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notes:physics:quantum-mechanics:angular-momentum

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notes:physics:quantum-mechanics:angular-momentum [2026/10/01 16:51] – [Using Operators] chrisnotes:physics:quantum-mechanics:angular-momentum [2026/10/01 19:57] (current) – [Example calculations of different quantities] chris
Line 408: Line 408:
 ===== Example calculations of different quantities ===== ===== Example calculations of different quantities =====
  
 +Calculating quantities is trivial if their operator can be described as
 +a function of the raising and lowering operators. Take for example the
 +$x$ and $y$ components of the angular momentum.
  
 +$$\hat L_x=\frac{\hat L_++\hat L_-}{2},\quad\hat L_y=\frac{\hat L_+-\hat L_-}{2}$$
  
 +For example below are the calculations for
 +$\left<\hat L_x\right>,\left<\hat L_y\right>,\left<\hat L_x^2\right>, \left<\hat L_y^2\right>$.
  
 +$$\begin{align}
 +\left<\hat L_x\right>
 +&= \left< l,m\left|\hat L_x\right|l,m\right> \\
 +&\propto \left< l,m\left|\hat L_+\right|l,m\right> + \left< l,m\left|\hat L_-\right|l,m\right> \\
 +&\propto \left< l,m|l,m+1\right> + \left< l,m|l,m-1\right> \\
 +&= 0
 +\end{align}$$
 +
 +The system is $\phi$-independent, since it has a central potential. This means symmetry around
 +the $z$ axis and thus $\left<\hat L_x\right>=\left<\hat L_y\right> =0$. Similarly
 +$\left<\hat L_x^2\right>=\left<\hat L_y^2\right>$. Then:
 +
 +$$
 +\left<\hat L^2\right>
 += \left<\hat L_x^2+\hat L_y^2+\hat L_z^2\right>
 += \left<\hat L_x^2\right> + \left<\hat L_y^2\right> + \left<\hat L_z^2\right>
 += 2\left<\hat L_x^2\right> + \left<\hat L_z^2\right>
 +$$
 +
 +$$\begin{align}
 +\left<\hat L_x^2\right>
 += \left<\hat L_y^2\right>
 +&= \frac12\left(\left<\hat L^2\right> - \left<\hat L_z^2\right> \right) \\
 +&= \frac12\left(\left< l,m\left|\hat L^2\right|l,m\right> -\left< l,m\left|\hat L_z^2\right|l,m\right>\right) \\
 +&= \frac12\Big(l(l+1)\left< l,m|l,m\right> - m^2\left< l,m|l,m\right>\Big) \\
 +&= \frac{l(l+1)-m^2}{2}
 +\end{align}$$
 +
 +This way we can find the dispersion around the expected value.
 +
 +$$\require{cancel}
 +\Delta L_x
 += \Delta L_y
 += \sqrt{\left<\hat L_x^2\right> - \cancelto{0}{\left<\hat L_x\right>^2}}
 += \sqrt{\frac{l(l+1)-m^2}{2}}
 +$$
  
  
notes/physics/quantum-mechanics/angular-momentum.1790873503.txt.gz · Last modified: by chris