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Orthogonality gives $⟨{\hat{a}}_{+}^2⟩=⟨{\hat{a}}_{-}^2⟩=0$, while the ladder coefficients give $⟨{\hat{a}}_{+}{\hat{a}}_{-}⟩=n$ and $⟨{\hat{a}}_{-}{\hat{a}}_{+}⟩=n+1$. Squaring the expressions for $\hat{x}$ and $\hat{p}$ therefore givesThusTheir equality is the harmonic-potential case of the quantum virial theorem, and their sum is the oscillator energy $E_n=\hslash \omega (n+1/2)$.

Take $m,\omega >0$ and use the usual common invariant domain for the position operator, momentum operator and their products. Since $X$ and $P$ are self-adjoint,Expanding the commutator and using $[P,X]=-i\hslash I$ givesThe two products areConsequently the quantum harmonic oscillator Hamiltonian operator isThe identity $[AB,C]=A[B,C]+[A,C]B$ now givesThus $H(a_{\pm }\psi )=(E\pm \hslash \omega )a_{\pm }\psi$. Each ladder image is an eigenstate if it is nonzero; a zero vector is not an eigenstate. The distinction is needed for lowering the ground state. Their squared norms areThese follow by moving the operator to the other side of the inner product with its adjoint operator. The first inequality implies $E\ge \hslash \omega /2$ for every eigenstate.

For a lowest-energy quantum state, $a_{-}{\psi }_0$ must vanish, since otherwise it would be an eigenstate with lower energy. The norm formula then forces $E_0=\hslash \omega /2$. Raising never vanishes and producesTo see that there are no other eigenvalues, repeatedly lower any eigenstate. If $E/(\hslash \omega )-1/2$ were not a nonnegative integer, there would be a final nonzero lowered quantum state with energy strictly between $\hslash \omega /2$ and $3\hslash \omega /2$. Its lower image would have positive norm but energy below the lower bound, a contradiction. Thus every eigenstate reaches the ground eigenspace after an integer number of lowerings. On a positive level, lowering is injective by its strictly positive norm factor. The assumed one-dimensional ground eigenspace therefore makes every eigenspace one-dimensional, so the displayed raised quantum states give all eigenstates up to nonzero multiples. If ${\psi }_0$ is normalized, their norms are $\Vert {\psi }_n{\Vert }^2=n!$.

For $k\ge 1$, $(a_{+}{)}^k\psi$ has eigenvalue $E+k\hslash \omega$, distinct from $E$. eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal: applying self-adjointness to their inner product gives $(E^{\prime }-E)⟨\psi ,\varphi ⟩=0$. ThereforeThe expectation of $(a_{-}{)}^k$ also vanishes, either by lowering and orthogonality or by conjugating the adjoint relation.

Solving the ladder definitions for $X$ and $P$ givesso $T-V=\frac{1}{2}\hslash \omega ((a_{+}{)}^2+(a_{-}{)}^2)$. Both terms have zero expectation by the preceding orthogonality. Since $T+V=H$, the oscillator form of the quantum virial theorem isExpectations here are divided by $\Vert \psi {\Vert }^2$ if the eigenstate is not normalized; no normalization assumption was used in the ladder argument.

Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.

Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.