# 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?

• If you are doing derivatives with respect to $x$ for the momentum operator, then you are in the position basis, not the momentum basis. Feb 28, 2020 at 20:08
• The momentum operator takes different form. In position basis, momentum is a derivative. In momentum basis, momentum is just multiplying by p. Feb 28, 2020 at 20:14
• oh I see, we use de Broglie's relation $p = \hbar k$ so $\hat T = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}$ and we just multiply by it, that's great, thanks to both of you!
– mpr
Feb 28, 2020 at 20:35

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 $$|\psi\rangle$$.

• Thanks Aaron, that's what I was looking for, I'd think in the script they meant $\langle k \vert \hat{T} \vert \psi \rangle$. Just one question, you say because $k$ is a variable it can't be an eigenvalue. Why not? What's in the definition of eigenvectors/eigenvalues that prohibits that?
– mpr
Mar 5, 2020 at 23:06
• @mpr Yeah I agree I was being pretty unclear on that first part. When I have time I'll edit it to be more clear, and I'll let you know when I do that. Mar 6, 2020 at 0:19
• @mpr I added a part at the end. Mar 6, 2020 at 9:59
• that clarifies it well, thank you very much!
– mpr
Mar 6, 2020 at 10:29