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GR8677 #28
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Quantum Mechanics$\Rightarrow$}Normalization

Recall that $\int |\psi(x)|^2 dx=1$ is the condition for a normalized function.

$\begin{eqnarray} |\psi|^2&=&\int^{2\pi}_0 |A|^2 d\phi\\ &=&2\pi|A|^2\\ &=&1\\ \Rightarrow A&=&\frac{1}{\sqrt{2\pi}},
\end{eqnarray}$

where the condition $|e^{im\phi}|^2=e^{im\phi}e^{-im\phi}=1$ is used.

This is choice (D).

Alternate Solutions
 Giubenez2014-10-22 02:42:06 The only way to solve this problem is to remember the first Spherical Harmonics, or, at least, their dependence to $m_z$. In fact, using the conventional coordinates where $\theta \in [0,\pi]$ is the polar angle and $\phi\in [0,2\pi]$ is the azimutal angle, they always have a term $Y^m_l \propto e^{i m \phi}$. The direction is to estimate A, we have thus to integrate on $d\phi$ that in the question is CAPITALIZED and we have to choose the $[0,2\pi]$ intervalReply to this comment
Giubenez
2014-10-22 02:42:06
The only way to solve this problem is to remember the first Spherical Harmonics, or, at least, their dependence to $m_z$.

In fact, using the conventional coordinates where $\theta \in [0,\pi]$ is the polar angle and $\phi\in [0,2\pi]$ is the azimutal angle, they always have a term
$Y^m_l \propto e^{i m \phi}$.
The direction is to estimate A, we have thus to integrate on $d\phi$ that in the question is CAPITALIZED and we have to choose the $[0,2\pi]$ interval
alemsalem
2009-09-23 19:47:52
i think the answer is correct but the it's missing smthing,, how do u know u should integrate over 2 PI u should integrate over one PI and the other half is accounted for by symmetry there are no additional probabilities,, that would compensate for the factor of 2 pauli mentioned
 flyboy6212010-11-14 19:33:17 The integration is over all possible values of $\phi$. Since the given function is periodic, you only need to integrate over the period, which is $2\pi$.
pauli568
2007-10-12 13:48:01
I think there is no correct answer given here.
The con dition of normalization goes as
\int{psi}dv=1 in which case theta coordinate should aso be taken care of and wich will result in an extra 2.
The result should be \frac{1}{2\sqrt{pi}}
 dean2008-10-09 21:40:08 The official solution is essentially correct, though I think it's better to think of the normalization as <$\psi$|$\psi$>=1, you get the conjugation for free.
 HaveSpaceSuit2008-10-17 17:50:12 They do not specify a theta dependence for the wave function so you can assume it is only a function of phi. Normalize the given function.

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