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Can anyone see how the solution to the differential equation $$\frac{dc(t)}{dt} = -c(t)\bigg\langle \psi(t) \left \lvert \frac{\partial \psi(t)}{\partial t } \bigg\rangle \right .$$ has the solution $c(t) = c(0)e^{i \gamma(t)}$ where $$\gamma(t) := i \int_{0}^{t} \bigg\langle \psi(t') \left \lvert \frac{\partial}{\partial t'} \psi(t') \bigg\rangle dt' \right. .$$

I get the solution $c(t) = c(0)e^{-\left \langle \psi \, \left \lvert \frac{\partial \psi}{\partial t} \right \rangle t \right .}$. Where $\langle \cdot, \cdot \rangle$ is the inner product of the functions.

Thanks for any assistance.

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    $\begingroup$ Differentiating your solution doesn't yield the original ODE, as $\psi$ is a function of $t$. This is, however, just a separable ODE. $\endgroup$ – Mattos Apr 29 '17 at 13:47
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Define

$$ f(t) = i\langle \psi(t) | \partial_t \psi(t) \rangle $$ Such that

$$ \frac{{\rm d}c(t)}{{\rm d}t} = if(t) c(t) $$

which can be rearranged as

$$ \frac{{\rm d}c(t)}{c(t)} = if(t){\rm d}t $$

And integrating at both sides

$$ \ln \frac{c(t)}{c(0)} = i\int_0^t {\rm d}t'f(t') = i\gamma(t) ~~~\Rightarrow~~~ c(t) = c(0)e^{i\gamma(t)} $$

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  • $\begingroup$ I just want to confirm that it would not make a difference that the inner product is define as $\langle \phi | \psi \rangle = \int\phi^*\psi dx$. That would just imply that $f(t)$ is a constant function, which doesn't alter the result, right? $\endgroup$ – Alex Apr 29 '17 at 13:55
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    $\begingroup$ @Alex It wouldn't even make a difference if you just replaced the inner product entirely with, say, $h(t)$. $\endgroup$ – J.G. Apr 29 '17 at 13:56
  • $\begingroup$ @Alex Not a constant function, but a function of only $t$ $\endgroup$ – caverac Apr 29 '17 at 13:57
  • $\begingroup$ @caverac No sorry it was a typo. I meant integral over $t$ not $x$ hence a constant function. $\endgroup$ – Alex Apr 29 '17 at 14:10
  • $\begingroup$ @Alex I do not know the context of this question, but my feeling is that $\psi = \psi(x,t)$, and the inner product is over $x$, hence a function of $t$. But this is just guessing. If you're sure that $\psi = \psi(t)$ then you are right, the inner product is just a number $\endgroup$ – caverac Apr 29 '17 at 14:15

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