Prefactor in Canonical Quantization of Scalar Field

In summary, the different normalizations for the ##\hat a## and ##\hat a^\dagger## in Peskin and Srednicki's books are causing confusion for me because they seem to conflict with each other. The commutators of the creation/destruction operators should be the same, but they are not.
  • #1
thatboi
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Hey all,
I am encountering an issue reconciling the choice of prefactors in the canonical quantization of the scalar field between Srednicki and Peskin's books. In Peskin's book (see equation (2.47)), there is a prefactor of ##\frac{1}{\sqrt{2E_{p}}}## whereas in Srednicki's book (see equation (3.18) and (3.19)), there is a prefactor of ##\frac{1}{2\omega}##. What concerns me is that if we take the derivative with respect to time of the field, then in Peskin's case, we are left with a ##\sqrt{E_{p}}## factor whereas in Srednicki's book, the ##\frac{1}{\omega}## prefactor completely disappears, so I fail to see how these 2 definitions can be equivalent.
Thanks.
 
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  • #2
Compare also the commutators of creation/destruction operators, they should be different too, so that at the end the commutators between the field and its time derivative are the same.
 
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  • #3
Yes, different books uses different normalizations for the ##\hat a## and ##\hat a^\dagger##
 
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  • #4
Great, thanks a lot. My confusion initially came from when I was looking through Itzykson's QFT book and came upon this discussion on pg. 521:
1679847504828.png

Specifically, looking at how the scalar field is quantized in equation (11.39), it seems to me that if we used Peskin's definition of normalization, then the energy term wouldn't cancel out after we take the derivative with respect to time and thus we couldn't evaluate the integral in equation (11.41) right (since the energy term is also necessarily a function of the spatial coordinates of momentum).
 
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  • #5
It depends, at which place you want to have it convenient. If you want the creation and annihilation operators to create momentum eigenstates normalized to 1, you need the ##1/\sqrt{2 E_{\vec{p}}}## factors in the mode decomposition. If you want manifestly covariant integrals in the mode decomposition you need the ##1/(2 E_{\vec{p}})## factors. That's because ##\mathrm{d}^2 p/(2 E_{\vec{p}})## is mainfestly Lorentz invariant.

The correct factor of the field is of course always uniquely defined by the equal-time (anti-)commutator relations,
$$[\hat{\Phi}(t,\vec{x}),\hat{\Pi}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
 
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