- #1
spaghetti3451
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The quantum Klein-Gordon field ##\phi({\bf{x}})## and its momentum density ##\pi({\bf{x}})## are given in Fourier space by
##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)## and
##\pi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} (-i) \sqrt{\frac{ \omega_{{\bf{p}}}}{2}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} - a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##.
[These are equations (2.25) and (2.26) from the Peskin and Schroeder.]
Now, I used the substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained
##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = a^{\dagger}_{{\bf{p}}}##.
On the other hand, I used the same substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained
##a_{{\bf{p}}} = - a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = - a^{\dagger}_{{\bf{p}}}##.
Can someone explain what's going on?
##\phi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2 \omega_{{\bf{p}}}}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} + a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)## and
##\pi({\bf{x}}) = \int \frac{d^{3}p}{(2\pi)^{3}} (-i) \sqrt{\frac{ \omega_{{\bf{p}}}}{2}} \big( a_{{\bf{p}}} e^{i{\bf{p}} \cdot {\bf{x}}} - a^{\dagger}_{{\bf{p}}} e^{-i{\bf{p}} \cdot {\bf{x}}} \big)##.
[These are equations (2.25) and (2.26) from the Peskin and Schroeder.]
Now, I used the substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained
##a_{{\bf{p}}} = a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = a^{\dagger}_{{\bf{p}}}##.
On the other hand, I used the same substitution ##{\bf{p}} \rightarrow {\bf{-p}}## in the expression for ##\phi({\bf{x}})## and obtained
##a_{{\bf{p}}} = - a^{\dagger}_{-{\bf{p}}}## and ##a_{-{\bf{p}}} = - a^{\dagger}_{{\bf{p}}}##.
Can someone explain what's going on?
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