Math Infinitum

Mapping the lesson.

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Intuition

The ladder argument decides the whole spectrum from the commutators alone. $\hat{J}^{2}$ can only take the values $\hbar^{2}j(j+1)$ , and for each $j$ the component $\hat{J}_{z}$ takes the $2j+1$ values $m\hbar$ with $m$ running from $-j$ to $j$ in steps of one. Since the top is reached from the bottom in whole steps, $2j$ is a whole number: $j$ is a whole number or a half-integer. The half-integers are not a curiosity — they are the spin of the electron.

A vector of fixed length that may only point at a few heights above the floor looks like a cone of possible directions around the vertical, with rungs where its tip may sit. That cone is the usual picture of quantum angular momentum.

The five states of $j=2$ in the usual picture: a vector of length $\sqrt{j(j+1)}=\sqrt6\approx2.45$ in units of , the vertical axis being , whose height can only be . It never points straight up: even for the length exceeds the height, because and are never both zero.

The eigenvalues of \hat{J}^{2} and of its z component

The simultaneous eigenstates of $\hat{J}^{2}$ and $\hat{J}_{z}$ are labelled by and . The commutation relations alone fix the possible values.

\begin{aligned}\hat{J}^{2}\ket{j,m}&=\hbar^{2}j(j+1)\ket{j,m}\\ \hat{J}_{z}\ket{j,m}&=\hbar m\ket{j,m}\\ j&=0,\tfrac12,1,\tfrac32,\dots\\ m&=-j,-j+1,\dots,j\end{aligned}

Properties

For each $j$ there are $2j+1$ values of $m$ : 1 for $j=0$ , 2 for $j =$ , 3 for , 4 for .

The spectrum of angular momentum

\begin{aligned}&\hat{J}^{2}=\hbar^{2}j(j+1),\quad \hat{J}_{z}=\hbar m\\ &m\in\{-j,\dots,j\},\quad 2j\in\{0,1,2,\dots\}\end{aligned}

At fixed $\hat{J}^{2}$ the eigenvalue of $\hat{J}_{z}$ is bounded, so raising must stop at a top rung and lowering at a bottom one. Applying to those rungs, written through and , fixes the top at and the bottom at , and whole steps from one to the other make a whole number.

Proof steps

\hat{J}^{2}-\hat{J}_{z}^{2}=\hat{J}_{x}^{2}+\hat{J}_{y}^{2}\ \Rightarrow\ \hbar^{2}m^{2}\le\langle\hat{J}^{2}\rangle

Squares of Hermitian operators have non-negative averages, so $m$ is bounded.

\begin{aligned}&\hat{J}_{+}\ket{m_{\max}}=0\ \Rightarrow\ \hat{J}^{2}\ket{m_{\max}}\\ &=\bigl(\hat{J}_{-}\hat{J}_{+}+\hat{J}_{z}^{2}+\hbar\hat{J}_{z}\bigr)\ket{m_{\max}}\\ &=\hbar^{2}m_{\max}(m_{\max}+1)\ket{m_{\max}}\end{aligned}

On the top rung the first term gives zero.

j\equiv m_{\max}\ \Rightarrow\ \hat{J}^{2}=\hbar^{2}j(j+1)

Name the top value $j$ .

\begin{aligned}&\hat{J}_{-}\ket{m_{\min}}=0\ \Rightarrow\ j(j+1)=m_{\min}(m_{\min}-1)\\ &\Rightarrow\ m_{\min}=-j\end{aligned}

Use $\hat{J}^{2}=\hat{J}_{+}\hat{J}_{-}+\hat{J}_{z}^{2}-\hbar\hat{J}_{z}$ on the bottom rung; the other root, , lies above the top.

j-(-j)=2j\in\{0,1,2,\dots\}

The bottom is reached from the top in whole steps of one.

Applications

Practice

2j+1 States

For a given $j$ , the $z$ component takes the $2j+1$ values from $-j$ to $j$ in steps of one, in units of .

Try it

How many states does $j=\tfrac32$ have?