Matrix elements of angular momentum ladder operators

From eq132 of the previous article, we have

$\small \hat{L}_z\vert l,m_l+1\rangle=(m_l+1)\hbar\vert l,m_l+1\rangle\; \; \; \; \; \; \; \; 134$

From eq113 where $k=1$ and eq130, we have

$\small \hat{L}_z\hat{L}_+\vert l,m_l\rangle=(m_l+1)\hbar\hat{L}_+\vert l,m_l\rangle\; \; \; \; \; \; \; \; 135$

Comparing eq134 and eq135, both eigenstates $\small \vert l,m_l+1\rangle$ and $\small \hat{L}_+\vert l,m_l\rangle$ are associated with the same eigenvalue $\small (m_l+1)\hbar$ . Therefore, one must be proportional to the other (e.g. the eigenstates $\small e^{-ikx}$ and $\small Ae^{-ikx}$ have the same eigenvalue), i.e.

$\small \hat{L}_+\vert l,m_l\rangle=c_+\vert l,m_l+1\rangle\; \; \; \; \; \; \; \; 136$

Similarly, for the lowering operator (using eq114), we have,

$\small \hat{L}_-\vert l,m_l\rangle=c_-\vert l,m_l-1\rangle\; \; \; \; \; \; \; \; 137$

where $\small c_+$ and $\small c_-$ are scalars; or in summary

$\small \hat{L}_\pm\vert l,m_l\rangle=c_\pm\vert l,m_l\pm 1\rangle\; \; \; \; \; \; \; \; 138$

If the complete set of eigenstates is normalised, multiplying the above equation on the left by $\small \langle l,m_l\pm1\vert$ gives the matrix elements of $\small \hat{L}_\pm$ :

$\small \langle l,m_l\pm 1\vert\hat{L}_\pm\vert l,m_l\rangle=c_\pm\; \; \; \; \; \; \; \; 139$

From eq116, we have $\small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=(\hat{L}^{2}-\hat{L}_z^{\; 2}-\hbar\hat{L}_z)\vert l,m_l\rangle$ . Using eq132 and eq133,

$\small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=[l(l+1)-m_l(m_l+1)]\hbar^{2}\vert l,m_l\rangle\; \; \; \; \; \; \; \; 140$

From eq136 and eq137

$\small \hat{L}_-\hat{L}_+\vert l,m_l\rangle=c_+\hat{L}_-\vert l,m_l+1\rangle=c_+c_-\vert l,m_l\rangle\; \; \; \; \; \; \; \; 141$

Comparing eq140 and eq141

$\small c_+c_-=[l(l+1)-m_l(m_l+1)]\hbar^{2}\; \; \; \; \; \; \; \; 142$

Question

Show that $\small c_+c_-=\vert c_+\vert^{2}$ .

Answer

From eq139, where we let $\small m_l=m_l+1$

$\small \langle l,m_l\vert\hat{L}_-\vert l,m_l+1\rangle=c_-$

Substituting eq109 in the above equation

$\small \langle l,m_l\vert\hat{L}_x\vert l,m_l+1\rangle-i\langle l,m_l\vert\hat{L}_y\vert l,m_l+1\rangle=c_-$

Since $\small \hat{L}_x$ and $\small \hat{L}_y$ are Hermitian,

$\small \left \{ \langle l,m_l+1\vert\hat{L}_x\vert l,m_l\rangle+i\langle l,m_l+1\vert\hat{L}_y\vert l,m_l\rangle\right \}^{*}=c_-$

Substituting eq108 in the above equation

$\small \langle l,m_l+1\vert\hat{L}_+\vert l,m_l\rangle^{*}=c_-$

Substituting eq139 in the above equation

$\small c_+^{\; *}=c_-\; \; \; \; \; \; \; \; 143$

Therefore, $\small c_+c_-=\vert c_+\vert^{2}$ .

Substituting $\small c_+c_-=\vert c_+\vert^{2}$ in eq142

$\small c_+=\hbar \sqrt{l(l+1)-m_l(m_l+1)}$

Substituting the above equation in eq136,

$\small \hat{L}_+\vert l,m_l\rangle=\hbar \sqrt{l(l+1)-m_l(m_l+1)}\; \vert l,m_l+1\rangle \; \; \; \; \; \; \; \; 144$

To find $\small c_-$ , we repeat the above steps for eq139 onwards and using eq115 to give

$\small \hat{L}_+\hat{L}_-\vert l,m_l\rangle=\left ( \hat{L}^{2}-\hat{L}_z^{\; 2} +\hbar\hat{L}_z\right )\vert l,m_l\rangle=[l(l+1)-m_l(m_l-1)]\hbar^{2}\vert l,m_l\rangle\; \; \; \; \; \; \; \; 145$