Momentum Eigenfunction

  • Thread starter Safinaz
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  • #1
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Homework Statement



## \psi_1 ## and ## \psi_2 ## are momentum eigenfunctions corresponding to
different momentum eigenvalues ## p_1 \not= p_2 ##. Is ## \psi_1 ## + ## \psi_2 ## also momentum eigenfunction ?

Homework Equations



Is the right answer[/B]

Yes
No
It Depends ?


The Attempt at a Solution



I think yes, because

$$ \frac{h}{i} \frac{d}{dx} \psi_1 = p_1 \psi_1, $$
$$ \frac{h}{i} \frac{d}{dx} \psi_2 = p_2 \psi_2, $$
Then
$$ \frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2) = p_1+p_2 (\psi_1+\psi_2), $$

is valid also
 

Answers and Replies

  • #2
blue_leaf77
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Think again how you should calculate
$$ \frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2)$$
from
$$ \frac{h}{i} \frac{d}{dx} \psi_1 = p_1 \psi_1, $$
$$ \frac{h}{i} \frac{d}{dx} \psi_2 = p_2 \psi_2, $$
 
  • #3
219
2
Ya ..

$$ \frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2) = (p_1 \psi_1+ p_2 \psi_2), $$
so ## (\psi_1+\psi_2) ## is not an Eigenfunction .
 
  • #4
blue_leaf77
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No, it's not.
 

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