Derivative of complex exponential differs by a sign

[Summer95]

I know this is probably the least of my worries at the moment but my quantum textbook solves $\frac{\mathrm{d}\phi (t) }{\mathrm{d} t}=\frac{iC}{h}\phi (t)$ as $\phi (t) = e^{-i(\frac{C}{h})t}$. Is this not off by a sign? Its really bugging me.

 

[Summer95]

And the h is really h bar but I don't know the latex code for it

 

[Mark44]

I know this is probably the least of my worries at the moment but my quantum textbook solves $\frac{\mathrm{d}\phi (t) }{\mathrm{d} t}=\frac{iC}{h}\phi (t)$ as $\phi (t) = e^{-i(\frac{C}{h})t}$. Is this not off by a sign? Its really bugging me.

I don't see how they got the minus sign. Can you post a picture of the textbook page?

And the h is really h bar but I don't know the latex code for it

\hbar -- $\hbar$

 

[Summer95]

again

 

[Geofleur]

Write $i\hbar\frac{1}{\phi}\frac{d\phi}{dt} = C$ as $\frac{d}{dt}\ln\phi = \frac{C}{i\hbar}$. Now multiply the RHS by one in disguise as $\frac{i}{i}\frac{C}{i\hbar} = \frac{iC}{i^2\hbar} = -\frac{iC}{\hbar}$. We get $\frac{d}{dt}\ln\phi = -\frac{iC}{\hbar}$. Integrate both sides to get $\ln\phi = -\frac{iCt}{\hbar} + A$, where $A$ is some constant. Finally, exponentiate both sides to get $\phi = Be^{-\frac{iCt}{\hbar}}$, where $B$ is a constant.

 

[Mark44]

Write $i\hbar\frac{1}{\phi}\frac{d\phi}{dt} = C$ as $\frac{d}{dt}\ln\phi = \frac{C}{i\hbar}$. Now multiply the RHS by one in disguise as $\frac{i}{i}\frac{C}{i\hbar} = \frac{iC}{i^2\hbar} = -\frac{iC}{\hbar}$. We get $\frac{d}{dt}\ln\phi = -\frac{iC}{\hbar}$. Integrate both sides to get $\ln\phi = -\frac{iCt}{\hbar} + A$, where $A$ is some constant. Finally, exponentiate both sides to get $\phi = Be^{-\frac{iCt}{\hbar}}$, where $B$ is a constant.

I don't think the problem was how to solve the differential equation, but rather how the text went from one DE to another. In the image of post #4, the text has

$$i\hbar\frac 1 {\phi(t)} \frac{d \phi(t)}{dt} = C \text{ or } \frac{d \phi(t)}{dt} = \frac{iC}{\hbar} \phi(t)$$

@Summer95, I believe the text is being a little sloppy here -- C in the left equation isn't the same as C in the right equation; the two differ by a sign. That's what I think is going on here.

 

[Ssnow]

If the $C$ is the same the second equation needs a minus in front the right side ... ( in any case the disjunction is true and also the solution given)

 

[Geofleur]

Oh I see! Oops...

 

[Summer95]

I don't think the problem was how to solve the differential equation, but rather how the text went from one DE to another. In the image of post #4, the text has@Summer95, I believe the text is being a little sloppy here -- C in the left equation isn't the same as C in the right equation; the two differ by a sign. That's what I think is going on here.

ok I see now, of course. thank you!