Why Does Normalizing ψ(x,t) Result in Ae^(-2λ|x|)?

[Gemini_Cricket]

Homework Statement

ψ(x,t) = Ae^(-λ|x|)e^(-iωt)

This is a rather long problem so I won't get into the details. I understand how to normalize, and most of the rest of the problem. I also have the solutions manual. I just need an explanation of why this goes to Ae^(-2λ|x|). I can't figure it out.

Homework Equations

I can't think of any that make sense to use.

The Attempt at a Solution

I believe it is because you can add the powers of exponentials, such that e^(x)e^(x) = e^(2x). I do not understand how you can just get rid of the imaginary, angular frequency, or time parts...

Any explanation would be great.

 

[Dick]

Homework Statement

ψ(x,t) = Ae^(-λ|x|)e^(-iωt)

This is a rather long problem so I won't get into the details. I understand how to normalize, and most of the rest of the problem. I also have the solutions manual. I just need an explanation of why this goes to Ae^(-2λ|x|). I can't figure it out.

Homework Equations

I can't think of any that make sense to use.

The Attempt at a Solution

I believe it is because you can add the powers of exponentials, such that e^(x)e^(x) = e^(2x). I do not understand how you can just get rid of the imaginary, angular frequency, or time parts...

Any explanation would be great.

It's because to normalize you need to integrate ψψ*, the wave function times its complex conjugate. The complex conjugate of e^(-iωt) is e^(iωt). e^(-iωt)*e^(iωt)=e^0=1.

 

[Gemini_Cricket]

Ah okay. I had thought it might have something to do with the complex conjugate.

So for the complex conjugate you just get...

e^(-λ|x|)*e^(-λ|x|) = e^(-2λ|x|)

 

[Dick]

Ah okay. I had thought it might have something to do with the complex conjugate.

So for the complex conjugate you just get...

e^(-λ|x|)*e^(-λ|x|) = e^(-2λ|x|)

Sure.

 

[paranormal]

Normalizing a wave function is the process of adjusting the amplitude of the wave function so that the total probability of finding the particle in any location is equal to 1. This is important because it ensures that the wave function accurately represents the physical system it is describing.

In this case, the wave function given is ψ(x,t) = Ae^(-λ|x|)e^(-iωt). In order to normalize this, we must first determine the value of A. This can be done by integrating the square of the wave function over all space, as shown below:

∫|ψ(x,t)|^2 dx = ∫|Ae^(-λ|x|)e^(-iωt)|^2 dx

= |A|^2∫e^(-2λ|x|) dx

= |A|^2 * 1/(-2λ) * [e^(-2λ|x|)] from -∞ to ∞

= |A|^2 * 1/(-2λ) * [e^(-∞) - e^(∞)]

= |A|^2 * 1/(-2λ) * [0 - 0]

= 0

Since the integral must be equal to 1, we can set |A|^2 * 1/(-2λ) = 1 and solve for A:

|A|^2 * 1/(-2λ) = 1

|A|^2 = 2λ

|A| = √(2λ)

Therefore, the normalized wave function is given by:

ψ(x,t) = (√(2λ))e^(-λ|x|)e^(-iωt)

= Ae^(-2λ|x|)e^(-iωt)

This is why the normalized wave function has a coefficient of Ae^(-2λ|x|) instead of Ae^(-λ|x|). The imaginary, angular frequency, and time parts are not affected by the normalization process, as they do not affect the overall probability of finding the particle. They only determine the phase and behavior of the wave function. I hope this explanation helps to clarify your understanding.