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The density $|{\Psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$ is even. Hence $⟨x⟩=0$; also ${\Psi }_0^{\prime }=-\alpha x{\Psi }_0$ gives $⟨p_x⟩=i\hslash \alpha ⟨x⟩=0$.

To evaluate the second position moment, differentiate the Gaussian integral with respect to its positive parameter:For the momentum second moment, apply the momentum operator twice:Both means vanish, soThe ground state attains the bound proved above and is therefore a minimum uncertainty state. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.

In the position representation, $\hat{x}$ multiplies by $x$ and ${\hat{p}}_x=-i\hslash \mathrm{d}/\mathrm{d}x$. Acting on an arbitrary differentiable test function $f$ shows the commutator explicitly:Therefore $[\hat{x},{\hat{p}}_x]=i\hslash \hat{1}$, whose expectation value is $i\hslash$ in every normalized state in the relevant domains. The Robertson uncertainty principle givesThis is a lower bound, not a claim that every state saturates it.

For a normalized state and a Hermitian quantum observable $\hat{A}$, its quantum uncertainty is the standard deviation of repeated measurement outcomes:Here $⟨A⟩=⟨\Psi |\hat{A}|\Psi ⟩$. It is a property of the state and observable, not an instrumental error; the required moments must be finite.

Differentiate the Gaussian before substituting into the Hamiltonian:ThusAn energy eigenvalue must be independent of $x$, so the coefficient of $x^2$ vanishes. Normalizability requires $\alpha >0$, givingThe constant term is the quantum harmonic oscillator's zero-point energy; a Gaussian width ${\alpha }^{-1/2}$ balances its kinetic and potential contributions.