- #1
yjc
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The short version of this question:
Suppose we have the differential equation ## \partial^2_t \phi (x,t) = -k \phi (x,t) ##. Is it ever true that
$$i \partial_t \phi (x,t) = \sqrt{k} \phi (x,t) $$
In the one-parameter case with ##\phi= \phi (t) ##, the general solution is
$$\phi = Ae^{-i\sqrt{k}t} + Be^{i\sqrt{k}t}$$
Then we get
$$i \partial_t \phi (t) = A\sqrt{k}e^{-i\sqrt{k}t} - B\sqrt{k}e^{i\sqrt{k}t} \neq \sqrt{k} \phi (t)$$
, so we end up with a sign change on the second term, so the equation never holds. However, a book I'm using suggests that it is true. Am I making a very simple mistake?
The full context of my question is this. We're given
1. ##\partial_t^2 \phi = (\vec{\nabla}^2 - m^2) \phi##
2. ##i \langle 0 \vert \partial_t \phi \vert \psi \rangle = ... = \langle 0 \vert \sqrt{m^2 - \vec{\nabla}^2} \phi \vert \psi \rangle ##
The omitted steps is when the text expands out the full form of the solution ##\phi = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{ipx} + a^{\dagger}_p e^{ipx})##
(In case you're wondering, the ##t##-dependence is inside the ##x = (t,x_1,x_2,x_3)##.
Suppose we have the differential equation ## \partial^2_t \phi (x,t) = -k \phi (x,t) ##. Is it ever true that
$$i \partial_t \phi (x,t) = \sqrt{k} \phi (x,t) $$
In the one-parameter case with ##\phi= \phi (t) ##, the general solution is
$$\phi = Ae^{-i\sqrt{k}t} + Be^{i\sqrt{k}t}$$
Then we get
$$i \partial_t \phi (t) = A\sqrt{k}e^{-i\sqrt{k}t} - B\sqrt{k}e^{i\sqrt{k}t} \neq \sqrt{k} \phi (t)$$
, so we end up with a sign change on the second term, so the equation never holds. However, a book I'm using suggests that it is true. Am I making a very simple mistake?
The full context of my question is this. We're given
1. ##\partial_t^2 \phi = (\vec{\nabla}^2 - m^2) \phi##
2. ##i \langle 0 \vert \partial_t \phi \vert \psi \rangle = ... = \langle 0 \vert \sqrt{m^2 - \vec{\nabla}^2} \phi \vert \psi \rangle ##
The omitted steps is when the text expands out the full form of the solution ##\phi = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{ipx} + a^{\dagger}_p e^{ipx})##
(In case you're wondering, the ##t##-dependence is inside the ##x = (t,x_1,x_2,x_3)##.