# Spring constant and dispersion relation

In order to calculate the dispersion relation (i.e $$w(k)$$) for the electrons and protons, I used the following relations:
$$E = ℏω$$, $$p = ℏk$$, and I substituted them in this formula for energy:
$$E = p^2/ 2m$$.

Therefore I got that $$ω(k) = ℏk^2/ 2m$$. However, here it is another exercise:

A particle of mass $$m$$ moves in the potential of a $$1D$$ harmonic oscillator. Calculate the spring constant $$k$$ of the oscillator if the zero point energy of the particle equals the zero point energy of the same particle in a $$1D$$ potential box of width $$a$$.

In order to solve this, I equated the expressions for zero-point energies, respectively. But for the formula for the zero-point energy of a harmonic oscillator, I have that:
$$E = ℏω/2$$, according to my book.
Now, should I substitute $$ω$$ = $$(k/m)$$^$$0.5$$ or the $$w(k)$$ I derived for the dispersion relation?

• You have to equate $E = \hbar\omega/2$ with the zero point energy of the particle in the 1D potential box, i.e. $E = \frac{h^2}{8mL^2}$. From here, find $\omega$! And once you have $\omega$, insert it to the equation $k = m\omega^2$. – rnels12 Dec 13 '18 at 13:12
• @rnels12 That is exactly what I did and was asking a confirmation for. Thank you! – physicist Dec 14 '18 at 18:04