Solve Spherical Harmonics: Y_{1,1} Eigenfunction of L^2 & L_z

Click For Summary
SUMMARY

The discussion confirms that the spherical harmonic function ##Y_{1,1}(\theta,\phi)## is an eigenfunction of the operators ##\hat{L}^2## and ##\hat{L_z}##, with corresponding eigenvalues of ##2\hbar^2## and ##\hbar##, respectively. The calculations demonstrate the application of the angular momentum operators in quantum mechanics, specifically using the definitions and properties of spherical harmonics. The discussion also highlights a best practice regarding the numbering of equations, emphasizing that only essential equations should be numbered for clarity.

PREREQUISITES
  • Understanding of quantum mechanics and angular momentum operators
  • Familiarity with spherical harmonics and their properties
  • Knowledge of partial derivatives and their application in quantum mechanics
  • Basic proficiency in LaTeX for mathematical notation
NEXT STEPS
  • Study the derivation of spherical harmonics in quantum mechanics
  • Learn about the mathematical properties of angular momentum operators ##\hat{L}^2## and ##\hat{L_z}##
  • Explore the significance of eigenvalues in quantum mechanics
  • Review best practices for presenting mathematical equations in academic writing
USEFUL FOR

Students and professionals in physics, particularly those focusing on quantum mechanics, as well as educators looking to enhance their understanding of spherical harmonics and angular momentum.

docnet
Messages
796
Reaction score
486
Homework Statement
psb
Relevant Equations
psb
Screen Shot 2021-02-20 at 12.16.10 AM.png

To show ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L}^2## we operate on ##Y_{1,1}(\theta,\phi)## with ##\hat{L}^2##
\begin{equation}
\hat{L}^2Y_{1,1}(\theta,\phi)=\hat{L}^2\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)
\end{equation}
\begin{equation}
=-\hbar^2\Big[\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\Big(sin\theta\frac{\partial}{\partial\theta}\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)\Big)+\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2}\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)\Big]
\end{equation}

\begin{equation}
=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[ \frac{1}{sin\theta}\frac{\partial}{\partial\theta}\Big(sin\theta cos\theta\Big)e^{i\phi}+\frac{1}{sin\theta}\frac{\partial^2}{\partial\phi^2}e^{i\phi}\Big]
\end{equation}
\begin{equation}
=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[\frac{1}{sin\theta}(cos^2\theta-sin^2\theta)-\frac{1}{sin\theta}\Big]e^{i\phi}
\end{equation}
\begin{equation}
=\hbar^2\sqrt{{\frac{3}{8\pi}}}\Big[\frac{cos^2\theta-sin^2\theta-1}{sin\theta}\Big]e^{i\phi}
\end{equation}
\begin{equation}
\hat{L}^2Y_{1,1}(\theta,\phi)=2\hbar^2\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)=2\hbar^2Y_{1,1}(\theta,\phi)
\end{equation}
so ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L}^2## with a corresponding eigenvalue of ##2\hbar^2##. Next we work out how ##\hat{L_z}## operates on ##Y_{1,1}(\theta,\phi)##
\begin{equation}
\hat{L_z}Y_{1,1}(\theta,\phi)=-i\hbar\frac{\partial}{\partial\phi}(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi})
\end{equation}
\begin{equation}
=i\hbar\sqrt{{\frac{3}{8\pi}}}sin\theta\frac{\partial}{\partial\phi}e^{i\phi}=\hbar\Big(-\sqrt{{\frac{3}{8\pi}}}sin\theta e^{i\phi}\Big)=\hbar Y_{1,1}(\theta,\phi)
\end{equation}
and we find that ##Y_{1,1}(\theta,\phi)## is an eigenfunction of ##\hat{L_z}## with a corresponding eigenvalue of ##\hbar##.
 
  • Like
Likes   Reactions: vanhees71 and PeroK
Physics news on Phys.org
This is all correct, nice job. A minor point is that (2)-(5) are not separate equations. Usually, equations are numbered only if it is necessary to refer to them in the text. Numbered equations are usually "bottom lines" after all the algebraic manipulation has been completed.
 
  • Like
Likes   Reactions: PhDeezNutz
kuruman said:
This is all correct, nice job. A minor point is that (2)-(5) are not separate equations. Usually, equations are numbered only if it is necessary to refer to them in the text. Numbered equations are usually "bottom lines" after all the algebraic manipulation has been completed.
that is a great and valid point that I will be sure to remember on my next assignment. Thank you
 
  • Like
Likes   Reactions: PhDeezNutz and kuruman

Similar threads

Replies
1
Views
2K
Replies
29
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
3K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K