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I'm very confused with this problem and I was looking for some guidance.

$$\psi(x) = Ae^{ikx}e^{-x^2/2a^2}$$ Use the normalization condition to find A.

So I understand that you use the normalization condition where $$\int_\infty^\infty|\psi(x)|^2 dx = 1$$

But the integral is really complicated and doesn't really lead me anywhere. Can anyone suggest a hint? Am I missing something?

$$\int_\infty^\infty (Ae^{ikx}e^{-x^2/2a^2})^2dx$$

$$\int_\infty^\infty (A^2 e^{2ikx}e^{-x^2/a^2})dx$$

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up vote 1 down vote accepted

The expression $|\psi(x)|^2$ is the complex modulus squared; $$ |\psi(x)|^2 = \psi(x)^*\psi(x) $$ where here the star means complex conjugation. It follows that for any wavefunction of the form $$ \psi(x) = Ae^{ikx} f(x) $$ where $A$ and $f(x)$ are real, one has $$ |\psi(x)|^2 = A^2 f(x)^2 $$ since $|e^{ikx}|^2 = e^{ikx} e^{-ikx} = 1$.

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Notice that $$|\psi|^2 =\psi^*\psi = |A|^2 e^{-x^2/a^2}.$$ We are left with a Gaussian integral.

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