Quantum harmonic oscillator ($H=p^2/(2m)+m{\omega }^2{x}^2/2$)

The quantum harmonic oscillator describes a particle of mass $m>0$ in a quadratic potential energy with angular frequency $\omega >0$. In the position representation its Quantum Hamiltonian isThe model is useful far beyond a literal spring: a Taylor series about a stable equilibrium with positive potential curvature has this quadratic term as its leading nonconstant approximation. The pure quartic oscillator has zero curvature at its minimum, so that particular approximation does not apply. ^[ref].

In units with $\hslash =1$, creation and annihilation operators giveHere $N$ is the number operator. Restoring the reduced Planck constant, the energy eigenvalues are $E_n=\hslash \omega (n+1/2)$ for $n=0,1,2,\dots$: neighboring levels are separated by $\hslash \omega$, and the ground-state energy is the zero-point energy $\hslash \omega /2$.

With length scale $\ell =\sqrt{\hslash /(m\omega )}$ and dimensionless coordinate $u=x/\ell$, the ground state is the normalized Gaussian wave packetFor $\hslash =m=\omega =1$, this is precisely the $a=1$ trial wavefunction in the quantum variational principle. The balance of kinetic energy and potential energy therefore recovers the exact ground-state energy using only Gaussian integrals.

For equal masses in identical quadratic traps with an interaction depending quadratically on $x_1-x_2$, the displayed orthogonal coordinates of the normal modes separate the Hamiltonian operator into a collective and a relative oscillator. orthogonality preserves both the kinetic Laplacian and the absolute coordinate Jacobian. Particle exchange fixes $y_{+}$ and negates $y_{-}$. Hence bosonic exchange symmetry selects even relative oscillator levels, and fermionic exchange symmetry selects odd ones. For a trap frequency $\omega$ and interaction $-m{\omega }^2{\sin }^2\alpha (x_1-x_2{)}^2/4$, the two frequencies are $\omega$ and $\omega \cos \alpha$; the relative oscillator length diverges when the latter approaches zero.

Multiplying the quantum harmonic oscillator ground state by $x^2$ produces a quantum superposition of only its $n=0$ and $n=2$ energy eigenstates. In the conventional phase choice for the creation and annihilation operators, $X=\sqrt{\hslash /(2m\omega )}(b+b^{†})$ and $b|0⟩=0$. ThereforeAfter wavefunction normalization, the energy measurement probabilities are $1/3$ at $\hslash \omega /2$ and $2/3$ at $5\hslash \omega /2$. Rephasing the second basis state can change the displayed relative coefficient's sign but cannot change these probabilities. For example, defining the unscaled creation operator as $a_{+}=P+im\omega X=i\sqrt{2m\hslash \omega }{b}^{†}$ reverses the sign of the $n=2$ basis function generated from $(a_{+}{)}^2|0⟩$.

The expansion of the relativistic kinetic energy at fixed rest mass gives the perturbation $-p^4/(8m^3{c}^2)$. In the harmonic-oscillator ground state, $⟨p^4⟩=3(m\hslash \omega /2{)}^2$, proving the displayed first-order energy correction. Its small parameter is $\hslash \omega /(m{c}^2)$. This correction lowers the nonrelativistic energy and does not include other relativistic interactions.

The diagonal matrix element $K_{\beta }(x,x)=⟨x|e^{-\beta H}|x⟩$ combines the equal-endpoint Euclidean oscillator path with its endpoint-independent Dirichlet oscillator fluctuation determinant. It is not itself a normalized probability density: its integral is the canonical partition function $[2\sinh (z/2){]}^{-1}$. The normalized position probability density isAt high temperature, the variance approaches $1/(\beta m{\omega }^2)$ as in the classical canonical ensemble. At zero temperature, it approaches $\hslash /(2m\omega )$, the ground state variance. Dividing $K_{\beta }(x,x)$ by $K_{\beta }(0,0)$ cancels its entire fluctuation prefactor.

A free particle whose kinetic energy equals the ground-state total energy $\hslash \omega /2$ has momentum $\hslash /a$ and de Broglie wavelength $\lambda =2\pi a$, where $a=\sqrt{\hslash /(m\omega )}$. Substitution into the normalized bound-state Gaussian yields the title density. The comparison wavelength is a useful length scale; it is not a claim that the bound state has a single free momentum or spatially uniform probability density.

For a one-dimensional quantum harmonic oscillator, let $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. The stationary Schrödinger equation gives the displayed curvature relation. Inflections occur at classical turning points $\xi =\pm \sqrt{2n+1}$ and at simple wavefunction wavefunction nodes, provided the curvature changes sign. Thus the first excited state has an inflection at its central wavefunction node as well as at both turning points. Restricting attention to where $V=E$ misses wavefunction node inflections.

For a one-dimensional oscillator eigenstate, parity gives $⟨x⟩=0$ and its coordinate variance is the displayed expression. A classical oscillator with that coordinate energy has turning amplitude $A=\sqrt{2E_x/(m{\omega }^2)}$, so $\Delta x=A/\sqrt{2}$. In several dimensions the total energy does not specify a given coordinate's classical amplitude until its energy allocation is fixed. zero-point energies in the other quantum coordinates must not silently be assigned to that one classical mode.

For a positive diagonal quadratic potential $V=m{\sum }_i{\omega }_i^2{x}_i^2/2$, coordinate Hamiltonians commute and wavefunctions separate into one-dimensional oscillator factors. Equality of two squared frequencies gives rotation symmetry in their plane; equality of all three gives full rotation symmetry. Equalities create energy degeneracies and extra conserved angular momentum. Merely finding an accidental equality between particular level energies does not establish a global rotational symmetry.

For a two-dimensional anisotropic quantum harmonic oscillator in the indicated range, the six lowest number states, counted with multiplicity, have occupations $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2)$. Their excitation energies, divided by $\hslash$, are $0,{\omega }_1,{\omega }_2,2{\omega }_1,{\omega }_1+{\omega }_2,2{\omega }_2$. These are ordered because ${\omega }_2<2{\omega }_1$ and $2{\omega }_2<3{\omega }_1$; every omitted state has at least three quanta and costs at least $3\hslash {\omega }_1$. At equal angular frequencies, the two-dimensional isotropic harmonic oscillator has energy eigenvalue $\hslash \omega (N+1)$ and quantum degeneracy $N+1$, since there are $N+1$ ordered nonnegative pairs with sum $N$.

The number of nonnegative triples with $n_x+n_y+n_z=N$ is the stars-and-bars count $\left(\genfrac{}{}{0ex}{}{N+2}{2}\right)$. Including a two-state electron spin gives capacity $(N+1)(N+2)$. Filling non-interacting shells therefore uses capacities $2,6,12,\dots$, with shell energy $(N+3/2)\hslash \omega$ per electron. A partly filled shell has degenerate many-body ground states; rearranging its occupants is not a positive excitation energy.

A classical oscillator of amplitude $A$ has energy $m{\omega }^2{A}^2/2$. Comparing with the oscillator spacing $\hslash \omega$ gives the large-occupation scale displayed. A coherent state with this classical amplitude has exactly this mean excitation occupation and a Poisson number distribution; its total mean energy also includes the zero-point term. An observed macroscopic amplitude does not, by itself, establish a pure coherent state.

At imaginary duration $\beta \hslash$, $K_E(q,q)=\sqrt{m\omega /[2\pi \hslash \sinh x]}\exp [-m\omega q^2\tanh (x/2)/\hslash ]$, $x=\beta \hslash \omega$. Integrating the common endpoint yields $[2\sinh (x/2){]}^{-1}$.

A canonical quantum harmonic oscillator has geometric number probabilities and mean occupation $(e^{\beta \hslash \omega }-1{)}^{-1}$. The position operator then gives the displayed variance. It interpolates between zero-point fluctuations $\hslash /(2m\omega )$ and the classical variance $k_{B}T/(m{\omega }^2)$. The kinetic and potential energies are equal at every temperature, although classical equipartition describes their values only at high temperature.

The ground-state turning points are $\pm \sqrt{\hslash /(m\omega )}$. In dimensionless position $y=x\sqrt{m\omega /\hslash }$, the probability outside them is $(2/\sqrt{\pi }){\int }_1^{\infty }{e}^{-y^2}dy=\mathrm{erfc}(1)\simeq 0.15730$. This tail demonstrates that a bound quantum state need not be confined to its classically allowed region.

For identical masses and bare frequencies with coupling $kxy$, coordinates $(x\pm y)/\sqrt{2}$ diagonalize the Hamiltonian. A stable normalizable ground state requires $|k|<m{\omega }^2$. Its wave function is the product of Gaussians of frequencies ${\omega }_{\pm }$ in these normal coordinates; a two-mode squeeze of the original bare oscillators alone is generally insufficient.

For $H=p^2/(2m)+k{x}^2/2-qEx$, completing the square translates the well by $x_0=qE/k$ and shifts every energy by $-q^2{E}^2/(2k)$. The wavefunction of the ground state is translated without changing its width. Consequently the induced electric dipole moment is exactly linear in $E$, even when a low-order perturbative approximation to the state is inaccurate.

Let $x_{\mathrm{zp}}=\sqrt{\hslash /(2m\omega )}$ and $\eta =x_0/(2x_{\mathrm{zp}})$. The normalized first-order truncation is $(|0⟩+\eta |1⟩)/\sqrt{1+{\eta }^2}$. Its mean displacement is $2\eta x_{\mathrm{zp}}/(1+{\eta }^2)$, agreeing with $x_0$ only at small $|\eta |$. The exact translated ground state is a coherent state whose expansion contains all number states and has mean occupation ${\eta }^2$. Correct linear response does not justify extrapolating a truncated state to large fields.

An energy eigenstate has $⟨x⟩=⟨p⟩=0$, $⟨x^2⟩=\hslash (n+1/2)/(m\omega )$ and $⟨p^2⟩=m\hslash \omega (n+1/2)$. Thus $\Delta x\Delta p=\hslash (n+1/2)$; only the ground state in this sequence minimizes the Heisenberg uncertainty principle.

With oscillator length $\ell =\sqrt{\hslash /(m\omega )}$, the eigenfunctions are proportional to $H_n(x/\ell )e^{-x^2/(2{\ell }^2)}$. They have $n$ real nodes, parity $(-1{)}^n$, and energy $\hslash \omega (n+1/2)$.

A jump from positive frequency $\omega$ to $\Omega$ keeps an oscillator mode and its first derivative continuous. A normalized positive-frequency incident solution has outgoing coefficients $\alpha =(\Omega +\omega )/(2\sqrt{\Omega \omega })$ and $\beta =(\Omega -\omega )/(2\sqrt{\Omega \omega })$, satisfying ${\alpha }^2-{\beta }^2=1$. This temporal matching is not a sum-one spatial-scattering probability rule.

For the normalized ground state of a reduced quantum harmonic oscillator, ${\varphi }_0(q)={\pi }^{-1/4}{e}^{-q^2/2}$ and $E_0=1/2$. Its classical turning points are $q=\pm 1$. The Born rule assigns probability $P=2{\pi }^{-1/2}{\int }_1^{\infty }{e}^{-q^2}dq=\mathrm{erfc}(1)\simeq 0.157299$ to the classically forbidden region. This is a stationary probability of detection outside the classical interval. It is not a transmitted flux or an escape rate: the real stationary wavefunction has zero probability current.

Consider $-y^{\prime \prime }-(n-1)y^{\prime }/x+x^{2}y=\lambda y$ for a positive integer $n$, with regular radial data at zero and square-integrable decay at infinity in the measure $x^{n-1}dx$. Its Sturm-Liouville form has $p=w=x^{n-1}$ and $q=x^{n+1}$. For the Gaussian trial $y=e^{a{x}^2}$, $a<0$, the Rayleigh quotient isThe Gamma function recurrence evaluates the ratio, and the minimum is $n$ at $a=-1/2$. It is the exact lowest eigenvalue: after the endpoint term $[x^n{y}^2{]}_0^{\infty }$ vanishes, integration by parts givesEquality gives the displayed positive Gaussian eigenfunction. For $n=1$, the radial regularity condition is the Neumann boundary condition at zero, not a vanishing value there. Changing the boundary condition changes the lowest eigenvalue and whether this trial is admissible.

A half-harmonic oscillator has $V=m{\omega }^2{x}^2/2$ on $x>0$ and an infinite wall at $x=0$. Its wavefunction vanishes at the wall. Only the odd full-oscillator energy eigenstates satisfy this condition; restricting them to the positive half-line and multiplying by $\sqrt{2}$ restores normalization. Thus $E_r=\hslash \omega (2r+3/2)$ for $r=0,1,\dots$. With $\ell =\sqrt{\hslash /(m\omega )}$, its normalized ground state isIt has no interior node on its allowed half-line, and energy $3\hslash \omega /2$.

For the dimensionless half-harmonic oscillator $-{\psi }^{\prime \prime }+x^2\psi =\lambda \psi$ on $x>0$ with $\psi (0)=0$, the trial family $x{e}^{-\alpha x}$, $\alpha >0$, is admissible for the Rayleigh-Ritz variational principle. The moments ${\int }_0^{\infty }{x}^n{e}^{-2\alpha x}dx=n!/(2\alpha {)}^{n+1}$ give norm squared $1/(4{\alpha }^3)$, derivative energy $1/(4\alpha )$, and potential energy $3/(4{\alpha }^5)$. Its Rayleigh quotient is ${\alpha }^2+3/{\alpha }^2$, minimized at $\alpha =3^{1/4}$. The resulting upper bound $2\sqrt{3}$ exceeds the exact lowest eigenvalue three; the exact ground-state shape is $x{e}^{-x^2/2}$.

Suddenly remove the wall of a half-harmonic oscillator in its ground state, leaving the same harmonic potential on the whole line. The sudden approximation keeps its wavefunction unchanged during removal. The new normalized ground state is ${\psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$, soThe integration is over the initially occupied half-line. The final first excited state instead has probability $1/2$, since the initial half-line state equals $\sqrt{2}$ times its positive-half restriction. This also checks that removing the wall does not simply create the full oscillator's odd state.

In the isotropic three-dimensional quantum harmonic oscillator, the first excited Cartesian energy eigenstates are proportional to $x,y,z$ times the common Gaussian $e^{-r^2/(2a_0^2)}$. If ${\varphi }_x,{\varphi }_y,{\varphi }_z$ are normalized with the same real phase convention, their combinations in the title are normalized angular momentum eigenstates with $l=1$ and $m=1,0,-1$.

Indeed, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$ identify a common radial factor times the spherical harmonics $Y_{10},Y_{1,\pm 1}$. The transformation is unitary, so it changes the basis rather than the threefold orbital degeneracy.

For a number operator eigenstate $|n⟩$, the annihilation operator and creation operator satisfy $⟨a⟩=⟨a^{†}⟩=⟨a^2⟩=⟨(a^{†}{)}^2⟩=0$, $⟨a^{†}a⟩=n$ and $⟨a{a}^{†}⟩=n+1$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$ gives the displayed second moments and zero means.

Thus the kinetic energy and potential energy each have mean $\hslash \omega (n+1/2)/2$, and the uncertainty product is $\Delta x\Delta p=\hslash (n+1/2)$. Among number states only the ground state saturates the Heisenberg uncertainty relation. A quantum superposition can have nonzero position and momentum means despite each constituent number state's zero means.

The oscillator length fixes the spatial scale of a quantum harmonic oscillator of mass $m$ and angular frequency $\omega$. Its normalized ground wavefunction is ${\phi }_0(x)={\pi }^{-1/4}{\ell }^{-1/2}\exp [-x^2/(2{\ell }^2)]$, so $|{\phi }_0(0){|}^2=1/(\sqrt{\pi }\ell )$ and $⟨x^2⟩={\ell }^2/2$. The definition of $\ell$ is distinct from the root-mean-square displacement $\ell /\sqrt{2}$.

The perturbation $\lambda \hslash \omega (a^r+(a^{†}{)}^r)$ couples the oscillator vacuum only to $|r⟩$ at first order. Its squared coupling is $(\hslash \omega {)}^{2}r!$ and the energy denominator is $-r\hslash \omega$, giving the displayed formal second-order coefficient. For $r>2$ and nonzero real $\lambda$, coherent-state expectations contain $R^2+2\lambda R^r\cos r\theta$; choosing its phase negative proves the Hamiltonian is unbounded below. The formal vacuum branch must therefore not be called an exact ground state. For $r=2$ boundedness requires $|\lambda |<1/2$, while the displaced $r=1$ oscillator is stable for every real strength.

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.

For one quantum harmonic oscillator of frequency $\Omega$, $Z=e^{-x/2}/(1-e^{-x})$ with $x=\hslash \Omega /(k_{B}T)$. Differentiating $\ln Z$ gives $U=\hslash \Omega [1/2+1/(e^x-1)]$ and $C_V=k_B{x}^2{e}^x/(e^x-1{)}^2$. Zero-point energy leaves $U\to \hslash \Omega /2$ as $T\to 0$ but contributes no heat capacity. Excitation freeze-out makes $C_V$ exponentially small, while high temperature gives $U\sim k_{B}T$ and $C_V\to k_B$.

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.

Displacing a one-dimensional quantum harmonic oscillator produces a linear force term and a scalar shift relative to the unshifted Hamiltonian. The scalar adds only a common phase. The linear term couples number states separated by one excitation. Arbitrary linear driving of the ground state creates a coherent state; a first-order expansion of that state contains only the vacuum and first excited state.^[ref]

A sinusoidal trap displacement with amplitude $\epsilon$ and duration $T$ produces a resonant first-order excitation amplitude proportional to $\epsilon \omega T/a_0$, where $a_0$ is the oscillator length. Keeping its square much smaller than one is the long-time condition for weak excitation. The small-displacement condition $\epsilon /a_0\ll 1$ alone does not ensure uniform validity as $T$ grows. Exact linear driving remains solvable as a coherent state after this first-order approximation fails.

The ladder identity $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ proves the matrix element in the title. In time-dependent perturbation theory, a linear force excites only $n=1$ directly from $n=0$. Parity alone rules out even excited states; the ladder identity also rules out odd $n>1$. Higher-order evolution can reach other levels, so the driven oscillator is not an exact two-level system.

For $H(t)=\hslash \omega (a^{†}a+1/2)-Fx{e}^{t/\tau }$ on $t\le 0$, start in the unforced ground state at $t=-\infty$. Define $x_{01}=\sqrt{\hslash /(2m\omega )}$ and $\eta =F{x}_{01}/(\hslash \omega )$. Time-dependent perturbation theory givesThus the first excited-state probability in the original oscillator basis tends to zero for a rapid turn-on and to ${\eta }^2$ for slow turn-on, within the weak-force approximation $|\eta |\ll 1$.

This finite adiabatic limit is not a failure of the adiabatic theorem: the instantaneous ground state is displaced and has a nonzero projection onto the original excited states. In fact a linear force preserves a coherent state, whose exact original-basis displacement at $t=0$ is $\alpha =i\eta \omega \tau /(1+i\omega \tau )$. It follows either from $\dot{\alpha }=-i\omega \alpha +iF{x}_{01}{e}^{t/\tau }/\hslash$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha {|}^2{e}^{-|\alpha {|}^2}$. The instantaneous oscillator basis has displacement $\eta$, so its mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/[1+(\omega \tau {)}^2]$, which vanishes adiabatically.

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.

Zero-point motion is the nonzero quantum fluctuation of position and momentum in a system's ground state. For a quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ and nonzero position and momentum variances even at zero temperature. It does not represent a thermal mixture and need not carry entropy. The large zero-point motion of light helium atoms helps explain why helium can remain a liquid at low pressure even as temperature approaches absolute zero.

A squeezed coherent state is obtained from the vacuum by a squeeze operator and a displacement operator. Its quadrature variances can differ while saturating the uncertainty product. With zero displacement it is a squeezed vacuum state.

The bosonic ladder operators satisfy

For the quantum harmonic oscillator with $a_{-}=P-im\omega X$, $a_{+}=a_{-}^{\ast }$ and $\gamma =2m\hslash \omega$, factorization gives $a_{+}{a}_{-}=2m(H-E_0)$, where $E_0=\hslash \omega /2$. For an energy eigenstate $\psi$ at $E_n=E_0+n\hslash \omega$,The first identity shows that lowering to the ground state is injective on the $E_n$ eigenspace; the second shows how to invert it by raising. If the ground eigenspace is one-dimensional, $(a_{-}{)}^n\psi =c{\psi }_0$ and therefore $\psi =c(a_{+}{)}^n{\psi }_0/({\gamma }^{n}n!)$. Thus every level is also nondegenerate. This proof explicitly rules out a second independent energy eigenstate at an excited level rather than merely constructing one ladder of states.

For real coefficients in $a_{\pm }=c_{\pm }X+i{d}_{\pm }P$, imposing $a_{-}^{†}=a_{+}$, $[a_{-},a_{+}]=I$ and $H=\hslash \omega (a_{+}{a}_{-}+\lambda I)$ givesIndeed the adjoint operator condition makes the two $c$ coefficients equal and the two $d$ coefficients opposite. The canonical commutation relation gives $2\hslash c_{-}{d}_{-}=1$, and matching the coefficients of $X^2$ and $P^2$ fixes their magnitudes. Both signs change together. The conventional creation and annihilation operators choose $\epsilon =1$; the other sign is an equivalent phase convention for the number states.

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

A real centered Gaussian wave packet with initial position variance $1/(2\alpha )$ and momentum variance ${\hslash }^2\alpha /2$ evolves under a quantum harmonic oscillator byThe covariance matrix obeys $\Sigma (t)=M\Sigma (0)M^{\mathsf{T}}$ and $\det \Sigma ={\hslash }^2/4$. Thus the Robertson-Schrödinger uncertainty relation stays saturated, although the position-momentum variance product can exceed ${\hslash }^2/4$ because of nonzero quantum covariance. The widths are stationary only for $\alpha =m\omega /\hslash$; otherwise they breathe with frequency $2\omega$.

photon addition to an optical squeezed vacuum gives $a^{†}S|0⟩=\cosh rS|1⟩$; subtraction gives $aS|0⟩=-\sinh rS|1⟩$. After successful-event normalization these are the same state up to phase, with mean occupation $1+3{\sinh }^{2}r$ and odd photon parity. At zero squeezing subtraction never succeeds; the normalized limiting state is not an output at that exact parameter value.

A single optical mode in a squeezed vacuum state. In principal optical quadratures, its variances are $e^{-2r}/4$ and $e^{2r}/4$, while its mean photon number is ${\sinh }^{2}r$. Its mean field vanishes. A degenerate optical parametric amplifier generates the state by pair creation with phase fixed by the pump. Loss admixes unsqueezed vacuum and limits the measured variance reduction.

The single-mode squeeze operator changes occupation in pairs. Consequently an optical squeezed vacuum has only even photon numbers, with mean $s={\sinh }^{2}r$ and variance $2s(s+1)$. Its large number fluctuations are compatible with quadrature squeezing: they concern different observables.

Here $s={\sinh }^{2}r$. The pair moment is $⟨N(N-1)⟩=3s^2+s$. These expressions follow from the single-mode Bogoliubov transformation or by differentiating the even-occupation probability generating function $G(t)=[\cosh r\sqrt{1-t^2{\tanh }^{2}r}{]}^{-1}$.