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I am given the following Hamiltonian: $$H = \frac{p^2}{2m} + \lambda|x|^3$$

where $\lambda$ is a positive constant.

Is there a relation between the ground state energy of $H$ and $\lambda$ i.e. is there any $f(\lambda)$ such that $E_{ground}\propto f(\lambda)$?

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  • $\begingroup$ Are you asking if the ground state energy depends on $\lambda$? Or are you asking if the dependency is some simple function of $\lambda$? $\endgroup$
    – Ghoster
    Commented Aug 10, 2023 at 8:02
  • $\begingroup$ I am asking if the dependency is some simple function of $\lambda$. Actually my notes say: $E_{ground} \propto \lambda^{2/5}$. But my notes do not explain why. $\endgroup$ Commented Aug 10, 2023 at 8:22
  • $\begingroup$ I would guess that if you solve the eigenvalue equation for the TISE with the given Hamiltonian symbolically in $\lambda$, you'll notice the power law dependence in the exponent for the smallest eigenvalue. $\endgroup$
    – r_phys
    Commented Aug 10, 2023 at 8:58

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It's simple dimensional analysis. Work out a variational estimate of $\langle E_{gnd}\rangle$ with a normalized Gaussian test function, $$ \psi ={e^{-x^2/(2a)}\over \sqrt[4]{a\pi} },\leadsto \\ \langle E_{gnd}\rangle = \int\!\! dx ~~ {e^{-x^2/a}\over \sqrt{a\pi} }\Bigl (-\frac{\hbar^2}{2m} (x^2/a^2- 1/a)+\lambda |x|^3\Bigr )\\ = \alpha /a + \beta \lambda a^{3/2}, $$ where α,β are positive constants you need not compute; after you scale the variables, you get simple Gaussian moments, uninteresting, as you are only interested in the functional dependence of a on λ !!

You now minimize the energy expectation w.r.t. a, so the vanishing variation yields $$ 0=-\alpha/a^2 + (3/2)\beta \lambda a^{1/2}, \leadsto a\propto \lambda^ {-2/5} , ~~~~ \leadsto \langle E_{gnd}\rangle\propto \lambda ^{2/5}. $$

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