Abstract
We begin this chapter with the consideration of the orbital angular momentum. This gives rise to a general definition of angular momenta. We derive the eigenvalue spectrum of the orbital angular momentum with an algebraic method. After a brief presentation of the eigenfunctions of the orbital angular momentum in the position representation, we outline some concepts for the addition of angular momenta.
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Notes
- 1.
Instead of \(\mathbf {J}\), one often finds \(\mathbf {j}\) (whereby this is of course not be confused with the probability current density). In this section, we denote the operator by \(\mathbf {J}\) and the eigenvalue by j.
- 2.
The factor \(\hbar \) is due to the choice of units, and would be replaced by a different constant if one were to choose different units. The essential factor is i.
- 3.
j is also called the angular momentum quantum number and m the magnetic or directional quantum number.
- 4.
For e.g. \(J_{+}\), \(\left| j,m\right\rangle \rightarrow \left| j, m+1\right\rangle \rightarrow \left| j, m+2\right\rangle \rightarrow \cdots \).
- 5.
We omit here the distinction between \(=\) and \(\cong \).
- 6.
The anticommutator is defined as usual by \(\left\{ A, B\right\} =AB+BA\).
- 7.
We recall the equality
$$\begin{aligned} {\varvec{\nabla }}^{2}=\frac{\partial ^{2}}{\partial r^{2}}+\frac{2}{r}\frac{ \partial }{\partial r}-\frac{{\mathbf {l}}^{2}}{\hbar ^{2}r^{2}} \end{aligned}$$See also Appendix D, Vol. 1.
- 8.
Other notations: \(Y_{l}^{m}\left( \vartheta ,\varphi \right) =Y_{l}^{m}\left( \hat{r}\right) =\left\langle \hat{r}\right. \left| l, m\right\rangle \). \(\hat{r}\) is the unit vector and is an abbreviation for the pair \(\left( \vartheta ,\varphi \right) \). Moreover, the notation \( Y_{lm} \) \(\left( \vartheta ,\varphi \right) \) is also common.
- 9.
Intuitively-clear reason: If the two angular momentum vectors are parallel to each other, then their angular-momentum quantum numbers are added to give \(j=j_{1}+j_{2}\). For other arrangements, j is smaller. The smallest value is \(\left| j_{1}-j_{2}\right| \), because of \(j\ge 0\).
- 10.
There are several different notations for these coefficients, e.g. \( C_{m_{1}m_{2};jm}^{j_{1}j_{2}}\). The so-called 3j-symbols are related coefficients:
$$\begin{aligned} \left( \begin{array}{ccc} j_{1} &{} j_{2} &{} j \\ m_{1} &{} m_{2} &{} m \end{array} \right) =\left( -1\right) ^{j_{1}-j_{2}-m} \frac{1}{\sqrt{2j+1}}\left\langle j_{1}j_{2}m_{1}m_{2}\right. \left| j-m\right\rangle . \end{aligned}$$The 3j-symbols are invariant against cyclic permutation:
$$\begin{aligned} \left( \begin{array}{ccc} j_{1} &{} j_{2} &{} j \\ m_{1} &{} m_{2} &{} m \end{array} \right) =\left( \begin{array}{ccc} j &{} j_{1} &{} j_{2} \\ m &{} m_{1} &{} m_{2} \end{array} \right) , \text { etc.} \end{aligned}$$ - 11.
The two conditions \(\left| j_{1}-j_{2}\right| \le j\le j_{1}+j_{2}\) and \(m=m_{1}+m_{2}\) must be met in order that a Clebsch-Gordan coefficient is nonzero. The CGC satisfy the orthogonality relations
$$\begin{aligned} \mathop {\displaystyle \sum }\limits _{m_{1}, m_{2}}\left\langle j_{1}j_{2}m_{1}m_{2}\right. \left| jm\right\rangle \left\langle j_{1}j_{2}m_{1}m_{2}\right. \left| j^{\prime }m^{\prime }\right\rangle&=\delta _{jj^{\prime }}\delta _{mm^{\prime }} \\ \mathop {\displaystyle \sum }\limits _{j, m}\left\langle j_{1}j_{2}m_{1}m_{2}\right. \left| jm\right\rangle \left\langle j_{1}j_{2}m_{1}^{\prime }m_{2}^{\prime }\right. \left| jm\right\rangle&=\delta _{m_{1}m_{1}^{\prime }}\delta _{m_{2}m_{2}^{\prime }} . \end{aligned}$$A particular CGC can in principle be calculated using the ladder operators \( j_{1\pm }+j_{2\pm }\), whereby one starts e.g. from \(\left\langle j_{1}j_{2}j_{1}j_{2}\right. \left| j_{1}+j_{2} j_{1}+j_{2}\right\rangle =1\). In this way, one obtains for example
$$\begin{aligned} \left( \begin{array}{ccc} A &{} B &{} A+B \\ a &{} b &{} c \end{array}\right) =\left( -1\right) ^{A-B-c}\left[ \frac{\left( 2A\right) !\left( 2B\right) !\left( A+B+c\right) !\left( A+B-c\right) !}{\left( 2A+2B+1\right) !\left( A+a\right) !\left( A-a\right) !\left( B+b\right) !\left( B-b\right) ! }\right] ^{\frac{1}{2}} . \end{aligned}$$ - 12.
Of course, all the calculations may also be performed representation-free.
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Pade, J. (2018). Angular Momentum. In: Quantum Mechanics for Pedestrians 2. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-030-00467-5_16
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DOI: https://doi.org/10.1007/978-3-030-00467-5_16
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