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I have to calculate an upper bound for the ground state energy $E_0$ given the Yukawa potential $$ V(r) = -\dfrac{g}{r} e^{-kr}\ ,\quad g,k > 0\ , $$ and a test function family $$ \phi_\lambda (r) = N_\lambda e^{- \lambda r / 2}\ ,\quad \lambda > 0\ . $$

I started by calculating the expectation value of the energy with the test functions as $$ \langle\phi_\lambda | \hat{H} | \phi_\lambda \rangle = \langle\phi_\lambda | \hat{T} | \phi_\lambda \rangle + \langle\phi_\lambda | \hat{V} | \phi_\lambda \rangle\ , $$ where $$ \langle\phi_\lambda | \hat{T} | \phi_\lambda \rangle = -\dfrac{\hbar^2}{2m} \int_0^\infty r^2 \phi_\lambda^* \dfrac{\partial^2 \phi_\lambda}{\partial r^2}\ dr $$ and $$ \langle\phi_\lambda | \hat{V} | \phi_\lambda \rangle = \int_0^\infty r^2 \phi_\lambda^* V \phi_\lambda\ dr\ . $$ This leads to the value $$ E(\lambda) = -N_\lambda^2 \left( \dfrac{\hbar^2}{4m\lambda} + \dfrac{g}{(\lambda + k)^2} \right)\ . $$ The factor $r^2$ in the previous integrals appears due to the definition of scalar product between radial wavefunctions, though I would like if someone can confirm me that this is correct.

Now, I can get the upper bound by solving $dE/d\lambda = 0$, but $$ \dfrac{dE}{d\lambda} = N_\lambda^2 \left( \dfrac{\hbar^2}{4m\lambda^2} + \dfrac{2g}{(\lambda + k)^3} \right) = 0 $$ does not have real solutions for $\lambda$, making it impossible to get the correponding energy.

What's the correct way of doing this?

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  • $\begingroup$ The spherically-symmetric part of the Laplacian in spherical coordinates is not $\partial^2/\partial r^2$. $\endgroup$ – G. Smith May 15 at 18:48
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I think you have a wrong sign on your energy equation. The terms in the parentheses should have different signs. Given the potential: \begin{equation} V(r)=-A \frac{e^{-\lambda r}}{r} \end{equation} If we work with the trial wave function \begin{equation} \psi(\mathbf{x} ; \alpha)=\sqrt{\frac{\alpha^{3}}{\pi}} e^{-\alpha r} \end{equation} Then the corresponding energy is calculated by the following formula: \begin{equation} E(\alpha)=\langle\psi(\alpha)|\hat{H}| \psi(\alpha)\rangle \end{equation} which is just the following integral over all of space: \begin{equation} \int\psi(\mathbf{x} ; \alpha)(T+V)\psi(\mathbf{x} ; \alpha)r^2 drd\Omega \end{equation} Where $T$ is the kinetic energy operator and $V$ the potential energy. Calculating this integral we get the following expression for the energy: \begin{equation} E(\alpha)=\frac{\hbar^{2} \alpha^{2}}{2 m}-\frac{4 A \alpha^{3}}{(\lambda+2 \alpha)^{2}} \end{equation} This energy has negative values if \begin{equation} \lambda<\frac{A m}{\hbar^{2}} \end{equation}

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  • $\begingroup$ Why is the trial function a function of $\mathbf{x}$ and not of $r$? $\endgroup$ – Gert May 15 at 17:53
  • $\begingroup$ you are right should be just of $r$ since i didn't include the angular part $\endgroup$ – Μπαμπης Ποζουκιδης May 15 at 18:07
  • $\begingroup$ Could you elaborate a few lines to show how to get from the trial function to the $E$ level? Thanks! (I'll upvote if you do) $\endgroup$ – Gert May 15 at 18:21
  • $\begingroup$ I did add it. Also i was mistaken before, the wavefunction is a function of $\mathbf{x}$. It is normalized in such a way that when you integrate over all of space (including the angular part) the integral is equal to 1 $\endgroup$ – Μπαμπης Ποζουκιδης May 15 at 19:49

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