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# 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 .

 $$m, 1/2$$ $$m+1, -1/2$$ $$m, m_s$$ [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 $$1/3$$.

 $$-1, -1/2$$ $$-1, 1/2$$ $$0,-1/2$$ $$0,1/2$$ $$1,-1/2$$ $$1,1/2$$ $$m,m_s$$ [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

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