Normalizing this wavefunction

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In summary, the conversation is about normalizing the wave function for a particle moving in one dimension, which involves finding the total probability of the particle being found anywhere. The equation for the wave function is given and the steps to multiply it by its complex conjugate are shown. The result is a standard gaussian integral, which can be solved using the formula listed in the link provided.
  • #1
czaroffishies
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I am very sorry that I did not use latex here. It didn't seem to be functioning properly, but I tried to make this readable.

Homework Statement



The wave function for a particle moving in one dimension is

Psi(x, t) = A x e^[-(sqrt(km)/2(hbar))*x^2] e^[-i*sqrt(k/m)*(3/2)*t]

Normalize this wave function.

Homework Equations



Psi(x,t)^2 = probability that particle is found at point x.
So, total probability of particle being found anywhere is integral from -inf to +inf of Psi*(x, t)*Psi(x,t) dx, where Psi*(x, t) is the complex conjugate of Psi(x, t), and must be equal to 1.

The Attempt at a Solution



First multiplied Psi(x, t) by complex conjugate:

Psi*(x, t) * Psi(x, t) =

(A x e^[-(sqrt(km)/2(hbar))*x^2] e^[i*sqrt(k/m)*(3/2)*t]) times
(A x e^[-(sqrt(km)/2(hbar))*x^2] e^[-i*sqrt(k/m)*(3/2)*t])

= A^2 x^2 e^2*[-(sqrt(km)/2(hbar))*x^2]

= A^2 x^2 e^[-(sqrt(km)/(hbar))*x^2]

This is somewhat embarrassing, but I am not sure I know how to integrate this. And if it can't be integrated by normal means, it probably wouldn't show up as the first problem in an introductory quantum mechanics book.

I'm not seeing where I went wrong... any help is appreciated!
 
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  • #2
It's a standard gaussian integral.
 
  • #3
Sweet, thanks.
 
  • #5


Hello,

Thank you for sharing your attempted solution. It seems like you are on the right track, but there are a few things that can be improved upon.

Firstly, when taking the complex conjugate of a function, you should also take the complex conjugate of any complex numbers within the function. In this case, the complex number in question is e^[-i*sqrt(k/m)*(3/2)*t]. The complex conjugate of this is e^[i*sqrt(k/m)*(3/2)*t]. So, the complex conjugate of Psi(x,t) is:

Psi*(x,t) = A x e^[-(sqrt(km)/2(hbar))*x^2] e^[i*sqrt(k/m)*(3/2)*t]

Next, when multiplying the complex conjugate of Psi(x,t) by Psi(x,t), you should use the distributive property to multiply each term within the parentheses by each term in the other parentheses. This will result in four terms:

Psi*(x,t) * Psi(x,t) = A^2 x^2 e^[-(sqrt(km)/hbar)*x^2] e^[i*sqrt(k/m)*(3/2)*t] e^[-i*sqrt(k/m)*(3/2)*t]

Simplifying this further, we get:

Psi*(x,t) * Psi(x,t) = A^2 x^2 e^[-(sqrt(km)/hbar)*x^2]

This is the same result you obtained, but with the additional term e^[i*sqrt(k/m)*(3/2)*t] e^[-i*sqrt(k/m)*(3/2)*t], which simplifies to 1.

Now, to integrate this function, we can use the substitution method. Let u = (sqrt(km)/hbar)*x^2. Then du = (sqrt(km)/hbar)*2x dx. Rearranging this, we get dx = (hbar/sqrt(km))*(1/2x) du. Substituting this into our integral, we get:

Integral of Psi*(x,t) * Psi(x,t) dx = Integral of A^2 x^2 e^[-u] * (hbar/sqrt(km))*(1/2x) du

= (A^2 hbar/sqrt(km)) * Integral of x e^[-u]
 

1. What does it mean to "normalize" a wavefunction?

Normalizing a wavefunction means to adjust the overall magnitude or size of the wavefunction so that its squared magnitude integrates to 1. This ensures that the total probability of finding the particle is equal to 1.

2. Why is normalizing a wavefunction important?

Normalizing a wavefunction is important because it ensures that the probability of finding a particle in a certain location is meaningful and accurate. It also allows for easier comparison between different wavefunctions.

3. How do you normalize a wavefunction?

To normalize a wavefunction, you take the square root of the integral of the squared magnitude of the wavefunction, and then divide the original wavefunction by this value. This will result in a normalized wavefunction.

4. Can any wavefunction be normalized?

Yes, any wavefunction can be normalized as long as it is square-integrable, meaning its squared magnitude can be integrated to a finite value.

5. What happens if a wavefunction is not normalized?

If a wavefunction is not normalized, the probability of finding the particle will not be meaningful. Additionally, it may not accurately represent the physical system it is describing. Normalization is an important step in the mathematical description of quantum systems.

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