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Ladder operators for real scalar field

  1. Aug 9, 2015 #1
    Puting a minus in front of the momentum in the field expansion gives

    ##\phi \left( {\bf{x}} \right) = \int {{d^3}\tilde p} \left( {{a_{\bf{p}}}{e^{i{\bf{p}} \cdot {\bf{x}}}} + a_{\bf{p}}^ + {e^{ - i{\bf{p}} \cdot {\bf{x}}}}} \right){\rm{ }}\phi \left( {\bf{x}} \right) = \int {{d^3}\tilde p} \left( {{a_{ - {\bf{p}}}}{e^{ - i{\bf{p}} \cdot {\bf{x}}}} + a_{ - {\bf{p}}}^ + {e^{i{\bf{p}} \cdot {\bf{x}}}}} \right)##.

    Is this implise that

    ##a_{ - {\bf{p}}}^ + = {a_{{\bf{p}}{\rm{ }}}}## ##{a_{ - {\bf{p}}}} = a_{\bf{p}}^ + ## ?

    Becuse if so

    ##\displaylines{
    \pi \left( {\bf{x}} \right) = - i\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}\sqrt {{{{\omega _{\bf{p}}}} \over 2}} \left( {{a_{\bf{p}}}{e^{i{\bf{p}} \cdot {\bf{x}}}} - a_{\bf{p}}^ + {e^{ - i{\bf{p}} \cdot {\bf{x}}}}} \right)} = \cr
    = - i\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}\sqrt {{{{\omega _{\bf{p}}}} \over 2}} \left( {{a_{\bf{p}}}{e^{i{\bf{p}} \cdot {\bf{x}}}} - a_{ - {\bf{p}}}^{}{e^{ - i{\bf{p}} \cdot {\bf{x}}}}} \right) = } \cr
    = - i\int {{{{d^3}p} \over {{{\left( {2\pi } \right)}^3}}}\sqrt {{{{\omega _{\bf{p}}}} \over 2}} \left( {{a_{\bf{p}}}{e^{i{\bf{p}} \cdot {\bf{x}}}} - a_{\bf{p}}^{}{e^{i{\bf{p}} \cdot {\bf{x}}}}} \right)} = 0 \cr} ##

    Wich is obviosly wrong, Where is the mistake?
     
  2. jcsd
  3. Aug 9, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    No, why should it?
    The first term on the left side gets transformed to the first one on the right side, and the second term gets transformed to the second one.
    Just a transformation p -> -p, no physics involved in that step. The integral is over the whole space anyway so integration limits don't change.
     
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