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Commutation relation of the creation/annihilation operators in a field

  1. Aug 23, 2013 #1
    Hello, I'm having trouble calculating this commutator, at the moment I've got:

    [itex]\left[a_{p},a_{q}^{\dagger}\right]=\left[\frac{i}{\sqrt{2\omega_{p}}}\Pi(p)+\sqrt{\frac{w_p}{2}}\Phi(p),\frac{-i}{\sqrt{2\omega_{p}}}\Pi(p)+\sqrt{\frac{w_p}{2}}\Phi(p)\right]=i\left[\Pi(p),\Phi(q)\right]=i\int d^{3}x d^{3}ye^{-i(\vec{p}\bullet\vec{x}+\vec{q}\bullet\vec{y})}\left[\Pi(x),\Phi(y)\right]=\int d^{3}x d^{3}ye^{-i(\vec{p}\bullet\vec{x}+\vec{q}\bullet\vec{y})}\delta^{3}(x-y)=\int d^{3}x e^{-i(\vec{p}+\vec{q})\bullet\vec{x}}=(2\pi)^{3}\delta^{3}(p+q)[/itex]

    But I should get [itex](2\pi)^{3}\delta^{3}(p-q)[/itex] instead. Where have I made the mistake?
     
    Last edited: Aug 23, 2013
  2. jcsd
  3. Aug 23, 2013 #2
    I'm a little puzzled by your expressions for a and a† in the second part of your equation. Where did you get those from? Typically, you define the creation and annihilation operators a and a† implicitly sort of like "fourier components" of Φ(x), and inverting the definitions to get an explicit expression for them is not nearly as pretty.

    I'm not even entirely sure exactly the problem you're doing: the 1D scalar field/Klein Gordon field? If I'm assuming correctly, then the computation you are trying to do is actually usually a bit more messy because you don't have such nice expressions for a and a†.

    Ryder does a good job of showing that calculation and the necessary techniques. I actually have never seen a better derivation than in Ryder (and I have seen many nice looking but actually incorrect derivations by my classmates). Attached is the excerpt from Ryder's "Quantum Field Theory" 2nd ed. where he does it right [the equations he mentions that aren't in that image, 4.9 and 4.10 are just the commutators of ∏ and Φ and their definition].

    Anyway it's totally possible that you're doing it some other way that I don't know about. I don't often see ∏(p) and Φ(p)--that's an unusual aspect of what you have written.
     

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    Last edited: Aug 23, 2013
  4. Aug 23, 2013 #3
    Sorry, yeah it's the Klein-Gordon field (in 3+1 dimensions).

    The expressions I got from the quantum harmonic oscillator, and the fact that the KG field can be considered as one of these per momentum. I haven't verified it in any literature actually, but it made sense, and gave me the right Hamiltonian.
     
  5. Aug 23, 2013 #4

    vanhees71

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    That's a nice idea, but it doesn't work that easily, because the KG equation is of 2nd order in time. You have to analyze the situation from the very beginning using the Hamiltonian. Let's take the neutral KG field. Then you have the Lagrangian
    [tex]\mathscr{L}=\frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi)-\frac{m^2}{2} \phi^2.[/tex]
    The canonical field momentum is
    [tex]\frac{\partial \mathscr{L}}{\partial \dot{\phi}}=\dot{\phi}.[/tex]
    The field equation is the Klein Gordon equation, and you have
    [tex](\Box+m^2) \phi=0,[/tex]
    implying that the normal modes are plane waves
    [tex]u_{\vec{p}}(x)=\left . \frac{1}{(2 \pi)^{3/2} \sqrt{2 E(\vec{p})}} \exp(-\mathrm{i} p_{\mu} x^{\mu}) \right|_{p^0=E(\vec{p})}[/tex]
    with the on-shell energy
    [tex]E(\vec{p})=+\sqrt{m^2+\vec{p}^2}.[/tex]
    The expansion in annihilation and creation operators wrt. to energy-momentum eigenmodes thus reads
    [tex]\phi(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [a(\vec{p}) u_{\vec{p}}(x) + a^{\dagger}(\vec{p}) u_{\vec{p}}^*(x)].[/tex]
    The inverse of this Fourier-like decomposition reads
    [tex]a(\vec{p})=\mathrm{i} \int_{\mathrm{R}^3} \mathrm{d}^3 \vec{x} [u_{\vec{p}}(x) \overleftrightarrow{\partial}_t \phi(x)],[/tex]
    where for any two functions one defines
    [tex]A(x) \overleftrightarrow{\partial}_t B(x)=A(x) \dot{B}(x)-\dot{A}(x) B(x).[/tex]
    Now you use the equal-time-commutation relations
    [tex][\phi(t,\vec{x}),\phi(t,\vec{y})]=[\dot{\phi}(t,\vec{x}),\dot{\phi}(t,\vec{y})]=0, \quad [\phi(t,\vec{x}),\dot{\phi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y})[/tex]
    to arrive at the commutation relations for [itex]a(\vec{p})[/itex] and [itex]a^{\dagger}(\vec{p})[/itex], i.e.,
    [tex][a(\vec{p}),a(\vec{k})]=0, \quad [a(\vec{p}),a(\vec{k})]=\delta^{(3)}(\vec{p}-\vec{k}).[/tex]
    The calculation is a bit lengthy, but I'm sure, you'll manage it :-).
     
  6. Aug 23, 2013 #5
    Lengthy is a good word to describe this calculation. It's not very hard if you have Ryder or vanhees' tips [which are almost identical], but as I recall it's about two pages to fill in all the steps [mostly since the equations get very long].
     
    Last edited: Aug 23, 2013
  7. Aug 26, 2013 #6
    Thank you for the help, and yeah it was quite lengthy/mistake-prone. I think I found my fundamental mistake. It was in not changing the sign of the i inside the exponential of the Fourier transform of Pi and Phi for the a_dagger. Incidentally, following your definition, I actually arrive to the same expressions I got in my first post, where [itex]\Pi(\textbf{p})=\int d^{3}x e^{-i(\textbf{p}\cdot \textbf{x})} \Pi(\textbf{x})[/itex]
    But, as I said my missing minus sign is in the fact that for the a_dagger, [itex]\Pi^{\dagger}(\textbf{p})=\int d^{3}x e^{i(\textbf{p}\cdot\textbf{x})} \Pi(\textbf{x})[/itex], which when integrated from -inf to inf, gives the same result, but when in an expression like this, doesn't! Also, you were using [itex]\mu[/itex] indices, which I think means that you are in the Heisenberg picture, while I am in the Schrödinger pitcture, but, of course, it doesn't change the result)
     
    Last edited: Aug 26, 2013
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