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Momentum Eigenfunction

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data

    ## \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 ?

    2. Relevant equations

    Is the right answer


    Yes
    No
    It Depends ?


    3. 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
     
  2. jcsd
  3. Sep 19, 2015 #2

    blue_leaf77

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    Think again how you should calculate
    $$ \frac{h}{i} \frac{d}{dx} (\psi_1+\psi_2)$$
    from
    $$ \frac{h}{i} \frac{d}{dx} \psi_1 = p_1 \psi_1, $$
    $$ \frac{h}{i} \frac{d}{dx} \psi_2 = p_2 \psi_2, $$
     
  4. Sep 19, 2015 #3
    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 .
     
  5. Sep 19, 2015 #4

    blue_leaf77

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    No, it's not.
     
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