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I Schrodinger equation and Heisenberg equation of motion

  1. Aug 13, 2017 #1
    My question is that how does the Schrodinger equation arise from the Heisenberg equation of motion in the quantum field formalism.

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    These are from Hatfield's book. So I'm having some difficulties to reproduce (2.36) by plugging (2.55) into (2.37) primarily because H is an integral.

    Thanks!
     
  2. jcsd
  3. Aug 13, 2017 #2

    vanhees71

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    Just use (2.54). Note that ##\hat{H}## is conserved in the Heisenberg picture, i.e., you can evaluate the integral at time ##t## (it doesn't depend on ##t## after all anyway). That's why you only need the canonical equal-time commutation relations (2.54).
     
  4. Aug 13, 2017 #3
    Hi vanhees, great to see you again! Sorry I don't really follow you this time. I was also trying to use (2.54) and I guess I was stuck because I'm not familiar with an operator that is an integral.

    So first of all, for the right side of (2.37) I'm not sure whether [tex] [H, \phi] = (\int dx \phi^* h \phi) \phi - \phi (\int dx \phi^* h \phi) [/tex] or [tex] \int dx \phi^* h \phi \phi - \int \phi dx \phi^* h \phi [/tex], where [tex] h = -\frac{1}{2} \partial^2_x + V(x) [/tex]. Furthermore, when the operators are in the form [tex] h \phi \phi [/tex], is it equal to simply [tex] (h\phi) \phi [/tex] or is it [tex] (h\phi)\phi + \phi (h\phi) [/tex]?
     
  5. Aug 13, 2017 #4
    Wait, I think I get something now.

    Is it like this?

    [tex] [H, \phi] = H\phi - \phi H = \int dx' \phi^*(x') h \phi(x') \phi(x) - \phi(x) \int dx' \phi^*(x') h \phi(x') [/tex]
    [tex] = \int dx' \phi^*(x') h \phi(x') \phi(x) - \int dx' \phi(x) \phi^*(x') h \phi(x') [/tex]
    the phi(x) can be taken inside the integral since it's independent of x', then by (2.54),
    [tex] = \int dx' \phi^*(x') \phi(x) h \phi(x') - \int dx' \phi(x) \phi^*(x') h \phi(x') = \int \delta (x' - x) h \phi(x') dx' = h \phi(x) [/tex]

    Did I make any mistakes in the math?
     
    Last edited: Aug 13, 2017
  6. Aug 14, 2017 #5

    vanhees71

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    I'm not familiar with your notation, and it's also good to write out the arguments. First of all you have
    $$\hat{H}=\int_{\mathbb{R}} \mathrm{d} x \hat{\varphi}^*(t,x) \left (-\frac{1}{2} \Delta +V(x) \right) \hat{\varphi}(t,x).$$
    Now ##\hat{H}## is not explicitly time dependent and that implies that it is conserved:
    $$\frac{\mathrm{d}}{\mathrm{d} t} \hat{H}=[\hat{H},\hat{H}]+\partial_t \hat{H}=0.$$
    That implies that you can use any ##t## in evaluating ##\hat{H}## since ##\hat{H}## doesn't depend on it. Now you have
    $$[\hat{H},\hat{\varphi}(t,x)]=\int_{\mathbb{R}} \mathrm{d} x' \left [\hat{\varphi}^*(t,x') \left (-\frac{1}{2} \Delta' +V(x') \right) \hat{\varphi}(t,x'),\hat {\varphi}(t,x) \right ].$$
    Now you can just use the equal-time commutator relations given in #1 to show (2.37). You also need the general formula
    $$[\hat{A},\hat{B} \hat{C}]=[\hat{A},\hat{C}] \hat{B}+\hat{C} [\hat{A},\hat{B}],$$
    valid for any three operators ##\hat{A}##, ##\hat{B}##, and ##\hat{C}##.

    It is allowed to use ##t## in the integral for ##\hat{H}## as the time argument of the fields since ##\hat{H}## doesn't depend on time as argued above, and that's why you can use the equal-time commutation relations to evaluate this commutator.
     
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