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Wave function simplification in relativistic coordinates?

  1. Feb 24, 2014 #1
    My text book says
    ψ(x,t)=exp(i(p[itex]_{0}[/itex]x[itex]^{0}[/itex] + p[itex]^{→}[/itex][itex]\cdot[/itex]x[itex]^{→}[/itex])/h)=exp(i*p[itex]\cdot[/itex]x/h)
    (note that by h I mean 'h-bar'...couldn't find the symbol).
    I don't recognize (like my text implies I should) how the first equation equals the second. Where did the p[itex]_{0}[/itex]x[itex]^{0}[/itex] go? Sorry for my stupidity here. Any hints appreciated. Thanks.
     
  2. jcsd
  3. Feb 24, 2014 #2

    jtbell

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    Staff: Mentor

    On the right side, he's considering p and x as relativistic 4-vectors. p⋅x is the 4-vector "dot product" which is defined as the expression in the inner parentheses on the left side.
     
  4. Feb 24, 2014 #3
    I can't believe I didn't think of that. Thank you!
     
  5. Feb 24, 2014 #4

    Bill_K

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    Science Advisor

    In TeX it's \ hbar. In UTF it's just ħ
     
  6. Feb 24, 2014 #5
    Thanks!
     
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