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A Complex scalar field - commutation relations

  1. Sep 18, 2016 #1
    I find it difficult to believe that the canonical commutation relations for a complex scalar field are of the form

    ##[\phi(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})##
    ##[\phi^{*}(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})##

    This seems to imply that the two scalar fields ##\phi## and ##\phi^{*}## are somehow coupled, even though they are not.

    Can you explain this?
     
  2. jcsd
  3. Sep 18, 2016 #2
    Where did you get these commutation relations from? The notes that I have seem to suggest otherwise.
     
  4. Sep 18, 2016 #3
    I assumed that these are the fundamental commutation relations for complex scalar field and using the commutation relations

    ##[a_{{\bf{p}}},a_{{\bf{p}}'}^{\dagger}]=(2\pi)^{3}\delta^{(3)}({\bf{p}}-{\bf{p}}')##
    ##[b_{{\bf{p}}},b_{{\bf{p}}'}^{\dagger}]=(2\pi)^{3}\delta^{(3)}({\bf{p}}-{\bf{p}}')##

    and the expansion of the field ##\phi## as

    ##\phi=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega_{\bf{p}}}}(a_{\bf{p}}e^{i{\bf{p}}\cdot{\bf{x}}}+b_{\bf{p}}^{\dagger}e^{-i{\bf{p}}\cdot{\bf{x}}})##

    I showed that the commutation relations in my first post hold true.

    What commutation relations do you have? Can you post them?

    I need to check my calculations using both mine and yours commutation realtions?
     
  5. Sep 18, 2016 #4
    Did you perhaps forget that [itex]\pi (x) = \dot{\phi^{*}}(x)[/itex] and not [itex]\dot{\phi}(x)[/itex] ?

    I have the commutation relations:
    ##[\phi(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})##
    ##[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})##

    from the same starting point as you have for the creation/annihilation operators
     
  6. Sep 18, 2016 #5
    I do have ##\pi=\dot{\phi}^{*}##.
     
  7. Sep 18, 2016 #6
    Hmm...might you have forgotten to take the Hermitian conjugate of the creation/annihilation operators when obtaining ##\phi^{*}##? Otherwise I think I will need to see the explicit expressions for your fields and conjugate momenta to figure out where you went wrong.
     
  8. Sep 18, 2016 #7
    I started with the two commutation relations ##[\phi^{i}(\vec{x}),\pi_{i}(\vec{y})]=i\delta^{3}(\vec{x}-\vec{y})## for the free real scalar fields ##\phi^{1}## and ##\phi^{2}##

    and plugged ##\phi = \frac{1}{\sqrt{2}}(\phi^{1}+i\phi^{2})## and the corresponding derivative into the two commutation relations to obtain a set of two equations in terms of ##\phi##, ##\phi^{\dagger}##, ##\pi## and ##\pi^{\dagger}##.

    I added these two equations and also subtracted the same two equations to obtain a new set of two equations as follows:

    ##[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi(\vec{x}),\pi(\vec{y})]=2i\delta^{3}(\vec{x}-\vec{y})##

    ##[\phi(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi^{\dagger}(\vec{x}),\pi(\vec{y})]=0##

    It's obvious now that your commutation relations are a possible solution to these equations, but I want to be sure that yours is the only solution.

    With that in mind, let me proceed.

    ##[\phi^{\dagger}(\vec{x}),\pi^{\dagger}(\vec{y})]+[\phi(\vec{x}),\pi(\vec{y})]=2i\delta^{3}(\vec{x}-\vec{y})##

    ##\implies [\phi(\vec{x}),\pi(\vec{y})]-[\phi(\vec{x}),\pi(\vec{y})]^{\dagger}=2i\delta^{3}(\vec{x}-\vec{y})##

    I'm now only slightly away from proving that ##[\phi(\vec{x}),\pi(\vec{y})]=i\delta^{3}(\vec{x}-\vec{y})##.

    How do I show that ##-[\phi(\vec{x}),\pi(\vec{y})]^{\dagger}=[\phi(\vec{x}),\pi(\vec{y})]##?
     
    Last edited: Sep 18, 2016
  9. Sep 19, 2016 #8
    This is a strange way to proceed; usually its done by the method you put in post #3 - starting from the expression of ##\phi##, generate ##\phi^{*}## and their conjugate momenta, and use the commutation relations of the creation and annihilation operators to simplify the integrals to get the desired results.

    Your alternative method still works, but why not consider ##[\phi(x), \pi(y)]## directly? That, for example, immediately yields
    [tex][\phi(x), \pi(y)] = \left[\frac{1}{\sqrt{2}} \left(\phi_{1}(x) + i \phi_{2}(x)\right) , \frac{1}{\sqrt{2}} \left(\pi_{1}(y) - i \phi_{2} (y)\right) \right] = \frac{1}{2} \left(
    \left[\phi_{1}(x),\pi_{1}(y) \right] + \left[\phi_{2}(x),\pi_{2}(y) \right]
    \right) = i \delta^{(3)} (x-y)[/tex]
     
  10. Sep 19, 2016 #9
    This method requires that I guess an expansion for ##\pi## in terms of the creation and annihilation operators and then check that it satisfies the commutation relations for ##\phi## with ##\pi## as well as for ##\phi^{*}## with ##\pi^{*}##.

    I was wanting to avoid this guesswork, so ...

    This is a better approach, and I will use it now. But, the commutators for real ##\phi^{i}## with real ##\pi_{i}## directly yield the non-zero commutators. If I start with the commutators for complex ##\phi## and ##\pi##, I have to check all the commutators one by one to find the non-zero ones.
     
  11. Sep 19, 2016 #10
    If you start with an expression for ##\phi## already, there's no need for guesswork - ##\phi^{*}## and the corresponding conjugate momenta can be obtained from it.
     
  12. Sep 19, 2016 #11

    vanhees71

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    It's all clearly defined. The (free) charged Klein-Gordon field has the Lagrangian (east-coast convention)
    $$\mathcal{L}=(\partial_{\mu} \phi^*)(\partial^{\mu} \phi)-m^2 \phi^* \phi.$$
    The canonical field momenta for the independent field degrees of freedom ##\phi## and ##\phi^*## (you can as well rewrite everything in terms of two real KG fields of course) are
    $$\Pi=\frac{\partial \mathcal{L}}{\partial \dot{\phi}}=\dot{\phi}^*, \quad \Pi^*=\frac{\partial \mathcal{L}}{\partial \dot{\phi}^*}=\dot{\phi},$$
    and the equal-time commutator relations are
    $$[\phi(t,\vec{x}),\Pi(t,\vec{y})]=\mathrm{i} \delta^{(3)}(x-y), \quad [\phi^{\dagger}(t,\vec{x}),\Pi^{\dagger}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(x-y).$$
     
  13. Sep 19, 2016 #12
    I know how to obtain ##\phi^{*}## from ##\phi##, but I'm not sure how to obtain ##\pi=\dot{\phi}^{*}## from ##\dot{\phi}^{*}##, since

    ##\displaystyle{\pi=\dot{\phi}^{*}=\int\ \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega_{{\bf{p}}}}}\Big(\dot{b}_{\vec{p}}\ e^{i\ \vec{p}\cdot{\vec{x}}} + \dot{a}_{\vec{p}}^{\dagger}\ e^{-i\ \vec{p}\cdot{\vec{x}}} \Big)}##.

    But what do you do next to obtain an expansion for ##\pi##?
     
  14. Sep 19, 2016 #13

    vanhees71

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    Don't forget the Heisenberg-picture equations of motion. The time dependence is ##\hat{a}_{\vec{p}}(t) = \hat{a}_{\vec{p}} \exp(-\mathrm{i} E_{\vec{p}} t)## and ##\hat{b}_{\vec{p}}(t) = \hat{b}_{\vec{p}} \exp(-\mathrm{i} E_{\vec{p}} t)##.
     
  15. Sep 19, 2016 #14
    I knew that this was exactly what I was supposed to do, but for some reason, I did not attempt it.

    Thanks!
     
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