How to Normalize the Basic Wave Equation

[Marthius]

This is a fairly simple question, but the first such question I have done. Inorder to check my work I was hoping somone could show me how to normalize the following.

\Psi(x,t) = Ae^{-a[(mx^{2}/\hbar)+it]
where m is the particles mass

And also that the expectation values of x and x² would be.

Don't wory, this is not for a class, I am studying this on my own

 

[Meir Achuz]

Psi^2 leads to a Gaussian integral, which is done by completing the square in the exponent.
<x> is zero bly symmetry.
<x^2>is found by integrating by parts.

 

[Marthius]

After playing with this I found

A = \sqrt[4]{\frac{2am}{\hbar\pi}}*e^{ait}

making

\Psi = \sqrt[4]{\frac{2am}{\hbar\pi}}*e^{-amx^{2}/\hbar}

Can annyone confirm this for me because I am really uncomfortable with my answer.

 

[tshafer]

that looks alright, but have you lost your time component along the way? when calculating \left|A\right|^{2} the time-dependence drops off, but you need to be sure to attach your value for A to the full wavefunction. i think it should look like this? \Psi\left(x,t\right)=\left(2ma/\pi\hbar\right)^{1/4}e^{-amx^{2}/\hbar}e^{-iat}

 

[Marthius]

that looks alright, but have you lost your time component along the way? when calculating \left|A\right|^{2} the time-dependence drops off, but you need to be sure to attach your value for A to the full wavefunction. i think it should look like this? \Psi\left(x,t\right)=\left(2ma/\pi\hbar\right)^{1/4}e^{-amx^{2}/\hbar}e^{-iat}

the only think was that e^{iat} from the second part of A canceld with e^{-iat} from the wave function, or is that wrong.

 

[tshafer]

You would be correct, but technically you're A is wrong. The e^{-iat} term cancels with its conjugate in the process of calculating A through normalization. A should be just \left(2ma/\pi\hbar\right)^{1/4}

 

[Marthius]

You would be correct, but technically you're A is wrong. The e^{-iat} term cancels with its conjugate in the process of calculating A through normalization. A should be just \left(2ma/\pi\hbar\right)^{1/4}

Looking back my mistake was simply squaring the wave function without taking the modulus first

 

[Marthius]

I was working with this a little more, and came up with a corisponding potential energy function of:

V(x) = 2a^{2}mx^{2}

Could anyone run it and verify that I have this right (My text has no answer key)?

Here is the wave function again.
\Psi(x,t) = Ae^{-a[(mx^{2}/\hbar)+it]
where m is the particles mass