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

AI Thread Summary
The discussion centers on determining the parity of a wavefunction described in spherical coordinates. The initial attempt concluded that the wavefunction exhibits odd parity, as substituting negative values for r resulted in a negative wavefunction. However, clarification is sought regarding the application of the parity operator, specifically how r should be treated during reflection. It is noted that in spherical coordinates, r remains positive, which leads to confusion about the implications for the wavefunction's parity. Ultimately, understanding the correct application of the parity operator is crucial for resolving the parity classification of the system.
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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