QM: Linear momentum of angular momentum eigenstate

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SUMMARY

The discussion focuses on calculating the commutation relations [Lz, Px] and [Lz, Py] in quantum mechanics, specifically for angular momentum eigenstates. It establishes that [Lz, Px] = iħPy and [Lz, Py] = -iħPx. This leads to the conclusion that the matrix element ⟨l'm'|Px|lm⟩ equals zero for m' ≠ m ± 1, utilizing the properties of angular momentum operators and their action on eigenstates.

PREREQUISITES
  • Understanding of angular momentum operators in quantum mechanics
  • Familiarity with commutation relations
  • Knowledge of matrix elements in quantum states
  • Basic concepts of quantum mechanics, particularly eigenstates
NEXT STEPS
  • Study the derivation of commutation relations in quantum mechanics
  • Learn about the action of angular momentum operators on eigenstates
  • Explore the significance of matrix elements in quantum mechanics
  • Investigate the implications of the ladder operators L± on angular momentum states
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Students and professionals in quantum mechanics, particularly those studying angular momentum, as well as educators preparing materials on operator theory and eigenstate analysis.

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Homework Statement


Find [Lz, Px] and [Lz,Py] and use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1.[/tex]


Homework Equations


[tex]L_z|lm\rangle = \hbar m |lm\rangle[/tex]
[tex]L^2|lm\rangle = \hbar^2 l(l+1)|lm\rangle[/tex]
[tex]L_{\pm}|lm\rangle = \hbar \sqrt{l(l+1)-m(m\pm 1)}|l,m\pm 1 \rangle[/tex]


The Attempt at a Solution


It was easy to show that [tex][L_z,P_x] = i \hbar P_y[/tex] and that [tex][L_z,P_y] = - i\hbar P_x[/tex] but how might I use this to show that [tex]\langle l'm'|P_x|lm\rangle = 0[/tex] for [tex]m' \neq m \pm 1?[/tex]
 
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Start with [itex]-i\hbar\langle l' m' \lvert P_x \rvert l m \rangle = \langle l' m' \lvert [L_z,P_y] \rvert l m \rangle[/itex].
 

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