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Eigenvalues of $L^2$

Consider the angular wavefunction $\psi(\theta,\phi) = L_+ Y_{l,m}(\theta,\phi)$. We know that
\oint \psi^\ast(\theta,\phi) \psi(\theta,\phi) d\Omega \geq 0,
\end{displaymath} (576)

since $\psi^\ast \psi\equiv \vert\psi\vert^2$ is a positive-definite real quantity. Hence, making use of Eqs. (194) and (539), we find that
$\displaystyle \oint (L_+ Y_{l,m})^\ast (L_+ Y_{l,m}) d\Omega$ $\textstyle =$ $\displaystyle \oint Y_{l,m}^{ \ast} (L_+)^\dag (L_+ Y_{l,m}) d\Omega$  
  $\textstyle =$ $\displaystyle \oint Y_{l,m}^{ \ast} L_- L_+ Y_{l,m} d\Omega\geq 0.$ (577)

It follows from Eqs. (541), and (556)-(558) that
$\displaystyle \oint Y_{l,m}^{ \ast} (L^2 -L_z^{ 2}-\hbar L_z) Y_{l,m} d\Omega$ $\textstyle =$ $\displaystyle \oint Y_{l,m}^{ \ast} \hbar^2\left[l (l+1) -m (m+1)\right]Y_{l,m} d\Omega$  
  $\textstyle =$ $\displaystyle \hbar^2 \left[l (l+1) -m (m+1)\right] \oint Y_{l,m}^{ \ast} Y_{l,m} d\Omega$  
  $\textstyle =$ $\displaystyle \hbar^2 \left[l (l+1) -m (m+1)\right]\geq 0.$ (578)

We, thus, obtain the constraint
l (l+1) \geq m (m+1).
\end{displaymath} (579)

Likewise, the inequality
\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} (580)

leads to a second constraint:
l (l+1) \geq m (m-1).
\end{displaymath} (581)

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 (579) and (581) are equivalent to the following constraint:

-l \leq m \leq l.
\end{displaymath} (582)

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. (560), 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 wavefunction, $\psi=0$. Thus, we must have

L_- Y_{l,m_-} = 0.
\end{displaymath} (583)

Now, from Eq. (540),
L^2 = L_+ L_-+L_z^{ 2} - \hbar L_z
\end{displaymath} (584)

L^2 Y_{l,m_-} = (L_+ L_-+L_z^{ 2} - \hbar L_z) Y_{l,m_-},
\end{displaymath} (585)

l (l+1) Y_{l,m_-} = m_- (m_- -1) Y_{l,m_-},
\end{displaymath} (586)

where use has been made of (556), (557), and (583). It follows that
l (l+1) = m_- (m_-1).
\end{displaymath} (587)

Assuming that $m_-$ is negative, the solution to the above equation is
m_- = - l.
\end{displaymath} (588)

We can similarly show that the largest possible value of $m$ is
m_+ =+ l.
\end{displaymath} (589)

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

-l, -l+1, \cdots  0,  \cdots,  l-1,  l.
\end{displaymath} (590)

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
Next: Spherical Harmonics Up: Orbital Angular Momentum Previous: Eigenvalues of
Richard Fitzpatrick 2010-07-20