What is the Parity of a System Described by a Wavefunction?

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Homework Help Overview

The discussion revolves around determining the parity of a quantum system described by a specific wavefunction in spherical coordinates. The original poster presents a wavefunction and attempts to analyze its parity by applying the parity operator.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the parity by substituting negative coordinates into the wavefunction and concludes it has odd parity. Other participants question the application of the parity operator, particularly regarding the treatment of the radial coordinate.

Discussion Status

Participants are exploring different interpretations of how the parity operator affects the coordinates, with some suggesting that the radial coordinate remains positive while others express confusion about the implications of reflection on vector direction. There is no explicit consensus yet on the correct application of the parity operator.

Contextual Notes

There is a discussion about the nature of the parity operator in spherical coordinates and how it relates to the wavefunction's behavior under reflection. Participants are also considering the implications of vector direction changes during reflections.

humanist rho
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Homework Statement


The wavefunction describing state of a system is,
\psi (r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}%<br /> \right) ^{3/2}\frac{r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi }
Find the parity of system in this state.


The Attempt at a Solution



\psi (-r,\theta ,\phi )=\frac{1}{8\sqrt{\pi }}\left( \frac{1}{a_{0}}%<br /> \right) ^{3/2}\frac{-r}{a_{0}}e^{-4/2a_{0}}\sin \theta e^{-i\phi }<br /> =-\psi (r,\theta ,\phi )
odd parity.


I'm wrong.But donno where. Pls help.
 
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In spherical coordinates, the parity operator takes
\begin{align*}
r &\to r \\
\theta &\to \pi-\theta \\
\phi &\to \phi+\pi
\end{align*}
 
you mean r →-r ?

Then that gives me the perfect answer.
Thank you very much dear friend.
 
No, r goes to +r.
 
How is that possible?
is it bcz r^2 = x^2+y^2+z^2?
But the reflection changes direction of vector also na?

Can you please explain?
 
humanist rho said:
But the reflection changes direction of vector also na?
What?
 

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