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I have a few issues with making the transition between these:

$\phi(x)=\int{\frac{d^3p}{2\pi^3}\frac{1}{\sqrt{2\omega_{\vec{p}}}}(a_{\vec{p}}e^{i \vec{p} \vec{x}}+ a^{\dagger}_{\vec{p}}e^{-i \vec{p} \vec{x}}})$

$\pi(x)=\int{\frac{d^3p}{2\pi^3}(-i)\sqrt{\frac{\omega_{\vec{p}}}{2}}(a_{\vec{p}}e^{i \vec{p} \vec{x}}- a^{\dagger}_{\vec{p}}e^{-i \vec{p} \vec{x}}})$.

So from the Lagrangian of the real scalar field, we get

$\pi(x)=\partial_0 \phi(x)$

which just means differentiating our $\phi(x)$ w.r.t time. But where does the $-i$ come from (as in the minus sign)? Also aren't the exponentials independent of time as $\vec{p} \vec{x}$ isn't a four vector? So now I'm wondering where the $\omega_{\vec{p}}$ has come from as well!

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You should write $p \cdot x$, instead of $\vec{p} \cdot \vec{x}$, because these are 4-vectors. –  Melquíades May 4 at 11:40

2 Answers 2

up vote 6 down vote accepted

The answer depends on whether you represent the operators in the Schrödinger or Heisenberg picture. In the Schrödinger picture, there is no explicit time-dependence of the operators. Hence, the conjugate momentum is not given by the time derivative of the field. The concrete form is rather given by analogy to the position and momentum operators of a harmonic oscillator:

$$x=\frac{1}{\sqrt{2\omega}}(a+a^\dagger)$$ $$p=-i\sqrt{\frac{\omega}{2}}(a-a^\dagger).$$

In the Heisenberg picture, the operators acquire an explicit time-dependence, i.e.

$$\phi(x)\rightarrow e^{iHt}\phi(x) e^{-iHt}.$$

In this case, taking the derivative with respect to time makes sense, and the relation between the field and its conjugate momentum can be shown to hold.

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But can't we just use $\pi(x)=\partial_0 \phi(x)$ like my questions asks? –  user13223423 May 4 at 11:21
    
No. At this point, the operator $\phi(x)$ does not depend explicitly on time. –  Frederic Brünner May 4 at 11:23
    
Ah ok I understand now! So when, for example again, Peskin and Schroeder just state $\phi(\vec{x})$ and $\pi(\vec{x})$, there isn't an easy way to get one from the other? –  user13223423 May 4 at 11:25
    
@user13223423: That is how I understood it. –  Frederic Brünner May 4 at 11:26

I think your confusion arises from a misunderstanding of the mode expansion of the field.

For a real scalar field we usually write the Fourier expansion as $$\phi(x)=\int \frac{d^3k}{(2\pi)^3}a^\dagger v_k e^{i\vec{k}\cdot\vec{x}}+a v_k^* e^{-i\vec{k}\cdot\vec{x}}$$ However, now we still have to find the mode functions $v_k$ and $v_k^*$! These are solutions to the Klein-Gordon equation in Fourier space: $$(\partial_t^2+\omega^2)\phi_k=0,\hspace{2.5cm}\omega^2=|\vec{k}|^2+m^2 $$ Now, you can easily check that $$v_k=\frac{1}{\sqrt{2\omega}}e^{-i\omega t}$$ is a solution (the constant is there because it will make the measure of the mode expansion Lorentz invariant). This leads to the full mode expansion $$\phi(x)=\int \frac{d^3k}{(2\pi)^3\sqrt{2\omega}}a^\dagger e^{ikx}+a e^{-ikx} $$ where $kx=k_\mu x^\mu=\vec{k}\cdot\vec{x}-\omega t$. As you see, your problem was not realizing that the exponents DO contain four-vectors even though it's just a Fourier expansion in the spatial dimension.

A very clear discussion of mode functions in flat space as well as the more general case when spacetime is curved is found in chapter 9 of Carroll's book on general relativity. I also suspect that most QFT books will have at least some sort of discussion of this result.

EDIT: As mentioned in another answer, I should have specified that I will be working in the Heisenberg picture, as is conventional in QFT.

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Thanks for the answer. But why do Peskin and Schroeder, for example, have $\vec{p}\vec{x}$ in the exponent rather than $px$ (dot product of course)? –  user13223423 May 4 at 11:16
    
@user13223423 I'm not quite sure what you are referring to, but it might be as simple as an issue with notation. –  Danu May 4 at 11:17
    
Well $px=Et-\vec{p}\vec{x}$, so it's important to know which we're using. –  user13223423 May 4 at 11:19
    
@user13223423: My answer is consistent with the notation in Peskin and Schroeder, where the distinction between the two pictures is made. –  Frederic Brünner May 4 at 11:19

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