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If $\left|\psi\right> \in L_2$, is $\left|\psi\right> ^2 \in L_2$, where $\left|\cdot\right>$ is Dirac notation.

Or for everyone to get what I am asking: $f(x) \in L_2$ if the following condition is met:

$$\int_{-\infty}^{\infty}|f|^2(x)dx<\infty$$

So my question consists in following: if $\psi(x) \in L_2$ will $\psi^2(x) \in L_2$ or in other words will the $\int_{-\infty}^{\infty}|\psi|^4(x)dx$ be less than $\infty$

The answer is not obvious to me, because if an integral converges it doesn't mean that it will do so with the squared integrand.

blitzar
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    Try to consider some example, like $\psi(x) = \frac{1}{|x|^\beta}$. – Siminore Nov 04 '16 at 15:16
  • @Siminore with this example it converges even more rapidly and it means that $\psi(x)^2\in L_2$. But is it true for any $\psi$? – blitzar Nov 04 '16 at 15:21
  • @Winther I tried to translate it in pure math. – blitzar Nov 04 '16 at 15:28
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    I think what @Siminore means is: consider $\psi(x) = 1/|x|^\beta$ on $|x| < 1$, $0$ elsewhere. $\int_{-\infty}^\infty |\psi(x)|^p; dx$ converges iff $p \beta < 1$. – Robert Israel Nov 04 '16 at 15:34
  • @hardmath Well OK then... – Parcly Taxel Nov 04 '16 at 15:35
  • Another example (with no poles): consider the function $\psi(x)$ which is zero everywhere apart from spikes of height $2^n$ and width $\frac{1}{4^n}$ around each integer $n$. We then have $\int_0^\infty \psi(x){\rm d}x = \sum_{n=1}^\infty 2^n\cdot \frac{1}{4^n} < \infty$, however $\int_0^\infty \psi^2(x){\rm d}x = \sum 4^n\cdot \frac{1}{4^n} = \infty$ – Winther Nov 04 '16 at 15:36
  • @RobertIsrael ok I got it. I need to check additional conditions. – blitzar Nov 04 '16 at 15:39
  • As I understood the correct answer for arbitrary $\psi$ is NO. – blitzar Nov 04 '16 at 15:41
  • Be careful with the definition of the $L_2$ space you have given, as $L_2$ norm is defined as $\int_{-\infty}^{\infty}|f(x)|^2dx<\infty$, it's important to notice that $|f(x)|^2 = f(x)^2$ is not true always, see complex functions for example. As it looks like you are working on quantum mechanics, lots of wave functions are complex, so you should be careful with your definintion. – Josu Etxezarreta Martinez Nov 04 '16 at 15:43
  • @JosuEtxezarretaMartinez I agree with you. Thank you, I have edited the question, even more your comment helped me to get what I wanted from my exercise. Because If I have a normed $\psi$ function, It means that the square of it belongs to $L_2$. – blitzar Nov 04 '16 at 15:46
  • @blitzar No probs man, we are here to help! – Josu Etxezarreta Martinez Nov 04 '16 at 15:53

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