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I am trying to solve a problem involving variational approximation, where the task is to calculate a value $C$ such that

$C > \frac{\int_{-\infty}^{\infty} |f'(y)|^2 dy}{\int_{-\infty}^{\infty} \frac{1}{y^2} |f(y)|^2 dy}$

for a given function $f$ that takes the form $f = \frac{y}{y^{2} + \alpha^{2}}$, where $\alpha$ is a variational parameter. This problem takes a potential of the form $V(x) = \frac{C}{x^{2}}$ and the expression for $C$ above gives a numerical value for the absolute lowest bound for which there is no ground state.

I have done simple plug and chug in the expression for $C$ above and got the relation that $C > \frac{1}{2}$. However, when I do some variational approach, I got $C > 1$, which I have done in this fashion:

Let $I_1$ be the integral expression in the numerator and $I_2$ be the expression for the denominator so that

$C > \frac{I_1}{I_2}$

Minimizing with respect to $\alpha$, I get

$\frac{d}{d\alpha} \left(\frac{I_1}{I_2}\right) = 0$

$\frac{I_2 \frac{dI_{1}}{d\alpha} - I_1 \frac{dI_2}{d\alpha}}{I_2 ^{2}} = 0$

which means that

$I_2 \frac{dI_{1}}{d\alpha} - I_1 \frac{dI_2}{d\alpha} = 0$

and can be written as

$I_2 I_1' -I_1I_2' = 0$

where the prime represents derivative with respect to $\alpha$. This can be arranged in a way that

$\frac{I'_1}{I_1} = \frac{I'_2}{I_2}$

and after integrating both sides, yields the relation $\ln(I_1) = \ln(I_2)$, and thus $I_1 = I_2$. If this is the case, then we get $C>1$. But that is not just right, I think.

Is there something wrong with the math right here? How can I extend this to functions with two variational parameters $\alpha$ and $\beta$, like for example $f(y) = \frac{y}{(y^2+\alpha^{2})^{\beta}}$?

Newbie
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  • The equation $\frac{I'_1}{I_1} = \frac{I'_2}{I_2}$ is only valid (as a function of $\alpha$) at the point where $I_1/I_2$ is extremized. You can't validly integrate it to get your final result, because at other values of $\alpha$ you don't have $\frac{I'_1}{I_1} = \frac{I'_2}{I_2}$. – Michael Seifert Dec 06 '23 at 12:49
  • @MichaelSeifert Thanks for the insights. What I am thinking of instead is both $I'{1}$ and $I'{2}$ should separately vanish, leading to a system with two equations. However, when I do that, I get $\alpha = 0$ and upon plugging it in to the original expression for $f(y)$ and carry out the integration, the result yields $C>0$, which is wrong I think. – Newbie Dec 06 '23 at 13:58
  • Yes, that's also incorrect. By that logic the function $f(x) = x/(x^2 + 1)$ has no extrema for $x > 0$ because neither the denominator nor the numerator have an extremum. But $f(x)$ is maximized when $x = 1$. – Michael Seifert Dec 06 '23 at 16:02

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