從位置空間中的Klein Gordon開始,\ begin {align *} \ left(\ frac {\ partial ^ 2} {\ partial t ^ 2}-\ nabla ^ 2 + m ^ 2 \ right)\ phi (\ mathbf {x},t)= 0 \ end {align *},然後使用傅立葉變換:$ \ displaystyle \ phi(\ mathbf {x},t)= \ int \ frac {d ^ 3p} {(2 \ pi)^ 3} e ^ {i \ mathbf {p} \ cdot \ mathbf {x}} \ phi(\ mathbf {p},t)$:\ begin {align *} \ int \ frac {d ^ 3p } {(2 \ pi)^ 3} \ left(\ frac {\ partial ^ 2} {\ partial t ^ 2}-\ nabla ^ 2 + m ^ 2 \ right)e ^ {i \ mathbf {p} \ cdot \ mathbf {x}} \ phi(\ mathbf {p},t)& = 0 \\\ int \ frac {d ^ 3p} {(2 \ pi)^ 3} e ^ {i \ mathbf {p} \ cdot \ mathbf {x}} \ left(\ frac {\ partial ^ 2} {\ partial t ^ 2} + | \ mathbf {p} | ^ 2 + m ^ 2 \ right)\ phi(\ mathbf {p },t)& = 0 \ end {align *}現在我不明白為什麼我們能夠擺脫積分,而只剩下\ begin {align *} \ left(\ frac {\ partial ^ 2 } {\ partial t ^ 2} + | \ mathbf {p} | ^ 2 + m ^ 2 \ right)\ phi(\ mathbf {p},t)= 0 \ end {align *}