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A Solution of the classical Klein-Gordon equation

  1. Mar 30, 2016 #1
    The classical Klein-Gordon equation is ##(\partial^{2}+m^{2})\varphi(t,\vec{x})=0##.

    To solve this equation, we need to Fourier transform ##\varphi(t,\vec{x})## with respect to its space coordinates to obtain

    ##\varphi(t,\vec{x}) = \int \frac{d^{3}\vec{k}}{(2\pi)^{3}}e^{i\vec{k}\cdot{\vec{x}}}\tilde{\varphi}(t,\vec{k})##.

    Plugging ##\varphi(t,\vec{x})## into Klein-Gordon equation is supposed to give us the solution

    ##\tilde{\phi}(t,\vec{k})=A(\vec{k})e^{-iE_{\vec{k}}t}+B(\vec{k})e^{iE_{\vec{k}}t}##.

    I am stuck in getting the solution. If I plug the Fourier transform of ##\varphi(t,\vec{x})## into the Klein-Gordon equation, I get ##(-\vec{k}^{2}+m^{2})\tilde{\phi}(t,\vec{k})=0##. I'm not sure how to proceed from there onwards. Can you help you me out?
     
  2. jcsd
  3. Mar 30, 2016 #2
    Your ##\partial^2## is not just the three dimensional Laplacian; it also contains a second-order derivative in time (with an appropriate sign depending on your choice of metric).
     
  4. Mar 31, 2016 #3

    vanhees71

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    Obviously, here it's the west-coast metric, i.e., ##(\eta_{\mu \nu})=\mathrm{diag}(+1,-1,-1,-1)##. Your symbol is quite uncommon usually one writes
    $$\Box=\partial_{\mu} \partial^{\mu}=\partial_t^2-\partial_x^2-\partial_y^2-\partial_z^2.$$
    The free Klein Gordon equation then reads
    $$(\Box+m^2)\varphi=0.$$
    Now plug in your Fourier ansatz
    $$(\Box+m^2) \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3} \tilde{\varphi}(t,\vec{p}) \exp(\mathrm{i} \vec{p} \cdot \vec{x}).$$
    Then you get
    $$(\partial_t^2 + E_{\vec{p}}^2) \tilde{\varphi}(t,\vec{p})=0, \quad E_{\vec{p}}=\sqrt{m^2+\vec{p}^2},$$
    which is easy to solve.
     
  5. Mar 31, 2016 #4
    Why is the energy ##E_{\vec{p}}## constrained to take on positive values only?

    In other words, why isn't the Klein-Gordon defined for ##E_{\vec{p}}=-\sqrt{m^2+\vec{p}^2}##?
     
  6. Mar 31, 2016 #5

    vanhees71

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    Sure it is, and you have to write down the complete solution!
     
  7. Mar 31, 2016 #6
    Well,

    ##(\partial_t^2 + E_{\vec{p}}^2) \tilde{\varphi}(t,\vec{p})=0, \quad E_{\vec{p}}=\sqrt{m^2+\vec{p}^2},##

    is the equation for simple harmonic equation. Therefore, it has the solution

    ##\tilde{\varphi}(t,\vec{p}) = A(\vec{p}) \exp{(-iE_{\vec{p}}t)} + B(\vec{p}) \exp{(iE_{\vec{p}}t)}##,

    where ##A(\vec{p})## and ##B(\vec{p})## are arbitrary constants of integration.

    But, isn't this solution only valid for ##E_{\vec{p}}=\sqrt{m^2+\vec{p}^2}##?
     
  8. Mar 31, 2016 #7

    vanhees71

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    No, why should it be? Of course, you are on the right track. It's one of the points that show that a wave-function interpretation in the same sense as for the non-relativistic Schrödinger equation is troublesome and that's why one switches to quantum field theory.

    The trouble is that the energy spectrum of the free Klein-Gordon equation would lead to the conclusion that energy is not bounded from below and thus that there's not a stable ground state. This problem occurs with any relativistic single-particle wave equation, and thus one switches to QFT, where the negative-frequency solutions are reinterpreted as positive-energy solutions for anti-particles, making ##A(\vec{p})## and ##B(\vec{p})## to annihilation and creation operators in the mode decomposition of the field operator:
    $$A(\vec{p}) \rightarrow \hat{a}(\vec{p}), \quad B(\vec{p}) \rightarrow \hat{b}^{\dagger}(-\vec{p}).$$
    This is the formal version of the socalled "Feynman Stückelberg trick", which clearly shows that it has nothing to do with something running "backwards in time" as is often written in popular-science books!
     
  9. Mar 31, 2016 #8
    Does the Klein-Gordon theory for a single real scalar field admit the existence of antiparticles? From what I've learnt, antiparticles arise only in the Klein-Gordon theory for two real scalar fields (or a complex scalar field).
     
  10. Mar 31, 2016 #9

    vanhees71

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    If the original field is real, you have neutral particles and ##\hat{b}=\hat{a}##.
     
  11. Mar 31, 2016 #10
    So, if the field is real, you can't really interpret the negative frequency solutions as positive energy solutions for antiparticles, can you?

    Does that mean that there no negative energy solutions for a real scalar field?
     
  12. Apr 1, 2016 #11

    vanhees71

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    No, but the Feynman-Stückelberg trick still works. Since you write a creation operator instead of a annihilation operator in front of the negative-frequency modes, which lead to an interpretation of the negative-frequency mode as a contribution to the field with positive energy,
    $$\hat{\phi}(t,\vec{x})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x)+\hat{a}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x)]|_{p^0=+E_{\vec{p}}}.$$
    The here chosen normalization condition leads to the most simple commutation relations for the annihilation and creation operators
    $$[\hat{a}(\vec{p}),\hat{a}(\vec{p}')]=0, \quad [\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=\delta^{(3)}(\vec{p}-\vec{p}').$$
     
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