By definition, the ground state of a quantum system is the state of lowest energy. What is the ground state of an atom in which $Z$ electrons orbit a nucleus with $Z$ protons? Because the $1s$ state is the lowest energy state in the independent particle approximation, it seems that the ground state should be one in which all $Z$ electrons are in the $1s$ state. However, this idea is not consistent with the experimental evidence.

In 1925, the young Austrian physicist Wolfgang Pauli hypothesized that no two electrons in a quantum system can be in the same quantum state. That is, no two electrons can have exactly the same set of quantum numbers $\left(n\text{,}l\text{,}n\text{,}{m}_s\right)$. If one electron is present in a state, it excludes all others. This statement, which is called the Pauli exclusion principle, turns out to be an extremely profound statement about the nature of matter.

The exclusion principle is not applicable to the hydrogen atom, which has only a single electron. But in helium, with $Z=2$ electrons, we must make sure that the two electrons are in different quantum states. This is not difficult. For a $1s$ state, with $l=0$, the only possible value of the magnetic quantum number is $m=0$. But there are two possible values of $m_s$, namely $+\frac{1}{2}$ and $-\frac{1}{2}$. If a first electron is in the spin-up $1s$ state $(1\text{,}0\text{,}0\text{,}+\frac{1}{2})$, a second $1s$ electron can still be added to the atom as long as it is in the spin-down state $(1\text{,}0\text{,}0\text{,}-\frac{1}{2})$. This is shown schematically in FIGURE 41.16a, where the dots represent electrons on the rungs of the “energy ladder” and the arrows represent spin-up or spin-down.

FIGURE 41.16 The ground state and the first excited state of helium.

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The Pauli exclusion principle does not prevent both electrons of helium from being in the $1s$ state as long as they have opposite values of $m_s$, so we predict this to be the ground state. A list of an atom’s occupied energy levels is called its electron configuration. The electron configuration of the helium ground state is written $1s^2$, where the superscript 2 indicates two electrons in the $1s$ energy level. An excited state of the helium atom might be the electron configuration $1s2s$. This state is shown in FIGURE 41.16b. Here, because the two electrons have different values of $n$, there is no restriction on their values of $m_s$.

The states $(1\text{,}0\text{,}0\text{,}+\frac{1}{2})$ and $(1\text{,}0\text{,}0\text{,}-\frac{1}{2})$ are the only two states with $n=1$. The ground state of helium has one electron in each of these states, so all the possible $n=1$ states are filled. Consequently, the electron configuration $1s^2$ is called a closed shell. Because the two electron magnetic moments point in opposite directions, we can predict that helium has no net magnetic moment and will be undeflected in a Stern-Gerlach apparatus. This prediction is confirmed by experiment.

The next element, lithium, has $Z=3$ electrons. The first two electrons can go into $1s$ states, with opposite values of $m_s$, but what about the third electron? The $1s^2$ shell is closed, and there are no additional quantum states having $n=1$. The only option for the third electron is the next energy state, $n=2$. The $2s$ and $2p$ states had equal energies in the hydrogen atom, but they do not in a multielectron atom. As Figure 41.14 showed, a lower-$l$ state has lower energy than a higher-$l$ state with the same $n$. The $2s$ state of lithium is lower in energy than $2p$, so lithium’s third ground-state electron will be $2s$. This requires $l=0$ and $m=0$ for the third electron, but the value of $m_s$ is not relevant because there is only a single electron in $2s$. FIGURE 41.17a shows the electron configuration with the $2s$ electron being spin-up, but it could equally well be spin-down. The electron configuration for the lithium ground state is written $1s^{2}2s$. This indicates two $1s$ electrons and a single $2s$ electron.

FIGURE 41.17 The ground state and the first excited state of lithium.

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FIGURE 41.18a shows the probability density of electrons in the $1s^{2}2s$ ground state of lithium. You can see the $2s$ electron shell surrounding the inner $1s^2$ core. For comparison, FIGURE 41.18b shows the first excited state of lithium, in which the $2s$ electron has been excited to the $2p$ energy level. This forms the $1s^{2}2p$ configuration, also shown in FIGURE 41.17b.

FIGURE 41.18 Electron clouds for two lithium electron configurations.

The Schrödinger equation accurately predicts the energies of the $1s^{2}2s$ and the $1s^{2}2p$ configurations of lithium, but the Schrödinger equation does not tell us which states the electrons actually occupy. The electron spin and the Pauli exclusion principle were the final pieces of the puzzle. Once these were added to Schrödinger’s theory, the initial phase of quantum mechanics was complete. Physicists finally had a successful theory for understanding the structure of atoms.