[proofplan] We first conjugate the physical oscillator by a unitary scaling to reduce the spectral problem to the dimensionless harmonic oscillator. The dimensionless operator is then factored using annihilation and creation operators, and the Gaussian ground state generates the Hermite eigenfunctions by repeated raising. Completeness of the Hermite functions identifies the self-adjoint closure with a diagonal operator, so the full spectrum and eigenspace dimensions can be read from the diagonal representation. The final step transfers the spectral statement back to the original Hamiltonian through the unitary scaling. [/proofplan]

[step:Scale the Hamiltonian to the dimensionless oscillator] Define the scaling constant $\alpha > 0$ by \begin{align*} \alpha = \left(\frac{m\omega}{\hbar}\right)^{1/2}. \end{align*} Define the unitary scaling map \begin{align*} U: L^2(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1) \to L^2(\mathbb{R}, \mathcal{B}(\mathbb{R}), \mathcal{L}^1), \qquad (U\psi)(y) = \alpha^{-1/2}\psi\left(\frac{y}{\alpha}\right). \end{align*} Indeed, by the change of variables $y = \alpha x$, so that $d\mathcal{L}^1(y) = \alpha\,d\mathcal{L}^1(x)$, \begin{align*} \int_{\mathbb{R}} |U\psi(y)|^2\,d\mathcal{L}^1(y) = \int_{\mathbb{R}} \alpha^{-1}\left|\psi\left(\frac{y}{\alpha}\right)\right|^2\,d\mathcal{L}^1(y) = \int_{\mathbb{R}} |\psi(x)|^2\,d\mathcal{L}^1(x). \end{align*} Thus $U$ is an isometry, and its inverse is \begin{align*} U^{-1}: L^2(\mathbb{R}) \to L^2(\mathbb{R}), \qquad (U^{-1}f)(x) = \alpha^{1/2}f(\alpha x). \end{align*} For $f \in \mathcal{S}(\mathbb{R})$, set $\psi = U^{-1}f$. The chain rule gives \begin{align*} \psi''(x) = \alpha^{5/2} f''(\alpha x). \end{align*} Since $y = \alpha x$, we compute \begin{align*} (UH_0U^{-1}f)(y) = \hbar\omega\left(-\frac{1}{2}f''(y) + \frac{1}{2}y^2f(y)\right). \end{align*} Therefore it is enough to analyze the closure $h$ of the dimensionless symmetric operator \begin{align*} h_0: \mathcal{S}(\mathbb{R}) \to L^2(\mathbb{R}), \qquad h_0 f = -\frac{1}{2}f'' + \frac{1}{2}y^2 f, \end{align*} because \begin{align*} H = \hbar\omega\, U^{-1}hU. \end{align*} Consequently $\lambda \in \sigma(h)$ iff $\hbar\omega\lambda \in \sigma(H)$, and eigenspace dimensions are preserved by $U^{-1}$. [/step]

[step:Factor the dimensionless oscillator by annihilation and creation operators]Define the annihilation operator on $\mathcal{S}(\mathbb{R})$ by \begin{align*} A: \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R}), \qquad Af = \frac{1}{\sqrt{2}}(f' + yf). \end{align*} Define the creation operator on $\mathcal{S}(\mathbb{R})$ by \begin{align*} A^*: \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R}), \qquad A^*f = \frac{1}{\sqrt{2}}(-f' + yf). \end{align*} Here $A^*$ denotes the formal adjoint of $A$ on the common domain $\mathcal{S}(\mathbb{R})$; it is not yet being asserted to be the Hilbert-space adjoint of a closed operator. The notation is justified because [integration by parts](/theorems/210) gives, for all $f,g \in \mathcal{S}(\mathbb{R})$, \begin{align*} (Af,g)_{L^2} = (f,A^*g)_{L^2}. \end{align*} The boundary term vanishes since Schwartz functions and all their derivatives decay faster than any inverse power. A direct calculation on $\mathcal{S}(\mathbb{R})$ gives \begin{align*} A^*Af = \frac{1}{2}(-f'' + y^2f - f). \end{align*} Hence \begin{align*} h_0 = A^*A + \frac{1}{2}I \end{align*} on $\mathcal{S}(\mathbb{R})$. The same calculation gives the commutator identity \begin{align*} AA^* - A^*A = I \end{align*} on $\mathcal{S}(\mathbb{R})$.[/step]

[guided]The point of introducing $A$ and $A^*$ is that the second-order operator $h_0$ factors into first-order pieces. We define \begin{align*} A: \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R}), \qquad Af = \frac{1}{\sqrt{2}}(f' + yf), \end{align*} and \begin{align*} A^*: \mathcal{S}(\mathbb{R}) \to \mathcal{S}(\mathbb{R}), \qquad A^*f = \frac{1}{\sqrt{2}}(-f' + yf). \end{align*} These maps preserve $\mathcal{S}(\mathbb{R})$ because differentiation and multiplication by the coordinate function $y$ preserve the Schwartz class. We must justify the notation $A^*$. For $f,g \in \mathcal{S}(\mathbb{R})$, [integration by parts](/theorems/2098) with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) gives \begin{align*} \int_{\mathbb{R}} f'(y)\overline{g(y)}\,d\mathcal{L}^1(y) = -\int_{\mathbb{R}} f(y)\overline{g'(y)}\,d\mathcal{L}^1(y). \end{align*} The boundary term is zero because $f$ and $g$ are Schwartz functions. Therefore \begin{align*} (Af,g)_{L^2} = \int_{\mathbb{R}} \frac{1}{\sqrt{2}}(f'(y)+yf(y))\overline{g(y)}\,d\mathcal{L}^1(y) = \int_{\mathbb{R}} f(y)\overline{\frac{1}{\sqrt{2}}(-g'(y)+yg(y))}\,d\mathcal{L}^1(y) = (f,A^*g)_{L^2}. \end{align*} Now compute the product $A^*A$. For $f \in \mathcal{S}(\mathbb{R})$, \begin{align*} A^*Af = \frac{1}{2}(-\partial_y + y)(\partial_y + y)f. \end{align*} Expanding and using $\partial_y(yf) = f + yf'$ gives \begin{align*} A^*Af = \frac{1}{2}(-f'' - f - yf' + yf' + y^2f) = \frac{1}{2}(-f'' + y^2f - f). \end{align*} Thus \begin{align*} A^*Af + \frac{1}{2}f = -\frac{1}{2}f'' + \frac{1}{2}y^2f = h_0f. \end{align*} So the dimensionless oscillator is \begin{align*} h_0 = A^*A + \frac{1}{2}I \end{align*} on the common core $\mathcal{S}(\mathbb{R})$. The commutator is just as important. Similarly, \begin{align*} AA^*f = \frac{1}{2}(\partial_y + y)(-\partial_y + y)f = \frac{1}{2}(-f'' + y^2f + f). \end{align*} Subtracting the formula for $A^*Af$ gives \begin{align*} (AA^* - A^*A)f = f. \end{align*} Hence \begin{align*} AA^* - A^*A = I \end{align*} on $\mathcal{S}(\mathbb{R})$. This identity is the algebraic source of the equally spaced spectrum.[/guided]

[step:Construct the Hermite eigenfunctions from the Gaussian ground state] Define the ground state \begin{align*} \phi_0: \mathbb{R} \to \mathbb{R}, \qquad \phi_0(y) = \pi^{-1/4}e^{-y^2/2}. \end{align*} Then $\phi_0 \in \mathcal{S}(\mathbb{R})$ and \begin{align*} A\phi_0 = 0. \end{align*} Since \begin{align*} \int_{\mathbb{R}} |\phi_0(y)|^2\,d\mathcal{L}^1(y) = \pi^{-1/2}\int_{\mathbb{R}} e^{-y^2}\,d\mathcal{L}^1(y) = 1, \end{align*} $\phi_0$ is normalized. For each integer $n \ge 0$, define \begin{align*} \phi_n: \mathbb{R} \to \mathbb{R}, \qquad \phi_n = \frac{1}{\sqrt{n!}}(A^*)^n\phi_0. \end{align*} Each $\phi_n$ lies in $\mathcal{S}(\mathbb{R})$. Using $AA^* = A^*A + I$ inductively gives \begin{align*} A\phi_n = \sqrt{n}\,\phi_{n-1} \end{align*} for $n \ge 1$, and \begin{align*} A^*\phi_n = \sqrt{n+1}\,\phi_{n+1} \end{align*} for $n \ge 0$. Therefore \begin{align*} A^*A\phi_n = n\phi_n \end{align*} and hence \begin{align*} h\phi_n = \left(n+\frac{1}{2}\right)\phi_n. \end{align*} The normalization follows by induction: \begin{align*} \|\phi_n\|_{L^2}^2 = \frac{1}{n}\|A^*\phi_{n-1}\|_{L^2}^2 = \frac{1}{n}(AA^*\phi_{n-1},\phi_{n-1})_{L^2} = \frac{1}{n}(n\phi_{n-1},\phi_{n-1})_{L^2} = 1. \end{align*} If $m \ne n$, then $h\phi_m = (m+\frac{1}{2})\phi_m$ and $h\phi_n = (n+\frac{1}{2})\phi_n$. Since $h$ is self-adjoint, \begin{align*} \left(m+\frac{1}{2}\right)(\phi_m,\phi_n)_{L^2} = \left(n+\frac{1}{2}\right)(\phi_m,\phi_n)_{L^2}. \end{align*} Thus $(\phi_m,\phi_n)_{L^2}=0$. Hence $(\phi_n)_{n=0}^{\infty}$ is an orthonormal sequence of eigenvectors. [/step]

[step:Use completeness of the Hermite functions to diagonalize the operator]We invoke the classical Hermite basis theorem: the functions $(\phi_n)_{n=0}^{\infty}$ constructed from the normalized Gaussian by repeated application of the raising operator form a complete [orthonormal basis](/page/Orthonormal%20Basis) of $L^2(\mathbb{R},\mathcal{B}(\mathbb{R}),\mathcal{L}^1)$, and the Hermite expansion of every $f \in \mathcal{S}(\mathbb{R})$ converges to $f$ in the Schwartz topology. In particular, for every integer $r \ge 0$ there is a constant $C_r(f)>0$ such that \begin{align*} |(f,\phi_n)_{L^2}| \le C_r(f)(1+n)^{-r} \end{align*} for all $n \ge 0$. Let $T$ be the operator on $L^2(\mathbb{R})$ defined by \begin{align*} D(T) = \left\{f \in L^2(\mathbb{R}) : \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)^2 |(f,\phi_n)_{L^2}|^2 < \infty\right\} \end{align*} and, for $f \in D(T)$, by \begin{align*} Tf = \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)(f,\phi_n)_{L^2}\phi_n. \end{align*} The coefficient map $W: L^2(\mathbb{R}) \to \ell^2(\mathbb{N}\cup\{0\})$ given by $Wf = ((f,\phi_n)_{L^2})_{n=0}^{\infty}$ is unitary by completeness and orthonormality. Under $W$, the operator $T$ is multiplication by the real sequence $(n+\frac{1}{2})_{n=0}^{\infty}$ on its maximal square-summability domain, so $T$ is self-adjoint. For $f \in \mathcal{S}(\mathbb{R})$, the Schwartz-topology convergence of the Hermite expansion permits applying the continuous differential expression $-\frac{1}{2}\frac{d^2}{dy^2}+\frac{1}{2}y^2$ term by term. Since $h_0\phi_n = (n+\frac{1}{2})\phi_n$ for every $n \ge 0$, we obtain \begin{align*} h_0f = \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)(f,\phi_n)_{L^2}\phi_n = Tf. \end{align*} Thus $T$ extends $h_0$. Since $h$ is the closure of $h_0$, this extension gives $h \subseteq T$. Both $h$ and $T$ are self-adjoint, and a self-adjoint operator cannot have a proper symmetric operator extension; therefore $h=T$.[/step]

[guided]The Hermite basis theorem is the analytic input that converts the ladder construction into a full spectral theorem. It applies here because the functions $(\phi_n)_{n=0}^{\infty}$ are exactly the normalized Hermite functions obtained from the normalized Gaussian by repeated application of the raising operator. The theorem gives two conclusions: first, $(\phi_n)_{n=0}^{\infty}$ is a complete orthonormal basis of $L^2(\mathbb{R},\mathcal{B}(\mathbb{R}),\mathcal{L}^1)$; second, if $f \in \mathcal{S}(\mathbb{R})$, then its Hermite expansion converges to $f$ in the Schwartz topology. In particular, for every integer $r \ge 0$ there is a constant $C_r(f)>0$ such that \begin{align*} |(f,\phi_n)_{L^2}| \le C_r(f)(1+n)^{-r} \end{align*} for all $n \ge 0$. This rapid decay is what justifies using the unbounded diagonal operator on Schwartz functions. We now define the diagonal operator associated with the eigenvalue list. Let $T$ be the operator on $L^2(\mathbb{R})$ with domain \begin{align*} D(T) = \left\{f \in L^2(\mathbb{R}) : \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)^2 |(f,\phi_n)_{L^2}|^2 < \infty\right\}. \end{align*} For $f \in D(T)$, define \begin{align*} Tf = \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)(f,\phi_n)_{L^2}\phi_n. \end{align*} The square-summability condition in $D(T)$ is exactly the condition that the displayed series converge in $L^2(\mathbb{R})$. The operator $T$ is self-adjoint. To see this, define the coefficient map \begin{align*} W: L^2(\mathbb{R}) \to \ell^2(\mathbb{N}\cup\{0\}), \qquad Wf = ((f,\phi_n)_{L^2})_{n=0}^{\infty}. \end{align*} Completeness and orthonormality of the Hermite functions imply that $W$ is unitary. Under this unitary map, $T$ becomes multiplication by the real sequence $(n+\frac{1}{2})_{n=0}^{\infty}$ on the maximal domain of all square-summable coefficient sequences for which the multiplied sequence is still square-summable. Multiplication by a real sequence on this maximal domain is self-adjoint, so $T$ is self-adjoint. We also need to verify that $T$ extends the differential operator $h_0$ on the Schwartz core. The ladder identities give \begin{align*} A\phi_n = \sqrt{n}\,\phi_{n-1} \end{align*} for $n \ge 1$, and \begin{align*} A^*\phi_n = \sqrt{n+1}\,\phi_{n+1} \end{align*} for $n \ge 0$. Hence \begin{align*} A^*A\phi_n = n\phi_n. \end{align*} Since $h_0 = A^*A + \frac{1}{2}I$ on $\mathcal{S}(\mathbb{R})$, this gives \begin{align*} h_0\phi_n = \left(n+\frac{1}{2}\right)\phi_n \end{align*} for every $n \ge 0$. Now let $f \in \mathcal{S}(\mathbb{R})$. The Hermite expansion of $f$ converges to $f$ in the Schwartz topology, and the map $g \mapsto -\frac{1}{2}g''+\frac{1}{2}y^2g$ is continuous from $\mathcal{S}(\mathbb{R})$ to itself. Therefore applying the differential expression term by term is justified in $\mathcal{S}(\mathbb{R})$, and hence also in $L^2(\mathbb{R})$. Using the eigenfunction identity just proved, we get \begin{align*} h_0f = \sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)(f,\phi_n)_{L^2}\phi_n = Tf. \end{align*} Thus $T$ extends $h_0$. Finally, because $h$ is the closure of $h_0$, every vector in $D(h)$ is a graph-norm limit of Schwartz functions for $h_0$. Since $T$ is closed and extends $h_0$, this gives $h \subseteq T$. Both $h$ and $T$ are self-adjoint. If the inclusion were proper, then the self-adjoint operator $T$ would be a proper symmetric extension of the self-adjoint operator $h$, which is impossible because [self-adjoint operators](/page/Self-Adjoint%20Operators) are maximal symmetric. Hence $h=T$. This is the step where completeness is used. Without completeness, the ladder construction would only produce eigenvectors inside some invariant subspace; it would not rule out additional spectral components orthogonal to all constructed eigenfunctions.[/guided]

[step:Read the spectrum and eigenspaces from the diagonal form] Since $h$ is the diagonal operator with diagonal entries $\lambda_n = n+\frac{1}{2}$ for $n \in \mathbb{N}\cup\{0\}$, each $\lambda_n$ belongs to $\sigma(h)$. Conversely, if $\lambda \in \mathbb{C}$ is not one of the numbers $\lambda_n$, then the sequence $(|\lambda-\lambda_n|^{-1})_{n=0}^{\infty}$ is bounded: for large $n$, $|\lambda-\lambda_n| \to \infty$, and only finitely many remaining terms must be checked. The diagonal operator with entries $(\lambda-\lambda_n)^{-1}$ is therefore a bounded inverse for $\lambda I-h$, so $\lambda$ lies in the resolvent set of $h$. Hence \begin{align*} \sigma(h) = \left\{n+\frac{1}{2} : n \in \mathbb{N}\cup\{0\}\right\}. \end{align*} Each $\lambda_n$ is an eigenvalue because $h\phi_n = \lambda_n\phi_n$. Let $f \in D(h)$ satisfy $hf = \lambda_n f$. Write \begin{align*} f = \sum_{k=0}^{\infty} c_k\phi_k, \end{align*} where $c_k = (f,\phi_k)_{L^2}$. Applying the diagonal formula for $h$ gives \begin{align*} \sum_{k=0}^{\infty}\left(k+\frac{1}{2}\right)c_k\phi_k = \left(n+\frac{1}{2}\right)\sum_{k=0}^{\infty}c_k\phi_k. \end{align*} Taking the [inner product](/page/Inner%20Product) with $\phi_k$ yields \begin{align*} \left(k-n\right)c_k = 0 \end{align*} for every $k \ge 0$. Hence $c_k = 0$ for $k \ne n$, and $f = c_n\phi_n$. Therefore \begin{align*} \ker\left(h-\left(n+\frac{1}{2}\right)I\right) = \operatorname{span}\{\phi_n\}, \end{align*} so each eigenspace is one-dimensional. [/step]

[step:Transfer the spectral statement back to the physical Hamiltonian] From the scaling identity \begin{align*} H = \hbar\omega\, U^{-1}hU \end{align*} and the unitarity of $U$, the spectrum satisfies \begin{align*} \sigma(H) = \hbar\omega\,\sigma(h). \end{align*} Therefore \begin{align*} \sigma(H) = \left\{\hbar\omega\left(n+\frac{1}{2}\right) : n \in \mathbb{N}\cup\{0\}\right\}. \end{align*} Moreover, if $\phi_n$ is the normalized dimensionless eigenfunction of $h$, then \begin{align*} \psi_n = U^{-1}\phi_n \end{align*} satisfies \begin{align*} H\psi_n = \hbar\omega\left(n+\frac{1}{2}\right)\psi_n. \end{align*} Because $U^{-1}$ is a linear isometric isomorphism, it maps one-dimensional eigenspaces of $h$ onto one-dimensional eigenspaces of $H$. Thus every spectral point of $H$ is a simple eigenvalue, and the eigenspace corresponding to $\hbar\omega(n+\frac{1}{2})$ is one-dimensional. This proves the theorem. [/step]