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 1st vector space and 2M_2+1 basis vectors in the 2nd 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}

or

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 2nd term on RHS of 2nd 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 3rd term on RHS of 2nd 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 0M_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.

 

 

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