Normalizing a wave function

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Homework Statement



A quantum mechanical wavefunction for a particle of mass m moving in one dimension where α and A are constants.

Normalize the function - that is find a value of A for which [tex]\int^{\infty}_{-\infty}|ψ|^2dx=1[/tex]

Homework Equations



[tex]ψ(x,t)= |Ae^{-α(x^2 + it\hbar/m)}|^2[/tex]

A useful integral: [tex]\int^{\infty}_{-\infty}e^{-z^2}dz = √\pi[/tex]

The Attempt at a Solution



[tex]ψ(x,t)= |Ae^{-α(x^2 + it\hbar/m)}|^2[/tex]

[tex]1= \int^{\infty}_{-\infty}|Ae^{-α(x^2 + it\hbar/m)}|^2 [/tex]

[tex]1= |A|^2\int^{\infty}_{-\infty}(e^{-α(x^2 + it\hbar/m)})(e^{α(x^2 + it\hbar/m)})[/tex]

I'm pretty sure the last line is incorrect. My reasoning was that since i is a complex number, for all complex numbers |z|^2≠|z^2z|. Before this, I tried changing the variable by letting [tex]z=√(2α(x^2 + it\hbar/m))[/tex]
 

Answers and Replies

  • #2
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Right, the last line is incorrect, since you do the following:

[tex]|z|^2 = z^2[/tex]

This is incorrect for complex numbers.

What you have here is basically something of the form

[tex]|e^{x^2 + it}| = |e^{x^2}| |e^{it}|[/tex]

Can you simplify this further?
 
  • #3
vanhees71
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First of all, one should clarify what's the wave function. I guess it's the expression without the modulus squared, i.e., a Gaussian wave packet
[tex]\psi(x,t)=A \exp \left [-\alpha \left (x^2+ \mathrm{i} \beta t \right ) \right ].[/tex]
Note that there is something fishy with the dimensions in the original expression. That's why I've introduced another real constant [itex]\beta[/itex]. I also guess [itex]\alpha>0[/itex].

Then just take the modulus squared using micromass's suggestion.
 
  • #4
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[tex]|e^{-αx^2}||e^{-iαt\hbar/m}|[/tex]

Take the complex conjugate.

[tex]|e^{-αx^2}||e^{-iαt\hbar/m}||e^{iαt\hbar/m}|[/tex]

And I'm left with this.

[tex]e^{-αx^2}[/tex]

Then,

[tex]1= |A|^2\int^{\infty}_{-\infty}e^{-αx^2}dx[/tex]

[tex]1= |A|^2√(\pi/α)[/tex]

[tex] A= a^{1/4}/\pi^{1/4}[/tex]

Did I get there?
 
  • #5
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[tex]|e^{-αx^2}||e^{-iαt\hbar/m}|[/tex]

Take the complex conjugate.

[tex]|e^{-αx^2}||e^{-iαt\hbar/m}||e^{iαt\hbar/m}|[/tex]

Not sure what you did here? Did you just multiply by

[tex]|e^{i\alpha t \hbar/m}|[/tex]

Why can you do this?

Aside, from this, all the rest (and the final answer) is ok.

And I'm left with this.

[tex]e^{-αx^2}[/tex]

Then,

[tex]1= |A|^2\int^{\infty}_{-\infty}e^{-αx^2}dx[/tex]

[tex]1= |A|^2√(\pi/α)[/tex]

[tex] A= a^{1/4}/\pi^{1/4}[/tex]

Did I get there?
 
  • #6
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I'm not sure what you're asking. I multiplied by [tex]|e^{i\alpha t \hbar/m}|[/tex] because to normalize a wave function I have to multiply by its complex conjugate and get [tex]e^0[/tex].

For the final answer, did I forget to square the [tex]e^{-αx^2}?[/tex] I just did right now and my final answer is [tex] A= (2a)^{1/4}/\pi^{1/4}[/tex]

I really appreciate your help, by the way.
 
  • #7
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I'm not sure what you're asking. I multiplied by [tex]|e^{i\alpha t \hbar/m}|[/tex] because to normalize a wave function I have to multiply by its complex conjugate and get [tex]e^0[/tex].

I know what you did, but I'm asking why you can multiply with some value like that? Doesn't that change the entire integral?

I mean, you have to calculate the integral of

[tex]|e^{\alpha x^2}||e^{-\alpha t \hbar / m}|[/tex]

And instead of that, you calculate

[tex]|e^{\alpha x^2}||e^{-\alpha t \hbar / m}||e^{\alpha t \hbar / m}|[/tex]

How do these two integrals relate?
 

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