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You get the solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ as follows

$$ \left(\frac{p^2}{2m}-E \right)\psi_E(p) = 0 $$

For this equation to hold it must be either $\frac{p^2}{2m}-E = 0$ or $\psi_E(p) = 0$. That means for every $p$ except for $p=\sqrt{2mE}$ it must be $\psi_E(p) = 0$. Only for $p=\sqrt{2mE}$ it is allowed that $\psi_E(p)$ is non-zero.

This is exactly the behavior of the Dirac delta function. So you can write (with an arbitrary constant $A$)

$$ \psi_E(p) = A \delta(p-\sqrt{2mE}) $$

You get the solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ as follows

$$ \left(\frac{p^2}{2m}-E \right)\psi_E(p) = 0 $$

For this equation to hold it must be either $\frac{p^2}{2m}-E = 0$ or $\psi_E(p) = 0$. That means for every $p$ except for $p=\pm\sqrt{2mE}$ it must be $\psi_E(p) = 0$. Only for $p=\sqrt{2mE}$ and $p=-\sqrt{2mE}$ it is allowed that $\psi_E(p)$ is non-zero.

So the most general solution to all this is (using the Dirac delta function)

$$ \psi_E(p) = A \delta(p-\sqrt{2mE}) + B \delta(p+\sqrt{2mE})$$ where $A$ and $B$ are arbitrary constants.

You get the solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ as follows

$$ \left(\frac{p^2}{2m}-E \right)\psi_E(p) = 0 $$

For this equation to hold it must be either $\frac{p^2}{2m}-E = 0$ or $\psi_E(p) = 0$. That means for every $p$ except for $p=\sqrt{2mE}$ it must be $\psi_E(p) = 0$. Only for $p=\sqrt{2mE}$ it is allowed that $\psi_E(p)$ is non-zero.

This is exactly the behavior of the [Dirac delta function][1]. So you can write (with an arbitrary constant $A$)

$$ \psi_E(p) = A \delta(p-\sqrt{2mE}) $$ [1]: https://en.wikipedia.org/wiki/Dirac_delta_function

You get the solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ as follows

$$ \left(\frac{p^2}{2m}-E \right)\psi_E(p) = 0 $$

For this equation to hold it must be either $\frac{p^2}{2m}-E = 0$ or $\psi_E(p) = 0$. That means for every $p$ except for $p=\sqrt{2mE}$ it must be $\psi_E(p) = 0$. Only for $p=\sqrt{2mE}$ it is allowed that $\psi_E(p)$ is non-zero.

This is exactly the behavior of the Dirac delta function. So you can write (with an arbitrary constant $A$)

$$ \psi_E(p) = A \delta(p-\sqrt{2mE}) $$

The solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ is $$ \psi_E(p) = \delta(p-\sqrt{2mE}) $$

It is zero for every $p$, except for $p=\sqrt{2mE}$ where it has a peak.
  See also the definition and visualization of the Dirac $\delta$ function.

You get the solution of $$ \frac{p^2}{2m}\psi_E(p) = E\psi_E(p) $$ as follows

$$ \left(\frac{p^2}{2m}-E \right)\psi_E(p) = 0 $$

For this equation to hold it must be either $\frac{p^2}{2m}-E = 0$ or $\psi_E(p) = 0$. That means for every $p$ except for $p=\sqrt{2mE}$ it must be $\psi_E(p) = 0$. Only for $p=\sqrt{2mE}$ it is allowed that $\psi_E(p)$ is non-zero.

This is exactly the behavior of the [Dirac delta function][1]. So you can write (with an arbitrary constant $A$)

$$ \psi_E(p) = A \delta(p-\sqrt{2mE}) $$ [1]: https://en.wikipedia.org/wiki/Dirac_delta_function