Finding most probable position for given wave function

[Adoniram]

Homework Statement

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

Homework Equations

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

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.

 

[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.

 

[Adoniram]

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.

 

[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.

 

[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.

 

[Adoniram]

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

 

[DEvens]

You took the derivative wrong.

 

[Adoniram]

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

 

[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)

 

[RUber]

I see, you canceled 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).