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Hamiltonian in terms of creation/annihilation operators

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider the free real scalar field [itex]\phi(x)[/itex] satisfying the Klein-Gordon equation, write the Hamiltonian in terms of the creation/annihilation operators.

    2. Relevant equations
    Possibly the definition of the free real scalar field in terms of creation/annihilation operators [tex]\phi(x)=\int \dfrac{d^3k}{(2\pi)^3} \dfrac{1}{\sqrt{2\omega_k}}(a(k) e^{-ik_\mu x^\mu}+a^\dagger(k)e^{ik_\mu x^\mu})[/tex] and also the commutation relations [tex][a(k),a^\dagger(k')]=(2\pi)^3\delta(k-k') \quad [a(k),a(k')]=[a^\dagger(k),a^\dagger(k')]=0.[/tex]

    3. The attempt at a solution
    First I've computed the Hamiltonian of the theory from the Lagrangian as
    [tex]H=\int d^3x \left(\dfrac{1}{2}\pi^2+\dfrac{1}{2}(\nabla \phi)^2+\dfrac{1}{2}m^2\phi^2 \right)[/tex]
    with this we know that we need [itex]\pi(x)=\partial_t \phi(x)[/itex] and [itex]\nabla \phi(x)[/itex]. I computed both of them from the definition of [itex]\phi(x)[/itex] and found
    [tex]\pi(x)=-i\int \dfrac{d^3 k}{(2\pi)^3}\sqrt{\dfrac{\omega_k}{2}}(a(k)e^{-i k_\mu x^\mu} - a^\dagger(k)e^{i k_\mu x^\mu})[/tex]
    and
    [tex]\nabla \phi(x)=i \int \dfrac{d^3k}{(2\pi)^3}\dfrac{1}{\sqrt{\omega_k}}\mathbf{k}(a(k)e^{-ik_\mu x^\mu}+a^\dagger(k)e^{ik_\mu x^\mu}).[/tex]

    Now I believe we just need to put everything into the formula and see what comes out, but it turns out this seems like a bad idea, as it didn't give much to work with. I found out (now I write [itex]kx = k_\mu x^\mu[/itex] for brevity)
    [tex]H=\int d^3x \dfrac{1}{2} \int \dfrac{d^3 k d^3 k'}{(2\pi)^6}\left( \sqrt{\dfrac{\omega_k \omega_{k'}}{4}}(a(k)e^{-ikx}-a^\dagger(k) e^{ikx})(a(k')e^{-ik'x}-a^\dagger(k')e^{ik'x})+\dfrac{\mathbf{k}\cdot \mathbf{k}'}{\sqrt{4\omega_k \omega_{k'}}}(a(k)e^{-ikx}-a^\dagger(k) e^{ikx})(a(k')e^{-ik'x}-a^\dagger(k')e^{ik'x})+\dfrac{m^2}{\sqrt{4\omega_k \omega_{k'}}}(a(k)e^{-ikx}+a^\dagger(k) e^{ikx})(a(k')e^{-ik'x}+a^\dagger(k')e^{ik'x}) \right)[/tex]

    after that I'm stuck. I know somehow I must use the commutation relations. I also know that the integral over [itex]x[/itex] and over [itex]k'[/itex] should disappear, leaving just one integral over [itex]k[/itex]. But I don't have any idea how to do this.
     
  2. jcsd
  3. Apr 13, 2017 #2
    You have to multiply out the different terms. Then, use that
    ##\frac{1}{(2\pi)^3}\int d^3 x e^{i(k-k')x}=\delta^{(3)}(k-k')##
     
  4. Apr 29, 2017 #3
    Actually some time after I've posted the thread I found out a simple way to do it. It was just a matter of rewriting the fields in a more convenient way:

    [tex]\phi(x)=\int \dfrac{d^3k}{(2\pi)^3} \dfrac{1}{2\omega_k}(a_k+a_{-k}^\dagger)e^{i\mathbf{k}\cdot \mathbf{x}}[/tex]

    where now the creation/annihilation operators carry the time dependency. Using this one can easily integrate over the exponentials to generate deltas and from the deltas perform one of the momentum integrations getting the result.
     
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