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In an atom, two electrons can have the same set of $n,\ell,m$ quantum numbers as long as they have opposite spins. My introductory physics and chemistry courses have all introduced this as two electrons occupying the same energy state; however, the spins of the electrons cause splitting of the energy levels they occupy. This leads me to think two electrons with opposite spins are only able to share the same set of $n,\ell,m$ quantum numbers because they actually occupy slightly different energy levels. My question is: if there were no splitting of energy levels due to spin, would two electrons still be able to occupy the now-exactly the same energy level or do they need to have different energies, no matter how small the difference?

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up vote 3 down vote accepted

If there are two states with the same energy, those two states can still accommodate two electrons. The Pauli exclusion principle says that two fermions can't be in the same state. However, if the states differ by a quantum number they are still distinguishable.

Taking your example, two electrons may have the same $n, l, m $ in an atom but one with spin up and another with spin down. However, even if there was no energy splitting they are not the same state since they have different angular momentum; the states in principle are distinct.

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This echoes JeffDror's (correct) answer but also includes some math/notation that might help illuminate things further.

Let's consider essentially your example. Let $H$ denote the hamiltonian for a single electron in a central potential (like when it moves around a nucleus. Then the state space (Hilbert space) of the system is spanned by energy eigenstates \begin{align} |n,\ell,m,m_s\rangle \end{align} where $m_s$ is the $z$-component of spin quantum number. Note that the energy of each of these states is determined entirely by the principal quantum number $n$, so there is a lot of degeneracy in the spectrum. Mathematically, when $H$ acts on any of these states, it yields a value $E_n$ that only depends on $n$; \begin{align} H|n,\ell,m,m_s\rangle = E_n|n,\ell,m,m_s\rangle. \end{align} In particular, a spin up electron, with $m_s = +1$, has the same energy as a spin down electron with $m_s=-1$; \begin{align} H|n,\ell,m,+1\rangle &= E_n|n,\ell,m,+1\rangle \\ H|n,\ell,m,-1\rangle &= E_n|n,\ell,m,-1\rangle \\ \end{align} There is a two-fold spin degeneracy in energy levels; different spins do not cause energy splitting. This situation would be different if, for example, there were an external magnetic field that interacted with the magnetic moment of the electron.

Now consider a system of two, non-interacting electrons in the central potential. The state space of this system is spanned by (tensor) product states that specify the individual state of each electron; \begin{align} |n_1,\ell_1,m_1,m_{s,1}\rangle|n_2,\ell_2,m_2,m_{s,2}\rangle \end{align} For convenience, let's suppress the orbital angular momentum quantum numbers in the notation for now, so that such a state would be written as \begin{align} |n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle. \end{align} Let $H_2$ denote the hamiltonian of this system, then the energy of such a state is given by the sum of the energies of the individual states; \begin{align} H_2|n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle = (E_{n_1}+E_{n_2})|n_1, m_{s,1}\rangle|n_2, m_{s,2}\rangle \end{align} However, states of this form are not allowed in this system because electrons are fermions, and the Pauli Exlusion Principle tells us that the for such a system, the only states that are allowed are those that are antisymmetric in the quantum numbers of the electrons. This means that if we write some two-electron state, and we switch the factors in each product state that appears, then the overall effect on the state should be that is gets multiplied by $-1$. As an example, consider the following state: \begin{align} |2,+1\rangle|2,-1\rangle. \end{align} Is this state allowed? Now, because when we flip the factors, we don't get negative of the same state back; \begin{align} |2,+1\rangle|2,-1\rangle \neq |2,-1\rangle|2,+1\rangle. \end{align} But we can fix this. Consider, instead, the following state: \begin{align} |\psi\rangle = \frac{1}{\sqrt 2}\Big(|2,+1\rangle|2,-1\rangle - |2,-1\rangle|2,+1\rangle\Big) \end{align} Notice what happens when we flip all the terms in this state; \begin{align} \mathrm{Flip}(|\psi\rangle) &= \frac{1}{\sqrt 2}\Big(|2,-1\rangle|2,+1\rangle - |2,+1\rangle|2,-1\rangle\Big)\\ &= -\frac{1}{\sqrt 2}\Big(|2,+1\rangle|2,-1\rangle - |2,-1\rangle|2,+1\rangle\Big) \\ &= -|\psi\rangle \end{align} This state works! When we exchange the quantum numbers of the two particles, the state changes by a negative sign. This state actually has a special name; it's called the ``spin singlet state." Notice that the two factors in each product state always have on electron having spin up, and the other electron having spin down. This antisymmetry property cannot be satisfied by a state in which the electrons have the same spin because flipping would just given the same state back again without the minus sign!

So the upshot is this:

Regardless of the form of the hamiltonian, and in particular whether or not different spin states lead to different energies, the allowed states of electrons are determined by symmetry properties of their states.

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I have never understood this. You have assumed that there is no external field: "This situation would be different if, for example, there were an external magnetic field that interacted with the magnetic moment of the electron." But if there is no external field then what makes spins to have well-defined $|+1\rangle$ or $|-1\rangle$ states? – hyportnex Jan 30 '14 at 12:24

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