So far, we have not seen any very spectacular consequences of electron spin and the antisymmetry requirement. In the hydrogen and helium atoms, the spin factors in the wave functions and the antisymmetry requirement simply affect the degeneracy of the levels but do not (except for very small effects to be considered later) affect the previously obtained energies. For lithium, the story is quite different.
Suppose we take the interelectronic repulsions in the Li atom as a perturbation on the remaining terms in the Hamiltonian. By the same steps used in the treatment of helium, the unperturbed wave functions are products of three hydrogenlike functions. For the ground state,
\(
\begin{equation}
\psi^{(0)}=1 s(1) 1 s(2) 1 s(3) \tag{10.31}
\end{equation}
\)
and the zeroth-order (unperturbed) energy is [Eq. (9.48) and the paragraph after (9.50)]
\(
E^{(0)}=-\left(\frac{1}{1^{2}}+\frac{1}{1^{2}}+\frac{1}{1^{2}}\right)\left(\frac{Z^{2} e^{2}}{8 \pi \varepsilon{0} a{0}}\right)=-27\left(\frac{e^{2}}{8 \pi \varepsilon{0} a{0}}\right)=-27(13.606 \mathrm{eV})=-367.4 \mathrm{eV}
\)
The first-order energy correction is $E^{(1)}=\left\langle\psi^{(0)}\right| \hat{H}^{\prime}\left|\psi^{(0)}\right\rangle$. The perturbation $\hat{H}^{\prime}$ consists of the interelectronic repulsions, so
\(
\begin{aligned}
& E^{(1)}=\int|1 s(1)|^{2}|1 s(2)|^{2}|1 s(3)|^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d v+\int|1 s(1)|^{2}|1 s(2)|^{2}|1 s(3)|^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{23}} d v \
& +\int|1 s(1)|^{2}|1 s(2)|^{2}|1 s(3)|^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{13}} d v
\end{aligned}
\)
The way we label the dummy integration variables in these definite integrals cannot affect their value. If we interchange the labels 1 and 3 on the variables in the second integral, it is converted to the first integral. Hence these two integrals are equal. Interchange of the labels 2 and 3 in the third integral shows it to be equal to the first integral also. Therefore
\(
E^{(1)}=3 \iint|1 s(1)|^{2}|1 s(2)|^{2} \frac{e^{2}}{4 \pi \varepsilon{0} r{12}} d v{1} d v{2} \int|1 s(3)|^{2} d v_{3}
\)
The integral over electron 3 gives 1 (normalization). The integral over electrons 1 and 2 was evaluated in the perturbation treatment of helium, and [Eqs. (9.52) and (9.53)]
\(
\begin{gathered}
E^{(1)}=3\left(\frac{5 Z}{4}\right)\left(\frac{e^{2}}{8 \pi \varepsilon{0} a{0}}\right)=\frac{45}{4}(13.606 \mathrm{eV})=153.1 \mathrm{eV} \
E^{(0)}+E^{(1)}=-214.3 \mathrm{eV}
\end{gathered}
\)
Since we can use the zeroth-order perturbation wave function as a trial variation function (recall the discussion at the beginning of Section 9.4), $E^{(0)}+E^{(1)}$ must be, according to the variation principle, equal to or greater than the true ground-state energy. The experimental value of the lithium ground-state energy is found by adding up the first, second, and third ionization energies, which gives [C. E. Moore, "Ionization Potentials and Ionization Limits," publication NSRDS-NBS 34 of the National Bureau of Standards (1970); available at www.nist.gov/data/nsrds/NSRDS-NBS34.pdf]
\(
-(5.39+75.64+122.45) \mathrm{eV}=-203.5 \mathrm{eV}
\)
We thus have $E^{(0)}+E^{(1)}$ as less than the true ground-state energy, which is a violation of the variation principle. Moreover, the supposed configuration $1 s^{3}$ for the Li ground state disagrees with the low value of the first ionization potential and with all chemical evidence. If we continued in this manner, we would have a $1 s^{Z}$ ground-state configuration for the element of atomic number $Z$. We would not get the well-known periodic behavior of the elements.
Of course, our error is failure to consider spin and the antisymmetry requirement. The hypothetical zeroth-order wave function $1 s(1) 1 s(2) 1 s(3)$ is symmetric with respect to interchange of any two electrons. If we are to have an antisymmetric $\psi^{(0)}$, we must multiply this symmetric space function by an antisymmetric spin function. It is easy to construct completely symmetric spin functions for three electrons, such as $\alpha(1) \alpha(2) \alpha(3)$. However, it is impossible to construct a completely antisymmetric spin function for three electrons.
Let us consider how we can systematically construct an antisymmetric function for three electrons. We shall use $f, g$, and $h$ to stand for three functions of electronic coordinates, without specifying whether we are considering space coordinates or spin coordinates or both. We start with the function
\(
\begin{equation}
f(1) g(2) h(3) \tag{10.32}
\end{equation}
\)
which is certainly not antisymmetric. The antisymmetric function we desire must be converted into its negative by each of the permutation operators $\hat{P}{12}, \hat{P}{13}$, and $\hat{P}_{23}$. Applying each of these operators in turn to $f(1) g(2) h(3)$, we get
\(
\begin{equation}
f(2) g(1) h(3), \quad f(3) g(2) h(1), \quad f(1) g(3) h(2) \tag{10.33}
\end{equation}
\)
We might try to construct the antisymmetric functions as a linear combination of the four functions (10.32) and (10.33), but this attempt would fail. Application of $\hat{P}_{12}$ to the last two functions in (10.33) gives
\(
\begin{equation}
f(3) g(1) h(2) \text { and } f(2) g(3) h(1) \tag{10.34}
\end{equation}
\)
which are not included in (10.32) or (10.33). We must therefore include all six functions (10.32) to (10.34) in the desired antisymmetric linear combination. These six functions are the six $(3 \cdot 2 \cdot 1)$ possible permutations of the three electrons among the three functions $f, g$, and $h$. If $f(1) g(2) h(3)$ is a solution of the Schrödinger equation with eigenvalue $E$, then, because of the identity of the particles, each of the functions (10.32) to (10.34) is also a solution with the same eigenvalue $E$ (exchange degeneracy), and any linear combination of these functions is an eigenfunction with eigenvalue $E$.
The antisymmetric linear combination will have the form
\(
\begin{align}
c{1} f(1) g(2) h(3)+c{2} f(2) g(1) h(3) & +c{3} f(3) g(2) h(1)+c{4} f(1) g(3) h(2) \
& +c{5} f(3) g(1) h(2)+c{6} f(2) g(3) h(1) \tag{10.35}
\end{align}
\)
Since $f(2) g(1) h(3)=\hat{P}{12} f(1) g(2) h(3)$, in order to have (10.35) be an eigenfunction of $\hat{P}{12}$ with eigenvalue -1 , we must have $c{2}=-c{1}$. Likewise, $f(3) g(2) h(1)=$ $\hat{P}{13} f(1) g(2) h(3)$ and $f(1) g(3) h(2)=\hat{P}{23} f(1) g(2) h(3)$, so $c{3}=-c{1}$ and $c{4}=-c{1}$. Since $f(3) g(1) h(2)=\hat{P}{12} f(3) g(2) h(1)$, we must have $c{5}=-c{3}=c{1}$. Similarly, we find $c{6}=c{1}$. We thus arrive at the linear combination
\(
\begin{align}
c_{1}[f(1) g(2) h(3)-f(2) g(1) h(3) & -f(3) g(2) h(1)-f(1) g(3) h(2) \
& +f(3) g(1) h(2)+f(2) g(3) h(1)] \tag{10.36}
\end{align}
\)
which is easily verified to be antisymmetric with respect to $1-2,1-3$, and $2-3$ interchange. [Taking all signs as plus in (10.36), we would get a completely symmetric function.]
Let us assume $f, g$, and $h$ to be orthonormal and choose $c_{1}$ so that (10.36) is normalized. Multiplying (10.36) by its complex conjugate, we get many terms, but because of the assumed orthogonality the integrals of all products involving two different terms of (10.36) vanish. For example,
\(
\begin{aligned}
& \int[f(1) g(2) h(3)]^{} f(2) g(1) h(3) d \tau \
&=\int f^{}(1) g(1) d \tau{1} \int g^{*}(2) f(2) d \tau{2} \int h^{*}(3) h(3) d \tau_{3}=0 \cdot 0 \cdot 1=0
\end{aligned}
\)
Integrals involving the product of a term of (10.36) with its own complex conjugate are equal to 1 , because $f, g$, and $h$ are normalized. Therefore,
\(
\begin{aligned}
1=\int|(10.36)|^{2} d \tau & =\left|c{1}\right|^{2}(1+1+1+1+1+1) \
c{1} & =1 / \sqrt{6}
\end{aligned}
\)
We could work with (10.36) as it stands, but its properties are most easily found if we recognize it as the expansion [Eq. (8.24)] of the following third-order determinant:
\(
\frac{1}{\sqrt{6}}\left|\begin{array}{lll}
f(1) & g(1) & h(1) \tag{10.37}\
f(2) & g(2) & h(2) \
f(3) & g(3) & h(3)
\end{array}\right|
\)
(See also Prob. 8.22.) The antisymmetry property holds for (10.37) because interchange of two electrons amounts to interchanging two rows of the determinant, which multiplies it by -1 .
We now use (10.37) to prove that it is impossible to construct an antisymmetric spin function for three electrons. The functions $f, g$, and $h$ may each be either $\alpha$ or $\beta$. If we take $f=\alpha, g=\beta, h=\alpha$, then (10.37) becomes
\(
\frac{1}{\sqrt{6}}\left|\begin{array}{lll}
\alpha(1) & \beta(1) & \alpha(1) \tag{10.38}\
\alpha(2) & \beta(2) & \alpha(2) \
\alpha(3) & \beta(3) & \alpha(3)
\end{array}\right|
\)
Although (10.38) is antisymmetric, we must reject it because it is equal to zero. The first and third columns of the determinant are identical, so (Section 8.3) the determinant vanishes. No matter how we choose $f, g$, and $h$, at least two columns of the determinant must be equal, so we cannot construct a nonzero antisymmetric three-electron spin function.
We now use (10.37) to construct the zeroth-order ground-state wave function for lithium, including both space and spin variables. The functions $f, g$, and $h$ will now involve both space and spin variables. We choose
\(
\begin{equation}
f(1)=1 s(1) \alpha(1) \tag{10.39}
\end{equation}
\)
We call a function like (10.39) a spin-orbital. A spin-orbital is the product of a one-electron spatial orbital and a one-electron spin function.
If we were to take $g(1)=1 s(1) \alpha(1)$, this would make the first and second columns of (10.37) identical, and the wave function would vanish. This is a particular case of the Pauli exclusion principle: No two electrons can occupy the same spin-orbital. Another way of stating this is to say that no two electrons in an atom can have the same values for all their quantum numbers. The Pauli exclusion principle is a consequence of the more general antisymmetry requirement for the wave function of a system of identical spin- $\frac{1}{2}$ particles and is less satisfying than the antisymmetry statement, since the exclusion principle is based on approximate (zeroth-order) wave functions.
We therefore take $g(1)=1 s(1) \beta(1)$, which puts two electrons with opposite spin in the $1 s$ orbital. For the spin-orbital $h$, we cannot use either $1 s(1) \alpha(1)$ or $1 s(1) \beta(1)$, since these choices make the determinant vanish. We take $h(1)=2 s(1) \alpha(1)$, which gives the familiar Li ground-state configuration $1 s^{2} 2 s$ and the zeroth-order wave function
\(
\psi^{(0)}=\frac{1}{\sqrt{6}}\left|\begin{array}{lll}
1 s(1) \alpha(1) & 1 s(1) \beta(1) & 2 s(1) \alpha(1) \tag{10.40}\
1 s(2) \alpha(2) & 1 s(2) \beta(2) & 2 s(2) \alpha(2) \
1 s(3) \alpha(3) & 1 s(3) \beta(3) & 2 s(3) \alpha(3)
\end{array}\right|
\)
Note especially that (10.40) is not simply a product of space and spin parts (as we found for H and He ), but is a linear combination of terms, each of which is a product of space and spin parts.
Since we could just as well have taken $h(1)=2 s(1) \beta(1)$, the ground state of lithium is, like hydrogen, doubly degenerate, corresponding to the two possible orientations of the spin of the $2 s$ electron. We might use the orbital diagrams
to indicate this. Each spatial orbital such as $1 s$ or $2 p_{0}$ can hold two electrons of opposite spin. A spin-orbital such as $2 s \alpha$ can hold one electron.
Although the $1 s^{2} 2 p$ configuration will have the same unperturbed energy $E^{(0)}$ as the $1 s^{2} 2 s$ configuration, when we take electron repulsion into account by calculating $E^{(1)}$ and
higher corrections, we find that the $1 s^{2} 2 s$ configuration lies lower for the same reason as in helium.
Consider some points about the Pauli exclusion principle, which we restate as follows: In a system of identical fermions, no two particles can occupy the same state. If we have a system of $n$ interacting particles (for example, an atom), there is a single wave function (involving $4 n$ variables) for the entire system. Because of the interactions between the particles, the wave function cannot be written as the product of wave functions of the individual particles. Hence, strictly speaking, we cannot talk of the states of individual particles, only the state of the whole system. If, however, the interactions between the particles are not too large, then as an initial approximation we can neglect them and write the zeroth-order wave function of the system as a product of wave functions of the individual particles. In this zeroth-order wave function, no two fermions can have the same wave function (state).
Since bosons require a wave function symmetric with respect to interchange, there is no restriction on the number of bosons in a given state.
In 1925, Einstein showed that in an ideal gas of noninteracting bosons, there is a very low temperature $T{c}$ (called the condensation temperature) above which the fraction $f$ of bosons in the ground state is negligible but below which $f$ becomes appreciable and goes to 1 as the absolute temperature $T$ goes to 0 . The equation for $f$ for noninteracting bosons in a cubic box is $f=1-\left(T / T{c}\right)^{3 / 2}$ for $T<T_{c}$ [McQuarrie (2000), Section 10-4]. The phenomenon of a significant fraction of bosons falling into the ground state is called Bose-Einstein condensation. Bose-Einstein condensation is important in determining the properties of superfluid liquid ${ }^{4} \mathrm{He}$ (whose atoms are bosons), but the interatomic interactions in the liquid make theoretical analysis difficult.
In 1995, physicists succeeded in producing Bose-Einstein condensation in a gas [Physics Today, August 1995, p. 17; C. E. Wieman, Am. J. Phys., 64, 847 (1996)]. They used a gas of ${ }_{37}^{87} \mathrm{Rb}$ atoms. $\mathrm{An}{ }^{87} \mathrm{Rb}$ atom has 87 nucleons and 37 electrons. With an even number (124) of fermions, ${ }^{87} \mathrm{Rb}$ is a boson. With a combination of laser light, an applied inhomogeneous magnetic field, and applied radiofrequency radiation, a sample of $10^{4}{ }^{87} \mathrm{Rb}$ atoms was cooled to $10^{-7} \mathrm{~K}$, thereby condensing a substantial fraction of the atoms into the ground state. The radiofrequency radiation was then used to remove most of the atoms in excited states, leaving a condensate of 2000 atoms, nearly all of which were in the ground state. Each Rb atom in this experiment was subject to a potential-energy function $V(x, y, z)$ produced by the interaction of the atom's total spin magnetic moment with the applied magnetic field (Sections 6.8 and 10.9). The inhomogeneous applied magnetic field was such that the potential energy $V$ was that of a three-dimensional harmonic oscillator (Prob. 4.20) plus a constant. The Rb atoms in the Bose-Einstein condensate are in the ground state of this harmonic-oscillator potential.