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(Someone may please change the title if they can think of a better one)

We have a Hilbert Space $\mathcal{H}$ that consists of all functions $\psi(x)$ such that

$\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $

Now I have to show that there are functions in $\mathcal{H}$ such that $Q\phi(x)=x\psi(x)$ is not in $\mathcal{H}$ . Does it suffice if I think of a counterexample.. but even then, the only two square integrables i can think of are the Gaussian and the dirac-delta "function". For both the cases, the corresponding $x\psi(x)$ is also L2. Could someone give me more examples of functions that are square integrable over the entire real axis?

Furthermore, if I now consider the function space $\Omega$ with the infinite set of conditions,

$\int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty$ for $n=0,1,2,\cdots$

I have to show now, that for any $\psi(x)$ in $\Omega$ the function $Q\phi(x)=x\psi(x)$ is also in $\Omega$. I can think of a weak argument like

$\int_{-\infty}^{\infty} |Q\psi(x)|^2 (1+|x|^n)dx = \int_{-\infty}^{\infty} |\psi(x)|^2 (x^2+|x|^{n+2})dx \approx \int_{-\infty}^{\infty} |\psi(x)|^2 (1+|x|^n)dx \lt \infty $

The last step because we would be concerned only with the behaviour of $x$ when it is very large, as the function would have to go to zero at large $x$ for it to be in $\Omega$.

I am not so sure about this, and face a basic conceptual difficulty. Can we conclude if $\int_{-\infty}^{\infty} |\psi(x)|^2 dx \lt \infty $ then $\psi(x)$ should fall off as $1/x^2$?

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Dear Approximist, the Dirac delta function does not belong to the Hilbert space $L^2$ (it is a so-called "distribution" or generalized function). Have you considered a function like $1/x$ (which has to be modified to get rid of the pole at the origin)? –  Akhil Mathew Mar 8 '11 at 19:41
    
@Akhil: Wikipedia says "The Dirac delta distribution is a densely defined unbounded linear functional on the Hilbert space L2 of square integrable functions." (I know it is a generalized function, that's why I called it a "function"- with quotation marks. I wished to avoid that debate) Also I know I can construct piecewise functions, but is there something better than that? –  Please Delete Account Mar 8 '11 at 19:45
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The function $1/x$ which is $0$ from $-\infty$ to $1$ is square integrable. And further, if you multiply it by $x$ it fails. Why would you want to avoid piecewise? –  Jonas Teuwen Mar 8 '11 at 19:47
    
@ Jonas neither $\int \frac{1}{x} dx$ nor $\int \frac{1}{x^2} dx$ converge from $-\infty$ to $\infty$ –  Please Delete Account Mar 8 '11 at 19:48
    
@Approximist: If you cut it off it does... –  Jonas Teuwen Mar 8 '11 at 19:49
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2 Answers 2

up vote 3 down vote accepted

Hint: Think along lines of Cauchy distribution function. Say $\psi(x) = \frac{1}{\sqrt{1+x^2}}$

For the second part of your question, you could define your function space $\Omega$ to be functions such that $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx < \infty$, for $n \in \mathbb{N}$

(Your function space and the function space defined in the above line are the same. Since if $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx < \infty$, for $n \in \mathbb{N}$, then $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 dx < \infty$ by taking $n=0$ and adding the two gives your definition and also the function satisfying your definition also fits in the definition given in this answer since $\displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 \left|x \right|^n dx \leq \displaystyle \int_{-\infty}^{\infty} \left| \psi(x) \right|^2 (1 + \left|x \right|^n) dx < \infty$)

The conclusion now is trivial since $2 \in \mathbb{N}$ and if $n \in \mathbb{N}$, then so does $n+2$

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Do I correctly infer that $0$ is in $\mathbb{N}$? (Edit: Your subsequent update answered my question.) –  Jonas Meyer Mar 8 '11 at 20:09
    
@Jonas: I tend to use $0 \in \mathbb{N}$ –  user17762 Mar 8 '11 at 20:12
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Also I know I can construct piecewise functions, but is there something better than that?

I don't know what's wrong with piecewise functions, but how about $f(x)=1/(1+|x|)$?

As Akhil mentioned, the Dirac delta function is not in $L^2$. A densely defined linear functional on a space is typically a very different object from an element of the space.

To show that $Q(\Omega)\subseteq\Omega$, note that $x^2+|x|^{n+2}\leq 1+|x|^{n+3}$ outside of a bounded interval, and $x^2+|x|^{n+2}$ is bounded on bounded intervals.

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According to physicists even $e^{i k x}$ is in $L^2$. –  Jonas Teuwen Mar 8 '11 at 20:01
    
$f(x)=1/(1+|x|)$ is not square integrable over the entire real axis. But I guess piecewise functions are good. A question: I must aplogize for my lack or rigour, but what is the difference between a densely defined linear functional from an element of the space? @Jonas I do not consider $e^{ikx}$ to be in $L^2$. Please do not generalize my naivety to all physicists. –  Please Delete Account Mar 8 '11 at 20:12
    
@Approximist: $f(x)=1/(1+|x|)$ actually is square integrable. For example, on $[1,\infty)$ or on $(-\infty,-1]$, $f(x)^2<1/x^2$, while on $[-1,1]$, $f^2\leq 1$. –  Jonas Meyer Mar 8 '11 at 20:16
    
@Approximist. A functional on a vector space is a linear map from the space to the scalar field. So if your scalars are complex, a functional on $L^2$ is a map $T:L^2\to\mathbb{C}$ satisfying $T(af+bg)=aT(f)+bT(g)$ for all scalars $a$ and $b$ and for all $f$ and $g$ in $L^2$. The functional $T$ is called bounded if there is a $C>0$ such that $|T(f)|\leq C\|f\|$ for all $f\in L^2$. A densely defined functional is functional defined on a dense linear subspace of $H$. ("Dense" meaning that all of $H$ is in the closure of the subspace.) –  Jonas Meyer Mar 8 '11 at 20:22
    
@Jonas Sorry, I made an error. The square integral of the function converges to two in the entire real axis +1. (Deleting that comment). And thanks for the clarification. I'll look it up more closely. –  Please Delete Account Mar 8 '11 at 20:33
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