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Comprehensive Study Notes

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Comprehensive Study Notes for the full course

The spin angular-momentum operators obey the general angular-momentum commutation relations of Section 5.4, and it is often helpful to use spin-angular-momentum ladder operators.

From (5.110) and (5.111), the raising and lowering operators for spin angular momentum are

\(
\begin{equation}
\hat{S}{+}=\hat{S}{x}+i \hat{S}{y} \text { and } \hat{S}{-}=\hat{S}{x}-i \hat{S}{y} \tag{10.63}
\end{equation}
\)

Equations (5.112) and (5.113) give

\(
\begin{align}
& \hat{S}{+} \hat{S}{-}=\hat{S}^{2}-\hat{S}{z}^{2}+\hbar \hat{S}{z} \tag{10.64}\
& \hat{S}{-} \hat{S}{+}=\hat{S}^{2}-\hat{S}{z}^{2}-\hbar \hat{S}{z} \tag{10.65}
\end{align}
\)

The spin functions $\alpha$ and $\beta$ are eigenfunctions of $\hat{S}{z}$ with eigenvalues $+\frac{1}{2} \hbar$ and $-\frac{1}{2} \hbar$, respectively. Since $\hat{S}{+}$is the raising operator, the function $\hat{S}{+} \beta$ is an eigenfunction of $\hat{S}{z}$ with eigenvalue $+\frac{1}{2} \hbar$. The most general eigenfunction of $\hat{S}_{z}$ with this eigenvalue is an arbitrary constant times $\alpha$. Hence

\(
\begin{equation}
\hat{S}_{+} \beta=c \alpha \tag{10.66}
\end{equation}
\)

where $c$ is some constant. To find $c$, we use normalization [Eq. (10.11)]:

\(
\begin{gather}
1=\sum{m{s}}\left[\alpha\left(m_{s}\right)\right] \alpha\left(m{s}\right)=\sum\left(\hat{S}{+} \beta / c\right) \left(\hat{S}{+} \beta / c\right) \
|c|^{2}=\sum\left(\hat{S}{+} \beta\right) \hat{S}{+} \beta=\sum\left(\hat{S}{+} \beta\right) \left(\hat{S}{x}+i \hat{S}{y}\right) \beta \
|c|^{2}=\sum\left(\hat{S}_{+} \beta\right) \hat{S}{x} \beta+i \sum\left(\hat{S}{+} \beta\right) \hat{S}_{y} \beta \tag{10.67}
\end{gather}
\)

We now use the Hermitian property of $\hat{S}{x}$ and $\hat{S}{y}$. For an operator $\hat{A}$ that acts on functions of the continuous variable $x$, the Hermitian property is

\(
\int{-\infty}^{\infty} f^{*}(x) \hat{A} g(x) d x=\int{-\infty}^{\infty} g(x)[\hat{A} f(x)] * d x
\)

For an operator such as $\hat{S}{x}$ that acts on functions of the variable $m{s}$, which takes on discrete values, the Hermitian property is

\(
\begin{equation}
\sum{m{s}} f^{}\left(m{s}\right) \hat{S}{x} g\left(m{s}\right)=\sum{m{s}} g\left(m{s}\right)\left[\hat{S}{x} f\left(m{s}\right)\right]^{} \tag{10.68}
\end{equation}
\)

Taking $f=\hat{S}_{+} \beta$ and $g=\beta$, we can write (10.67) as

\(
c^{} c=\sum \beta\left(\hat{S}{x} \hat{S}{+} \beta\right)^{}+i \sum \beta\left(\hat{S}{y} \hat{S}{+} \beta\right)^{*}
\)

Taking the complex conjugate of this equation and using (10.63) and (10.65), we have

\(
\begin{aligned}
& c c^{}=\sum \beta^{} \hat{S}{x} \hat{S}{+} \beta-i \sum \beta^{} \hat{S}{y} \hat{S}{+} \beta \
&|c|^{2}=\sum \beta^{}\left(\hat{S}{x}-i \hat{S}{y}\right) \hat{S}{+} \beta=\sum \beta^{*} \hat{S}{-} \hat{S}{+} \beta \
&|c|^{2}=\sum \beta^{*}\left(\hat{S}^{2}-\hat{S}{z}^{2}-\hbar \hat{S}_{z}\right) \beta \
&|c|^{2}=\sum \beta^{}\left(\frac{3}{4} \hbar^{2}-\frac{1}{4} \hbar^{2}+\frac{1}{2} \hbar^{2}\right) \beta=\hbar^{2} \sum \beta^{} \beta=\hbar^{2} \
& \quad|c|=\hbar
\end{aligned}
\)

Choosing the phase of $c$ as zero, we have $c=\hbar$, and (10.66) reads

\(
\begin{equation}
\hat{S}_{+} \beta=\hbar \alpha \tag{10.69}
\end{equation}
\)

A similar calculation gives

\(
\begin{equation}
\hat{S}_{-} \alpha=\hbar \beta \tag{10.70}
\end{equation}
\)

Since $\alpha$ is the eigenfunction with the highest possible value of $m{s}$, the operator $\hat{S}{+}$acting on $\alpha$ must annihilate it [Eq. (5.135)]:

\(
\hat{S}_{+} \alpha=0
\)

Likewise,

\(
\hat{S}_{-} \beta=0
\)

From these last four equations, we get

\(
\begin{equation}
\left(\hat{S}{+}+\hat{S}{-}\right) \beta=\hbar \alpha, \quad\left(\hat{S}{+}-\hat{S}{-}\right) \beta=\hbar \alpha \tag{10.71}
\end{equation}
\)

Use of (10.63) in (10.71) gives

\(
\begin{equation}
\hat{S}{x} \beta=\frac{1}{2} \hbar \alpha, \quad \hat{S}{y} \beta=-\frac{1}{2} i \hbar \alpha \tag{10.72}
\end{equation}
\)

Similarly, we find

\(
\begin{equation}
\hat{S}{x} \alpha=\frac{1}{2} \hbar \beta, \quad \hat{S}{y} \alpha=\frac{1}{2} i \hbar \beta \tag{10.73}
\end{equation}
\)

Matrix representatives of the spin operators are considered in Prob. 10.28.


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