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Consider the following operator $\hat{x}=i\hbar \frac{\partial}{\partial p}$.

I am trying to show that the eigenfunctions of $\hat{x}$ are not square-normalizable. I am interested in doing so since theoretically, we notice that the eigenfunction of the momentum operator $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$ is not square-normalizable. How do we set up the equation for the $\hat{x}$ operator?

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    $\begingroup$ You can get a bit of intuition by noting that the eigenstate of $\hat{x}$ are Delta functions, which are not square-normalizable. $\endgroup$ – DumpsterDoofus Feb 24 '14 at 14:31
  • $\begingroup$ @DumpsterDoofus Thank you for your quick response. Could you elaborate on that intuition ? $\endgroup$ – Carpediem Feb 24 '14 at 14:32
  • $\begingroup$ How can I express your idea in a more formal way. $\endgroup$ – Carpediem Feb 24 '14 at 14:51
  • $\begingroup$ Just take the (inverse) Fourier transform of position eigenfunction in momentum representation - it'll be the eigenfunction in position representation. $\endgroup$ – Ruslan Feb 24 '14 at 15:00
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The eigenvalue equation

$$\tag{1} \hat{x}\psi(x)~=~x_0\psi(x)$$

in the standard Schrödinger position representation

$$\tag{2} \hat{x}~=~x, \qquad \hat{p}~=~-i\hbar\frac{\partial}{\partial x},$$

reads

$$\tag{3} (x-x_0)\psi(x)~=~0,$$

which has general solution

$$\tag{4} \psi(x) ~\propto~ \delta(x-x_0). $$

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