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Using Minkowski signature $(+,-,-,-)$, for the Lagrangian density

$${\cal L}=\partial_{\mu}\phi\partial^{\mu}\phi^{\dagger}-m^2\phi \phi^{\dagger}$$

of the complex scalar field, we have the field

$$\phi(x)=\int{\frac{d^3 \vec{p}}{2(2\pi)^3\omega_\vec{p}}}(a(\vec{p})e^{-ipx}+b^{\dagger}(\vec{p})e^{+ipx}).$$

I'm trying to now find an equation for the $a(\vec{p})$ and $b(\vec{p})$ (with the final goal of finding an expression for $[a(\vec{p}),b^{\dagger}(\vec{q})]$ using $[\phi(x),\Pi(y)]$).

What I should note is that we are considering all of this in the Schrodinger picture (t=0) so I suppose the very first thing to do is change all the $x$'s to $\vec{x}$ right?

The strategy I'm struggling to implement and failing at many points along the way:

  1. Find the momentum $\Pi^{\phi}(x)=\frac{\partial L}{\partial \dot{\phi}}=\dot{\phi^{\dagger}}$.

  2. Add some combination of $\phi(x)$ and $\Pi(x)$ to get rid of one of the creation/annihilation operators.

  3. Do an inverse Fourier transform to find $a(\vec{p})$ in terms of $\phi(x)$, for example.

None of the major text books seem to actually carry this through, and instead write something like "and it's easy to show…". However I don't find it too easy, especially part 3. as I'm not a Fourier transform expert.

Could anyone either direct me somewhere where the above is computed explicitly (in more than 2/3 lines), or help me understand each of the 3 steps above?

(I realise it is easy to find a reference where this is done for the real scalar field, in which case we have $a(\vec{p})$ and $a^{\dagger}(\vec{p})$. Even still, I find it hard to follow parts.)

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    $\begingroup$ So this seems like a very broad question to me, it might be useful to break it into smaller parts. For example, what do you have for part 1 (computing the momentum)? You should find that the momentum conjugate to $\phi^\dagger$ is $\dot{\phi}$ and that the momentum conjugate to $\phi$ is $\dot{\phi}^\dagger$, is that what you find? $\endgroup$
    – Andrew
    Commented Apr 14, 2014 at 17:41
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    $\begingroup$ Yes but you can easily get that just from looking at the Lagrangian. Perhaps I'll add it in then $\endgroup$
    – Phibert
    Commented Apr 14, 2014 at 17:43
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    $\begingroup$ Good :) Just trying to figure out exactly where you are stuck. OK so you can get the momentum, and you know that $[\phi,\Pi] = i \delta$ on equal time slices. Can you express $\Pi$ as a mode expansion of $a,b$ and their daggers? If so what do you get when you plug the mode expansions of $\phi$ and $\Pi$ into $[\phi,\Pi]=i\delta$? Again it's useful for giving an answer to have a clearer idea of exactly what you tried and where things aren't working. $\endgroup$
    – Andrew
    Commented Apr 14, 2014 at 17:51

1 Answer 1

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The fields satisfy the wave equation. We can therefore write \begin{equation} \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + a^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \end{split} \end{equation} where $\omega_{\bf p} = \sqrt{ {\bf p}^2 + m^2 } $. Inverting this, we find (VERIFY THIS) \begin{equation} \begin{split} a^\dagger ( {\bf p} ) &= - i \int d^3 x e^{i p \cdot x} \overleftrightarrow{\partial_0} \phi^\dagger(x) \\ b^\dagger ( {\bf p} ) &= - i \int d^3 x e^{i p \cdot x} \overleftrightarrow{\partial_0} \phi(x) \\ a ( {\bf p} ) &= i \int d^3 x e^{-i p \cdot x} \overleftrightarrow{\partial_0} \phi(x) \\ b ( {\bf p} ) &= i \int d^3 x e^{-i p \cdot x} \overleftrightarrow{\partial_0} \phi^\dagger(x) \\ \end{split} \end{equation} where $A \overleftrightarrow{\partial_0} B = A \partial_0 B - (\partial_0 A ) B$.

The conjugate momenta can be determined from the Lagrangian as $$ \pi = \partial_0 \phi^\dagger,~~ \pi^\dagger = \partial_0 \phi $$ The commutation relations in terms of the fields are $$ [ \phi(t, {\bf x}), \pi(t, {\bf y}) ] = i \delta^3 ( {\bf x} - {\bf y} ) $$ Using this information, you should be able to compute the brackets of the mode coefficients.

PS - I should add that I'm using the $(-+++)$ signature for the metric.

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    $\begingroup$ Nice answer; most likely the OP will also need the well-known identity: \begin{equation} \delta^n(k) = \int \frac{\mathrm{d}^n x }{(2 \pi)^n} \; e^{i k \cdot x} \end{equation} $\endgroup$
    – Hunter
    Commented Apr 14, 2014 at 18:05
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    $\begingroup$ The conceptually easy but 'algebraically hard' way that I did it in my QFT class was simply plug in the Fourier expansion of the field in the commutation relation. Using all field/momentum commutation relations will give you the right relation between all creationg/annihilation operators, no inverting necessary. $\endgroup$
    – Danu
    Commented Apr 14, 2014 at 18:12
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    $\begingroup$ For example, in the case of Minkowski space, it is standard to choose $\phi_k^\pm = e^{ \mp i k \cdot x }$. The creation and annihilation operators are then defined as $a^\dagger_k = (\phi^+_k, \phi)$ and $a_k = (\phi_k^-, \phi)$. Working out all these explicitly, one obtains the formula that I have described. $\endgroup$
    – Prahar
    Commented Apr 15, 2014 at 1:55
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    $\begingroup$ @newtothis - the factor of $2\omega_p$ is a matter of convention. Some authors (like Weinberg and Tong) like to do the field expansion with $\frac{1}{\sqrt{2 \omega_{\bf p}}}$ instead of what I have. I like this convention (used by Srednicki) because then $a({\bf p})$ transforms nicely under Lorentz transformations, whereas in Weinberg's conventions there are square roots in all the formulas. The bigger problem with your formula however is the sign. It should be $-2 \omega_{\bf p}$. The sign is physically important, the overall factor is a convention. $\endgroup$
    – Prahar
    Commented Jul 12, 2022 at 9:46
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    $\begingroup$ yes! you can easily see this by simply rescaling $a({\bf p}) \to \sqrt{2\omega_{\bf p}} a({\bf p})$. Note that you are also missing a factor of $(2\pi)^3$ somewhere (and the previously mentioned sign which is far more important!) $\endgroup$
    – Prahar
    Commented Jul 12, 2022 at 9:52

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