,???Department of Physics, Beijing Jiaotong University, Beijing 100044, ChinaCorresponding authors: ? ? E-mail:fmchen@bjtu.edu.cn
Received:2018-12-17Online:2019-06-1
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Ling-Di Yang, Fa-Min Chen. A Spinor Approach to the SU(2) Clebsch-Gordan Coefficients *. [J], 2019, 71(6): 663-669 doi:10.1088/0253-6102/71/6/663
1 Introduction
It is well known that the finite dimensional (irreducible) representation of an angular momentum vector operator can be constructed by employing a pair of spinors.[1-2] The CG coefficients1(1 The analytic expression for CG coefficients were first derived by Wigner,[3] then by Racah,[4] using a different approach.) couple the individual eigenstates of two independent angular momentum operators $J_1$ and $J_2$ to form the eigenstates of the total angular momentum operator $J=J_1+J_2$. So, if we use two independent pairs of spinors to construct the individual states of $J_1$ and $J_2$, and take care of the coupling of these two pairs of spinors, it is possible to work out the CG coefficients. Our goal is to derive all CG coefficients using this spinor approach. It seems that this approach is simpler than the conventional approaches[5-6] for calculating the CG coefficients.This paper is organized as follows. In Sec. 2, we briefly review the irreducible representation of the set of SU(2) generators or angular momentum vector operator using the spinor approach. The reader who is familiar with it may skip this section. In Sec. 3, we present our detailed calculations on CG coefficients utilizing the spinor algebra. In Sec. 4, we work out some simple physical examples to illustrate the spinor approach. Section 5 is devoted to discussions. In Appendix A, we sketch two conventional methods for calculating the ${\rm SU}(2)$ CG coefficients.
2 Review of Representation of SU(2)
In this section, we briefly review the spinor construction of the irreducible representation of the Lie algebra of SU(2).[1] The three generators of SU(2) or the components of angular momentum $J$ satisfy the usual commutation relation2(2 For simplicity, we set the planck constant $\hbar=1$.)We denote the "spin up" and "spin down" spinors as $\xi_{1}$ and $\xi_{2}$, respectively. The action of $J_i$ on $\xi_{a}$ ($a=1,2$) is as follows
where $\sigma_i{}^\beta{}_a$ is the set of pauli matrices
And the Leibnitz rule
is assumed. Equation (2) is equivalent to
where $J_{\pm}=J_1\pm i J_2$ or $J_x\pm i J_y$ are the usual raising and lowering operators.
The (irreducible) ($2j+1$)-dimensional representation of $J_i$ can be constructed by introducing the set of basis state vectors
with $j=0,1/2,1,3/2,2,\ldots\,$, and $m=-j, -j+1,\ldots\,$, $j-1, j$. The physical interpretation of Eq. (6) is that it describes a state of $2j$ spin $1/2$ particles; specifically, it describes the state of $(j+m)$ "spin up'' particles and $(j-m)$ "spin down" particles.
Using Eqs. (2) and (4), we see that the corresponding eigenvalues of $J^2$ and $J_3$ are $j(j+1)$ and $m$, respectively, i.e.,
and it is easy to show that
The irreducible representation matrices $J^{(j)}_{m'm}$, defined by $J\psi_{jm}=\sum_{m'}J^{(j)}_{m'm}\psi_{jm'},$ can be read off from Eqs. (7) and (8). By Eq. (8), we see that the matrix elements of $J^{(j)}_{\pm m'm}$ are real and positive3(If we use the following "$\tau$-pauli" matrices
to replace the $\sigma$-pauli matrices in Eqs. (2) and (3), Eqs. (7) remain unchanged, but the matrix elements (9) turn out to be $ J^{(j)}_{\pm m'm}=e^{i\delta}\sqrt{(j\pm m+1) (j\mp m)}\delta_{m',m\pm1}\,. $ However, we follow the convention of Refs. [5-7] by setting the phase factor $e^{i\delta}=1$.), i.e.,Also, the matrix elements of the raising and lowering operators defined via Eqs. (14), (15), and (40) are real and positive.
3 SU(2) CG Coefficients and Spinors
We now consider two independent angular momentum vector operators $J_1$ and $J_2$ and their coupling. Following the idea of the last section, it is not difficult to construct the simultaneous eigenstate of $J^2_1$, $J^2_2$, $J_{1z}$, and $J_{2z}$:Here $\eta_a$ is another independent pair of spinors. The action of $J_1$ and $J_2$ on the spinors is as follows
and the Leibnitz rule, similar to Eq. (4), is assumed. We see that Eq. (10) is an obvious generalization of Eq. (6). In summary, the state vector $\psi_{j_1j_2;m_1m_2}$ satisfies
Here $J_{A\pm}=J_{Ax}\pm i J_{Ay}$. Notice that $\psi_{j_1j_2;m_1m_2}$ is also an eigenstate of $J_z=J_{1z}+J_{2z}$, the $z$-component of the total angular momentum
It is useful to consider the general ${\rm SU}(2)$ transformation generated by $J$. It can be defined as follows. Let $U=exp(i{\theta}\cdot J)$, where $\theta$ is a vector of parameters. Using Eq. (11), we learn that
where $u^{\beta}{}_a=(e^{i\theta\cdot\sigma/2})^{\beta}{}_a$, obeying $u^\dagger u =\mathbf{1}_{2\times2}$ and $\det u=1$, namely
Here $\epsilon^{a\beta}$ ($\epsilon^{12}=-\epsilon^{21}=-1$) is the antisymmetric tensor. Later we will see that $\epsilon^{a\beta}$ plays the key role for constructing the eigenstates of the total angular momentum.
and read off the CG coefficients from the above unitary transformation. As usual, we use $j$ and $m$ to label the quantum numbers of $J^2$ and $J_z$, respectively.
The basic strategy is to construct $\psi_{j_1j_2;jj}$ first, with $|j_2-j_1|\leq j\leq j_1+j_2$, then use the lowering operator $J_-=J_{1-}+J_{2-}$ to act on it $(j-m)$ times to get the general state $\psi_{j_1j_2;jm}$.
We begin by considering the special case $j_1=j_2$. In this case, the minimum $j$ is $0$. The essential observation is that the state $\psi_{j_1j_1;j=0,m=0}$ must take the following form:
(up to a normalization constant). In fact, due to the constraint of the antisymmetric tensor $\epsilon^{a\beta}$, the number of "spin up" particles is exactly the same as the number of "spin down" particles, so Eq. (20) must be the spin zero state of the total angular momentum $J$; Or in other words, due to the second equation of Eqs. (18), (20) must transform as a scalar under the SU(2) transformation (17). This can be proved as follows: Using
and Eq. (11), a short calculation gives
Taking account of the equation
we see that it is indeed the spin zero state, namely,
Finally, notice that Eq. (20) is also the simultaneous eigenstate of $J_1^2$ and $J_2^2$, belonging to the same eigenvalue $j_1(j_1+1)$. This completes the proof. The next step is to construct $\psi_{j_1j_1;j=1,m=1}$; it must take the following form:
More generally
where $0\leq j\leq j_1+j_2=2j_1$. We are going to prove Eq. (26) a little later.
We are now ready to construct $\psi_{j_1j_1;jj}$, without assuming $j_1=j_2$. Following the idea for constructing Eqs. (20) and (26), it is natural to propose4(4 Similarly, one may construct $\psi_{j_1j_2;j(-j)}$ as follows: $ \psi_{j_1j_2;j(-j)}=c\overbrace{(\xi_2)^{[j-(j_2-j_1)]} (\eta_2)^{[j+(j_2-j_1)]}}^{{\rm spin}\ j\ {\rm part}\ (m=-j)} \,\overbrace{\sum_{a,\beta}(\epsilon^{a\beta}\xi_{a}\eta_{\beta})^{[(j_1+j_2)-j]}}^{\rm spin\ zero\ part}\,. $)
with $c$ a constant. If $j_1=j_2$, Eq. (27) is reduced to Eq. (26); So it is sufficient to prove Eq. (27). In Eq. (27), the part constrained by the antisymmetric tensor $\epsilon^{a\beta}$ is obviously the "spin zero" part, while the rest part, containing $[j-(j_2-j_1)]+[j+(j_2-j_1)]=2j$ "spin up" particles, must be the "spin $j$" part with $m=j$. It is not difficult to prove this claim: Using Eqs. (21) and (11), we see immediately that the right-hand side of Eq. (27) is annihilated by $J_+$,
while it is an eigenstate of $J_z$, belonging to the eigenvalue $j$,
On account of Eq. (23), we learn that the right-hand side of Eq. (27) is indeed the eigenstate of $J^2$, and the corresponding eigenvalue is $j(j+1)$.
On other hand, in Eq. (27), we have $2j_1$ $\xi$-type spinors and $2j_2$ $\eta$-type spinors, so Eq. (27) must be the simultaneous eigenstate of $J_1^2$ and $J_2^2$. To see this, let us expand the right-hand side of Eq. (27):
Using Eq. (10), the above expression can be converted into
In the second line, we have set $j_1-r=m_1$ and $m_2=j-j_1+r$; And
is the normalization constant, to be given by Eq. (37). Now it is manifest that the second line of Eq. (31) is the simultaneous eigenstate of $J_1^2$, $J_2^2$, and $J_z$, and the corresponding eigenvalues are $j_1(j_1+1)$, $j_2(j_2+1)$, and $m_1+m_2$, respectively.
The range of $j$ can be determined by noting that the powers of the spinors in Eq. (27) must be non-negative integers, i.e.,
which are nothing but
The constant $\tilde c$ can be determined (up to a phase factor) by normalizing the first line of Eq. (31),
which can be evaluated using the identity5(5 It can be proved by Taylor expanding the equation $(x+y)^{-n}(x+y)^{-m}=(x+y)^{-(n+m)}\,,$ where $m$ and $n$ are positive real numbers.)
where $n_i\geq0$ ($i=1,2,3$). A short computation gives
with $e^{i\gamma}$ the phase factor; We choose our phase convention by setting
Now $\tilde c$ is a real and non-negative number. As a result, the CG coefficients must also be real. By Eq. (32), we have
i.e., the normalization constant $c$ of Eq. (27) depends on the sign of $(-1)^{j_1+j_2-j}.$
Now the general state $\psi_{j_1j_2;jm}$ can be worked out in the standard way. Using the lower sign of the equations
we learn that
Using Eq. (14), we learn that
and $(J_{2-})^{j-m-r}\psi_{j_1j_2;m_1m_2}$ has a similar expression. Substituting Eqs. (31), (37), and (42) into Eq. (41), and relabeling the indices, then comparing the expression with Eq. (19), we obtain
which is in agreement with the result in Ref. [7].
Let us comment on our phase convention. With the choice of phase factor (38), our phase convention is identical with that of Ref. [7]. In Ref. [7], the phase factor of $\psi_{j_1j_2;jj}$ is fixed by requiring that
$\bullet$ $C^{j_1j_2}_{j_1j_2;(j_1+j_2)(j_1+j_2)}=1$ and the matrix element $(\psi_{j_1j_2;jm}, J_{1z}\psi_{j_1j_2;(j+1)m})$ is real and non-negative,
$\bullet$ or the non-vanishing CG coefficient $C^{j_1j_2}_{m_1=j_1,m_2;jj}$ is positive6(6 In Ref. [7], $C^{j_1j_2}_{m_1=j_1,m_2;jj}$ is written as $(j_1j_1j_2m_2|j_1j_2jj)$, and (44) reads $(j_1j_1j_2m_2|j_1j_2jj)={\rm arg}(j_1j_1j_2m_2|j_1j_2jj)|(j_1j_1j_2m_2|j_1j_2jj) | $ $=|(j_1j_1j_2m_2|j_1j_2jj)|$, i.e., the phase convention is ${\rm arg}(j_1j_1j_2m_2|j_1j_2jj)=1.$ (See (3.5.11) of Ref. [7]). (In Re. [7], the phase factor $e^{i\delta}$ in Eq. (44) is denoted as ${\rm arg}(j_1j_1j_2m_2|j_1j_2jj)$ to emphasize it may depend on the quantum numbers $j_1$, $j_2$, and $j$.)), i.e.,
The above two requirements are equivalent (see Sec. 3.4 and 3.5 of Ref. [7]). One can easily read off our $C^{j_1j_2}_{m_1=j_1,m_2;jj}$ from Eq. (31): $ C^{j_1j_2}_{m_1=j_1,(j-j_1);jj}=\tilde c\sqrt{(2j_1)!(j+j_2-j_1)!/(j_1+j_2-j)!}\,, $ which is obviously positive.
We see that using the antisymmetric tensor $\epsilon_{a\beta}$ properly, we have been able to construct $\psi_{j_1j_2;jj}$ in terms of $\psi_{j_1j_2;m_1m_2}$ with very little calculations (see Eqs. (27) and (31)), leading to a simple calculation of the explicit expression of CG coefficients (43). However, in the most conventional approach,[5] it is not that easy to construct $\psi_{j_1j_2;jj}$ when $j\neq j_1+j_2$: One has to work out $\psi_{j_1j_2;(j_1+j_2)(j_1+j_2)}$, $\psi_{j_1j_2;(j_1+j_2-1)(j_1+j_2-1)}$, …, and $\psi_{j_1j_2;|j_1-j_2||j_1-j_2|}$ step by step7(7 Alternatively, one has to work out $\psi_{j_1j_2;(j_1+j_2)(-j_1-j_2)}$, $\psi_{j_1j_2;(j_1+j_2-1)(-j_1-j_2+1)}$, …, and $\psi_{j_1j_2;|j_1-j_2|(-|j_1-j_2|)}$ step by step.).
In another conventional approach,[6] the strategy is to derive recursion relations for the CG coefficients. Using the recursion relations and the normalization conditions of CG coefficients, in principle one can work out all non-vanishing CG coefficients. However, the calculation is so involved that the textbook[6] says that "With enough patience we can obtain the Clebsch-Gordan coefficient of every site in terms of the coefficient of the starting site, A."
In summary, it is not that simple to use the conventional approaches to work out the general expression of the SU(2) CG coefficients (43). See Appendix A for a quick review for these two conventional methods of computing the CG coefficients.
4 Simple Physical Examples
In this section, we work out some simple physical examples using the method developed in Sec. 3. We begin by considering the familiar case of the coupling of two spin $1/2$ particles, such as two electrons. For $j_1=j_2=1/2$, according to Eq. (34), $j$ has two possible values: $1$ or $0$. If $j=0$, using Eq. (20) or Eq. (27), we obtain the spin singletimmediately. Here we have omitted the quantum numbers $j_1$ and $j_2$. By Eqs. (37)-(39), the normalization constant is $c=-1/\sqrt2$. Following the convention of Ref. [6], we will use "$\pm$" to denote "spin up" and "spin down", respectively. Defining $\psi_{1+}\equiv \xi_1$, $\psi_{1-}\equiv \xi_2$, $\psi_{2+}\equiv \eta_1$, and $\psi_{2-}\equiv \eta_2$, the singlet Eq. (45) takes the familiar form,
Also, using Eq. (27), we learn that the spin triplet with quantum numbers $j=m=1$ takes the following form (the normalization constant $c$ is obviously equal to $1$):
And the rest state vectors of spin triplet can be determined in the standard way Eq. (41); For instance,
Similarly,
It is a simple matter to read off the CG coefficients from the above equations.
We now turn to the case of $j_1=j_2=1$, i.e., the coupling of two identical spin $1$ particles. According to Eq. (34), $j=0,\ 1$, or $2$. If $j=0$, using Eqs. (27) and (10), or using Eq. (31) directly, we obtain
By Eqs. (37)-(39), the normalization constant is $\tilde c=2c=1/\sqrt 3$. Comparing the above equation with Eq. (19) determines the corresponding CG coefficients:
which are in agreement with the results in the textbook.[5] Similarly, by Eqs. (27) and (10), or Eq. (31), we have
The normalization constant can be determined by using Eqs. (37)-(39); It is $\tilde c=-c=1/ 2$ by Eqs. (37)-(39). And
where the normalization constant $c$ is $1/2$ by our convention defined by Eqs. (37)-(39). One can use the standard way (41) to work out $\psi_{11,(j=1)(m=0,-1)}$ and $\psi_{11,(j=2)(m=0,\pm1,-2)}$, and read off the CG coefficients. Since these calculations are quite standard and straightforward, we do not present them here.
Since the quantum numbers discussed in this section are small ($j_1=j_2=1/2$ or $j_1=j_2=1$), it is also not difficult to work out the coupling of (spin) state vectors of two identical particles by using the conventional approaches described in Appendix A. However, for arbitrary $j_1$ and $j_2$, the spinor approach for working out the general expression (43) of CG coefficients is more convenient and efficient.
5 Discussions
In summary, we have derived the expression of SU(2) CG coefficients coupling two independent angular momentum vector operators utilizing the spinor algebra, and presented two simple physical examples to illustrate the spinor approach. Clearly, one can generalize this spinor approach to work out the expressions for $6j$ and $9j$ coefficients coupling three and four angular momentum vector operators, respectively.Appendix A: Sketches of Conventional Calculations of SU(2) CG Coefficients
In this appendix, we briefly sketch the two conventional methods for calculating the SU(2) CG coefficients in Refs. [5] and [6], respectively. In Refs. [5-6], the overall phase factor for the CG coefficients is not specified; We shall fix the phase factor by adopting the convention of Ref. [7] (see (44)).In the most conventional approach to CG coefficients, say for example, in Ref. [5], the first step is to assume that both $j$ and $m$ take the maximum value $j=j_1+j_2=m$ and to construct the state vector8(8 For convenience, we still use $\psi$ to denote the wave functions, though they do not necessarily take exactly the same forms as that of (6) and (10).)
The only CG coefficient is
where we have fixed the phase factor by using the convention Eq. (44).
Evaluating the following equation
gives
Comparing the above equation with Eq. (19), one can read off the corresponding CG coefficients. Similarly, one can obtain $\psi_{j_1j_2;(j_1+j_2)m}$ by evaluating $(J_{-})^{(j_1+j_2)-m} \psi_{j_1j_2;(j_1+j_2)(j_1+j_2)}$, and then one can read of the CG coefficients $C^{j_1j_2}_{m_1m_2;(j_1+j_2)m}$.
The next step is to consider
The state vector $\psi_{j_1j_2;(j_1+j_2-1)(j_1+j_2-1)}$ must be orthogonal to $\psi_{j_1j_2;(j_1+j_2)(j_1+j_2-1)}$. (Recall that $J^2$ is a hermitian operator, so its two eigenstates corresponding to two different eigenvalues must be orthogonal.) Equation (A4) suggests that
Let us fix above overall phase factor $e^{i\delta}$ by using the convention (44). Comparing Eq. (A6) with Eq. (19), we can read off the CG coefficient
Comparing (A7) with (44) determines the phase factor $e^{i\delta}=1$. However, in Ref. [5], the phase factor in Eq. (59) is $e^{i\delta}=-1$. This is not a problem: According to chapter 4 of Ref. [5], the state vector Eq. (A6) is defined "apart from an arbitrary choice of a phase factor".
One then derives $\psi_{j_1j_2;(j_1+j_2-1)m}$ by calculating $(J_{-})^{(j_1+j_2-1)-m}\psi_{j_1j_2;(j_1+j_2-1)(j_1+j_2-1)}$, and read of the CG coefficients $C^{j_1j_2}_{m_1m_2;(j_1+j_2-1)m}$.
Similarly, one can construct
by noting that it must be orthogonal to both $\psi_{j_1j_2;(j_1+j_2-1)(j_1+j_2-2)}$ and $\psi_{j_1j_2;(j_1+j_2)(j_1+j_2-2)}$; And the overall phase factor can be fixed by using the convention (44). Then one can derive $\psi_{j_1j_2;(j_1+j_2-2)m}$ by using the lowering operator $J_-$ to act on Eq. (A8) $[(j_1+j_2-2)-m]$ times.
Continuing in this way until $j$ reaches the minimum value $j=|j_1-j_2|$, one can work out all CG coefficients in principle. However, the calculation for working out the general expression of SU(2) CG coefficients (43) is lengthy and complicated.
Another conventional way is based on the recursion relations for CG coefficients.[6] We now briefly outline this method. For simplicity, we will omit the quantum numbers $j_1j_2$; For instance, we will write $|j_1j_2;jm\rangle$ as $|jm\rangle$. The recursion relations can be derived by calculating the matrix elements
in two different ways. Using $(J_{1\pm}+J_{2\pm})=(J_{1\mp}^\dagger+J_{2\mp}^\dagger)$, a short computation gives the recursion relations:
In terms of our notation, the CG coefficients read
The textbook[6] derives Eq. (63) in a slightly different way. Similarly, by evaluating the matrix elements
we obtain conditions for non-vanishing CG coefficients
The recursion relations (A10) and the normalization conditions
determine all CG coefficients (up to an overall phase factor). To see this, we consider the lower signs of Eq. (A10). Note that if we set $m_1=j_1$, the first term of the right-hand side of Eq. (A10) vanishes, i.e.,
For convenience, we use "A" and "B" to stand for $\langle j_1(m_2+1)|jm\rangle$ and $\langle j_1m_2|j(m-1)\rangle$, respectively9(9 They are the CG coefficients in the sites "A" and "B" of Fig.3.9 of Ref. [6].). We see that "B" can be expressed in terms of "A". Here we treat "A" as a starting coefficient. Later we will see, the rest of CG coefficients can be expressed in terms of "A" as well. For instance, if we take the upper signs of Eq. (A10), and do the replacements:
Eq. (A10) becomes
Here the key point is that we have used "A" and "B" to generate a new CG coefficient "D". By Eqs. (A15) and (A17), we see that "D" can be also expressed in terms of "A".
$\bullet$ Similarly, we can use "B" and "D" to generate a new CG coefficient "E", and express "E" in terms of "A";
$\bullet$ And then it is possible to use "B" and "E" to generate a new CG coefficient "C", and express "C" in terms of "A".
Continuing in this way, with "enough patience", as the textbook[6] pointed out, we can express the rest of CG coefficients in terms of the CG coefficient "A". And the CG coefficient "A" can be determined (up to an overall phase factor) by the normalization conditions (A14). However, the calculation is quite involved, as the textbook[6] noticed.
We shall fix the overall phase factor by adopting the convention (44). In Eq. (A15), if we set $m=j$, and replace $m_2+1$ by $m_2$, then CG coefficient "A" reads $ \langle j_1m_2|jj\rangle$. According to Eq. (44), it satisfies
namely, the above overall phase factor is $e^{i\delta}=1$.
For more detailed discussions and calculations of the CG coefficients using the approach of recursion relations, see Refs. [6] and [7].
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