Harmonic Oscillator Ladder Operators

Purpose

For the one-dimensional oscillator

$$H = \frac{p^2}{2m} + \frac12m\omega^2x^2, \qquad [x,p]=i\hbar,$$

define the oscillator length

$$\ell = \sqrt{\frac{\hbar}{m\omega}}$$

and the dimensionless ladder operators

$$a = \frac{1}{\sqrt2} \left( \frac{x}{\ell} + \frac{i\ell p}{\hbar} \right),$$ $$a^\dagger = \frac{1}{\sqrt2} \left( \frac{x}{\ell} - \frac{i\ell p}{\hbar} \right).$$

They satisfy

$$[a,a^\dagger]=I,$$

and factor the Hamiltonian as

$$H = \hbar\omega \left( N+\frac12I \right), \qquad N=a^\dagger a.$$

The full algebraic proof of the spectrum is at Ladder-Operator Solution: First Encounter. This card collects the identities needed after the construction is known.

At a glance

Task	Formula
Lower a number state	$a\lvert n\rangle=\sqrt n\lvert n-1\rangle$
Raise a number state	$a^\dagger\lvert n\rangle=\sqrt{n+1}\lvert n+1\rangle$
Build a number state	$\lvert n\rangle=(a^\dagger)^n\lvert0\rangle/\sqrt{n!}$
Count quanta	$N\lvert n\rangle=n\lvert n\rangle$
Write the spectrum	$E_n=\hbar\omega(n+1/2)$
Recover position	$x=\ell(a+a^\dagger)/\sqrt2$
Recover momentum	$p=\hbar(a-a^\dagger)/(i\ell\sqrt2)$
Evolve a ladder operator	$a_H(t)=e^{-i\omega(t-t_0)}a_H(t_0)$
Ground-state condition	$a\lvert0\rangle=0$

The number label starts at $n=0$. The lowering formula gives zero on the vacuum because its coefficient is $\sqrt0$.

Definitions in dimensional form

The equivalent expressions are

$$a = \sqrt{\frac{m\omega}{2\hbar}}\,x + \frac{i}{\sqrt{2m\hbar\omega}}\,p,$$ $$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\,x - \frac{i}{\sqrt{2m\hbar\omega}}\,p.$$

Every term is dimensionless. The inverse relations are

$$x = \frac{\ell}{\sqrt2}(a+a^\dagger),$$ $$p = \frac{\hbar}{i\ell\sqrt2}(a-a^\dagger) = -i\sqrt{\frac{m\hbar\omega}{2}} (a-a^\dagger).$$

Changing the sign of the momentum term exchanges the roles of the displayed $a$ and $a^\dagger$. A different convention is acceptable only when every commutator, vacuum condition, and ladder action is changed consistently.

Core algebra

From $[x,p]=i\hbar$,

$$[a,a^\dagger]=I.$$

Consequently,

$$aa^\dagger=N+I, \qquad a^\dagger a=N.$$

The number-operator commutators are

$$[N,a]=-a, \qquad [N,a^\dagger]=a^\dagger.$$

Since $H=\hbar\omega(N+I/2)$,

$$[H,a] = -\hbar\omega a,$$ $$[H,a^\dagger] = \hbar\omega a^\dagger.$$

Thus $a$ and $a^\dagger$ are spectrum-generating operators. They do not commute with $H$ and are not conserved observables.

Number states and repeated actions

Choose normalized number states with phases fixed so that the ladder coefficients are positive real numbers:

$$a\lvert n\rangle = \sqrt n\,\lvert n-1\rangle,$$ $$a^\dagger\lvert n\rangle = \sqrt{n+1}\,\lvert n+1\rangle.$$

Repeated lowering gives

$$a^r\lvert n\rangle = \sqrt{\frac{n!}{(n-r)!}} \lvert n-r\rangle, \qquad 0\leq r\leq n,$$

and

$$a^r\lvert n\rangle=0, \qquad r\gt n.$$

Repeated raising gives

$$(a^\dagger)^r\lvert n\rangle = \sqrt{\frac{(n+r)!}{n!}} \lvert n+r\rangle.$$

Starting from the vacuum,

$$\lvert n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} \lvert0\rangle.$$

The basis is orthonormal and complete:

$$\langle m\vert n\rangle=\delta_{mn}, \qquad \sum_{n=0}^{\infty} \lvert n\rangle\langle n\rvert=I.$$

Hamiltonian and zero-point energy

Multiplication in the stated order gives

$$a^\dagger a = \frac12 \left( \frac{x^2}{\ell^2} + \frac{\ell^2p^2}{\hbar^2} -I \right).$$

Therefore

$$H = \hbar\omega \left( a^\dagger a+\frac12I \right).$$

On $\lvert n\rangle$,

$$H\lvert n\rangle = \hbar\omega \left(n+\frac12\right) \lvert n\rangle.$$

The $1/2$ term comes from operator noncommutativity. Replacing $a^\dagger a$ by $aa^\dagger$ without compensating for their commutator shifts every energy by $\hbar\omega$.

Position and momentum matrix elements

The inverse relations and ladder actions give

$$\begin{aligned} \langle m\rvert x\lvert n\rangle &= \frac{\ell}{\sqrt2} \left[ \sqrt n\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1} \right], \end{aligned}$$

and

$$\begin{aligned} \langle m\rvert p\lvert n\rangle &= \frac{\hbar}{i\ell\sqrt2} \left[ \sqrt n\,\delta_{m,n-1} - \sqrt{n+1}\,\delta_{m,n+1} \right]. \end{aligned}$$

Both connect only adjacent number states:

$$\Delta n=\pm1.$$

This is the ideal-oscillator selection rule for a perturbation linear in $x$ or $p$. A selection rule identifies zeros of matrix elements; whether a nonzero transition is resonant also depends on the perturbation’s time and frequency content.

Quadratic matrix elements

Squaring the position operator gives

$$x^2 = \frac{\ell^2}{2} \left[ a^2+(a^\dagger)^2+2N+I \right].$$

Therefore

$$\begin{aligned} \langle m\rvert x^2\lvert n\rangle &= \frac{\ell^2}{2} \bigl[{} \sqrt{n(n-1)}\,\delta_{m,n-2}\\ &\quad+(2n+1)\delta_{mn}\\ &\quad+\sqrt{(n+1)(n+2)}\,\delta_{m,n+2} \bigr]. \end{aligned}$$

Likewise,

$$p^2 = \frac{\hbar^2}{2\ell^2} \left[ -a^2-(a^\dagger)^2+2N+I \right],$$

so

$$\begin{aligned} \langle m\rvert p^2\lvert n\rangle &= \frac{\hbar^2}{2\ell^2} \bigl[{} -\sqrt{n(n-1)}\,\delta_{m,n-2}\\ &\quad+(2n+1)\delta_{mn}\\ &\quad-\sqrt{(n+1)(n+2)}\,\delta_{m,n+2} \bigr]. \end{aligned}$$

A quadratic perturbation therefore has

$$\Delta n=0,\pm2.$$

Number-state moments

For every number state,

$$\langle x\rangle_n=0, \qquad \langle p\rangle_n=0.$$

The second moments are

$$\langle x^2\rangle_n = \ell^2 \left(n+\frac12\right),$$ $$\langle p^2\rangle_n = \frac{\hbar^2}{\ell^2} \left(n+\frac12\right) = m\hbar\omega \left(n+\frac12\right).$$

Thus

$$(\Delta x)_n(\Delta p)_n = \hbar \left(n+\frac12\right).$$

Only the vacuum reaches the lower bound:

$$(\Delta x)_0 = \frac{\ell}{\sqrt2}, \qquad (\Delta p)_0 = \frac{\hbar}{\ell\sqrt2},$$ $$(\Delta x)_0(\Delta p)_0 = \frac{\hbar}{2}.$$

The kinetic and potential expectations are equal:

$$\langle T\rangle_n = \langle V\rangle_n = \frac{E_n}{2}.$$

This is the oscillator virial theorem in the number basis.

Ground-state wavefunction

In position representation,

$$p=-i\hbar\frac{d}{dx},$$

so

$$a = \frac{1}{\sqrt2} \left( \frac{x}{\ell} + \ell\frac{d}{dx} \right).$$

The vacuum condition $a\lvert0\rangle=0$ becomes

$$\left( \frac{x}{\ell} + \ell\frac{d}{dx} \right) \psi_0(x)=0.$$

Its normalized solution is

$$\psi_0(x) = \left( \frac{1}{\pi\ell^2} \right)^{1/4} e^{-x^2/(2\ell^2)}.$$

Repeated application of $a^\dagger$ generates the normalized Hermite-function eigenstates. Their explicit formula is collected in Number States.

Heisenberg evolution

For the time-independent oscillator Hamiltonian,

$$\frac{da_H}{dt} = \frac{i}{\hbar}[H,a_H] = -i\omega a_H.$$

Hence

$$a_H(t) = e^{-i\omega(t-t_0)}a_H(t_0),$$ $$a_H^\dagger(t) = e^{i\omega(t-t_0)}a_H^\dagger(t_0).$$

Substituting into $x$ and $p$ gives

$$x_H(t) = x_H(t_0)\cos(\omega\tau) + \frac{p_H(t_0)}{m\omega} \sin(\omega\tau),$$ $$p_H(t) = p_H(t_0)\cos(\omega\tau) - m\omega x_H(t_0) \sin(\omega\tau),$$

where $\tau=t-t_0$. These are exact operator identities.

Coherent-state bridge

A coherent state is an eigenstate of $a$:

$$a\lvert\alpha\rangle = \alpha\lvert\alpha\rangle.$$

Its number-basis expansion is

$$\lvert\alpha\rangle = e^{-\lvert\alpha\rvert^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \lvert n\rangle.$$

Under oscillator evolution, the eigenvalue rotates:

$$\alpha(t)=\alpha(0)e^{-i\omega t}$$

up to the common state phase. The dedicated coherent-state formula card is the next entry in this category; the canonical treatment is at Coherent States.

Interpretation and domains

For a single particle in an external quadratic potential, $a^\dagger$ raises the oscillator excitation and $a$ lowers it. They do not create or destroy the particle. In field and many-body applications, each normal mode has the same oscillator algebra, and its ladder operators may acquire a particle or quasiparticle interpretation.

$x$, $p$, $a$, and $a^\dagger$ are unbounded. Their products and formal commutators require a common domain. The Schwartz space and finite linear combinations of number states provide standard invariant cores for the displayed algebra. A normalized vector need not belong to the domain of every power of a ladder operator.

Units and convention checks

Quantity	Dimensions
$a$, $a^\dagger$, $N$	dimensionless
$\ell$	length
$x$	length
$p$	momentum
$H$, $\hbar\omega$	energy

Fast checks are

$$[a,a^\dagger]=I,$$ $$\lVert a\lvert n\rangle\rVert^2=n, \qquad \lVert a^\dagger\lvert n\rangle\rVert^2=n+1,$$

and Hermiticity of the reconstructed $x$ and $p$.

Common mistakes

• Dropping a factor of $m$, $\omega$, or $\hbar$ so that $a$ is not dimensionless.
• Reversing the sign of the momentum term in only one formula.
• Treating $a$ and $a^\dagger$ as commuting numbers.
• Replacing $a^\dagger a$ by $aa^\dagger$ without the identity term.
• Omitting the zero-point energy.
• Starting number-state labels at $n=1$.
• Dropping the square-root coefficients in ladder actions.
• Applying the factorial lowering formula when $r\gt n$ instead of returning the zero vector.
• Calling $a$ or $a^\dagger$ Hermitian observables.
• Calling $a^\dagger$ a particle-creation operator in the single-particle oscillator without qualification.
• Using $\Delta n=\pm1$ for an $x^2$ perturbation.
• Ignoring domains for unbounded operator products.

Related formulas

• Harmonic Oscillator Spectrum
• Canonical Commutation Relations
• Commutator Identities
• Heisenberg Equation
• Number States
• Hermite Functions
• Zero-Point Energy
• Harmonic Oscillator to Fields

Exercises

1. Verify the ladder commutator

Starting from the dimensional definitions of $a$ and $a^\dagger$, show that $[a,a^\dagger]=I$.

Solution

Write

$$a=\alpha x+i\beta p, \qquad a^\dagger=\alpha x-i\beta p,$$

where

$$\alpha=\sqrt{\frac{m\omega}{2\hbar}}, \qquad \beta=\frac{1}{\sqrt{2m\hbar\omega}}.$$

Then

$$\begin{aligned} [a,a^\dagger] &= -i\alpha\beta[x,p] +i\alpha\beta[p,x]\\ &= 2\alpha\beta\hbar I. \end{aligned}$$

Since $2\alpha\beta\hbar=1$, the result is

$$[a,a^\dagger]=I.$$

2. Derive the position matrix element

Compute $\langle m\rvert x\lvert n\rangle$ and state the selection rule.

Solution

Use

$$x=\frac{\ell}{\sqrt2}(a+a^\dagger).$$

The two terms give

$$\langle m\rvert a\lvert n\rangle = \sqrt n\,\delta_{m,n-1},$$ $$\langle m\rvert a^\dagger\lvert n\rangle = \sqrt{n+1}\,\delta_{m,n+1}.$$

Therefore

$$\langle m\rvert x\lvert n\rangle = \frac{\ell}{\sqrt2} \left[ \sqrt n\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1} \right].$$

It vanishes unless $m=n\pm1$.

3. Quadratic selection rule

Show that $x^2$ connects only $\Delta n=0,\pm2$ and compute $\langle n\rvert x^2\lvert n\rangle$.

Solution

Square $x=\ell(a+a^\dagger)/\sqrt2$:

$$x^2 = \frac{\ell^2}{2} \left[ a^2+aa^\dagger+a^\dagger a+(a^\dagger)^2 \right].$$

Using $aa^\dagger=N+I$ gives

$$x^2 = \frac{\ell^2}{2} \left[ a^2+(a^\dagger)^2+2N+I \right].$$

The first and second terms change $n$ by $-2$ and $+2$, while $N$ and $I$ are diagonal. Hence $\Delta n=0,\pm2$. On a number state,

$$\langle n\rvert x^2\lvert n\rangle = \frac{\ell^2}{2}(2n+1) = \ell^2\left(n+\frac12\right).$$

4. Heisenberg evolution

Derive $a_H(t)=e^{-i\omega(t-t_0)}a_H(t_0)$ and use it to recover $x_H(t)$.

Solution

The Heisenberg equation and $[H,a]=-\hbar\omega a$ give

$$\dot a_H = \frac{i}{\hbar}[H,a_H] = -i\omega a_H.$$

Therefore

$$a_H(t) = e^{-i\omega\tau}a_H(t_0), \qquad \tau=t-t_0.$$

Taking the adjoint gives

$$a_H^\dagger(t) = e^{i\omega\tau}a_H^\dagger(t_0).$$

Substitute these into $x_H=\ell(a_H+a_H^\dagger)/\sqrt2$ and rewrite the initial ladder operators in terms of $x_H(t_0)$ and $p_H(t_0)$. The result is

$$x_H(t) = x_H(t_0)\cos(\omega\tau) + \frac{p_H(t_0)}{m\omega}\sin(\omega\tau).$$

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Ch. 7.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018, Ch. 2.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, Sec. 2.3.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, Wiley, 1977.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.