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Eigenvalues of $L^2$

Consider the angular wave-function $\psi(\theta,\phi) = L_+\,Y_{l,m}(\theta,\phi)$. We know that
\begin{displaymath}
\oint \psi^\ast(\theta,\phi)\,\psi(\theta,\phi)\,d\Omega \geq 0,
\end{displaymath} (555)

since $\psi^\ast\,\psi\equiv \vert\psi\vert^2$ is a positive-definite real quantity. Hence, making use of Eqs. (176) and (521), we find that
$\displaystyle \oint (L_+\,Y_{l,m})^\ast\,(L_+\,Y_{l,m})\,d\Omega$ $\textstyle =$ $\displaystyle \oint Y_{l,m}^{\,\ast}\,(L_+)^\dag\,(L_+\,Y_{l,m})\,d\Omega$  
  $\textstyle =$ $\displaystyle \oint Y_{l,m}^{\,\ast}\,L_-\,L_+\,Y_{l,m}\,d\Omega\geq 0.$ (556)

It follows from Eqs. (523), and (535)-(537) that
$\displaystyle \oint Y_{l,m}^{\,\ast}\,(L^2 -L_z^{\,2}-\hbar\,L_z)\,Y_{l,m}\,d\Omega$ $\textstyle =$ $\displaystyle \oint Y_{l,m}^{\,\ast}\,\hbar^2\left[l\,(l+1) -m\,(m+1)\right]Y_{l,m}\,d\Omega$  
  $\textstyle =$ $\displaystyle \hbar^2\,\left[l\,(l+1) -m\,(m+1)\right]\,\oint Y_{l,m}^{\,\ast}\,Y_{l,m}\,d\Omega$  
  $\textstyle =$ $\displaystyle \hbar^2\,\left[l\,(l+1) -m\,(m+1)\right]\geq 0.$ (557)

We, thus, obtain the constraint
\begin{displaymath}
l\,(l+1) \geq m\,(m+1).
\end{displaymath} (558)

Likewise, the inequality
\begin{displaymath}
\oint (L_-\,Y_{l,m})^\ast\,(L_-\,Y_{l,m})\,d\Omega
=\oint Y_{l,m}^{\,\ast}\,L_+\,L_-\,Y_{l,m}\,d\Omega\geq 0
\end{displaymath} (559)

leads to a second constraint:
\begin{displaymath}
l\,(l+1) \geq m\,(m-1).
\end{displaymath} (560)

Without loss of generality, we can assume that $l\geq 0$. This is reasonable, from a physical standpoint, since $l\,(l+1)\,\hbar^2$ is supposed to represent the magnitude squared of something, and should, therefore, only take non-negative values. If $l$ is non-negative, then the constraints (558) and (560) are equivalent to the following constraint:

\begin{displaymath}
-l \leq m \leq l.
\end{displaymath} (561)

We, thus, conclude that the quantum number $m$ can only take a restricted range of integer values.

Well, if $m$ can only take a restricted range of integer values, then there must exist a lowest possible value it can take. Let us call this special value $m_-$, and let $Y_{l,m_-}$ be the corresponding eigenstate. Suppose we act on this eigenstate with the lowering operator $L_-$. According to Eq. (539), this will have the effect of converting the eigenstate into that of a state with a lower value of $m$. However, no such state exists. A non-existent state is represented in quantum mechanics by the null wave-function $\psi=0$. Thus, we must have

\begin{displaymath}
L_-\,Y_{l,m_-} = 0.
\end{displaymath} (562)

Now, from Eq. (522),
\begin{displaymath}
L^2 = L_+\,L_-+L_z^{\,2} - \hbar\,L_z
\end{displaymath} (563)

Hence,
\begin{displaymath}
L^2\,Y_{l,m_-} = (L_+\,L_-+L_z^{\,2} - \hbar\,L_z)\,Y_{l,m_-},
\end{displaymath} (564)

or
\begin{displaymath}
l\,(l+1)\,Y_{l,m_-} = m_-\,(m_- -1)\,Y_{l,m_-},
\end{displaymath} (565)

where use has been made of (535), (536), and (562). It follows that
\begin{displaymath}
l\,(l+1) = m_-\,(m_--1).
\end{displaymath} (566)

Assuming that $m_-$ is negative, the solution to the above equation is
\begin{displaymath}
m_- = - l.
\end{displaymath} (567)

We can similarly show that the largest possible value of $m$ is
\begin{displaymath}
m_+ =+ l.
\end{displaymath} (568)

The above two results imply that $l$ is an integer, since $m_-$ and $m_+$ are both constrained to be integers.

We can now formulate the rules which determine the allowed values of the quantum numbers $l$ and $m$. The quantum number $l$ takes the non-negative integer values $0, 1, 2, 3, \cdots$. Once $l$ is given, the quantum number $m$ can take any integer value in the range

\begin{displaymath}
-l,\,-l+1,\,\cdots\, 0, \,\cdots,\, l-1,\, l.
\end{displaymath} (569)

Thus, if $l=0$ then $m$ can only take the value $0$, if $l=1$ then $m$ can take the values $-1, 0, +1$, if $l=2$ then $m$ can take the values $-2,-1,0,+1,+2$, and so on.


next up previous contents
Next: Spherical harmonics Up: Orbital angular momentum Previous: Eigenvalues of   Contents
Richard Fitzpatrick 2006-12-12