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Consider a particle in a box $\Lambda = [0, L]$. The wavefunction $\psi \in L_D^2(\Lambda)$ where $D$ denotes a Dirichlet Condition $\psi(0)=0=\psi(L)$. We have, then

$$ - \frac{\hbar^{2}}{2m} \frac{d^{2}\psi}{dx^{2}} = E \psi$$

inside de box. Solving this equation,

$$ \psi_{m}(x) = \sqrt{2/L} \sin(n\pi x/L).$$

Now consider a weak formulation of this problem:

$$\frac{\hbar^{2}}{2m} \int_{0}^{L} \frac{d\psi}{dx} \frac{d\varphi}{dx} dx = E \int_{0}^L \psi \varphi dx$$

so that $\varphi \in H^1_0(\Lambda)$. Is it possible to find an analytical solution for this equation? If so, how can one proceed to find solutions of weak formulations in general?

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  • $\begingroup$ What do you mean for $L^2_D(\Lambda)$? Is it different from $L^2(\Lambda)$? Boundary conditions do not change the Hilbert space... $\endgroup$
    – Valter Moretti
    Mar 2, 2021 at 18:54
  • $\begingroup$ It's just that the wavefunction is in $L^2(\Lambda)$ and $\psi$ has Dirichlet Conditions $\psi(0)=0=\psi(L)$. So I'm just saying that $\psi$ is in $L^{2}$ in the interval $\Lambda$ and $\psi(0)=0=\psi(L)$. $\endgroup$
    – amith
    Mar 2, 2021 at 18:56
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    $\begingroup$ Since $\psi$ is defined up to zero measure set it does not mean much. There is no a Hilbert space made of functions vanishing at the boundary of an interval. Boundary conditions instead can be used to define the domain of your Hamiltonian... $\endgroup$
    – Valter Moretti
    Mar 2, 2021 at 19:40

1 Answer 1

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Elliptic regularity implies that $\psi$ admits weak derivatives of every order which are $L^2$ locally. Now, Sobolev's lemma implies that these derivatives must be standard derivatives. In summary, every weak solution is just the standard solution $\psi_m$.

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  • $\begingroup$ But what if my problem only has weak solutions and I get a weak formulation like that? Is it possible to solve those kind of problems analytically? $\endgroup$
    – amith
    Mar 3, 2021 at 19:44
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    $\begingroup$ Usually the solutions are only weak because the domain of operators are not made of smooth functions when they are taken selfadjoint. However, if the potential is relatively regular, there are known results (essentially a theorem by Weyl) establishing that the wavefunctions are smooth outside the set of singularities of the potential. $\endgroup$
    – Valter Moretti
    Mar 3, 2021 at 20:10
  • $\begingroup$ @Valter Moretti: Would you mind posting a reference to that theorem, please? I have only found the one for Laplace's equation. $\endgroup$
    – user510186
    Dec 12, 2022 at 20:20

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