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A Canonical quantization of scalar fields

  1. Mar 19, 2016 #1
    In the srednicki notes he goes from

    $$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to
    $$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p) $$

    Where $$\tilde{a}(p) = \int \frac{d^{3}x}{(2\pi)^{\frac{3}{2}}}e^{-ipx}a(x)$$

    Is this as simple as substituting or isn't there a commutator to get to
    $$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p)$$ ?

    Do $$e^{ipx}$$ and $$P^{2}$$ commute? What about $$\nabla$$ and $$ a(x)$$?

    This is really bugging me any help would be greatly appreciated!
     
  2. jcsd
  3. Mar 19, 2016 #2

    vanhees71

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    This is obviously non-relativistic QFT. Then the scalar field (quantized Schrödinger field) in the Heisenberg picture has a mode decomposition
    $$\hat{\psi}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^{3/2}} \exp[-\mathrm{i} (\omega_{\vec{p}}t-\vec{x} \cdot \vec{p})] \hat{a}(\vec{p}), \quad \omega_{\vec{p}}=\frac{\vec{p}^2}{2m}.$$
    The free Hamiltonian is indeed given by
    $$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \hat{\psi}^{\dagger}(t,\vec{x}) \left (-\frac{\Delta}{2m} \right ) \hat{\psi}(t,\vec{x}).$$
    Now you can plug in the mode decomposition for ##\hat{\psi}(t,\vec{x})##. After some simple but lengthy manipulations, using
    $$\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \exp(\mathrm{i} \vec{p} \vec{x})=(2 \pi)^3 \delta^{(3)}(\vec{p})$$
    you indeed find that
    $$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \omega_{\vec{p}} \hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}).$$
    It's important, particularly in the beginning to somehow distinguish between usual real or complex numbers and operators. Above, I've written a hat above operators. Of course the operators are all linear operators on Hilbert space and thus commute with all usual real or complex numbers!
     
  4. Mar 19, 2016 #3
    But does $$P^{2}$$ commute with$$e^{ipx}$$ or how can I perform this commutation?
     
  5. Mar 19, 2016 #4

    vanhees71

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    In the above equation ##P^2## is obviously ##\vec{p}^2## and thus a real number, which commutes with all operators. Unfortunately I don't have my copy of Srednicky's book here; so I can't look at the context right now.
     
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