but everywhere. From this we conclude that particles with this negative exchange symmetry (fermions) cannot exist together in a common bound state with the same quantum numbers. These states are "excluded" from what is possible, and this is known as the exclusion principle. This is more than a mere curiosity – it has extremely far-reaching consequences. For example, it is commonly pointed out in particle physics circles that all matter is comprised of fermions, and this is necessary, as the exclusion principle prevents the catastrophic collapse of atoms. We will discuss this in the next section. Two bosons (particles that satisfy positive exchange symmetry) do not bring about a zero wave function when they have the same quantum number, so they don't satisfy an exclusion principle.

    Spin

    We have not yet discussed spin. This is largely because it was covered rather extensively in Physics 9HC. The time has come to reintroduce this iconic quantum mechanical phenomenon, and put into the context of the current discussion.

    Spin is also known as "intrinsic angular momentum." As we saw for the hydrogen atom, hamiltonians with potentials that have radial symmetry give rise to two quantum numbers (\(l\) and \(m_l\)), which give a measure of the magnitude of the angular momentum and \(z\)-component of the angular momentum:

    \[ L^2=l\left(l+1\right)\hbar\;,\;\;\;\;\;\;\;\;L_z=m_l\hbar\;,\;\;\;\;\; m_l=0,\;\pm 1,\; \dots,\; \pm l \]

    This angular momentum comes from the electron's motion relative to the origin (which is why it is called orbital angular momentum). We even wrote the operator for the \(z\)-component of angular momentum in terms of spherical coordinates (Equation 4.4.6).

    Spin is also angular momentum, but it doesn't count on any physical motion relative to a reference point. It is just an intrinsic property of the particle. It still behaves like angular momentum – a charged particle with spin has a magnetic moment, it counts in the accounting of angular momentum conservation of a system, and only its magnitude and \(z\)-component can be measured at one time (the uncertainty in the \(x\) and \(y\) components still exists).

    The important thing about acknowledging that spin is intrinsic for our current discussion is that the intrinsic angular momentum is additional information that needs to be logged by the quantum state. Remember the quantum state knows everything about the particle. So while an electron in a hydrogen atom has orbital angular momentum, it also has spin. This is not accounted-for in the (\(n,l,m_l\)) quantum numbers, so in fact we need to add an additional quantum number into the mix, bringing the full lineup to (\(n,l,m_l,m_s\)), the final quantum number indicating the \(z\)-component of the electron's spin. One might ask why there is not another quantum number for the magnitude of spin. Well, in fact there is, but that information is contained in the fact that we are talking about an electron. Fundamental particles like the electron have only one possible state of total spin, which cannot be altered by circumstances. It should be noted that when we do use the total spin quantum number (shockingly called "\(s\)"), the same rules apply for spin as for orbital angular momentum, with the possible exception of the allowed values of these numbers, as noted below.

    \[ S^2=s\left(s+1\right)\hbar\;,\;\;\;\;\;\;\;\;S_z=m_s\hbar \]

    Besides the fact that it is intrinsic, there is one other aspect to spin that distinguishes it from orbital angular momentum. Unlike orbital angular momentum, whose \(z\)-component must be some integer multiple of \(\hbar\) for all particles, for some types of particles the \(z\)-component of spin can come in integer multiples of \(\frac{\hbar}{2}\).

    It turns out that particles with "integer spin" (\(m_s = 0,\; \pm 1,\;\dots\)) and particles with "half-integer spin" (\(m_s = \pm\frac{1}{2},\; \pm\frac{3}{2},\;\dots\)) are fundamentally very different. It turns out that particles with integer spin all exhibit positive exchange symmetry (i.e. they are bosons), while particles with half-integer spin exhibit negative exchange symmetry (i.e. they are fermions). We know that exchange symmetry divides particles into two types, and we know that the nature of spin divides particles into two types, but it is not at all obvious is that this partitioning of particle types turns out to be equivalent.

    As was stated earlier, particles that comprise what we call matter (electrons, protons, and neutrons) are fermions, and they happen to all be spin-\(\frac{1}{2}\) particles. [Technically, protons and neutrons are composite particles, constructed from quarks, but they too are spin-\(\frac{1}{2}\).] Our discussions will primarily focus on electrons (as was the case for the hydrogen atom, even though a proton was also involved), though we have also made passing remarks about photons (which are spin-1 bosons), in particular the role they play in selection rules for energy level transitions for emission and absorption.

    Populating Energy States

    When we moved to quantum mechanics in three dimensions, we found that the additional quantum numbers led to degeneracies in energy levels. We concluded that additional quantum numbers come along anytime we increase the number of degrees of freedom, and that as long as the hamiltonian doesn't depend upon one or more of the quantum numbers (or even if it does, in the case of accidental degeneracy), there can be multiple quantum states associated with the same energy level.

    Intrinsic spin introduces a new degree of freedom, and with it, more degeneracy. There are only two spin states for the electron, but that doubles the degeneracy of all the states of the hydrogen atom, whose wave function we now know should be written as "\(\psi_{nlm_lm_s}\)."

    Degeneracy addresses the number of unique states for the same energy level, but where the addition of this additional quantum number really gets interesting is when we put several particles together into a common potential. Again, we continue to assume that the particles do not interact with each other.

    Let's take as an example a one-dimensional infinite square well (we aren't very creative, this is the first example we use for everything). Let's place three identical particles into this "box," such that they reach an energy eigenstate. [Keep in mind we are talking about a multi-particle quantum state here.] The question we want to answeris, what is the ground state energy of this configuration?

    It turns out that we can't answer this without knowing what types of particles we are putting into the box. We know from our separation of variables work (and frankly, just from energy conservation), that the total energy of the multi-particle state will be the sum of the energies of the individual particles, so we might think that the lowest total energy occurs when each particle is in its individual ground state. But not so fast!

    What if the particles are electrons? These are fermions, which means that there is zero probability of having two of them with the same quantum numbers. There are two quantum numbers present here – one for the single dimension, and one for the intrinsic spin degree of freedom. So we can "fit" two electrons into their lowest individual energy states if they have opposite spins, but we can't get the third electron into the ground state without violating the exclusion principle, so it must reside in its first excited state.

    If the particles are bosons, then no problem, all three particles can reside in the ground state at once, and that will be the lowest total energy state for the system. Have we solved this problem for all particles? No! Recall that not all fermions are spin-\(\frac{1}{2}\); higher spin quantum numbers are also possible. If we use spin-\(\frac{3}{2}\) fermions, then there are four different spin states available: \(m_s = \pm \frac{1}{2},\; \pm\frac{3}{2}\). With four different states available for a single energy level, there is plenty of room to fit all of these fermions in their individual ground states.