I am given $$\Psi(x)=A_0 e^{-kx^2} e^{i\alpha x}$$

Here, $\large A_0=[\frac{1}{\pi \sigma_0^2}]^\frac{1}{4}$, $\large k=\frac{1}{2\sigma_0^2}$, and $\large \alpha=\frac{p_0}{\hbar}$

I want to find the modulus and the complex conjugate.

As long as $A_0$ is real, the complex conjugate looks like: $$\Psi(x)=A_0 e^{-kx^2} e^{-i\alpha x}$$

Now, I have to find the modulus. I know that is $z=re^{i\theta}$, then $|z|=r$. Going by that, $$|\Psi|=A_0e^{-kx^2}$$

(Is there any restriction on what $r$ can be? The reason I ask this question is because, in the back of my mind, I have $r=\sqrt{a^2+b^2}$ and $\theta=tan^{-1}\frac{b}{a}$. It seems rather odd that we can take $r$ to be whatever multiplies $e^{i\theta}$, given that $r$ and $\theta$ are both determined by the values of $a$ and $b$, where $z=a+ib$.)

My main question is: Given a complex function like the one above, what's the general method for finding its modulus? I'm a little confused, and the value of $|\Psi|$ mentioned above is taken from my notes. I'm not completely sure how it's arrived at.

  • $\begingroup$ Since your normalizing constant $A_0$ and the Gaussian factor $\exp(-kx^2)$ are positive, then you're completely justified in taking their product as the modulus... FWIW, the identity $|c a|=c|a|,\quad c>0$ is always a useful thing to remember... $\endgroup$ – J. M. is a poor mathematician Nov 22 '12 at 4:56
  • $\begingroup$ Can I do $|\Psi|=[\Psi^* \Psi]^{\frac{1}{2}}$? $\endgroup$ – Joebevo Nov 22 '12 at 5:13
  • $\begingroup$ Sure, try it out... $\endgroup$ – J. M. is a poor mathematician Nov 22 '12 at 5:22
  • $\begingroup$ It seems to work, and I'll make a note of it in my notes, but I just wanted to make sure that it's the most general way to find the modulus. Somehow, I seem to have missed the class where they extended the idea of a modulus to cover complex functions. It's like a secret that everybody is in on except me! $\endgroup$ – Joebevo Nov 22 '12 at 5:36

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