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If $\psi(x)$ represents the wavefunction of a 1D quantum system, it satisfies the Schrodinger equation, has a unit norm, and $\lim_{x\rightarrow \infty }\psi(x)=0.$ Then is it true that $\lim_{x\rightarrow \infty }\dfrac{d\psi(x)}{dx}$ also zero or another constant?

Qmechanic
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Purushothaman
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1 Answers1

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In a formal mathematical sense, no. Consider

$$ \psi(x)=\sin(x^2)/x $$ This wavefunction is smooth and square integrable, and goes to zero at infinity. However, it's derivative does not go to zero at infinity, as you can verify.

However, for convenience we often assume all wavefunction have vanishing derivative at infinity. This is required to make $P^2$ Hermitian.

Jahan Claes
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    That's not really an assumption. Remember that $P^2$ is Hermitian if $\langle \psi, P^2\varphi\rangle = \langle P^2 \psi, \varphi\rangle$ for all $\psi,\varphi$ in the domain of $P^2$.The domain of $P^2$ is not the entire Hilbert space - it is restricted to twice-differentiable functions whose second derivative is also in $L^2(\mathbb R)$, which means that $\lim_{x\rightarrow \infty} \psi'' = 0$. This, along with the fact that $\lim_{x\rightarrow \infty}\psi = 0$, implies that $\lim_{x\rightarrow \infty} \psi' = 0$, meaning that $P^2$ is Hermitian on its domain of definition. – J. Murray Oct 12 '18 at 05:26
  • @J.Murray I think all I'm saying is that we assume our wavefunction is in the domain off $P^2$, which is not the same as all square integrable functions. – Jahan Claes Oct 12 '18 at 15:44
  • What I'm saying is that that assumption isn't actually necessary. You can do QM with a wavefunction which isn't even continuous if you want. The domain of $P^2$ is a dense subspace of the full Hilbert space $\mathcal H$, so even if some $\psi\in \mathcal H$ does not lie in the domain of $P^2$, we define the time evolution of $\psi$ by the time evolution of a sequence $\psi_n\in Dom(P^2)$ which converges to $\psi$. – J. Murray Oct 12 '18 at 17:16
  • @J.Murray Fair enough. I'm not going to edit my answer, because I don't think anything I said was explicitly incorrect, but it's a fair point that most things you prove by assuming $\psi$ is in the domain of $P^2$ can be generalized to the uniform limit of wavefunctions in the domain of $P^2$. – Jahan Claes Oct 12 '18 at 18:30
  • I agree. It's a small and technical point, but one that is particularly useful in resolving annoying questions which only arise when you look a bit too closely. – J. Murray Oct 12 '18 at 18:35