There is this problem, it is not for evaluation, I mean, it is not homework.

There is a 3-dimension potential $V(r)=C . r$, where $r$ is the radial coordinate and $C$ is a real constant (I think it has something to do with quarks separation, not sure).

Using separation of variables $\left<r,\theta,\varphi|\Psi\right>=\psi(r,\theta,\varphi)=R(r)Y(\theta,\varphi)$, and focusing in the radial equation only:

$$ \frac{1}{R(r)} \left( \frac{d}{dr}r^2\frac{dR(r)}{dr}+\frac{2mr^2}{\hbar^2}(E-C.r)R(r) \right)=l(l+1) $$

It is possible to make a (tricky) change in the $r$ variable such that:

$$ \left(\frac{d^2}{d \rho^2}-\frac{l(l+1)}{\rho^2}-\rho \right) u(\rho)=\lambda u(\rho) $$

which is a new eigenvalue/eigenvector problem. I am looking for a solution that could be general for $l=0,1,2,\dots$

Using the variational method, I've tried some test functions without success:

  • $u(\rho)=\rho \exp (-a \rho)$
  • $u(\rho)=\rho^{l+1} \exp (-a \rho)$

Any idea about what could be a good test function?

The "exact" ground state eigenvalues are known to be:

  • $l=0 \implies \lambda_0= 2.338$
  • $l=1 \implies \lambda_0= 3.361$
  • $l=2 \implies \lambda_0= 4.248$

Edit regarding to AHusain's comment

For instance, let's test $u(\rho)=\rho \exp(-a \rho)$, $a \in \mathbb{R}$.

Since the cumbersome transformation is:

$$ \rho=\left( \frac{2 m C}{\hbar^2} \right)^{1/3} \left( r-\frac{E}{C} \right) \,\,\, \dots or \dots \,\,\, r = \left( \frac{\hbar^2}{2mC} \right)^{1/3} \rho+\frac{E}{C} $$

The radial integral changes as:

$$ \int_0^{\infty} dr \, r^2 \rightarrow \left( \frac{\hbar^2}{2mC} \right)^{1/3} \int_{\rho_0}^{\infty} d\rho \left[ \left( \frac{\hbar^2}{2mC} \right)^{1/3} \rho+\frac{E}{C} \right]^2 \sim \int_{\rho_0}^{\infty} d\rho \, (a_0 \rho^2+a_1 \rho + a_2) $$


$$ \rho_0 = -\left( \frac{2mE^3}{\hbar^2 C^2} \right)^{1/3} $$

Using the above integral variable change and the variational method we have:

$$ \frac{\left<u|H|u\right>}{\left<u|u\right>}=\frac{\int_{\rho_0}^{\infty} d\rho (a_0 \rho^2+a_1 \rho + a_2) u^*(\rho) \left(\frac{d^2}{d \rho^2}-\frac{l(l+1)}{\rho^2}-\rho \right) u(\rho)}{\int_{\rho_0}^{\infty} d\rho (a_0 \rho^2+a_1 \rho + a_2) |u(\rho)|^2} $$

The second derivative in $\rho$ over $u(\rho)$ is:

$$ \frac{d^2 u(\rho)}{d\rho^2} = (a^2 \rho-2a) \exp(-a \rho) $$

Using it:

$$ \frac{\left<u|H|u\right>}{\left<u|u\right>}=\frac{\int_{\rho_0}^{\infty} d\rho (a_0 \rho^2+a_1 \rho + a_2) \left( a^2 \rho^2-2a\rho- l(l+1)-\rho^3 \right) \exp(-2a\rho)}{\int_{\rho_0}^{\infty} d\rho (a_0 \rho^2+a_1 \rho + a_2) \rho^2 \exp(-2a\rho)} $$

Now, we have integrals up to $\rho^5 \exp(-2a\rho)$, according to WolframAlpha:

  • $\int_{\rho_0}^{\infty} d\rho \exp(-2a\rho)=\frac{e^{-2a\rho_0}}{2a}$
  • $\int_{\rho_0}^{\infty} d\rho \rho \exp(-2a\rho)=\frac{e^{-2a\rho_0}(2a \rho_0+1)}{4a^2}$
  • $\int_{\rho_0}^{\infty} d\rho \rho^2 \exp(-2a\rho)=\frac{e^{-2a\rho_0}(2a \rho_0(a \rho_0+1)+1)}{4a^3}$
  • $\int_{\rho_0}^{\infty} d\rho \rho^3 \exp(-2a\rho)=\frac{e^{-2a\rho_0}(2a \rho_0(a \rho_0(2a \rho_0+3)+3)+3)}{8a^4}$
  • $\int_{\rho_0}^{\infty} d\rho \rho^4 \exp(-2a\rho)=\frac{e^{-2a\rho_0}(2a \rho_0(a \rho_0(a \rho_0(a \rho_0+2)+3)+3)+3)}{4a^5}$
  • $\int_{\rho_0}^{\infty} d\rho \rho^5 \exp(-2a\rho)=\frac{e^{-2a\rho_0}(2a \rho_0(a \rho_0(a \rho_0(a \rho_0(2a \rho_0+5)+10)+15)+15)+15)}{8a^6}$

It is so huge that makes me wonder whether it is the correct path.

  • 1
    $\begingroup$ Could you put the values that you get as $\langle u \mid E \mid u \rangle \geq \lambda_0$ from the variational principle? How far are the inequalities from being saturated when you minimize over $a$? $\endgroup$ – AHusain Jun 14 '18 at 3:28
  • 2
    $\begingroup$ OK, I assume you are not asking why your hydrogenic test functions are throughly inappropriate, and how they contrast with the Airy functions that solve the 1D problem to start with, no? $\endgroup$ – Cosmas Zachos Jun 14 '18 at 14:15
  • 2
    $\begingroup$ Might start from here. $\endgroup$ – Cosmas Zachos Jun 14 '18 at 14:17
  • $\begingroup$ Dear @AHusain, I am not getting the $\lambda_0$ values, the calculation becomes too tough, so I feel like I am doing something wrong. Would you mind in take a look in those changes I made? $\endgroup$ – Thiago Melo Jun 14 '18 at 18:01
  • 1
    $\begingroup$ Can one get anywhere if one thinks of it as $d_\rho d_\rho = l(l+1)\rho^{-2} +\lambda \rho^0 +\rho^1 $ ? Perhaps one can look at laurent series expansions for inspiration of the rhs? Or perhaps one can try to do the liebniz rule backwards? $\endgroup$ – Emil Jun 14 '18 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.