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Verbatim question for GR8677 #28
Quantum Mechanics}Normalization

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

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

This is choice (D).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
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
Alternate Solution - Unverified
Comments
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
Alternate Solution - Unverified
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
flyboy621
2010-11-14 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.


NEC
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}}
dean
2008-10-09 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.
HaveSpaceSuit
2008-10-17 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.
NEC

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