[proofplan] We first compute the ground state explicitly and record the algebraic identities satisfied by the ladder operators. The commutator relation gives the raising and lowering formulae, which prove normalisation and the eigenvalue equation by induction. Orthogonality follows directly from the lowering formula and the fact that repeated lowering eventually reaches $\psi_0$. Completeness is proved by identifying $\psi_n$ with the classical Hermite functions and using the generating function to show that any $L^2$ function orthogonal to all of them has vanishing [Fourier transform](/page/Fourier%20Transform), hence is zero. [/proofplan]

[step:Compute the normalised ground state and factor the Hamiltonian]Throughout the proof, $m>0$, $\omega>0$, and $\hbar>0$ are the constants from the theorem statement. Let $\mathcal{S}(\mathbb{R})$ denote the [Schwartz space](/page/Schwartz%20Space) on $\mathbb{R}$, namely the space of smooth functions whose derivatives decay faster than every polynomial. On $\mathcal{S}(\mathbb{R})$, define the ladder operators $a:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ and $a^*:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ by \begin{align*} au=\sqrt{\frac{m\omega}{2\hbar}}xu+\sqrt{\frac{\hbar}{2m\omega}}u' \end{align*} and \begin{align*} a^*u=\sqrt{\frac{m\omega}{2\hbar}}xu-\sqrt{\frac{\hbar}{2m\omega}}u'. \end{align*} Let $H$ denote the self-adjoint harmonic-oscillator operator from the theorem statement. The theorem statement defines the Hermite functions $(\psi_n)_{n\ge 0}$ as follows: $\psi_0$ is the positive $L^2(\mathbb{R})$-normalised solution of $a\psi_0=0$, and for $n\ge 0$ the functions $\psi_n$ are given by \begin{align*} \psi_n=\frac{1}{\sqrt{n!}}(a^*)^n\psi_0. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Set \begin{align*} \alpha := \sqrt{\frac{m\omega}{\hbar}}. \end{align*} The equation $a\psi_0=0$ is the first-order ordinary differential equation \begin{align*} \sqrt{\frac{m\omega}{2\hbar}}\,x\psi_0(x)+\sqrt{\frac{\hbar}{2m\omega}}\,\psi_0'(x)=0. \end{align*} Equivalently, \begin{align*} \psi_0'(x)=-\alpha^2 x\psi_0(x). \end{align*} Solving gives \begin{align*} \psi_0(x)=C_0 e^{-\alpha^2x^2/2} \end{align*} for a constant $C_0>0$. The normalisation condition $\|\psi_0\|_{L^2(\mathbb{R})}=1$ gives \begin{align*} 1=C_0^2\int_{\mathbb{R}}e^{-\alpha^2x^2}\,d\mathcal{L}^1(x)=C_0^2\frac{\sqrt{\pi}}{\alpha}, \end{align*} so \begin{align*} C_0=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}. \end{align*} Thus \begin{align*} \psi_0(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}e^{-\alpha^2x^2/2}. \end{align*} Since a Gaussian multiplied by any polynomial is a Schwartz function, $\psi_0\in\mathcal{S}(\mathbb{R})$. Since $a^*$ maps $\mathcal{S}(\mathbb{R})$ into itself, the defining formula $\psi_n=(n!)^{-1/2}(a^*)^n\psi_0$ gives $\psi_n\in\mathcal{S}(\mathbb{R})$ for every $n\ge 0$. For $u\in\mathcal{S}(\mathbb{R})$, direct differentiation gives \begin{align*} a^*au=-\frac{\hbar}{2m\omega}u''+\frac{m\omega}{2\hbar}x^2u-\frac12u. \end{align*} Let $I:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ denote the identity operator. Multiplying by $\hbar\omega$ yields \begin{align*} Hu=\hbar\omega\left(a^*a+\frac12 I\right)u. \end{align*} Similarly, \begin{align*} aa^*u=-\frac{\hbar}{2m\omega}u''+\frac{m\omega}{2\hbar}x^2u+\frac12u, \end{align*} and hence \begin{align*} [a,a^*]u=(aa^*-a^*a)u=u. \end{align*}[/step]

[guided]We first make the operators concrete. Let $\mathcal{S}(\mathbb{R})$ denote the [Schwartz space](/page/Schwartz%20Space) on $\mathbb{R}$, the space of smooth functions whose derivatives decay faster than every polynomial. The annihilation operator $a:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ and creation operator $a^*:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ are the first-order differential operators \begin{align*} au=\sqrt{\frac{m\omega}{2\hbar}}xu+\sqrt{\frac{\hbar}{2m\omega}}u' \end{align*} and \begin{align*} a^*u=\sqrt{\frac{m\omega}{2\hbar}}xu-\sqrt{\frac{\hbar}{2m\omega}}u'. \end{align*} The Hermite functions in the theorem are defined in the statement: the ground state $\psi_0$ is the positive $L^2(\mathbb{R})$-normalised solution of $a\psi_0=0$, and the excited states are obtained by repeated application of $a^*$. Thus $a\psi_0=0$ means \begin{align*} \sqrt{\frac{m\omega}{2\hbar}}x\psi_0(x)+\sqrt{\frac{\hbar}{2m\omega}}\psi_0'(x)=0. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. With $\alpha=\sqrt{m\omega/\hbar}$, this is \begin{align*} \psi_0'(x)=-\alpha^2x\psi_0(x). \end{align*} Solving this first-order ordinary differential equation gives \begin{align*} \psi_0(x)=C_0e^{-\alpha^2x^2/2} \end{align*} for a positive constant $C_0$. The normalisation condition determines $C_0$ because \begin{align*} 1=C_0^2\int_{\mathbb{R}}e^{-\alpha^2x^2}\,d\mathcal{L}^1(x)=C_0^2\frac{\sqrt{\pi}}{\alpha}. \end{align*} Therefore \begin{align*} C_0=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}, \end{align*} and hence \begin{align*} \psi_0(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}e^{-\alpha^2x^2/2}. \end{align*} This also shows $\psi_0\in\mathcal{S}(\mathbb{R})$, and since $a^*$ preserves $\mathcal{S}(\mathbb{R})$, every $\psi_n=(n!)^{-1/2}(a^*)^n\psi_0$ lies in $\mathcal{S}(\mathbb{R})$. Now compute the algebra. For $u\in\mathcal{S}(\mathbb{R})$, differentiating the products in $a^*au$ gives \begin{align*} a^*au=-\frac{\hbar}{2m\omega}u''+\frac{m\omega}{2\hbar}x^2u-\frac12u. \end{align*} Let $I:\mathcal{S}(\mathbb{R})\to\mathcal{S}(\mathbb{R})$ denote the identity operator. Multiplying by $\hbar\omega$ gives the factorisation \begin{align*} Hu=\hbar\omega\left(a^*a+\frac12 I\right)u \end{align*} on $\mathcal{S}(\mathbb{R})$. Similarly, \begin{align*} aa^*u=-\frac{\hbar}{2m\omega}u''+\frac{m\omega}{2\hbar}x^2u+\frac12u, \end{align*} so subtracting the two identities gives \begin{align*} [a,a^*]u=(aa^*-a^*a)u=u. \end{align*}[/guided]

[step:Derive the raising and lowering relations]For $n\ge 0$, the definition of $\psi_n$ gives \begin{align*} a^*\psi_n=\frac{1}{\sqrt{n!}}(a^*)^{n+1}\psi_0=\sqrt{n+1}\,\psi_{n+1}. \end{align*} We next prove that \begin{align*} a\psi_n=\sqrt{n}\,\psi_{n-1} \end{align*} for $n\ge 1$, while $a\psi_0=0$ by definition. The commutator identity $[a,a^*]=I$ implies, by induction on $n$, \begin{align*} a(a^*)^n=(a^*)^n a+n(a^*)^{n-1} \end{align*} on $\mathcal{S}(\mathbb{R})$. Applying this identity to $\psi_0$ and using $a\psi_0=0$ gives \begin{align*} a\psi_n=\frac{1}{\sqrt{n!}}a(a^*)^n\psi_0 =\frac{n}{\sqrt{n!}}(a^*)^{n-1}\psi_0 =\sqrt{n}\,\psi_{n-1}. \end{align*}[/step]

[guided]The purpose of this step is to turn the formal ladder-operator picture into exact identities. The raising formula is immediate from the definition: \begin{align*} a^*\psi_n=\frac{1}{\sqrt{n!}}(a^*)^{n+1}\psi_0. \end{align*} Since \begin{align*} \psi_{n+1}=\frac{1}{\sqrt{(n+1)!}}(a^*)^{n+1}\psi_0, \end{align*} we have \begin{align*} (a^*)^{n+1}\psi_0=\sqrt{(n+1)!}\,\psi_{n+1}, \end{align*} and therefore \begin{align*} a^*\psi_n=\frac{\sqrt{(n+1)!}}{\sqrt{n!}}\psi_{n+1}=\sqrt{n+1}\,\psi_{n+1}. \end{align*} The lowering formula requires the commutator. Since $[a,a^*]=I$, moving $a$ past one copy of $a^*$ produces one extra identity operator. Repeating this $n$ times gives \begin{align*} a(a^*)^n=(a^*)^n a+n(a^*)^{n-1}. \end{align*} This is proved by induction: the case $n=1$ is exactly $aa^*=a^*a+I$, and if it holds for $n$, then \begin{align*} a(a^*)^{n+1}=\left((a^*)^n a+n(a^*)^{n-1}\right)a^*. \end{align*} Using $aa^*=a^*a+I$, this becomes \begin{align*} a(a^*)^{n+1}=(a^*)^n(a^*a+I)+n(a^*)^n. \end{align*} Collecting the two copies of $(a^*)^n$ gives \begin{align*} a(a^*)^{n+1}=(a^*)^{n+1}a+(n+1)(a^*)^n. \end{align*} Applying the identity to $\psi_0$ eliminates the first term because $a\psi_0=0$. Hence, for $n\ge 1$, \begin{align*} a\psi_n =\frac{1}{\sqrt{n!}}a(a^*)^n\psi_0 =\frac{n}{\sqrt{n!}}(a^*)^{n-1}\psi_0. \end{align*} Since \begin{align*} \psi_{n-1}=\frac{1}{\sqrt{(n-1)!}}(a^*)^{n-1}\psi_0, \end{align*} we obtain \begin{align*} a\psi_n=\frac{n\sqrt{(n-1)!}}{\sqrt{n!}}\psi_{n-1}=\sqrt{n}\,\psi_{n-1}. \end{align*}[/guided]

[step:Prove normalisation and orthogonality]We use the complex $L^2$ [inner product](/page/Inner%20Product) convention \begin{align*} (u,v)_{L^2(\mathbb{R})}=\int_{\mathbb{R}}u(x)\overline{v(x)}\,d\mathcal{L}^1(x), \end{align*} which is linear in the first argument. We first prove $\|\psi_n\|_{L^2(\mathbb{R})}=1$ by induction. The case $n=0$ is the chosen normalisation. If $\|\psi_n\|_{L^2(\mathbb{R})}=1$, then \begin{align*} \|\psi_{n+1}\|_{L^2(\mathbb{R})}^2=\frac{1}{n+1}(a^*\psi_n,a^*\psi_n)_{L^2(\mathbb{R})}. \end{align*} For $u,v\in\mathcal{S}(\mathbb{R})$, [integration by parts](/theorems/210) on $[-R,R]$ and passage to the limit $R\to\infty$ give $(a^*u,v)_{L^2(\mathbb{R})}=(u,av)_{L^2(\mathbb{R})}$, because the Schwartz decay makes the boundary term vanish. Applying this adjointness identity to $u=\psi_n$ and $v=a^*\psi_n$ gives \begin{align*} \|\psi_{n+1}\|_{L^2(\mathbb{R})}^2=\frac{1}{n+1}(aa^*\psi_n,\psi_n)_{L^2(\mathbb{R})}. \end{align*} Since $aa^*=a^*a+I$ and $a\psi_n=\sqrt{n}\psi_{n-1}$ for $n\ge 1$, while $a\psi_0=0$, we have \begin{align*} (aa^*\psi_n,\psi_n)_{L^2(\mathbb{R})}=(a\psi_n,a\psi_n)_{L^2(\mathbb{R})}+(\psi_n,\psi_n)_{L^2(\mathbb{R})}. \end{align*} The induction hypothesis and the lowering relation give \begin{align*} (a\psi_n,a\psi_n)_{L^2(\mathbb{R})}=n. \end{align*} Therefore \begin{align*} (aa^*\psi_n,\psi_n)_{L^2(\mathbb{R})}=n+1. \end{align*} Thus $\|\psi_{n+1}\|_{L^2(\mathbb{R})}=1$. Now let $m,n\in\mathbb{N}\cup\{0\}$ with $m<n$. We prove $(\psi_n,\psi_m)_{L^2(\mathbb{R})}=0$ by induction on $m$. For $m=0$, the raising relation gives $\psi_n=n^{-1/2}a^*\psi_{n-1}$ for $n\ge 1$. Since $\psi_{n-1},\psi_0\in\mathcal{S}(\mathbb{R})$, the adjointness identity just proved on $\mathcal{S}(\mathbb{R})$ gives \begin{align*} (\psi_n,\psi_0)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{n}}(a^*\psi_{n-1},\psi_0)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{n}}(\psi_{n-1},a\psi_0)_{L^2(\mathbb{R})}=0. \end{align*} This proves orthogonality to $\psi_0$. Assume now that $m\ge 1$ and that $(\psi_r,\psi_{m-1})_{L^2(\mathbb{R})}=0$ for every $r>m-1$. Since $\psi_m=m^{-1/2}a^*\psi_{m-1}$, adjointness gives \begin{align*} (\psi_n,\psi_m)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{m}}(\psi_n,a^*\psi_{m-1})_{L^2(\mathbb{R})}. \end{align*} Moving $a^*$ to the first slot gives \begin{align*} (\psi_n,\psi_m)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{m}}(a\psi_n,\psi_{m-1})_{L^2(\mathbb{R})}. \end{align*} Using $a\psi_n=\sqrt{n}\psi_{n-1}$, we obtain \begin{align*} (\psi_n,\psi_m)_{L^2(\mathbb{R})}=\sqrt{\frac{n}{m}}(\psi_{n-1},\psi_{m-1})_{L^2(\mathbb{R})}. \end{align*} Because $n-1>m-1$, the induction hypothesis makes the right-hand side zero. Hence $(\psi_n,\psi_m)_{L^2(\mathbb{R})}=0$ for all $m<n$. Symmetry of the $L^2$ inner product gives the case $n<m$, and the normalisation already proved gives the diagonal case. Thus $(\psi_n)_{n\ge 0}$ is orthonormal.[/step]

[guided]We use the complex $L^2$ inner product convention \begin{align*} (u,v)_{L^2(\mathbb{R})}=\int_{\mathbb{R}}u(x)\overline{v(x)}\,d\mathcal{L}^1(x), \end{align*} which is linear in the first argument. We prove orthonormality in two parts: first each vector has norm one, then distinct vectors have inner product zero. The base case for normalisation is the defining normalisation $\|\psi_0\|_{L^2(\mathbb{R})}=1$. Suppose $\|\psi_n\|_{L^2(\mathbb{R})}=1$. Since every $\psi_k$ lies in $\mathcal{S}(\mathbb{R})$, [integration by parts](/theorems/2098) on $[-R,R]$ and passage to the limit $R\to\infty$ show that $a$ and $a^*$ are adjoint on the vectors used below; the boundary term vanishes because Schwartz functions decay faster than every polynomial. The raising relation gives $\psi_{n+1}=(n+1)^{-1/2}a^*\psi_n$, so \begin{align*} \|\psi_{n+1}\|_{L^2(\mathbb{R})}^2=\frac{1}{n+1}(a^*\psi_n,a^*\psi_n)_{L^2(\mathbb{R})}. \end{align*} Using adjointness and $aa^*=a^*a+I$, we get \begin{align*} \|\psi_{n+1}\|_{L^2(\mathbb{R})}^2=\frac{1}{n+1}\left((a\psi_n,a\psi_n)_{L^2(\mathbb{R})}+(\psi_n,\psi_n)_{L^2(\mathbb{R})}\right). \end{align*} The lowering relation gives $a\psi_n=\sqrt{n}\psi_{n-1}$ for $n\ge 1$, and gives $a\psi_0=0$ for $n=0$. Hence the numerator is $n+1$, and $\|\psi_{n+1}\|_{L^2(\mathbb{R})}=1$. For orthogonality, fix $m<n$. When $m=0$, the raising relation gives $\psi_n=n^{-1/2}a^*\psi_{n-1}$, and adjointness gives \begin{align*} (\psi_n,\psi_0)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{n}}(\psi_{n-1},a\psi_0)_{L^2(\mathbb{R})}=0. \end{align*} For $m\ge 1$, use $\psi_m=m^{-1/2}a^*\psi_{m-1}$ and move $a^*$ by adjointness: \begin{align*} (\psi_n,\psi_m)_{L^2(\mathbb{R})}=\frac{1}{\sqrt{m}}(a\psi_n,\psi_{m-1})_{L^2(\mathbb{R})}=\sqrt{\frac{n}{m}}(\psi_{n-1},\psi_{m-1})_{L^2(\mathbb{R})}. \end{align*} Induction on $m$ makes the final inner product zero because $n-1>m-1$. Symmetry of the inner product handles $n<m$, and the normalisation gives the diagonal. Therefore $(\psi_n)_{n\ge 0}$ is orthonormal.[/guided]

[step:Compute the eigenvalue equation]For each $n\ge 0$, the function $\psi_n$ belongs to $\mathcal{S}(\mathbb{R})$ because $a^*$ maps $\mathcal{S}(\mathbb{R})$ into itself. For the standard self-adjoint harmonic-oscillator realisation, the domain is \begin{align*} \mathcal{D}(H)=\left\{u\in L^2(\mathbb{R}): -\frac{\hbar^2}{2m}u''+\frac12m\omega^2x^2u\in L^2(\mathbb{R})\right\}, \end{align*} with derivatives understood distributionally. Since every Schwartz function and its derivatives decay faster than any polynomial, $\mathcal{S}(\mathbb{R})\subset\mathcal{D}(H)$; hence $\psi_n\in\mathcal{D}(H)$. Using the factorisation $H=\hbar\omega(a^*a+\frac12 I)$ on $\mathcal{S}(\mathbb{R})$ and the lowering relation, \begin{align*} a^*a\psi_n=a^*(\sqrt{n}\psi_{n-1}). \end{align*} Using the raising relation, this becomes \begin{align*} a^*a\psi_n=n\psi_n. \end{align*} for $n\ge 1$. For $n=0$, $a^*a\psi_0=0$. Therefore, for every $n\ge 0$, \begin{align*} H\psi_n=\hbar\omega\left(a^*a+\frac12 I\right)\psi_n =\hbar\omega\left(n+\frac12\right)\psi_n. \end{align*}[/step]

[guided]The domain point comes first. We already proved that $\psi_n\in\mathcal{S}(\mathbb{R})$ for every $n\ge 0$. The standard harmonic-oscillator domain is \begin{align*} \mathcal{D}(H)=\left\{u\in L^2(\mathbb{R}): -\frac{\hbar^2}{2m}u''+\frac12m\omega^2x^2u\in L^2(\mathbb{R})\right\}, \end{align*} where derivatives are distributional. A Schwartz function and all its derivatives decay faster than every polynomial, so the expression defining $Hu$ lies in $L^2(\mathbb{R})$ for $u\in\mathcal{S}(\mathbb{R})$. Therefore $\psi_n\in\mathcal{D}(H)$. On $\mathcal{S}(\mathbb{R})$ we proved the factorisation \begin{align*} H=\hbar\omega\left(a^*a+\frac12 I\right). \end{align*} For $n\ge 1$, the lowering relation gives $a\psi_n=\sqrt{n}\psi_{n-1}$, and the raising relation gives $a^*\psi_{n-1}=\sqrt{n}\psi_n$. Hence \begin{align*} a^*a\psi_n=a^*(\sqrt{n}\psi_{n-1})=n\psi_n. \end{align*} For $n=0$, the defining equation $a\psi_0=0$ gives $a^*a\psi_0=0$. Combining these cases, for every $n\ge 0$, \begin{align*} H\psi_n=\hbar\omega\left(a^*a+\frac12 I\right)\psi_n=\hbar\omega\left(n+\frac12\right)\psi_n. \end{align*}[/guided]

[step:Identify the functions with the classical Hermite functions]Let $H_n:\mathbb{R}\to\mathbb{R}$ denote the physicists' Hermite polynomial \begin{align*} H_n(y):=(-1)^n e^{y^2}\frac{d^n}{dy^n}e^{-y^2}. \end{align*} We prove by induction that \begin{align*} \psi_n(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}\frac{1}{\sqrt{2^n n!}}H_n(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} The case $n=0$ is the already computed formula for $\psi_0$. For the induction step, assume the formula holds for $n$. Since \begin{align*} a^*=\frac{1}{\sqrt{2}}\left(\alpha x-\alpha^{-1}\frac{d}{dx}\right) \end{align*} on $\mathcal{S}(\mathbb{R})$, the product rule gives \begin{align*} a^*\left(H_n(\alpha x)e^{-\alpha^2x^2/2}\right)=\frac{1}{\sqrt{2}}\left(2\alpha xH_n(\alpha x)-H_n'(\alpha x)\right)e^{-\alpha^2x^2/2}. \end{align*} Differentiating the Rodrigues formula defining $H_n$ gives the recurrence \begin{align*} H_{n+1}(y)=2yH_n(y)-H_n'(y). \end{align*} Using this identity with $y=\alpha x$, the preceding display becomes \begin{align*} a^*\left(H_n(\alpha x)e^{-\alpha^2x^2/2}\right)=\frac{1}{\sqrt{2}}H_{n+1}(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} Because $\psi_{n+1}=(n+1)^{-1/2}a^*\psi_n$, the induction hypothesis gives \begin{align*} \psi_{n+1}(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}\frac{1}{\sqrt{2^{n+1}(n+1)!}}H_{n+1}(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} Thus $(\psi_n)_{n\ge 0}$ is the rescaled classical Hermite family.[/step]

[guided]We need the exact normalising constants because completeness will use the fact that orthogonality to $\psi_n$ is the same as orthogonality to $H_n(y)e^{-y^2/2}$ after rescaling. Define $H_n:\mathbb{R}\to\mathbb{R}$ by \begin{align*} H_n(y):=(-1)^n e^{y^2}\frac{d^n}{dy^n}e^{-y^2}. \end{align*} We claim that \begin{align*} \psi_n(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}\frac{1}{\sqrt{2^n n!}}H_n(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} For $n=0$, $H_0(y)=1$, so this is exactly the previously computed formula for $\psi_0$. Assume the formula holds for some $n\ge 0$. On Schwartz functions, the creation operator can be rewritten as \begin{align*} a^*=\frac{1}{\sqrt{2}}\left(\alpha x-\alpha^{-1}\frac{d}{dx}\right). \end{align*} Applying this to the product $H_n(\alpha x)e^{-\alpha^2x^2/2}$ and using the product rule gives \begin{align*} a^*\left(H_n(\alpha x)e^{-\alpha^2x^2/2}\right)=\frac{1}{\sqrt{2}}\left(2\alpha xH_n(\alpha x)-H_n'(\alpha x)\right)e^{-\alpha^2x^2/2}. \end{align*} The expression in parentheses is exactly $H_{n+1}(\alpha x)$. Indeed, differentiating the Rodrigues formula defining $H_n$ gives the recurrence \begin{align*} H_{n+1}(y)=2yH_n(y)-H_n'(y). \end{align*} Therefore \begin{align*} a^*\left(H_n(\alpha x)e^{-\alpha^2x^2/2}\right)=\frac{1}{\sqrt{2}}H_{n+1}(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} Now use the raising definition with its normalising factor: \begin{align*} \psi_{n+1}=(n+1)^{-1/2}a^*\psi_n. \end{align*} Substituting the induction hypothesis and the last displayed identity yields \begin{align*} \psi_{n+1}(x)=\left(\frac{\alpha}{\sqrt{\pi}}\right)^{1/2}\frac{1}{\sqrt{2^{n+1}(n+1)!}}H_{n+1}(\alpha x)e^{-\alpha^2x^2/2}. \end{align*} This closes the induction and identifies $\psi_n$ with the rescaled classical Hermite functions.[/guided]

[step:Show that no nonzero function is orthogonal to all Hermite functions]Let $f\in L^2(\mathbb{R})$ satisfy \begin{align*} (f,\psi_n)_{L^2(\mathbb{R})}=0 \end{align*} for every $n\ge 0$. Define the rescaled function $F:\mathbb{R}\to\mathbb{C}$ by \begin{align*} F(y)=\alpha^{-1/2}f(y/\alpha). \end{align*} The change of variables $y=\alpha x$, with $d\mathcal{L}^1(y)=\alpha\,d\mathcal{L}^1(x)$, gives $F\in L^2(\mathbb{R})$. Since the $L^2$ inner product is linear in the first argument and each $\psi_n$ is real-valued, the condition $(f,\psi_n)_{L^2(\mathbb{R})}=0$ becomes \begin{align*} \int_{\mathbb{R}}F(y)H_n(y)e^{-y^2/2}\,d\mathcal{L}^1(y)=0 \end{align*} for every $n\ge 0$, after multiplication by the nonzero real normalising constant. Since the polynomials $H_n$ have exact degree $n$, their linear span is the space of all real polynomials. Hence \begin{align*} \int_{\mathbb{R}}F(y)y^k e^{-y^2/2}\,d\mathcal{L}^1(y)=0 \end{align*} for every $k\ge 0$. Define $G:\mathbb{R}\to\mathbb{C}$ by \begin{align*} G(y)=F(y)e^{-y^2/2}. \end{align*} By the [Cauchy-Schwarz inequality](/theorems/432), \begin{align*} \int_{\mathbb{R}}|G(y)|\,d\mathcal{L}^1(y) \le \|F\|_{L^2(\mathbb{R})}\left(\int_{\mathbb{R}}e^{-y^2}\,d\mathcal{L}^1(y)\right)^{1/2}<\infty. \end{align*} Moreover, for every $k\ge 0$, \begin{align*} \int_{\mathbb{R}}y^kG(y)\,d\mathcal{L}^1(y)=0. \end{align*} Define the transform $\Phi:\mathbb{C}\to\mathbb{C}$ by \begin{align*} \Phi(z)=\int_{\mathbb{R}}G(y)e^{-izy}\,d\mathcal{L}^1(y). \end{align*} For each $R>0$, the estimate \begin{align*} |G(y)e^{-izy}|\le |F(y)|e^{-y^2/2}e^{R|y|} \end{align*} holds whenever $|z|\le R$. The right-hand side is integrable by Cauchy-Schwarz, since \begin{align*} \int_{\mathbb{R}}e^{-y^2+2R|y|}\,d\mathcal{L}^1(y)<\infty. \end{align*} For each integer $k\ge 0$, the derivative of the integrand is $(-iy)^kG(y)e^{-izy}$. If $|z|\le R$, then \begin{align*} |(-iy)^kG(y)e^{-izy}|\le |y|^k|F(y)|e^{-y^2/2}e^{R|y|}. \end{align*} The right-hand side is integrable by Cauchy-Schwarz, since \begin{align*} \int_{\mathbb{R}}|y|^{2k}e^{-y^2+2R|y|}\,d\mathcal{L}^1(y)<\infty. \end{align*} Therefore differentiation under the integral is justified on each compact subset of $\mathbb{C}$, and \begin{align*} \Phi^{(k)}(0)=\int_{\mathbb{R}}(-iy)^kG(y)\,d\mathcal{L}^1(y)=0 \end{align*} for every $k\ge 0$. Since all Taylor coefficients of the entire function $\Phi$ at $0$ vanish, the Taylor series of $\Phi$ is identically zero in a neighbourhood of $0$. The identity theorem for holomorphic functions applied successively on overlapping discs then gives $\Phi\equiv 0$ on $\mathbb{C}$. Restricting to the real axis, the Fourier transform of the $L^1$ function $G$ vanishes identically. Let $\mathcal{S}'(\mathbb{R})$ denote the space of [tempered distributions](/page/Tempered%20Distributions) on $\mathbb{R}$. Define the regular tempered distribution $T_G\in\mathcal{S}'(\mathbb{R})$ by \begin{align*} T_G(\varphi)=\int_{\mathbb{R}}G(y)\varphi(y)\,d\mathcal{L}^1(y) \end{align*} for $\varphi\in\mathcal{S}(\mathbb{R})$. The vanishing of the Fourier transform of $G$ says exactly that the Fourier transform of $T_G$ is zero. Since the [Fourier transform on $\mathcal{S}'$](/theorems/230) is injective, $T_G=0$. Thus \begin{align*} \int_{\mathbb{R}}G(y)\varphi(y)\,d\mathcal{L}^1(y)=0 \end{align*} for every $\varphi\in\mathcal{S}(\mathbb{R})$. A locally integrable function whose associated [regular distribution](/page/Regular%20Distribution) is zero is zero almost everywhere; applying this to $G\in L^1(\mathbb{R})\subset L^1_{\mathrm{loc}}(\mathbb{R})$ gives $G=0$ for $\mathcal{L}^1$-almost every $y\in\mathbb{R}$. Since $e^{-y^2/2}>0$ for every $y\in\mathbb{R}$, it follows that $F=0$ in $L^2(\mathbb{R})$, and therefore $f=0$ in $L^2(\mathbb{R})$.[/step]

[guided]We prove completeness by proving the equivalent orthogonal-complement statement: if a function is orthogonal to every $\psi_n$, then it must be the zero element of $L^2(\mathbb{R})$. The scaling parameter in the oscillator is $\alpha=\sqrt{m\omega/\hbar}$. To remove it from the Hermite functions, define $F$ is the map $F:\mathbb{R}\to\mathbb{C}$ given by \begin{align*} F(y)=\alpha^{-1/2}f(y/\alpha). \end{align*} This is the unitary rescaling of $f$. Indeed, using the substitution $y=\alpha x$, so that $d\mathcal{L}^1(y)=\alpha\,d\mathcal{L}^1(x)$, we obtain \begin{align*} \int_{\mathbb{R}}|F(y)|^2\,d\mathcal{L}^1(y) =\int_{\mathbb{R}}\alpha^{-1}|f(y/\alpha)|^2\,d\mathcal{L}^1(y) =\int_{\mathbb{R}}|f(x)|^2\,d\mathcal{L}^1(x). \end{align*} Thus $F\in L^2(\mathbb{R})$. Because the $L^2$ inner product is linear in the first argument and each $\psi_n$ is real-valued, the orthogonality condition $(f,\psi_n)_{L^2(\mathbb{R})}=0$ becomes \begin{align*} \int_{\mathbb{R}}F(y)H_n(y)e^{-y^2/2}\,d\mathcal{L}^1(y)=0 \end{align*} for every $n\ge 0$, after multiplying by the nonzero real normalising constants. Since $H_n$ is a polynomial of degree exactly $n$, the family $(H_n)_{n\ge 0}$ spans all polynomials by triangular elimination. Therefore \begin{align*} \int_{\mathbb{R}}F(y)y^k e^{-y^2/2}\,d\mathcal{L}^1(y)=0 \end{align*} for every integer $k\ge 0$. Now define $G:\mathbb{R}\to\mathbb{C}$ by \begin{align*} G(y)=F(y)e^{-y^2/2}. \end{align*} This is the Gaussian-weighted version of $F$. The weight is chosen so that the Hermite orthogonality conditions become ordinary moment conditions. The function $G$ is integrable because \begin{align*} \int_{\mathbb{R}}|G(y)|\,d\mathcal{L}^1(y) \le \left(\int_{\mathbb{R}}|F(y)|^2\,d\mathcal{L}^1(y)\right)^{1/2} \left(\int_{\mathbb{R}}e^{-y^2}\,d\mathcal{L}^1(y)\right)^{1/2}<\infty. \end{align*} The moment identities become \begin{align*} \int_{\mathbb{R}}y^kG(y)\,d\mathcal{L}^1(y)=0 \end{align*} for every $k\ge 0$. We package all moments into one analytic function. Define $\Phi:\mathbb{C}\to\mathbb{C}$ by \begin{align*} \Phi(z)=\int_{\mathbb{R}}G(y)e^{-izy}\,d\mathcal{L}^1(y). \end{align*} This integral is absolutely convergent for every complex $z$. If $|z|\le R$, then \begin{align*} |G(y)e^{-izy}|\le |F(y)|e^{-y^2/2}e^{R|y|}. \end{align*} The right-hand side is integrable, because Cauchy-Schwarz gives \begin{align*} \int_{\mathbb{R}}|F(y)|e^{-y^2/2}e^{R|y|}\,d\mathcal{L}^1(y) \le \|F\|_{L^2(\mathbb{R})} \left(\int_{\mathbb{R}}e^{-y^2+2R|y|}\,d\mathcal{L}^1(y)\right)^{1/2}<\infty. \end{align*} For each integer $k\ge 0$, differentiating the integrand $k$ times with respect to $z$ produces the factor $(-iy)^k$. If $|z|\le R$, then \begin{align*} |(-iy)^kG(y)e^{-izy}|\le |y|^k|F(y)|e^{-y^2/2}e^{R|y|}. \end{align*} This majorant is integrable, because Cauchy-Schwarz gives \begin{align*} \int_{\mathbb{R}}|y|^k|F(y)|e^{-y^2/2}e^{R|y|}\,d\mathcal{L}^1(y) \le \|F\|_{L^2(\mathbb{R})} \left(\int_{\mathbb{R}}|y|^{2k}e^{-y^2+2R|y|}\,d\mathcal{L}^1(y)\right)^{1/2}<\infty. \end{align*} This domination is uniform on compact subsets of $\mathbb{C}$, so differentiating under the integral is justified. For every $k\ge 0$, \begin{align*} \Phi^{(k)}(0)=\int_{\mathbb{R}}(-iy)^kG(y)\,d\mathcal{L}^1(y)=0. \end{align*} Hence every Taylor coefficient of the entire function $\Phi$ at $0$ is zero. Therefore the Taylor series of $\Phi$ is identically zero in a neighbourhood of $0$, and the identity theorem for holomorphic functions applied successively on overlapping discs gives $\Phi\equiv 0$. For real $\xi$, $\Phi(\xi)$ is exactly the Fourier transform of $G$ at frequency $\xi$ under the convention used in the proof. Since $\Phi(\xi)=0$ for every $\xi\in\mathbb{R}$, we pass to distributions. Let $\mathcal{S}'(\mathbb{R})$ denote the space of tempered distributions on $\mathbb{R}$, and define the regular tempered distribution $T_G\in\mathcal{S}'(\mathbb{R})$ by \begin{align*} T_G(\varphi)=\int_{\mathbb{R}}G(y)\varphi(y)\,d\mathcal{L}^1(y) \end{align*} for $\varphi\in\mathcal{S}(\mathbb{R})$. The vanishing of $\Phi$ on the real axis says that the Fourier transform of $T_G$ is zero. The [Fourier transform on $\mathcal{S}'$](/theorems/230) is injective, so $T_G=0$. Equivalently, \begin{align*} \int_{\mathbb{R}}G(y)\varphi(y)\,d\mathcal{L}^1(y)=0 \end{align*} for every $\varphi\in\mathcal{S}(\mathbb{R})$. A locally integrable function whose associated regular distribution is zero is zero almost everywhere; since $G\in L^1(\mathbb{R})\subset L^1_{\mathrm{loc}}(\mathbb{R})$, this gives $G=0$ for $\mathcal{L}^1$-almost every $y$. Finally, the Gaussian factor never vanishes: \begin{align*} e^{-y^2/2}>0 \end{align*} for every $y\in\mathbb{R}$. Therefore $F=0$ almost everywhere, and undoing the unitary scaling gives $f=0$ in $L^2(\mathbb{R})$.[/guided]

[step:Conclude that the orthonormal family is a basis]We have shown that $(\psi_n)_{n\ge 0}$ is orthonormal and that its orthogonal complement in $L^2(\mathbb{R})$ is $\{0\}$. Let $M:=\overline{\operatorname{span}}\{\psi_n:n\ge 0\}\subset L^2(\mathbb{R})$. Since $M$ is closed by definition, the Hilbert-space orthogonal-complement identity gives $(M^\perp)^\perp=M$; this identity follows from the [orthogonal projection theorem](/theorems/4916) for closed subspaces of Hilbert spaces, because any vector orthogonal to $M^\perp$ must equal its projection onto $M$. As $M^\perp=\{0\}$, we get $M=L^2(\mathbb{R})$. Thus the closed linear span of $(\psi_n)_{n\ge 0}$ is all of $L^2(\mathbb{R})$. Hence $(\psi_n)_{n\ge 0}$ is an [orthonormal basis](/page/Orthonormal%20Basis) of $L^2(\mathbb{R})$. Together with the eigenvalue computation and the inclusion $\psi_n\in\mathcal{D}(H)$ for every $n\ge 0$, this proves the theorem.[/step]

[guided]The last step converts the orthogonal-complement calculation into the basis statement. Let $M:=\overline{\operatorname{span}}\{\psi_n:n\ge 0\}\subset L^2(\mathbb{R})$. We proved that every $f\in L^2(\mathbb{R})$ satisfying $(f,\psi_n)_{L^2(\mathbb{R})}=0$ for all $n\ge 0$ must be zero, so $M^\perp=\{0\}$. In a [Hilbert space](/page/Hilbert%20Space), the orthogonal-complement identity gives $(M^\perp)^\perp=\overline{M}$: by the [orthogonal projection](/theorems/437) theorem, a vector orthogonal to $M^\perp$ has zero component in the orthogonal complement of $\overline{M}$ and therefore lies in $\overline{M}$. Since $M$ is already closed by definition, $M=(M^\perp)^\perp=L^2(\mathbb{R})$. Thus the orthonormal family $(\psi_n)_{n\ge 0}$ has dense closed span equal to the whole Hilbert space, so it is an orthonormal basis. The previous eigenvalue step also proved $\psi_n\in\mathcal{D}(H)$ and $H\psi_n=\hbar\omega(n+1/2)\psi_n$, completing every assertion in the theorem.[/guided]