born's rule quantum mechanics

[physica]

The question asks to apply born's rule. I know what born's rule is. Squaring the wave equation will give a probability distribution. Integrating this over a distance with an arbitrary constant and summing the whole equation to one (normalizing the constant) will give me the normalized function. This question then give the normalized function and also gives 3 integrals and asks me to prove that the function is normalized. I know that you have to pull the squared constant out of the integral, do the definite integral. divide one by the definite integral and the take the square root of this to get the normalized constant. However, I have no idea how to use the definite integrals that they have given me in the question. They are slightly different from the integral I need the integral is too hard for me to solve. Can someone give me any clues?

 

Many thanks

 

 

[elfmotat]

For the wave function they give you, see what happens if you change variables: [math]x \to x / \sqrt{\alpha}[/math]. Then compare the result to the third integral they give you.

Edited October 15, 2014 by elfmotat

[physica]

I'm trying. There are two issues I still have.

 

1. Why do we do this substitution in a physical sense?

 

2. I'm starting to see a solution to this with the substitution but I can't get rid of the e^(-b^2/4) in the result

[elfmotat]

1. Why do we do this substitution in a physical sense?

 

My hint was meant to suggest that you can use u-substitution to solve the integral. Let [math]u= \sqrt{\alpha} x [/math]. It doesn't have any physical meaning.

 

 

 

 

2. I'm starting to see a solution to this with the substitution but I can't get rid of the e^(-b^2/4) in the result

 

Indeed. It seems the function isn't actually normalized.

Edited October 16, 2014 by elfmotat

[swansont]

For the first integral: Do you understand what is meant by an odd function?

[physica]

Odd function means that it's not symmetrical. But if the function is squared it should be symmetrical.

 

Will try the U substitution thank you

[elfmotat]

Well, I feel like an idiot. I was integrating the wavefunction instead of the square of the wavefunction. The square of it is just [math]\sqrt{\frac{2 \alpha}{\pi}} e^{-2\alpha x^2}[/math], which you can use u-substitution to get it into the form of integral (2). Everything works out, there are no extra factors.

[physica]

It's ok. I'm still stuck though. How did the e raised to the imaginary function disappear?

[elfmotat]

I'll drop the coefficient to simplify things:

 

[math]\psi^* \psi =(e^{ikx} e^{-\alpha x^2})(e^{-ikx} e^{-\alpha x^2})=e^{ikx-ikx-\alpha x^2-\alpha x^2}=e^{-2 \alpha x^2}[/math]

[physica]

Oh yes I remember this now when doing measurement and uncertainty. Thank you the rest is straightforward. Thank you for being patient with me.

[physica]

This may seem silly. I understand the mechanical way of doing the maths but do you know why the wave function star has a positive imaginary component whilst the non-star wave function has a negative imaginary component? I feel like I've gone through it but I can't remember why. Looking back at it this is why I was struggling with the question in the first place.

 

Many thanks

[Fuzzwood]

http://en.wikipedia.org/wiki/Complex_conjugate