Eigenvalues of quantum orbital angular momentum operators

As mentioned in an earlier article, if \small \left [ \hat{L}^{2},\hat{L}_z \right ]=0, a common complete set of eigenfunctions can be selected for the two operators. Let \small Y be the common set of normalised eigenfunctions with eigenvalues \small b and \small c for \small \hat{L}^{2} and \small \hat{L}_z respectively.

From eq75,

\small \hat{L}^{2}-\hat{L}_z^{\; 2} =\hat{L}_x^{\; 2}+\hat{L}_y^{\; 2}

Multiplying the above equation by \small Y on the right, \small Y^{*} on the left and integrating over all space, we have

\small \langle L^{2}\rangle-\langle L_z^{\; 2}\rangle=\langle L_x^{\; 2}\rangle+\langle L_y^{\; 2}\rangle

\small b-c^{2}=\langle L_x^{\; 2}\rangle+\langle L_y^{\; 2}\rangle

Even though \small Y is not an eigenfunction of \small \hat{L}_x^{\; 2}, we can still find the expectation value of \small \hat{L}_x^{\; 2}, which is

\small \langle L_x^{\; 2}\rangle=\int Y^{*}\hat{L}_x(\hat{L}_xY)d\tau=\int (\hat{L}_xY)(\hat{L}_xY)^{*}d\tau=\int \vert \hat{L}_xY\vert^{2}d\tau\geq 0

Note that we have used the fact that \small \hat{L}_x is Hermitian for the 2nd equality (see eq37). Similarly, \small \langle L_y^{\; 2}\rangle\geq 0. So,

\small b-c^{2}\geq 0\; \; \; \; \; or\; \; \; \; \; -\sqrt{b}\leq c\leq \sqrt{b}

Therefore, \small c has an upper bound and a lower bound. Since the eigenvalues of \small \hat{L}_z has an upper bound and a lower bound, from eq113 and eq114 we have

\small \hat{L}_zY_{max}=c_{max}Y_{max}\; \; \; \; \; \; \; \; 122

\small \hat{L}_zY_{min}=c_{min}Y_{min}\; \; \; \; \; \; \; \; 123

where \small Y_{max}=\hat{L}_+^{\; k_{max}}Y, \small c_{max}=c+k_{max}\hbar, \small Y_{min}=\hat{L}_-^{\; k_{min}}Y and \small c_{min}=c-k_{min}\hbar.

If we operate on eq122 with \small \hat{L}_+, we supposedly have

\small \hat{L}_z(\hat{L}_+Y_{max})=(c_{max}+\hbar)(\hat{L}_+Y_{max})

However, the above equation contradicts the upper bound eigenvalue of \small \hat{L}_z of \small c_{max}. This implies that we must truncate the ladder beyond eq122. Since \small c_{max}+\hbar\neq 0, we must have

\small \hat{L}_+Y_{max}=0\; \; \; \; \; \; \; \; 124

Using the same argument when operating on eq123 with \small \hat{L}_-, we have

\small \hat{L}_-Y_{min}=0 \; \; \; \; \; \; \; \; 125

If we further operate on eq124 with \small \hat{L}_-, we have

\small \hat{L}_-\hat{L}_+Y_{max}=0

Substitute eq116 in the above equation,

\small (\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z)Y_{max}=0

Using eq119 where \small k=k_{max}, and eq122,

\small (b-c_{max}^{\; \; \; \; \; \; 2}-\hbar c_{max})Y_{max}=0

Since \small Y_{max}\neq 0

\small b-c_{max}^{\; \; \; \; \; \; 2}-\hbar c_{max}=0 \; \; \; \; \; \; \; \; 126

Similarly, operating on eq125 with \small \hat{L}_+ and using eq115, eq119 and eq123, we have

\small b-c_{min}^{\; \; \; \; \; \; 2}+\hbar c_{min}=0 \; \; \; \; \; \; \; \; 127

Subtracting eq127 from eq126, we have \small (c_{max}+c_{min})(c_{min}-c_{max}-\hbar)=0. Since \small c_{max}> c_{min} and \small (c_{min}-c_{max}-\hbar)< 0, we have \small c_{max}+c_{min}=0 or

\small c_{max}=-c_{min}\; \; \; \; \; \; \; \; 128

As we know from eq122 and eq123, the value of \small c_{max}-c_{min} of a particular system is dependent on the number of consecutive operations on \small \hat{L}_zY by \small \hat{L}_+ or \small \hat{L}_-, with each operation raising or lowering the eigenvalue of \small \hat{L}_z by \small \hbar. Therefore,

\small c_{max}-c_{min}=0,\hbar,2\hbar,\cdots=2l\hbar\; \; \; \; \; \; \; \; 129

where \small l=0,\frac{1}{2},1,\frac{3}{2},\cdots

Substituting eq128 in eq129, we have \small c_{max}=l\hbar and \small c_{min}=-l\hbar. Therefore, the eigenvalues of \small \hat{L}_z are

\small c=-l\hbar,-l\hbar+\hbar,-l\hbar+2\hbar,\cdots,l\hbar=-l\hbar,(-l+1)\hbar,(-l+2)\hbar,\cdots,l\hbar

\small c=m_l\hbar\; \; \; \; \; \; \; \; 130

where \small m_l=-l,-l+1,-l+2,\cdots,l.

Substituting \small c_{min}=-l\hbar in eq127,

\small b=l(l+1)\hbar^{2}\; \; \; \; \; \; \; \; 131

where \small l=0,\frac{1}{2},1,\frac{3}{2},\cdots.

As mentioned in the previous article, the raising and lowering operators also apply to the spin angular momentum \small \boldsymbol{\mathit{S}} and the total angular momentum \small \boldsymbol{\mathit{J}}. We would therefore expect the eigenvalues of \small \hat{S}^{2} and \small \hat{S}_z to be \small s(s+1)\hbar^{2} and \small m_s\hbar respectively, and the eigenvalues of \small \hat{J}^{2} and \small \hat{J}_z to be \small j(j+1)\hbar^{2} and \small m_j\hbar respectively. However, the quantum numbers \small l and \small m_l for the orbital angular momentum \small \boldsymbol{\mathit{L}}, but not the quantum numbers for \small \boldsymbol{\mathit{S}} and \small \boldsymbol{\mathit{J}}, are restricted to integers. Therefore,


\small \boldsymbol{\mathit{L}} \small m_l\in \mathbb{Z} \small l\in \mathbb{Z}
\small \boldsymbol{\mathit{S}} \small m_s=-s,-s+1,-s+2,\cdots,s \small s=0,\frac{1}{2},1,\frac{3}{2},\cdots
\small \boldsymbol{\mathit{J}} \small m_j=-j,-j+1,-j+2,\cdots,j \small j=0,\frac{1}{2},1,\frac{3}{2},\cdots



Why are the quantum numbers for  restricted to integers?


The eigenvalue equation for \small \hat{L}_z (see eq95) is:

\small \frac{\hbar}{i}\frac{\partial}{\partial\phi}\psi=m_l\hbar\psi

where \small \psi=Ae^{im_l\phi}.

Since \small \psi must be single-valued,

\small Ae^{im_l\phi}=Ae^{im_l(\phi+2\pi)}

\small e^{im_l2\pi}=1

\small cosm_l2\pi + isinm_l2\pi=1

The solution to the above equation is \small m_l\in \mathbb{Z}. Furthermore, \small l is also an integer because \small m_l=-l,-l+1,-l+2,\cdots,l.


We would arrive at the same results (eq130 and eq131) if we have chosen \small \hat{L}_x or \small \hat{L}_y instead of \small \hat{L}_z. The significance of eq130 and eq131 is that we can simultaneously assign eigenvalues of \small \hat{L}^{2} and \small \hat{L}_z (or \small \hat{L}^{2} and \small \hat{L}_x or \small \hat{L}^{2} and \small \hat{L}_y) if the 2 operators commute. This, together with the fact that any pair of component angular momentum operators does not commute, implies that we cannot simultaneously specify eigenvalues of \small \hat{L}^{2} and more than one component angular momentum operators.

In conclusion, substituting eq130 in eq112 and eq131 in eq117

\small \hat{L}_zY=m_l\hbar Y

\small \hat{L}^{2}Y=l(l+1)\hbar^{2} Y

Since \small Y is a function of \small l and \small m_l, we can express the above eigenvalue equations as:

\small \hat{L}_z\vert l,m_l\rangle=m_l\hbar \vert l,m_l\rangle \; \; \; \; \; \; \; \; 132

\small \hat{L}^{2}\vert l,m_l\rangle=l(l+1)\hbar^{2} \vert l,m_l\rangle \; \; \; \; \; \; \; \; 133


Next article: matrix elements of angular momentum operators
Previous article: quantum orbital angular momentum ladder operators
Content page of quantum mechanics
Content page of advanced chemistry
Main content page

Leave a Reply

Your email address will not be published. Required fields are marked *