# Estimating ground state of Yukawa potential using a variational method [closed]

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?

• The spherically-symmetric part of the Laplacian in spherical coordinates is not $\partial^2/\partial r^2$. – G. Smith May 15 at 18:48

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 $$\begin{equation} \lambda<\frac{A m}{\hbar^{2}} \end{equation}$$
• Why is the trial function a function of $\mathbf{x}$ and not of $r$? – Gert May 15 at 17:53
• you are right should be just of $r$ since i didn't include the angular part – Μπαμπης Ποζουκιδης May 15 at 18:07
• 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) – Gert May 15 at 18:21
• 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 – Μπαμπης Ποζουκιδης May 15 at 19:49