GREPhysics.NET: GR8677 #28

GR8677 #28

Problem

GREPhysics.NET Official Solution

Quantum Mechanics $\Rightarrow$ }Normalization

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

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

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

This is choice (D).

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

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Comments

fredluis
2019-08-09 04:33:09

This problem is deceptively simple, people. Remember that if the charge is moving towards the wire its velocity vector would be negative. tree removal

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joshuaprice153
2019-08-09 02:42:01

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Giubenez
2014-10-22 02:42:06

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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 $2\pi$ .

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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.

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