Quantum harmonic oscillator

In units with $\hslash =1$, a harmonic oscillator of angular frequency $\omega$ has

For action $\frac{m}{2}\int ({\dot{q}}^2-{\omega }^2{q}^2)dt$, the kernel is $\sqrt{m\omega /(2\pi i\sin \omega T)}\exp \{im\omega [(q_f^2+q_i^2)\cos \omega T-2q_f{q}_i]/(2\sin \omega T)\}$. The Dirichlet oscillator determinant ratio supplies its prefactor, while the classical boundary action supplies its exponent. The Feynman i-epsilon prescription specifies its continuation through caustics.

For a quantum harmonic oscillator with $\omega >0$, a complex mode satisfying the Wronskian normalization $q{\dot{q}}^{\ast }-\dot{q}{q}^{\ast }=i$ can be written $q=r{e}^{is}$ with $\dot{s}=-1/(2r^2)$. Its vacuum energy is ${\dot{r}}^2/2+1/(8r^2)+{\omega }^2{r}^2/2$. This is at least $\omega /2$, with equality exactly when $\dot{r}=0$ and $r^2=1/(2\omega )$. The same argument fixes the subhorizon positive-frequency normalization in the Bunch-Davies vacuum of a canonical cosmological mode.

This Hilbert space has norm $\Vert u{\Vert }_X^2=\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2+\Vert xu{\Vert }_2^2$. Completeness follows from completeness of H1 space and closedness of the multiplication operator $u\mapsto xu$. Smooth cutoffs followed by convolution with a mollifier show that smooth compactly supported functions are dense. The space is the natural energy domain of the dimensionless quantum harmonic oscillator, whereas the operator domain additionally requires $u^{\prime \prime }$ and $x^{2}u$ to lie in L2 space.

For $c>0$ and smooth compactly supported $u$, integration by parts givesDropping the nonnegative weighted derivative term proves the displayed bound. For a weak resolvent solution in the one-dimensional harmonic oscillator form domain, apply the bound to ${\chi }_{R}u$, where ${\chi }_R$ is a smooth cutoff. The cutoff error $-2{\chi }_R^{\prime }{u}^{\prime }-{\chi }_R^{\prime \prime }u$ is bounded in L2 space and tends to zero. The Fatou lemma then shows that $u^{\prime \prime }$ and $x^{2}u$ belong to L2 space, identifying the strong operator domain.

For $Au=u^{\prime \prime }-(1+x^2)u$ on $D(A)=\{u\in H^2(\mathbb{R}):x^{2}u\in L^2\}$, the form $a_{\lambda }(u,v)=\int u^{\prime }\overline{v^{\prime }}+(1+\lambda +x^2)u\bar{v}$ is coercive on the one-dimensional harmonic oscillator form domain when $\lambda >0$. The Lax-Milgram theorem gives a unique weak solution of $(\lambda -A)u=g$. The graph estimate for the harmonic oscillator puts it in $D(A)$, and $a_{\lambda }(u,u)\ge (1+\lambda )\Vert u{\Vert }_2^2$ gives the displayed resolvent bound. The domain is dense and the bounded resolvent proves closedness, so the Hille-Yosida theorem gives a contraction semigroup. An exponentially shifted semigroup with generator $A+I$ solves the heat equation with oscillator potential.

For a quantum harmonic oscillator with $\hslash =1$, $\omega >0$ and inverse temperature $\beta >0$, the canonical partition function is ${\sum }_{r\ge 0}{e}^{-\beta \omega (r+1/2)}$. The geometric series gives the displayed formula. Keeping the zero-point energy is essential to its prefactor.

Adding $-Fx$ to the quantum harmonic oscillator potential gives $\frac{1}{2}m{\omega }^2(x-d{)}^2-F^2/(2m{\omega }^2)$, where $d=F/(m{\omega }^2)$. Translation preserves the oscillator spectrum, apart from the displayed constant shift. Eigenfunctions are translated by $d$ and the energy spacing is unchanged.

In the sudden approximation, an initial oscillator ground state overlaps the displaced ground state by $\exp [-m\omega d^2/(4\hslash )]$. Completing the square in the product of the two normalized Gaussian wavefunctions and using the Gaussian integral proves this amplitude. Squaring it gives the displayed probability. Only an irrelevant overall phase evolves before the sudden switch.

Zero-point energy is the nonzero ground-state energy left by quantum fluctuations. One harmonic oscillator contributes $\omega /2$ before any regularization or normal-ordering subtraction.

The bosonic ladder operators satisfy

An annihilation operator lowers the occupation number operator eigenvalue by one and annihilates the Fock vacuum.

A creation operator raises the occupation number operator eigenvalue of a quantum harmonic oscillator by one. Repeated application builds Fock states.

For a real squeezing parameter $\gamma$, the single-mode squeeze operator isIts exponent is skew-adjoint, so $S(\gamma )$ is a unitary operator, and it implements the Bogoliubov transformation

The squeezed vacuum is $|\gamma ⟩=S(\gamma )|0⟩$. Its position and momentum variances are rescaled in opposite directions,so it remains an minimum-uncertainty state with $\Delta X\Delta P=\hslash /2$.

For an invertible finite-mode Bogoliubov transformation $a=A{a}^{\prime }+B{a}^{\prime }{}^{†}$, the symmetric squeezing matrix is $M=-A^{-1}B$. The canonical commutation relations make its singular values strictly less than one. The normalized state has $|C|=\det (I-M{M}^{†}{)}^{1/4}$. Infinitely many modes additionally require bosonic mode mixing implementability.

The number operator $N=A^{†}A$ satisfiesIt is self-adjoint and positive semidefinite.

A number state is a normalized eigenstate of the number operator:For a single quantum harmonic oscillator, it is also an energy eigenstate with energy $\hslash \omega (n+1/2)$.

If $N|\nu ⟩=\nu |\nu ⟩$, then $A$ lowers the eigenvalue by one and $\Vert A|\nu ⟩{\Vert }^2=\nu \Vert \nu ⟩{\Vert }^2$. Positivity forces the lowering ladder to terminate at eigenvalue zero, so every eigenvalue is a nonnegative integer.

For independent frequencies $1$ and $2$,so the eigenspace at energy $k+3/2$ has one basis state $|n,m⟩$ for each nonnegative solution of $n+2m=k$.