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 $E_{ground}\propto f(\lambda)$?

  • $\begingroup$ 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

1 Answer 1


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 $$ 0=-\alpha/a^2 + (3/2)\beta \lambda a^{1/2}, \leadsto a\propto \lambda^ {-2/5} , ~~~~ \leadsto \langle E_{gnd}\rangle\propto \lambda ^{2/5}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.