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: The general problem : lecture_8 : lecture_8

A simple example of angular momentum addition

Given two spin-1/2 angular momenta, ${\bf S}_1$ and ${\bf S}_2$, we define

\begin{displaymath}
{\bf S}= {\bf S}_1 + {\bf S}_2
\end{displaymath}

The problem is to find the eigenstates of the total total spin operators $S^2$ and $S_z$ and identify the allowed total spin states.

In order to solve this problem, we recognize that the eigenvectors we seek will be expressible in terms of tensor products of the spin eigenstates of the individual spins. There will be four such vectors:

$\displaystyle {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle {\left\vert{1 \over 2}\;\;\;{1 \over 2}\right>}\bigotimes {\left\vert{1 \over 2}\;\;\;{1 \over 2}\right>}$  
$\displaystyle {\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle {\left\vert{1 \over 2}\;\;\;-{1 \over 2}\right>}\bigotimes {\left\vert{1 \over 2}\;\;\;{1 \over 2}\right>}$  
$\displaystyle {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle {\left\vert{1 \over 2}\;\;\;{1 \over 2}\right>}\bigotimes {\left\vert{1 \over 2}\;\;\;-{1 \over 2}\right>}$  
$\displaystyle {\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle {\left\vert{1 \over 2}\;\;\;-{1 \over 2}\right>}\bigotimes {\left\vert{1 \over 2}\;\;\;-{1 \over 2}\right>}$  

Although this is a valid set of basis vectors for the full spin Hilbert space, these are not eigenvectors of $S^2$ or $S_z$. Thus, we seek a unitary transformation among these vectors that generates a new set of four vectors that are eigenvectors of $S^2$ and $S_z$.

The four new vectors will correspond to adding the spins in either a parallel or anti-parallel fashion:

$\displaystyle {\rm Parallel}\;\;\;\;\;$ $\textstyle \uparrow\uparrow$ $\displaystyle \;\;\;\;\;\;\;\;
{\rm Total\;Spin}=1,\;\;\;\;\;\;\;\;m_s=-1,0,1\;\;\; = \;\;\;3\;{\rm states}$  
$\displaystyle {\rm Antiparallel}\;\;\;\;\;$ $\textstyle \uparrow\downarrow$ $\displaystyle \;\;\;\;\;\;\;\;
{\rm Total\;Spin}=0,\;\;\;\;\;\;\;\;m_s=0\;\;\;\;\;\;\;\;\;\;\;\;\; = \;\;\;1\;{\rm state}$  

for a total of four states, as expected.

To find these states, begin by noting that the state

\begin{displaymath}
{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}
\end{displaymath}

is, in fact, an eigenstate of $S^2$ and $S_z$. To see this note that
$\displaystyle S_z{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle (S_{1z}+S_{2z}){\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle {\hbar \over 2}{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2...
... 2}{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle \hbar{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  

Also, since
$\displaystyle S^2$ $\textstyle =$ $\displaystyle ({\bf S}_1 + {\bf S}_2)^2$  
  $\textstyle =$ $\displaystyle S_1^2 + S_2^2 + 2{\bf S}_1\cdot{\bf S}_2$  
  $\textstyle =$ $\displaystyle S_1^2 + S_2^2 + 2(S_{1x} S_{2x} + S_{1y}S_{2y} + S_{1z}S_{2z})$  
  $\textstyle =$ $\displaystyle S_1^2 + S_2^2 + 2S_{1z}S_{2z} + \left(S_{1+}S_{2-} + S_{1-}S_{2+}\right)$  

The action of $S^2$ on ${\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ is

\begin{displaymath}
S^2{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;...
...\over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}
\end{displaymath}

However, the second term on the right vanishes because each involves a raising operator acting on a state ${\left\vert{1 \over 2}\;\;\;{1 \over 2}\right>}$ for either the first or second spin, which annihilates the state. Thus,
$\displaystyle S^2{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ $\textstyle =$ $\displaystyle \left[S_1^2 + S_2^2 + 2S_{1z}S_{2z}\right]{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle \left[{1 \over 2}\left({1 \over 2}+1\right)\hbar^2 +
{1 \over 2}\...
...ht]{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle \hbar^2\left({3 \over 4}+{3 \over 4} + {1 \over 2}\right){\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle 2\hbar^2{\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  

Thus, the eigenvalue of $S^2$ is $2\hbar^2 = 1(1+1)\hbar^2$, corresponding to a value of $s=1$ with a corresponding $S_z$ value of $\hbar$. These facts make it clear that ${\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$ is a total spin-1 state $\vert s\;\;\;m_s\rangle = \vert 1\;\;1\rangle$ of the total spin:

\begin{displaymath}
\vert 1\;\;1\rangle = {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}
\end{displaymath}

In order to find the the states corresponding to $m_s = 0$ and $m_s=-1$, we can use the total lowering operator:

\begin{displaymath}
S_- = S_{1-} + S_{2-}
\end{displaymath}

Recall the general relation for the action of raising and lowering operators on general spin states:

\begin{displaymath}
S_{\pm} \vert s\;\;m_s\rangle =
\hbar\sqrt{s(s+1)-m_s(m_s\pm 1)}\vert s\;\;m_s\pm 1\rangle
\end{displaymath}

The procedure is, then, to act on both sides of

\begin{displaymath}
\vert 1\;\;1\rangle = {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}
\end{displaymath}

with $S_-$:

\begin{displaymath}
S_-\vert 1\;\;1\rangle = S_-{\left\vert{1 \over 2}\;\;\;{1 \...
...\over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}
\end{displaymath}

On the left, we obtain $\sqrt{2}\hbar \vert 1\;\;0\rangle$ so that
$\displaystyle \sqrt{2}\hbar\vert 1\;\;0\rangle$ $\textstyle =$ $\displaystyle (S_{1-} + S_{2-}){\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}$  
  $\textstyle =$ $\displaystyle \hbar{\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1...
...r {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$  
$\displaystyle \vert 1\;\;0\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}
\left[{\left\vert{1 \over 2}\;\;\;-{1 \over 2}...
...t\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}\right]$  

Finally, the state with $m_s=-1$ is obtained by acting again with the lowering operator:
$\displaystyle S_- \vert 1\;\;0\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}S_-
\left[{\left\vert{1 \over 2}\;\;\;-{1 \over...
...t\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}\right]$  
$\displaystyle \sqrt{2}\hbar \vert 1\;\;-1\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}\left[S_{1-}{\left\vert{1 \over 2}\;\;\;{1 \ove...
...t\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}\right]$  
$\displaystyle 2\hbar\vert 1\;\;-1\rangle$ $\textstyle =$ $\displaystyle \hbar{\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{...
...r{\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$  
$\displaystyle \vert 1\;\;-1\rangle$ $\textstyle =$ $\displaystyle {\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$  

The last state is $\vert\;\;0\rangle$ and is obtained by recognizing that it must be composed of those states that have opposite values of $m_{s_1}$ and $m_{s_2}$. So we include these states with arbitrary coefficients:

\begin{displaymath}
\vert\;\;0\rangle = \alpha {\left\vert{1 \over 2}\;\;\;-{1 \...
...over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}
\end{displaymath}

The coefficients $\alpha$ and $\beta$ are then determined by the conditions that $\vert\;\;0\rangle$ must be normalized and that it must be orthogonal to the other three states. The first condition yields:

\begin{displaymath}
\langle 0\;\;0\vert\;\;0\rangle = \alpha^2 + \beta^2 = 1
\end{displaymath}

In order to fulfill the second, we recognize that $\vert\;\;0\rangle$ is manifestly orthogonal to $\vert 1\;\;1\rangle$ and $1\;\;-1\rangle$. However, orthogonality to $\vert 1\;\;0\rangle$ needs to be enforced:
$\displaystyle \langle 1\;\;0\vert\;\;0\rangle$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle {1 \over \sqrt{2}}\left({\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \...
...t<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}\right)$   $\displaystyle \left(\alpha{\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\...
...rt{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}\right) = 0$  
$\displaystyle {1 \over \sqrt{2}}(\alpha + \beta)$ $\textstyle =$ $\displaystyle 0$  
$\displaystyle \Rightarrow\;\;\;\;\;\alpha$ $\textstyle =$ $\displaystyle -\beta$  

From the normalization condition, it is clear, then, that $\alpha = \pm 1/\sqrt{2}$ so that $\beta = \mp 1/\sqrt{2}$. The choice of sign is arbitrary, so we choose $\alpha = -1/\sqrt{2}$ and $\beta = 1/\sqrt{2}$ so that

\begin{displaymath}
\vert\;\;0\rangle = {1 \over \sqrt{2}}\left({\left\vert{1 \o...
...\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}\right)
\end{displaymath}

Thus, the four new basis vectors are:

$\displaystyle {\left\vert{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>},\;\;\;\;\;\;$   $\displaystyle {1 \over \sqrt{2}}\left({\left\vert{1 \over 2}\;\;\;{1 \over 2};{...
...
{\left\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}$  
    $\displaystyle {1 \over \sqrt{2}}\left({\left\vert{1 \over 2}\;\;\;{1 \over 2};{...
...t\vert{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right>}\right)$  

We can write the transformation from the old basis to the new basis as a matrix equation:

\begin{displaymath}
\left(\matrix{\vert 1\;\;1\rangle \cr \cr \vert 1\;\;0\rangl...
...\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right>}}\right)
\end{displaymath}

The matrix

\begin{displaymath}
{\rm U} =
\left(\matrix{1 & 0 & 0 & 0 \cr\cr
0 & {1 \over ...
...r\cr
0 & {1 \over \sqrt{2}} & -{1 \over \sqrt{2}} & 0}\right)
\end{displaymath}

can easily be shown to be unitary. Note that the elements of the matrix can be computed from the following overlaps:
$\displaystyle {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}1\;\;1\rangle$ $\textstyle =$ $\displaystyle 1$  
$\displaystyle {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}1\;\;0\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}1\;\;0\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}0\;\;0\rangle$ $\textstyle =$ $\displaystyle {1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;{1 \over 2}\right\vert}0\;\;0\rangle$ $\textstyle =$ $\displaystyle -{1 \over \sqrt{2}}$  
$\displaystyle {\left<{1 \over 2}\;\;\;-{1 \over 2};{1 \over 2}\;\;\;-{1 \over 2}\right\vert}1\;\;-1\rangle$ $\textstyle =$ $\displaystyle 1$  

These are examples of what are known as Clebsch-Gordan coefficients. In the next section, we will study the general forms of these coefficients.

Finally, note that the new basis vectors are, in fact, eigenvectors of $S^2$ and $S_z$. Thus, they satisfy

$\displaystyle S^2\vert 1\;\;1\rangle$ $\textstyle =$ $\displaystyle 2\hbar^2\vert 1\;\;1\rangle \;\;\;\;\;\;\;\;\;\;
S_z\vert 1\;\;1\rangle = \hbar\vert 1\;\;1\rangle$  
$\displaystyle S^2\vert 1\;\;0\rangle$ $\textstyle =$ $\displaystyle 2\hbar^2\vert 1\;\;0\rangle \;\;\;\;\;\;\;\;\;\;
S_z\vert 1\;\;0\rangle = 0$  
$\displaystyle S^2\vert 1\;\;-1\rangle$ $\textstyle =$ $\displaystyle 2\hbar^2\vert 1\;\;-1\rangle \;\;\;\;\;\;
S_z\vert 1\;\;-1\rangle = -\hbar\vert 1\;\;-1\rangle$  
$\displaystyle S^2\vert\;\;0\rangle$ $\textstyle =$ $\displaystyle 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
S_z\vert\;\;0\rangle = 0$  

The states $\vert 1\;\;1\rangle$, $\vert 1\;\;0\rangle$, $\vert 1\;\;-1\rangle$ form what are called the triplet states, corresponding to $s=1$. Note that these are symmetric with respect to exchange of the two spins. The state $\vert\;\;0\rangle$ is known as the singlet state as it is antisymmetric (changes sign) upon exchange of the spins.


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: The general problem : lecture_8 : lecture_8
Mark Tuckerman 平成17年2月18日