10.2: Angular Momentum in Hydrogen Atom

In a hydrogen atom, the wavefunction of an electron in a simultaneous eigenstate of $L^2$ and $L_z$ has an angular dependence specified by the spherical harmonic $Y_{l,m}(\theta,\phi)$. (See Section [sharm].) If the electron is also in an eigenstate of $S^{\,2}$ and $S_z$ then the quantum numbers $s$ and $m_s$ take the values $1/2$ and $\pm 1/2$, respectively, and the internal state of the electron is specified by the spinors $\chi_\pm$. (See Section [spauli].) Hence, the simultaneous eigenstates of $L^2$, $S^{\,2}$, $L_z$, and $S_z$ can be written in the separable form $$\label{e11.28} \psi^{(1)}_{l,1/2;m,\pm 1/2} = Y_{l,m}\,\chi_\pm.$$ Here, it is understood that orbital angular momentum operators act on the spherical harmonic functions, $Y_{l,m}$, whereas spin angular momentum operators act on the spinors, $\chi_\pm$.

Because the eigenstates $\psi^{(1)}_{l,1/2;m,\pm 1/2}$ are (presumably) orthonormal, and form a complete set, we can express the eigenstates $\psi^{(2)}_{l,1/2;j,m_j}$ as linear combinations of them. For instance,

$$\label{e11.29} \psi^{(2)}_{l,1/2;j,m+1/2} = \alpha\,\psi^{(1)}_{l,1/2;m,1/2} + \beta\, \psi^{(1)}_{l,1/2;m+1,-1/2},$$ where $\alpha$ and $\beta$ are, as yet, unknown coefficients. Note that the number of $\psi^{(1)}$ states that can appear on the right-hand side of the previous expression is limited to two by the constraint that $m_j=m+m_s$ [see Equation ([e11.23])], and the fact that $m_s$ can only take the values $\pm 1/2$. Assuming that the $\psi^{(2)}$ eigenstates are properly normalized, we have

$$\label{e11.30} \alpha^{\,2} + \beta^{\,2} = 1.$$

Now, it follows from Equation ([e11.26]) that

$$\label{e11.31} J^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2}= j\,(j+1)\,\hbar^{\,2}\,\psi^{(2)}_{l,1/2;j,m+1/2},$$ where [see Equation ([e11.12])] $$\label{e11.32} J^{\,2} = L^2+S^{\,2} +2\,L_z\,S_z+ L_+\,S_-+L_-\,S_+.$$ Moreover, according to Equations ([e11.28]) and ([e11.29]), we can write $$\label{e11.33} \psi^{(2)}_{l,1/2;j,m+1/2} = \alpha\,Y_{l,m}\,\chi_+ + \beta\, Y_{l,m+1}\,\chi_-.$$ Recall, from Equations ([eraise]) and ([elow]), that $$\begin{aligned} \label{e11.34} L_+\,Y_{l,m} &= [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar\,Y_{l,m+1},\\[0.5ex] L_-\,Y_{l,m} &= [l\,(l+1)-m\,(m-1)]^{1/2}\,\hbar\,Y_{l,m-1}.\end{aligned}$$ By analogy, when the spin raising and lowering operators, $S_\pm$, act on a general spinor, $\chi_{s,m_s}$, we obtain $$\begin{aligned} S_+\,\chi_{s,m_s} &= [s\,(s+1)-m_s\,(m_s+1)]^{1/2}\,\hbar\,\chi_{s,m_s+1},\\[0.5ex] S_-\,\chi_{s,m_s} &= [s\,(s+1)-m_s\,(m_s-1)]^{1/2}\,\hbar\,\chi_{s,m_s-1}.\end{aligned}$$ For the special case of spin one-half spinors (i.e., $s=1/2, m_s=\pm 1/2$), the previous expressions reduce to $$S_+\,\chi_+=S_-\,\chi_- = 0,$$ and $$\label{e11.39} S_\pm\,\chi_\mp = \hbar\,\chi_\pm.$$

It follows from Equations ([e11.32]) and ([e11.34])–([e11.39]) that $$\begin{aligned} J^{\,2}\,Y_{l,m}\,\chi_+&= [l\,(l+1)+3/4+m]\,\hbar^{\,2}\,Y_{l,m}\,\chi_+\nonumber\\[0.5ex] &\phantom{=}+ [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar^{\,2}\,Y_{l,m+1}\,\chi_-,\end{aligned}$$ and $$\begin{aligned} J^{\,2}\,Y_{l,m+1}\,\chi_-&= [l\,(l+1)+3/4-m-1]\,\hbar^{\,2}\,Y_{l,m+1}\,\chi_-\nonumber\\[0.5ex] &\phantom{=}+ [l\,(l+1)-m\,(m+1)]^{1/2}\,\hbar^{\,2}\,Y_{l,m}\,\chi_+.\end{aligned}$$ Hence, Equations ([e11.31]) and ([e11.33]) yield

$$\begin{aligned} \label{e11.42} (x - m)\,\alpha - [l\,(l+1)-m\,(m+1)]^{1/2}\,\beta &= 0,\\[0.5ex] -[l\,(l+1)-m\,(m+1)]^{1/2}\,\alpha +(x+m+1)\,\beta&= 0,\label{e11.43}\end{aligned}$$ where $$x = j\,(j+1) - l\,(l+1) - 3/4.$$ Equations ([e11.42]) and ([e11.43]) can be solved to give $$x\,(x+1) = l\,(l+1),$$ and

$$\label{e11.45} \frac{\alpha}{\beta} = \frac{[(l-m)\,(l+m+1)]^{1/2}}{x-m}.$$ It follows that $x=l$ or $x=-l-1$, which corresponds to $j=l+1/2$ or $j=l-1/2$, respectively. Once $x$ is specified, Equations ([e11.30]) and ([e11.45]) can be solved for $\alpha$ and $\beta$. We obtain

$$\label{e11.47} \psi^{(2)}_{l+1/2,m+1/2} = \left(\frac{l+m+1}{2\,l+1}\right)^{1/2} \psi^{(1)}_{m,1/2} + \left(\frac{l-m}{2\,l+1}\right)^{1/2}\psi^{(1)}_{m+1,-1/2},$$ and

$$\label{e11.48} \psi^{(2)}_{l-1/2,m+1/2} = \left(\frac{l-m}{2\,l+1}\right)^{1/2} \psi^{(1)}_{m,1/2} -\left(\frac{l+m+1}{2\,l+1}\right)^{1/2}\psi^{(1)}_{m+1,-1/2}.$$ Here, we have neglected the common subscripts $l,1/2$ for the sake of clarity: that is, $\psi^{(2)}_{l+1/2,m+1/2}\equiv \psi^{(2)}_{l,1/2;l+1/2,m+1/2}$, et cetera. The previous equations can easily be inverted to give the $\psi^{(1)}$ eigenstates in terms of the $\psi^{(2)}$ eigenstates:

$$\begin{aligned} \psi^{(1)}_{m,1/2} &= \left(\frac{l+m+1}{2\,l+1}\right)^{1/2}\!\psi^{(2)}_{l+1/2,m+1/2} + \left(\frac{l-m}{2\,l+1}\right)^{1/2}\!\psi^{(2)}_{l-1/2,m+1/2},\\[0.5ex] \!\!\!\!\!\!\psi^{(1)}_{m+1,-1/2}&= \left(\frac{l-m}{2\,l+1}\right)^{1/2}\!\psi^{(2)}_{l+1/2,m+1/2} - \left(\frac{l+m+1}{2\,l+1}\right)^{1/2}\! \psi^{(2)}_{l-1/2,m+1/2}.\label{e11.50}\end{aligned}$$ The information contained in Equations ([e11.47])–([e11.50]) is neatly summarized in Table [t2]. For instance, Equation ([e11.47]) is obtained by reading the first row of this table, whereas Equation ([e11.50]) is obtained by reading the second column. The coefficients in this type of table are generally known as Clebsch-Gordon coefficients .

Clebsch-Gordon coefficients for adding spin one-half to spin $l$.

	$m, 1/2$	$m+1, -1/2$	$m, m_s$
[0.5ex] $l+1/2, m+1/2$	${\scriptstyle\sqrt{(l+m+1)/(2\,l+1)}}$	${\scriptstyle\sqrt{(l-m)/(2\,l+1)}}$
[0.5ex] $l-1/2, m+1/2$	${\scriptstyle\sqrt{(l-m)/(2\,l+1)}}$	${\scriptstyle-\sqrt{(l+m+1)/(2\,l+1)}}$
[0.5ex] $j, m_j$

As an example, let us consider the $l=1$ states of a hydrogen atom. The eigenstates of $L^2$, $S^{\,2}$, $L_z$, and $S_z$, are denoted $\psi^{(1)}_{m,m_s}$. Because $m$ can take the values $-1,0,1$, whereas $m_s$ can take the values $\pm 1/2$, there are clearly six such states: that is, $\psi^{(1)}_{1,\pm 1/2}$, $\psi^{(1)}_{0,\pm 1/2}$, and $\psi^{(1)}_{-1,\pm 1/2}$. The eigenstates of $L^2$, $S^{\,2}$, $J^{\,2}$, and $J_z$, are denoted $\psi^{(2)}_{j,m_j}$. Because $l=1$ and $s=1/2$ can be combined together to form either $j=3/2$ or $j=1/2$ (see previously), there are also six such states: that is, $\psi^{(2)}_{3/2,\pm 3/2}$, $\psi^{(2)}_{3/2,\pm 1/2}$, and $\psi^{(2)}_{1/2,\pm 1/2}$. According to Table [t2], the various different eigenstates are interrelated as follows:

$$\begin{aligned} \label{ecgs} \psi^{(2)}_{3/2,\pm 3/2} &= \psi^{(1)}_{\pm 1, \pm 1/2},\\[0.5ex] \psi^{(2)}_{3/2,1/2} &= \sqrt{\frac{2}{3}}\,\psi^{(1)}_{0,1/2} + \sqrt{ \frac{1}{3}}\,\psi^{(1)}_{1,-1/2},\label{e11.52}\\[0.5ex] \psi^{(2)}_{1/2,1/2} &= \sqrt{\frac{1}{3}}\,\psi^{(1)}_{0,1/2} - \sqrt{ \frac{2}{3}}\,\psi^{(1)}_{1,-1/2},\\[0.5ex] \psi^{(2)}_{1/2,-1/2} &= \sqrt{\frac{2}{3}}\,\psi^{(1)}_{-1,1/2} -\sqrt{ \frac{1}{3}}\,\psi^{(1)}_{0,-1/2},\\[0.5ex] \psi^{(2)}_{3/2,-1/2} &= \sqrt{\frac{1}{3}}\,\psi^{(1)}_{-1,1/2} + \sqrt{ \frac{2}{3}}\,\psi^{(1)}_{0,-1/2}, \end{aligned}$$ and

$$\begin{aligned} \psi^{(1)}_{\pm 1,\pm 1/2} &= \psi^{(2)}_{3/2, \pm 3/2},\\[0.5ex] \psi^{(1)}_{1,-1/2} &= \sqrt{\frac{1}{3}}\,\psi^{(2)}_{3/2,1/2} - \sqrt{ \frac{2}{3}}\,\psi^{(2)}_{1/2,1/2},\\[0.5ex] \psi^{(1)}_{0,1/2} &= \sqrt{\frac{2}{3}}\,\psi^{(2)}_{3/2,1/2} + \sqrt{ \frac{1}{3}}\,\psi^{(2)}_{1/2,1/2},\label{e11.57}\\[0.5ex] \psi^{(1)}_{0,-1/2} &= \sqrt{\frac{2}{3}}\,\psi^{(2)}_{3/2,-1/2} -\sqrt{ \frac{1}{3}}\,\psi^{(2)}_{1/2,-1/2},\label{ecge}\\[0.5ex] \psi^{(1)}_{-1,1/2} &= \sqrt{\frac{1}{3}}\,\psi^{(2)}_{3/2,-1/2} + \sqrt{ \frac{2}{3}}\,\psi^{(2)}_{1/2,-1/2}, \end{aligned}$$ Thus, if we know that an electron in a hydrogen atom is in an $l=1$ state characterized by $m=0$ and $m_s=1/2$ [i.e., the state represented by $\psi^{(1)}_{0,1/2}$] then, according to Equation ([e11.57]), a measurement of the total angular momentum will yield $j=3/2$, $m_j=1/2$ with probability $2/3$, and $j=1/2$, $m_j=1/2$ with probability $1/3$. Suppose that we make such a measurement, and obtain the result $j=3/2$, $m_j=1/2$. As a result of the measurement, the electron is thrown into the corresponding eigenstate, $\psi^{(2)}_{3/2,1/2}$. It thus follows from Equation ([e11.52]) that a subsequent measurement of $L_z$ and $S_z$ will yield $m=0$, $m_s=1/2$ with probability $2/3$, and $m=1$, $m_s=-1/2$ with probability $1/3$.

Clebsch-Gordon coefficients for adding spin one-half to spin one. Only non-zero coefficients are shown.

	$-1, -1/2$	$-1, 1/2$	$0,-1/2$	$0,1/2$	$1,-1/2$	$1,1/2$	$m,m_s$
[0.5ex] $3/2, -3/2$	$\scriptstyle{1}$
[0.5ex] $3/2, -1/2$		${\scriptstyle\sqrt{1/3}}$	${\scriptstyle\sqrt{2/3}}$
[0.5ex] $1/2, -1/2$		${\scriptstyle\sqrt{2/3}}$	${\scriptstyle -\sqrt{1/3}}$
[0.5ex] $3/2, 1/2$				${\scriptstyle\sqrt{2/3}}$	${\scriptstyle\sqrt{1/3}}$
[0.5ex] $1/2, 1/2$				${\scriptstyle\sqrt{1/3}}$	${\scriptstyle - \sqrt{2/3}}$
[0.5ex] $3/2, 3/2$						${\scriptstyle 1}$
$j, m_j$

The information contained in Equations ([ecgs])–([ecge]) is neatly summed up in Table [t3]. Note that each row and column of this table has unit norm, and also that the different rows and different columns are mutually orthogonal. Of course, this is because the $\psi^{(1)}$ and $\psi^{(2)}$ eigenstates are orthonormal.

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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