How can the normalization of a wave function be achieved?

[Numnum]

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 $$\int^{\infty}_{-\infty}|ψ|^2dx=1$$

Homework Equations

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

A useful integral: $$\int^{\infty}_{-\infty}e^{-z^2}dz = √\pi$$

The Attempt at a Solution

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

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

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

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 $$z=√(2α(x^2 + it\hbar/m))$$

 

[micromass]

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

$$|z|^2 = z^2$$

This is incorrect for complex numbers.

What you have here is basically something of the form

$$|e^{x^2 + it}| = |e^{x^2}| |e^{it}|$$

Can you simplify this further?

 

[vanhees71]

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
$$\psi(x,t)=A \exp \left [-\alpha \left (x^2+ \mathrm{i} \beta t \right ) \right ].$$
Note that there is something fishy with the dimensions in the original expression. That's why I've introduced another real constant $\beta$. I also guess $\alpha>0$.

Then just take the modulus squared using micromass's suggestion.

 

[Numnum]

$$|e^{-αx^2}||e^{-iαt\hbar/m}|$$

Take the complex conjugate.

$$|e^{-αx^2}||e^{-iαt\hbar/m}||e^{iαt\hbar/m}|$$

And I'm left with this.

$$e^{-αx^2}$$

Then,

$$1= |A|^2\int^{\infty}_{-\infty}e^{-αx^2}dx$$

$$1= |A|^2√(\pi/α)$$

$$A= a^{1/4}/\pi^{1/4}$$

Did I get there?

 

[micromass]

$$|e^{-αx^2}||e^{-iαt\hbar/m}|$$

Take the complex conjugate.

$$|e^{-αx^2}||e^{-iαt\hbar/m}||e^{iαt\hbar/m}|$$

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

$$|e^{i\alpha t \hbar/m}|$$

Why can you do this?

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

And I'm left with this.

$$e^{-αx^2}$$

Then,

$$1= |A|^2\int^{\infty}_{-\infty}e^{-αx^2}dx$$

$$1= |A|^2√(\pi/α)$$

$$A= a^{1/4}/\pi^{1/4}$$

Did I get there?

 

[Numnum]

I'm not sure what you're asking. I multiplied by $$|e^{i\alpha t \hbar/m}|$$ because to normalize a wave function I have to multiply by its complex conjugate and get $$e^0$$.

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

I really appreciate your help, by the way.

 

[micromass]

I'm not sure what you're asking. I multiplied by $$|e^{i\alpha t \hbar/m}|$$ because to normalize a wave function I have to multiply by its complex conjugate and get $$e^0$$.

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

$$|e^{\alpha x^2}||e^{-\alpha t \hbar / m}|$$

And instead of that, you calculate

$$|e^{\alpha x^2}||e^{-\alpha t \hbar / m}||e^{\alpha t \hbar / m}|$$

How do these two integrals relate?