# Finding most probable position for given wave function

In summary: Just solved for x and it was x=ln(2).In summary, the particle is most likely to be at the position (x,y) where the probability density's derivative is zero.

## 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.

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.

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.

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.

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.

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

You took the derivative wrong.

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

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)

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).

## 1. What is a wave function?

A wave function is a mathematical representation of the quantum state of a particle or system. It describes the probability amplitude of the particle or system being in a certain state or position.

## 2. How do you find the most probable position for a given wave function?

The most probable position for a given wave function can be found by calculating the square of the wave function, known as the probability density. The position with the highest probability density is considered the most probable position for the particle or system.

## 3. Can the most probable position change over time?

Yes, the most probable position of a particle or system can change over time as the wave function evolves. This is known as wave function collapse, where the wave function collapses to a single state or position upon measurement.

## 4. Are there any limitations to finding the most probable position for a given wave function?

There are limitations to finding the most probable position for a given wave function, such as the Heisenberg uncertainty principle. This principle states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa.

## 5. How is the most probable position related to the energy of a particle?

The most probable position is not directly related to the energy of a particle. However, the energy of a particle can affect its wave function and therefore its most probable position. For example, a higher energy state may have a larger spread in position, resulting in a less defined most probable position.

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