Physics Asked by rm213 on May 8, 2021

As we all know there is the normal quantum harmonic oscillator with potential $V= frac{momega^2x^2}{2}$ and we get $E_n =hbaromega(n+1/2)$

What is my $E_n$ when the potential $V= 2momega^2x^2$ or how can I calculate it?

As suggested by @Frank $$V'=2momega'^2x^2=frac{1}{2}m(2omega')^2x^2=frac{1}{2}momega^2x^2$$ Thus $$E'_n=hbaromega'left(n'+frac{1}{2}right)=2hbaromega left(n'+frac{1}{2}right)=2E_n$$

Correct answer by Young Kindaichi on May 8, 2021

The hamiltonian $H = {p^2}/{2m} + momega^2x^2/2$ has the spectrum $E_n=hbaromega(n+1/2)$. Replacing $omega to 2omega$ you get the hamiltonian $H = {p^2}/{2m} + 2momega^2x^2$, which then must have the same spectrum with $omega to 2omega$, so $E_n to E_n = 2hbaromega(n+1/2)$

Answered by pp.ch.te on May 8, 2021

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