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).

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Comments

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

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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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