# Finding most probable position for given wave function

1. Aug 23, 2015

1. The problem statement, all variables and given/known data
A particle has a given wavefunction:
$ψ(x) = C e^{-x}(1-e^{-x})$

(many steps in between)
...
Find the most probable position of the particle

2. Relevant equations
Most probable is where the probability density's derivative = 0. Right?
$P(x) = |ψ(x)|^{2}$

3. The attempt at a solution
I've solved for C, which 2√3.

Then I take the derivative of $P(x)$, and set it to 0:
$-1-2e^{-2x}+3e^{-x}=0$

I've tried factoring this multiple ways and just can't seem to solve for x. I can do it in Mathematica no problem, but I could swear I've done this before and just can't remember the right tools.

2. Aug 23, 2015

### Henriamaa

it is not possible to have the constant in the equation all the terms in the wave functions have the exponential, the probability density has all exponentials so will its derivative.

3. Aug 23, 2015

Um I'm not sure what you meant by all that. The constant C was solved (described in first post).

I figured it out though, and I'm surprised I forgot this:
Replace $e^{x}$ with $y$, and $e^{-x}$ with $1/y$ etc, solve for y (it's a quadratic equation). Doing that, I found $y=1,2$

Then, solving for $x$ gives me $x=Ln(2),0$, but the only logical answer is $x=Ln(2)$ which agrees with what I found in Mathematica.

4. Aug 23, 2015

### blue_leaf77

Are you sure you were copying the wavefunction correctly from the original problem? That form of wavefunction is not square integrable and consequently you shouldn't be able to find $C$, possibilities might be that it's defined within certain space interval only or the $x$'s are actually absolute valued.

5. Aug 23, 2015

### Henriamaa

I meant that if you calculate the probability density there can't be a term that is not multiplied by an exponential function hence your derivative can't have a constant all by itself. You have a -1 all by itself in your expression for the derivative.

6. Aug 23, 2015

The original function was 0 from -Infinity to 0. I was being lazy when I copied it.

7. Aug 24, 2015

### DEvens

You took the derivative wrong.

8. Aug 24, 2015

You can factor that expression multiple ways. I promise it's right (just verified with Mathematica)

9. Aug 26, 2015

### RUber

You said there was more involved that you didn't write out.

$\psi(x)=Ce^{ −x} (1−e^{ −x} )$
Is always positive.
$P(x) = | \psi(x) | ^2 = \psi(x)^2$
$\frac{d}{dx} P(x) = \frac{d}{dx} \psi(x)^2 = 2 \psi(x) \frac{d}{dx} \psi(x)$
$\psi(x)=Ce^{ −x} (1−e^{ −x} ) = C(e^{ −x} - e^{ −2x})$
What is $\frac{d}{dx} \psi(x)$?
Like the others have said, I don't see any way to get a -1 in the expansion for P'(x)

10. Aug 26, 2015

### RUber

I see, you cancelled out the $2C^2 e^{-2x}$ factor from the derivative to leave what you replaced with a quadratic. I also come up with ln(2).