Odd Parity of States: 2p m=0, 2p m=1, 26f m=0, 2s

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SUMMARY

The discussion focuses on determining the parity of specific quantum states: 2p m=0, 2p m=1, 26f m=0, and 2s. It is concluded that the 2p m=0, 2p m=1, and 2s states exhibit odd parity, while the parity of the 26f m=0 state requires further analysis. Participants recommend writing down the quantum numbers (n, m, and ℓ) for each state and applying the parity transformation to ascertain the wavefunctions' behavior.

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Homework Statement



Which of the following states have odd parity?

a. 2p m=0
b. 2p m=1
c. 26f m=0
d. 2s


The Attempt at a Solution



Knowing the wavefunctions for a,b, and d, i think that all of them have odd parity. What about the 26f m=0 state?
 
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I'd recommend you write down the values of [itex]n[/itex], [itex]m[/itex] and [itex]\ell[/itex] for each of those orbitals, and then use those to deduce the form of each wavefunction, and then see what happens to each one under the parity transformation.

[tex]\psi_{n\ell m}(r,\vartheta,\varphi) \propto e^{- \rho / 2} \rho^{\ell} L_{n-\ell-1}^{2\ell+1}(\rho) \cdot Y_{\ell}^{m}(\vartheta, \varphi )[/tex]
 

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