Kinetic Energy Operator in Momentum Basis I'm having a little trouble with something that's 'easy to check' according to the script I'm using. I consider the kinetic energy operator
$$\hat T = \frac{p^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} $$
and the wave function in momentum space, which is obtained using the Fourier transform
$$\tilde \psi(k) = \mathcal{F}[\psi(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \psi(x) e^{-ikx} dx \ .$$
According to the script it's easy to see that
$$\hat T \tilde \psi(k) = \frac{\hbar^2 k^2}{2m} \tilde \psi(k)\ ,$$
i.e. that $\tilde \psi(k)$ is an eigenstate of the kinetic energy operator. I don't manage to replicate that result. In particular, using integration by parts a few times I get
$$\begin{align}
\hat T \tilde \psi(k) &= -\frac{\hbar^2}{2m} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \frac{d^2}{dx^2} ( \psi(x) e^{-ikx} ) dx \\
&= … \\
&= -\frac{\hbar^2}{2m} \tilde \psi(k) (-k^2 + 2k^2 - k^2) \\
&= 0 \end{align}$$
I see that $\mathcal{F}[\hat T \psi(x)] = \frac{\hbar^2 k^2}{2m} \mathcal{F}[\psi(x)]$ but $\mathcal{F}[\hat T \psi(x)] \neq \hat T \tilde \psi(k)$, right? So I'm confused ...
Can anybody show me where I go wrong?
 A: First, the expression
$$\hat T\bar\psi (k)=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$
does not mean $\bar\psi(k)$ is an eigenvalue of $\hat T$ because $\hbar^2k^2/2m$ is not a constant ($k$ is a variable now).
Second, $\hat T\bar\psi(k)$ is not an expression that makes much sense. I think you actually mean $\langle k|\hat T|\psi\rangle$. To be more formal, you are starting with
$$\bar\psi(k)=\langle k|\psi\rangle=\int_{-\infty}^\infty\langle k|x\rangle\langle x|\psi\rangle\,\text dx=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-ikx}\psi(x)\,\text dx$$ 
But you cannot "operate" $\hat T$ on $\langle k|\psi\rangle$, since this is technically just a number for some value of $k$. This is why I think you actually mean $\langle k|\hat T|\psi\rangle$ because we can compute this in the momentum basis
$$\langle k|\hat T|\psi\rangle=\int_{-\infty}^\infty\langle k|\hat T|k'\rangle\langle k'|\psi\rangle \,\text dk'=\int_{-\infty}^\infty\frac{\hbar^2k^2}{2m}\delta(k-k')\bar\psi(k') \,\text dk'=\frac{\hbar^2k^2}{2m}\bar\psi(k)$$
This also reveals why $\bar\psi(k)$ is not, in general, an eigenfunction of $\hat T$. Note that really what we want is the basis-independent statement $\hat T|\psi\rangle=c|\psi\rangle$ to be true for $|\psi\rangle$ to be an eigenstate of $\hat T$. Now, at first glance this seems to be the case here, but the issue is that we needed to move into the $k$-basis in order to get this expression. In other words, $\frac{\hbar^2k^2}{2m}$ isn't an eigenvalue, it is just what is multiplied by $\bar\psi(k)$ when looking at $\hat T$ being applied to $|\psi\rangle$ specifically in the $k$-basis. In other other words, the reason that we get $\frac{\hbar^2k^2}{2m}$ is because of the operator and the chosen basis, not because of $|\psi\rangle$.
