# Derivative of complex exponential differs by a sign

1. Sep 30, 2015

### 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.

2. Sep 30, 2015

### Summer95

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

3. Sep 30, 2015

### Staff: Mentor

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

\hbar -- $\hbar$

4. Sep 30, 2015

### Summer95

again

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5. Sep 30, 2015

### 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.

6. Oct 1, 2015

### Staff: Mentor

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.

7. Oct 1, 2015

### 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)

8. Oct 1, 2015

### Geofleur

Oh I see! Oops...

9. Oct 1, 2015

### Summer95

ok I see now, of course. thank you!