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.
Quantum Harmonic Oscillator – Proof of Average Kinetic and Potential Energy Equality
energyharmonic-oscillatorpotential energyquantum mechanicsvirial-theorem
Related Solutions
I) It depends on how abstract OP wants it to be. Say that we discard any reference to 1D geometry, and position and momentum operators $\hat{q}$ and $\hat{p}$. Say that we only know that
$$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1}, \qquad\qquad \nu\in\mathbb{R},$$ $$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, $$ $$\tag{3} [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1}, \qquad\qquad[{\bf 1}, \cdot]~=~0.$$
(Since we have cut any reference to geometry, there is no longer any reason why $\nu$ should be a half, so we have generalized it to an arbitrary real number $\nu\in\mathbb{R}$.)
II) Next assume that the physical states live in an inner product space $(V,\langle \cdot,\cdot \rangle )$, and that $V$ form a non-trivial irreducible unitary representation of the Heisenberg algebra,
$$\tag{4} {\cal A}~:=~ \text{associative algebra generated by $\hat{a}$, $\hat{a}^{\dagger}$, and ${\bf 1}$}.$$
The spectrum of a semi-positive operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ is always non-negative,
$$\tag{5} {\rm Spec}(\hat{N})~\subseteq~ [0,\infty[.$$
In particular, the spectrum ${\rm Spec}(\hat{N})$ is bounded from below. Since the operator $\hat{N}$ commutes with the Hamiltonian $\hat{H}$, we can use $\hat{N}$ to classify the physical states. Let us sketch how the standard argument goes. Say that $|n_0\rangle\neq 0$ is a normalized eigenstate for $\hat{N}$ with eigenvalue $n_0\in[0,\infty[$. We can use the lowering ladder (annihilation) operator $\hat{a}$ repeatedly to define new eigenstates
$$\tag{6} |n_0- 1\rangle,\quad |n_0- 2\rangle, \quad\ldots$$
which however could have zero norm. Since the spectrum ${\rm Spec}(\hat{N})$ is bounded from below, this lowering procedure (6) must stop in finite many steps. There must exists an integer $m\in\mathbb{N}_0$ such that zero-norm occurs
$$\tag{7} \hat{a}|n_0 - m\rangle~=~0.$$
Assume that $m$ is the smallest of such integers. The norm is
$$\tag{8} 0 ~=~ || ~\hat{a}|n_0 - m\rangle ~||^2 ~=~ \langle n_0 - m|\hat{N}|n_0 - m\rangle ~=~ ( n_0 - m) \underbrace{||~|n_0 - m\rangle~||^2}_{>0},$$
so the original eigenvalue is an integer
$$\tag{9} n_0 ~=~ m\in\mathbb{N}_0,$$
and eq. (7) becomes
$$\tag{10} \hat{a}|0\rangle ~=~0,\qquad\qquad \langle 0 |0\rangle ~\neq~0.$$
We can next use the raising ladder (creation) operator $\hat{a}^{\dagger}$ repeatedly to define new eigenstates
$$\tag{11} |1\rangle,\quad |2\rangle,\quad \ldots.$$
By a similar norm argument, one may see that this raising procedure (11) cannot eventually create a zero-norm state, and hence it goes on forever/doesn't stop. Inductively, at stage $n\in\mathbb{N}_0$, the norm remains non-zero,
$$\tag{12} || ~\hat{a}^{\dagger}|n\rangle ~||^2 ~=~ \langle n|\hat{a}\hat{a}^{\dagger}|n\rangle~=~ \langle n|(\hat{N}+1)|n\rangle ~=~ (n+1) ~\langle n|n\rangle~>~0. $$
So $V$ contains at least one full copy of the standard Fock space. On the other hand, by the irreducibility assumption, the vector space $V$ cannot be bigger, and $V$ is hence just a standard Fock space (up to isomorphism).
III) Finally, if $V$ is not irreducible, then $V$ could be a direct sum of several Fock spaces. In the latter case, the ground state energy-level is degenerate.
Short answer: By the uncertainty principle, the harmonic oscillator can't be localized at the minimum value of potential energy, i.e., $x=0$, because, by the uncertainty principle, it's momentum would become large (strictly speaking, the expectation value of $p^2$, and thereby it's kinetic energy, becomes large). The lowest energy state of the harmonic oscillator is a compromise between minimizing potential energy (i.e., $x^2$) and kinetic energy (i.e., $p^2$), which cannot be done simultaneously, because $\langle x^2\rangle \langle p^2 \rangle \ge \frac{\hbar^2}{4}$.
Best Answer
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.