Composite systems – beyond spin operators

In the previous article, we showed that the total $z$ -component spin angular momentum operator $\hat{S}_z^{\; (T)}$ is $\hat{S}_z^{\; (T)}=\hat{S}_z^{\; (1)}\otimes I+I\otimes\hat{S}_z^{\; (2)}$ , which is a special case of the general form:

$\hat{J}_i=\hat{M}_i^{\; (1)}\otimes I+I\otimes\hat{M}_i^{\; (2)}\; \; \; \; \; \; \; \; 205$

$\hat{J}_i$ (where $i=x,y,z$ ) is the total angular momentum component operator. $\hat{M}_i^{\; (1)}$ and $\hat{M}_i^{\; (2)}$ are component operators of $\hat{M}^{ (1)}$ and $\hat{M}^{ (2)}$ respectively. $\hat{M}^{ (1)}$ and $\hat{M}^{ (2)}$ are operators of two sources of angular momentum, where they may be: 1) the orbital angular momentum operator of particle 1 and orbital angular momentum operator of particle 2 respectively; 2) the spin angular momentum operator of particle 1 and spin angular momentum operator of particle 2 respectively; 3) the orbital angular momentum operator and spin angular momentum operator respectively of a particle.

Also mentioned in the previous article is that $L_z=\sum_{i=1}^{n}l_{z,i}$ and $S_z=\sum_{i=1}^{n}s_{z,i}$ . It is therefore easy to accept the validity of points 1) and 2). For point 3, the proposal that $j_z=l_z+s_z$ may seem untenable. However, spin angular momentum, like orbital angular momentum, is a form of angular momentum. In fact, the total angular momentum $\boldsymbol{\mathit{J}}$ of a system is defined as the vector sum $\boldsymbol{\mathit{J}}=\boldsymbol{\mathit{L}}+\boldsymbol{\mathit{S}}$ . If point 3) is valid, $\hat{J}_i$ must satisfy the same commutation relations as described by eq99, eq100 and eq101.

Question

Show that $\hat{J}_i$ satisfies the same commutation relations as described by eq99, eq100 and eq101.

Answer

$\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I+I\otimes\hat{M}_x^{\; (2)},\hat{M}_y^{\; (1)}\otimes I+I\otimes\hat{M}_y^{\; (2)}\right ]$

Expanding the RHS of the above equation and noting that $\hat{M}_i^{\; (1)}$ and $\hat{M}_i^{\; (2)}$ commute because they act on different vector spaces, we have

$\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)}\otimes I,\hat{M}_y^{\; (1)}\otimes I\right ] +\left [I\otimes\hat{M}_x^{\; (2)},I\otimes\hat{M}_y^{\; (2)}\right ]$

$\left [ \hat{J}_x,\hat{J}_y \right ]=\left [\hat{M}_x^{\; (1)} ,\hat{M}_y^{\; (1)}\right ]\otimes I +I\otimes\left [\hat{M}_x^{\; (2)},\hat{M}_y^{\; (2)}\right ]$

With reference to eq99, eq100, eq101, eq165, eq166 and eq167, $\left [\hat{M}_i^{\; (l)} ,\hat{M}_j^{\; (l)}\right ]=i\hbar\epsilon_{ijk}\hat{M}_k^{\; l}$ , where $l=1,2$ and $\epsilon_{ijk}$ is the Levi-Civita symbol. So,

$\left [ \hat{J}_x,\hat{J}_y \right ]=i\hbar\hat{M}_z^{\; (1)} \otimes I +i\hbar I\otimes\hat{M}_z^{\; (2)}=i\hbar\hat{J}_z$

Similarly, we have $\left [ \hat{J}_y,\hat{J}_z \right ]=i\hbar\hat{J}_x$ and $\left [ \hat{J}_z,\hat{J}_x \right ]=i\hbar\hat{J}_y$ .

Since the total angular momentum component operators satisfy the form of commutation relations as described by eq99, eq100 and eq101, the raising and lowering operators also apply to the total angular momentum operator $\hat{J}$ . We would therefore expect

$\hat{J}^2\psi=j(j+1)\hbar^2\psi\; \; \; \; and\; \; \; \; \; \hat{J}_z\psi=m_j\hbar\psi\; \; \; \; \; \; \; \; \; 205a$

You’ll realise from the workings of the above Q&A that we can simplify the notation $\hat{J}_z$ of eq205 as

$\hat{J}_z=\hat{M}_{z,1}+\hat{M}_{z,2}\; \; \; \; \; \; \; \; 206$

To show that $\hat{J}_z$ commutes with $\hat{M}_{z,i}$ , where $\hat{M}_{z,i}$ can be either $\hat{l}_{z,i}$ or $\hat{s}_{z,i}$ , we have $\left [ \hat{J}_z,\hat{l}_z \right ]=\left [ \hat{l}_z,\hat{l}_z \right ]+\left [ \hat{s}_z,\hat{l}_z \right ]=0$ and $\left [ \hat{J}_z,\hat{s}_z \right ]=\left [ \hat{l}_z,\hat{s}_z \right ]+\left [ \hat{s}_z,\hat{s}_z \right ]=0$ . Therefore, the eigenstate of $\hat{J}_z$ is simultaneously the eigenstates of $\hat{M}_{z,1}$ and $\hat{M}_{z,2}$ . This implies that the eigenvalues of $\hat{J}_z$ are the sum of the eigenvalues of $\hat{M}_{z,1}$ and $\hat{M}_{z,2}$ , i.e. $m_j\hbar=m_1\hbar+m_2\hbar$ , or

$m_j=m_1+m_2\; \; \; \; \; \; \; \; 207$

In other words, the allowed values of the total magnetic quantum number $m_j$ are the sum of the allowed values of the two contributing magnetic quantum numbers. As for the allowed values of the total angular momentum quantum number $j$ , let’s further define the eigenvalues of $\hat{M}_1^{\; 2}$ and $\hat{M}_2^{\; 2}$ as $M_1(M_1+1)\hbar^{2}$ and $M_2(M_2+1)\hbar^{2}$ respectively. This allows us to work with the quantum numbers $M_1$ and $M_2$ .

Now, the maximum value of $m_j$ in eq207 is $m_{j,max}=m_{1,max}+m_{2,max}$ . Since the maximum value of a magnetic quantum number is the angular momentum quantum number (i.e. $m_{j,max}=j_{max}$ and $m_{i,max}=M_{i}$ , where $i=1,2$ ), the highest value of $j$ is

$j_{max}=M_1+M_2\; \; \; \; \; \; \; \; 208$

Furthermore, for a particular value of $\boldsymbol{\mathit{j}}$ in the coupled representation, there are $2j+1$ values of $m_j$ and therefore $2j+1$ states. So $j_{max}$ has $2j_{max}+1=2M_1+2M_2+1$ states. These states are $\vert j_{max},m_{j,max}\rangle,\vert j_{max},m_{j,max}-1\rangle,\vert j_{max},m_{j,max}-2\rangle,\cdots$ . The states for the next lower value of $j$ (denoted by $j'$ ) are $\vert j',m'_{j,max}\rangle,\vert j',m'_{j,max}-1\rangle,\vert j',m'_{j,max}-2\rangle,\cdots$ . The same logic applies for states all the way to the lowest value of $j$ .

Question

Show that the total number of states in the uncoupled representation is $(2M_1+1)(2M_2+1)$ .

Answer

In eq193, the total number of states in the uncoupled representation $\vert M_1,m_1;M_2,m_2\rangle$ is the number of ways to form Kronecker products of basis vectors from each vector space. Since there are $2M_1+1$ basis vectors in the 1^st vector space and $2M_2+1$ basis vectors in the 2^nd vector space,

$total\: states\: in\: uncoupled\: representation=(2M_1+1)(2M_2+1)\; \; \; \; \; \; \; \; 209$

To determine the lower values of $j$ , we consider the lower values of $m_{j,max}$ , the first being $m_{j,max}-1=m'_{j,max}$ . There are two possible ways to obtain this value, with $m_{1,max}-1$ and $m_{2,max}$ , or $m_{1,max}$ and $m_{2,max}-1$ . Since each state is characterised by a unique value of $m_j$ for a particular value of $j$ , one of the two possibilities is accounted for by the state $\vert j_{max},m_{j,max}-1\rangle$ . The remaining possibility must be due to $\vert j',m'_{j,max}\rangle$ . Since $m_{j,max}=j_{max}$ , we must have $m'_{j,max}=j'$ . Furthermore, because $m_{j,max}-1=m'_{j,max}$ , we have $j'=m_{j,max}-1=j_{max}-1$ . The state $\vert j',m'_{j,max}\rangle$ is therefore $\vert j_{max}-1,m_{j,max}-1\rangle$ .

For $m_{j,max}-2$ , there are three possible ways to obtain it. Again, one of the possible ways is accounted for by $\vert j_{max},m_{j,max}-2\rangle$ and the second way by $\vert j',m'_{j,max}-1\rangle=\vert j_{max}-1,m_{j,max}-2\rangle$ . The remaining possibility must be due to the state $\vert j'',m''_{j,max}\rangle=\vert j'-1,m'_{j,max}-1\rangle=\vert j_{max}-2,m_{j,max}-2\rangle$ .

Therefore, the allowed values of are

$j=j_{max},j_{max}-1,j_{max}-2,\cdots,j_{min}$

$j=M_1+M_2,M_1+M_2-1,M_1+M_2-2,\cdots,j_{min}$

To determine $j_{min}$ , we note that the total number of states for the system can be written as $\sum_{j=j_{min}}^{j_{max}}(2j+1)$ because there are $2j+1$ states associated with each value of $j$ . Since $j_{min}\geq 0$ , we can further split the sum as:

$total\: states=\sum_{j=0}^{j_{max}}(2j+1)-\sum_{j=0}^{j_{min}-1}(2j+1)\; \; \; \; \; \; \; \; 210$

Question

Show that $\sum_{j=0}^{n-1}(2j+1)=n^{2}$ and hence $\sum_{j=0}^{x}(2j+1)=(x+1)^{2}$ .

Answer

$\sum_{j=0}^{n-1}(2j+1)=1+\sum_{j=1}^{n-1}(2j+1)=1+2\sum_{j=1}^{n-1}j+\sum_{j=1}^{n-1}1\; \; \; \; \; \; \; \; 211$

For the 2^nd term on RHS of 2^nd equality of eq211, $\sum_{j=1}^{n-1}j=1+2+\cdots+(n-2)+(n-1)$ , which if written in the reverse order becomes $\sum_{j=1}^{n-1}j=(n-1)+(n-2)+\cdots+2+1$ . Adding the two sums, we have

$2\sum_{j=1}^{n-1}j=n(n-1)\; \; \; \; \; \; \; \; 214$

For the 3^rd term on RHS of 2^nd equality of in eq211

$\sum_{j=1}^{n-1}1=\sum_{j=1}^{n-1}j^{0}=1+1+\cdots+1=n-1\; \; \; \; \; \; \; \; 215$

Substitute eq214 and eq215 back in eq211, we have $\sum_{j=0}^{n-1}(2j+1)=n^{2}$ . Let $x=n-1$ , we have

$\sum_{j=0}^{n-1}(2j+1)=(x+1)^{2}\; \; \; \; \; \; \; \; 216$

Using eq216, where $x=j_{max}$ for $\sum_{j=0}^{j_{max}}(2j+1)$ , and $x=j_{min}-1$ for $\sum_{j=0}^{j_{min}-1}(2j+1)$ , eq210 becomes

$total\: states=(j_{max}+1)^{2}-j_{min}^{\; \;\; \; \; \; 2}=j_{max}^{\; \;\; \; \; \; 2}+2j_{max}+1-j_{min}^{\; \;\; \; \; \; 2}\; \; \; \; \; \; \; \; 217$

The total number of states (energy levels) of a system must be independent of the chosen representation. Substituting eq209 in LHS of eq217 and eq208 in RHS of eq217 and simplifying,

$j_{min}=\pm(M_1-M_2)\; \; \; \; \; \; \; \; 218$

Eq218 is equivalent to $j_{min}=\vert M_1-M_2\vert$ because $j_{min}\geq 0$ , $M_1\geq 0$ , $M_2\geq 0$ , and $M_2$ may be a larger value than $M_1$ . Therefore, for a given value of $M_1$ and a given value of $M_2$ , the allowed values of the total angular momentum quantum number $j$ are:

$j=M_1+M_2,M_1+M_2-1,\cdots,\vert M_1-M_2\vert\; \; \; \; \; \; \; \; 219$

which is called the Clebsch-Gordan series.

Question

Write all the eigenstates (in the form of $\vert j,m_j\rangle$ ) and basis states (in the form of $\vert m_1,m_2\rangle$ ) of a system with two sources of angular momentum, $M_1=2$ and $M_2=1$ .

Answer

There are a total of 15 eigenstates and also 15 basis states. The allowed values of $j$ are 3, 2 and 1. The eigenstates are $\vert 3,3\rangle$ , $\vert 3,2\rangle$ , $\vert 3,1\rangle$ , $\vert 3,0\rangle$ , $\vert 3,-1\rangle$ , $\vert 3,-2\rangle$ , $\vert 3,-3\rangle$ , $\vert 2,2\rangle$ , $\vert 2,1\rangle$ , $\vert 2,0\rangle$ , $\vert 2,-1\rangle$ , $\vert 2,-2\rangle$ , $\vert 1,1\rangle$ , $\vert 1,0\rangle$ and $\vert 1,-1\rangle$ . The basis states are $\vert 2,1\rangle$ , $\vert 1,1\rangle$ , $\vert 0,1\rangle$ , $\vert -1,1\rangle$ , $\vert -2,1\rangle$ , $\vert 2,0\rangle$ , $\vert 2,-1\rangle$ , $\vert 1,0\rangle$ , $\vert 1,-1\rangle$ , $\vert 0,0\rangle$ , $\vert 0,-1\rangle$ , $\vert -1,0\rangle$ , $\vert -1,-1\rangle$ , $\vert -2,0\rangle$ and $\vert -2,-1\rangle$ . Each spin eigenstate of the system is a linear combination of the 15 basis states.

What we have described so far pertains to a system with two sources of angular momentum. If the system has more than two sources of angular momentum, the Clebsch-Gordan series is applied repeatedly, i.e. a first series is written with $M_1$ and $M_2$ , and then the Clebsch-Gordan procedure is again applied to each value of this series with $M_3$ to form a second resultant series, and the procedure is repeated until a final resultant series is developed with $M_{final}$ . For example, a system with three sources of angular momentum, $s_1=\frac{1}{2}$ , $s_2=\frac{1}{2}$ and $s_3=\frac{1}{2}$ , has the following allowed values of $S$ :

1st series using $s_1$ and $s_2$ , $S=1,0$

2nd and final series using $1,0$ and $s_3$ , $S=\frac{3}{2},\frac{1}{2},\frac{1}{2}$

For this system, there are 8 basis states, whose explicit forms can be expressed as follows:

$\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 1\\0 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}$

$\small \begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 1\\0 \end{pmatrix} \; \; \; \;\begin{pmatrix} 1\\0 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix} \; \; \; \;\begin{pmatrix} 0\\1 \end{pmatrix}\otimes\begin{pmatrix} 0\\1 \end{pmatrix} \otimes\begin{pmatrix} 0\\1 \end{pmatrix}$

Question

What are the allowed angular momenta of a system with three sources of angular momentum, $\small l_1=1$ , $\small l_2=1$ and $\small l_3=1$ , and how many basis states are there in total?

Answer

$\small j=3,2,2,1,1,1,0$ . The total number of basis states is 27.

Composite systems – beyond spin operators

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