Question on discrete commutation relation in QFT

In summary, the given commutation relation states that the Fourier transforms of the operators ##\phi(t,\vec{x})## and ##\pi(t,\vec{x})## are equal to each other. This can be written as ##\left[\tilde{\phi}(\vec{k}),\tilde{\pi}(\vec{k}')\right]=i\delta_{\vec{k},-\vec{k}'}=i\delta_{-\vec{k},\vec{k}'}##, where the Kronecker deltas are interchangeable. It is important to distinguish between the operators ##\phi(t,\vec{x})## and ##\tilde{\phi}(\vec{k})## to avoid confusion.
  • #1
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Homework Statement
The statement is in title
Relevant Equations
The equations are given below
Given the commutation relation

$$\left[\phi\left(t,\vec{x}\right),\pi\left(t,\vec{x}'\right)\right]=i\delta^{n-1}\left(\vec{x}-\vec{x}'\right)$$

and define the Fourier transform as

$$\tilde{\phi}(t,\vec{k})=\frac{1}{\sqrt{L^{n-1}}}\int_{\,0}^{\,L}\phi(t,\vec{x})e^{-i\vec{k}\cdot\vec{x}}\,\mathrm{d}^{n-1}\vec{x}$$

$$\tilde{\pi}(t,\vec{k})=\frac{1}{\sqrt{L^{n-1}}}\int_{\,0}^{\,L}\pi(t,\vec{x})e^{-i\vec{k}\cdot\vec{x}}\,\mathrm{d}^{n-1}\vec{x}$$

Is it then correct to say the following?

$$\left[\tilde{\phi}(\vec{k}),\tilde{\pi}(\vec{k}')\right]=i\delta_{\vec{k},-\vec{k}'}=i\delta_{-\vec{k},\vec{k}'}$$

i.e. can I use both Kronecker deltas interchangeably?
 
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  • #2
Have you tried to derive the commutator? You also should destinguish the operators ##\phi(t,\vec{x})## from ##\tilde{\phi}(\vec{k})##, because otherwise it leads to confusion. If so, where is your problem. Of course ##\delta_{\vec{k},-\vec{k}'}=\delta_{-\vec{k},\vec{k}'}##.
 

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