Is there a simple way to explain why the expectation of the kinetic energy equals the one of the potential energy in the quantum harmonic oscillator? I would like to find a simpler explanation than saying it is a consequence of the virial theorem, since that one is not so easy to prove.
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asked Dec 9, 2021 at 10:00
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3 Answers
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I don't know if this what you are looking for, but one can go to dimensionless position and momentum operators ($\tilde{p} = p/\sqrt{\hbar m \omega}$, $\tilde{x} = \sqrt{ m \omega/\hbar}\;x$) and have the Hamiltonian as $H = \frac{\hbar \omega}{2} \left(\tilde{p}^2 + \tilde{x}^2\right)$ and then you can see that there is a nice symmetry here where you can rotate in the $(\tilde{x}, \tilde{p})$ space and leave the Hamiltonian unchanged. The symmetry under changing $\tilde{x}$ with $\tilde{p}$ and vice-versa is just replacing kinetic and potential energy, but as this is a symmetry it is clear they need to be identical for the eigenstates of the Hamiltonian.
As a nice side-note, this is also why the eigenfunctions of the Harmonic oscillator are eigenfunctions of the Fourier transform. Because FT just goes from momentum to position basis, but for the Harmonic oscillators they are the same.
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To add to the answer by @yyy:
Once we have written Hamiltonian in dimensionless representation,
$$H=\frac{\hbar\omega}{2}(\bar{p}^2+\bar{x}^2),$$
one can see that the equation is the same in position and momentum representation, where position and momentum (and hence potential and kinetic energies) exchange with each other.
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As long as you are in an energy eigenstate $\vert n\rangle$ you can prove the equality by writing both position and momentum operator in terms of the ladder operators $a,a^\dagger$, see here. Because $x$ is proportional to $a+a^\dagger$ and $p$ to $i(a-a^\dagger)$ you get by multiplying out $a^2+(a^\dagger)^2$ and $aa^\dagger+a^\dagger a$ for $x^2$ (up to constants) and $-a^2-(a^\dagger)^2$ and $aa^\dagger+a^\dagger a$ for $p^2$. Now just apply these operators to $\vert n\rangle$, use orthonormality of the energy eigenstates and you see that only the $aa^\dagger+a^\dagger a$ term contributes, which is the same for both, whereas the $\pm(a^2+(a^\dagger)^2)$ term vanishes by orthogonality.
This approach also shows that the expectation values are not the same for general states. Try for example $\frac{1}{\sqrt{2}}(\vert n\rangle+\vert n+2\rangle) $ where the contribution of $\pm(a^2+(a^\dagger)^2)$ does not vanish and creates a difference between the expectation values of kinetic energy and potential energy.
