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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?

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    $\begingroup$ If you are doing derivatives with respect to $x$ for the momentum operator, then you are in the position basis, not the momentum basis. $\endgroup$ Commented Feb 28, 2020 at 20:08
  • $\begingroup$ The momentum operator takes different form. In position basis, momentum is a derivative. In momentum basis, momentum is just multiplying by p. $\endgroup$
    – FGSUZ
    Commented Feb 28, 2020 at 20:14
  • $\begingroup$ 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! $\endgroup$
    – mpr
    Commented Feb 28, 2020 at 20:35

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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 $|\psi\rangle$.

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  • $\begingroup$ 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? $\endgroup$
    – mpr
    Commented Mar 5, 2020 at 23:06
  • $\begingroup$ @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. $\endgroup$ Commented Mar 6, 2020 at 0:19
  • $\begingroup$ @mpr I added a part at the end. $\endgroup$ Commented Mar 6, 2020 at 9:59
  • $\begingroup$ that clarifies it well, thank you very much! $\endgroup$
    – mpr
    Commented Mar 6, 2020 at 10:29

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