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Perturbation Theory for the Harmonic Oscillator

The harmonic oscillator is the most transparent laboratory for stationary perturbation theory. Its spectrum is known, every energy denominator is an integer multiple of ℏω\hbar\omega, and ladder operators turn polynomial perturbations into finite-band matrices with visible parity selection rules.

That simplicity lets several logically different outcomes be compared in one place:

  • a linear perturbation has a vanishing first-order energy yet is exactly removable by displacement;
  • a quadratic perturbation changes the frequency and is exactly solvable by choosing a new oscillator basis;
  • a cubic perturbation has a symmetry-forced zero diagonal term but exposes an important stability caveat;
  • a positive quartic perturbation defines a stable non-Gaussian problem with a useful asymptotic series.

This page owns that comparative sandbox. The exact oscillator, displaced oscillator, squeezed states, durable anharmonic model, and full quartic worked calculation retain their own canonical homes.

Take

H0=p22m+12mω2x2,ω>0.H_0 = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2, \qquad \omega\gt0.

Its number states obey

H0∣n⟩=En(0)∣n⟩,En(0)=ℏω(n+12).\begin{aligned} H_0\lvert n\rangle &= E_n^{(0)}\lvert n\rangle, \\ E_n^{(0)} &= \hbar\omega \left(n+\frac{1}{2}\right). \end{aligned}

Define the zero-point position scale

xzpf≡ℏ2mωx_{\mathrm{zpf}} \equiv \sqrt{\frac{\hbar}{2m\omega}}

and the dimensionless coordinate

q≡a+a†.q\equiv a+a^\dagger.

Then

x=xzpfq,x=x_{\mathrm{zpf}}q,

with

a∣n⟩=n ∣n−1⟩,a†∣n⟩=n+1 ∣n+1⟩.\begin{aligned} a\lvert n\rangle &= \sqrt n\,\lvert n-1\rangle, \\ a^\dagger\lvert n\rangle &= \sqrt{n+1}\,\lvert n+1\rangle. \end{aligned}

The factor of 22 in xzpfx_{\mathrm{zpf}} is part of the convention. Some sources instead define an oscillator length ℓ=ℏ/(mω)\ell=\sqrt{\hbar/(m\omega)}, for which x=ℓ(a+a†)/2x=\ell(a+a^\dagger)/\sqrt2. Mixing the two definitions is a common source of incorrect coefficients.

Each factor of q=a+a†q=a+a^\dagger changes nn by one. Consequently, xrx^r can connect only states satisfying

∣Δn∣≤r,Δn≡r(mod2).\lvert\Delta n\rvert\le r, \qquad \Delta n\equiv r\pmod 2.

Parity gives the same result from another direction. Number states have parity (−1)n(-1)^n, while xrx^r is even for even rr and odd for odd rr.

PerturbationParityAllowed changesDiagonal element
xxoddΔn=±1\Delta n=\pm1zero
x2x^2evenΔn=0,±2\Delta n=0,\pm2generally nonzero
x3x^3oddΔn=±1,±3\Delta n=\pm1,\pm3zero
x4x^4evenΔn=0,±2,±4\Delta n=0,\pm2,\pm4generally nonzero

Allowed oscillator number-state couplings generated by the first four powers of position

Potentially nonzero matrix elements of xrx^r in the number basis. Polynomial degree fixes the matrix bandwidth, while parity fixes whether Δn\Delta n is even or odd.

The selection rule should be applied before expanding operator products. It predicts which energy corrections vanish, which intermediate states can appear, and how wide a numerical Hamiltonian matrix will be.

Because aa and a†a^\dagger do not commute, an ordinary binomial expansion of (a+a†)r(a+a^\dagger)^r is invalid. Normal ordering collects the commutator corrections. With colons denoting normal order,

q2=:q2:+1,q^2=:q^2:+1, q3=:q3:+3q,q^3=:q^3:+3q,

and

q4=:q4:+6:q2:+3.q^4=:q^4:+6:q^2:+3.

For a diagonal number-state matrix element, only terms with equal numbers of aa and a†a^\dagger survive. For example,

⟨n∣q4∣n⟩=6n(n−1)+12n+3=6n2+6n+3.\begin{aligned} \langle n\vert q^4\vert n\rangle ={}& 6n(n-1)+12n+3 \\ ={}& 6n^2+6n+3. \end{aligned}

These identities are the one-mode precursor of the normal-ordering and contraction bookkeeping used in many-body theory and quantum field theory.

Use λ\lambda only as an order-counting parameter and write

H(λ)=H0+λV.H(\lambda)=H_0+\lambda V.
VVFirst energy shiftFirst nonzero energy orderExact or large-order lesson
FxFxzerosecondenergy series terminates; exact displacement
κx2\kappa x^2nonzerofirstexact frequency change; finite Taylor radius
γx3\gamma x^3zerosecondreal pure cubic is unbounded below
gx4g x^4, g>0g\gt0nonzerofirststable model; perturbative series is asymptotic

The same unperturbed basis therefore supports four distinct lessons. A zero first-order diagonal element can mean exact cancellation, delayed response, or an ill-posed global model; the order label alone does not decide which.

Let

V=Fx,V=Fx,

where FF has units of force. Parity immediately gives

En(1)=F⟨n∣x∣n⟩=0.E_n^{(1)} = F\langle n\vert x\vert n\rangle =0.

The first-order state correction has only neighboring levels:

∣n(1)⟩=Fxzpfℏω[n ∣n−1⟩−n+1 ∣n+1⟩].\begin{aligned} \lvert n^{(1)}\rangle = \frac{F x_{\mathrm{zpf}}}{\hbar\omega} \Bigl[ &\sqrt n\,\lvert n-1\rangle \\ &- \sqrt{n+1}\,\lvert n+1\rangle \Bigr]. \end{aligned}

At second order,

En(2)=F2xzpf2[nℏω−n+1ℏω]=−F22mω2.\begin{aligned} E_n^{(2)} ={}& F^2x_{\mathrm{zpf}}^2 \left[ \frac{n}{\hbar\omega} - \frac{n+1}{\hbar\omega} \right] \\ ={}& -\frac{F^2}{2m\omega^2}. \end{aligned}

The cancellation of nn is a strong check. Every level receives the same shift, so the spacing remains ℏω\hbar\omega.

Completing the square gives the exact answer:

H(λ)=p22m+12mω2(x+λFmω2)2−λ2F22mω2.\begin{aligned} H(\lambda) ={}& \frac{p^2}{2m} + \frac{1}{2}m\omega^2 \left( x+\frac{\lambda F}{m\omega^2} \right)^2 \\ &- \frac{\lambda^2F^2}{2m\omega^2}. \end{aligned}

Hence

En(λ)=ℏω(n+12)−λ2F22mω2.E_n(\lambda) = \hbar\omega \left(n+\frac{1}{2}\right) - \frac{\lambda^2F^2}{2m\omega^2}.

The energy expansion terminates at second order, even though the translated eigenstate contains all orders when expanded in the original number basis. Displaced Oscillator owns the exact coordinate, translation-operator, and coherent-state descriptions. First-Order State Corrections owns the detailed perturbative mixing analysis.

Quadratic Perturbation: Frequency Renormalization

Section titled “Quadratic Perturbation: Frequency Renormalization”

Now take

V=κx2.V=\kappa x^2.

Applying x2=xzpf2q2x^2=x_{\mathrm{zpf}}^2q^2 to a number state gives

x2∣n⟩=xzpf2[n(n−1) ∣n−2⟩+(2n+1)∣n⟩+(n+1)(n+2) ∣n+2⟩].\begin{aligned} x^2\lvert n\rangle ={}& x_{\mathrm{zpf}}^2 \Bigl[ \\[-0.25em] &\sqrt{n(n-1)}\,\lvert n-2\rangle \\ &+ (2n+1)\lvert n\rangle \\ &+ \sqrt{(n+1)(n+2)}\,\lvert n+2\rangle \Bigr]. \end{aligned}

Therefore

En(1)=κxzpf2(2n+1)=κℏmω(n+12).\begin{aligned} E_n^{(1)} &= \kappa x_{\mathrm{zpf}}^2(2n+1) \\ &= \frac{\kappa\hbar}{m\omega} \left(n+\frac{1}{2}\right). \end{aligned}

The first-order state correction is

∣n(1)⟩=κxzpf22ℏω[n(n−1) ∣n−2⟩−(n+1)(n+2) ∣n+2⟩].\begin{aligned} \lvert n^{(1)}\rangle ={}& \frac{\kappa x_{\mathrm{zpf}}^2} {2\hbar\omega} \Bigl[ \\[-0.25em] &\sqrt{n(n-1)}\,\lvert n-2\rangle \\ &- \sqrt{(n+1)(n+2)}\,\lvert n+2\rangle \Bigr]. \end{aligned}

Only the off-diagonal terms enter the second-order energy:

En(2)=κ2xzpf42ℏω[n(n−1)−(n+1)(n+2)]=−κ2ℏ2m2ω3(n+12).\begin{aligned} E_n^{(2)} ={}& \frac{\kappa^2x_{\mathrm{zpf}}^4} {2\hbar\omega} \Bigl[ n(n-1) \\ &\qquad{} - (n+1)(n+2) \Bigr] \\ ={}& -\frac{\kappa^2\hbar} {2m^2\omega^3} \left(n+\frac{1}{2}\right). \end{aligned}

This perturbation is exactly another harmonic oscillator:

Ω(λ)2=ω2+2λκm.\Omega(\lambda)^2 = \omega^2+\frac{2\lambda\kappa}{m}.

Thus

En(λ)=ℏΩ(λ)(n+12).E_n(\lambda) = \hbar\Omega(\lambda) \left(n+\frac{1}{2}\right).

Expanding

Ω(λ)=ω1+2λκmω2\Omega(\lambda) = \omega \sqrt{ 1+\frac{2\lambda\kappa}{m\omega^2} }

reproduces En(1)E_n^{(1)} and En(2)E_n^{(2)}. The exact state has a changed width and is a squeezed number state relative to the original oscillator basis. Squeezed States: First Encounter supplies the state-space language.

Stability requires

ω2+2λκm>0.\omega^2+\frac{2\lambda\kappa}{m}\gt0.

The Taylor series about λ=0\lambda=0 has a branch point where the effective frequency vanishes. For the binomial expansion, the nearest such point sets the radius

∣2λκmω2∣<1.\left| \frac{2\lambda\kappa}{m\omega^2} \right| \lt1.

The exact oscillator can remain stable at some parameter values outside that Taylor disk; exact solvability and convergence of one chosen expansion are different questions.

Cubic Perturbation: Symmetry and Stability

Section titled “Cubic Perturbation: Symmetry and Stability”

Let

V=γx3.V=\gamma x^3.

Because x3x^3 is odd,

En(1)=γ⟨n∣x3∣n⟩=0.E_n^{(1)} = \gamma\langle n\vert x^3\vert n\rangle =0.

The action of q3q^3 is

q3∣n⟩=n(n−1)(n−2)∣n−3⟩+3nn ∣n−1⟩+3(n+1)n+1 ∣n+1⟩+(n+1)(n+2)(n+3)∣n+3⟩.\begin{gathered} q^3\lvert n\rangle = \\[-0.25em] \begin{aligned} & \sqrt{n(n-1)(n-2)} \lvert n-3\rangle \\ &+ 3n\sqrt n\,\lvert n-1\rangle \\ &+ 3(n+1)\sqrt{n+1}\, \lvert n+1\rangle \\ &+ \sqrt{(n+1)(n+2)(n+3)} \lvert n+3\rangle. \end{aligned} \end{gathered}

The four allowed channels give the formal second-order coefficient

En(2)=−γ2xzpf6ℏω(30n2+30n+11).\begin{aligned} E_n^{(2)} = -\frac{ \gamma^2x_{\mathrm{zpf}}^6 }{ \hbar\omega } \bigl( 30n^2+30n+11 \bigr). \end{aligned}

For the ground state,

E0(2)=−11γ2xzpf6ℏω.E_0^{(2)} = -\frac{ 11\gamma^2x_{\mathrm{zpf}}^6 }{ \hbar\omega }.

The vanishing first-order shift therefore does not mean that the cubic term has no spectral effect.

There is, however, a global caveat. The real potential

12mω2x2+λγx3\frac{1}{2}m\omega^2x^2+\lambda\gamma x^3

is unbounded below on one side for every nonzero real λγ\lambda\gamma. It does not define a stable bound-state oscillator on the full line. The coefficients above are formal local perturbative data and can be connected to resonance problems under additional analytic prescriptions; they should not be advertised as convergent corrections to a stable bound spectrum.

A physically stable asymmetric well needs stabilizing higher powers, a restricted effective domain, or a more complete potential. This is a case where symmetry algebra can be correct while the global Hamiltonian model is incomplete.

Quartic Perturbation: Stable Anharmonicity

Section titled “Quartic Perturbation: Stable Anharmonicity”

Take

V=gx4,g>0.V=gx^4, \qquad g\gt0.

Normal ordering gives

⟨n∣x4∣n⟩=xzpf4(6n2+6n+3).\langle n\vert x^4\vert n\rangle = x_{\mathrm{zpf}}^4 \bigl( 6n^2+6n+3 \bigr).

Hence

En(1)=3gxzpf4(2n2+2n+1)=3gℏ24m2ω2(2n2+2n+1).\begin{aligned} E_n^{(1)} ={}& 3g x_{\mathrm{zpf}}^4 \bigl( 2n^2+2n+1 \bigr) \\ ={}& \frac{3g\hbar^2} {4m^2\omega^2} \bigl( 2n^2+2n+1 \bigr). \end{aligned}

For the ground state, the first two coefficients are

E0(λ)=ℏω2+3λgxzpf4−42λ2g2xzpf8ℏω+O(λ3).\begin{aligned} E_0(\lambda) ={}& \frac{\hbar\omega}{2} + 3\lambda g x_{\mathrm{zpf}}^4 \\ &- \frac{ 42\lambda^2g^2x_{\mathrm{zpf}}^8 }{ \hbar\omega } + O(\lambda^3). \end{aligned}

The number 4242 comes from the two off-diagonal channels ∣2⟩\lvert2\rangle and ∣4⟩\lvert4\rangle. The complete derivation belongs to Anharmonic Oscillator by Perturbation Theory.

Unlike the pure real cubic model, the potential is bounded below for g>0g\gt0. Yet stability does not imply a convergent Rayleigh–Schrödinger series. The quartic-oscillator coefficients grow factorially at high order; low orders are useful for weak coupling, while the full series requires asymptotic and resummation ideas.

Anharmonic Oscillator is the canonical multi-method comparison among perturbative, variational, semiclassical, and numerical descriptions.

For a monomial

Vr=crxr,V_r=c_r x^r,

define the low-state dimensionless coupling

gr≡crxzpfrℏω.g_r \equiv \frac{ c_r x_{\mathrm{zpf}}^r }{ \hbar\omega }.

At number nn, matrix elements of qrq^r grow parametrically as nr/2n^{r/2} along allowed channels. A conservative mixing estimate is therefore

∣λgr∣nr/2≪1,\lvert\lambda g_r\rvert n^{r/2} \ll1,

up to order-one coefficients and the allowed value of Δn\Delta n. This estimate is not a universal error bound, but it makes one failure mode visible: a coupling that is perturbative for the ground state may not be perturbative high in the spectrum.

Different observables can have different effective criteria:

  • eigenvector mixing compares off-diagonal matrix elements with level gaps;
  • energy accuracy compares omitted corrections with the requested energy scale;
  • transition amplitudes can be sensitive to small state admixtures;
  • a near cancellation can make relative error misleading even when absolute error is small.

The exactly solvable linear and quadratic cases should be used to calibrate any proposed error diagnostic.

In the number basis, a polynomial of degree rr produces a banded matrix:

⟨m∣xr∣n⟩=0if ∣m−n∣>r.\langle m\vert x^r\vert n\rangle=0 \qquad \text{if } \lvert m-n\rvert\gt r.

Parity sharpens the structure:

  • even potentials preserve even and odd sectors separately;
  • odd potentials couple the two parity sectors;
  • x4x^4 requires only the five bands Δn=0,±2,±4\Delta n=0,\pm2,\pm4;
  • a sum of monomials has the union of their allowed bands.

For a basis truncated at n≤Nmax⁡n\le N_{\max}, test:

  1. convergence of target eigenvalues as Nmax⁡N_{\max} increases;
  2. residual norms of the computed eigenpairs;
  3. stability of perturbative coefficients fitted at small λ\lambda;
  4. parity-block consistency when the Hamiltonian is even;
  5. sensitivity to the reference frequency used to define the basis.

A basis with an optimized reference frequency can converge much faster for a strongly broadened or narrowed state. That is a basis choice, not a change in the physical Hamiltonian.

The oscillator examples clarify what a Gaussian trial family can and cannot do.

  • Linear force: a displaced Gaussian is the exact ground state.
  • Quadratic change: a Gaussian with adjustable width is the exact ground state of the new frequency.
  • Stable quartic term: an optimized Gaussian gives an upper bound but cannot reproduce the exact non-Gaussian tails.
  • Pure real cubic term: no variational ground-state minimum exists because the potential is unbounded below.

For the quartic oscillator, a trial frequency Ω\Omega resums selected effects of gg into the reference state. Expanding the optimized result at weak coupling can reproduce low-order trends, while its finite-coupling behavior answers a different variational question. Variational Perturbation Theory promotes that frequency to an order-dependent reference scale and derives the formal δ re-expansion. Compare methods at equal physical parameters, not merely by the number of algebraic terms retained.

See Variational Principle for the bound and stationarity logic.

Each free-field normal mode is a harmonic oscillator, and a field operator is a sum of mode coordinates proportional to ak+ak†a_{\mathbf k}+a_{\mathbf k}^\dagger. Polynomial field interactions therefore generate products of creation and annihilation operators much as x3x^3 and x4x^4 do here.

The analogy teaches several durable habits:

  • normal order noncommuting products rather than using a classical binomial expansion;
  • use selection rules before summing intermediate states;
  • distinguish diagonal shifts from state-changing terms;
  • expect disconnected and contraction terms at higher order;
  • check whether the perturbative series is convergent or only asymptotic.

One oscillator does not contain spatial locality, momentum conservation among modes, ultraviolet divergences, renormalization, or infinitely many degrees of freedom. The bridge is structural, not an equivalence. Harmonic Oscillator to Quantum Fields and Normal Ordering continue the comparison.

  1. Fix the oscillator-length convention. Write xx in terms of a+a†a+a^\dagger before computing coefficients.
  2. Classify parity and degree. List allowed Δn\Delta n values.
  3. Identify the leading nonzero order. A vanishing diagonal term shifts attention to state response and second order.
  4. Check global stability. A locally small perturbation can make the full potential unbounded.
  5. Exploit exact reorganizations. Displace a linear term or redefine the frequency for a quadratic term.
  6. Form a dimensionless coupling. Include the oscillator width and the state number.
  7. Cross-check coefficients. Expand an exact result, use parity, or compare with numerical diagonalization.
  8. State the series meaning. Distinguish terminating, convergent, formal, and asymptotic expansions.
  • Using xzpf=ℏ/(mω)x_{\mathrm{zpf}}=\sqrt{\hbar/(m\omega)} while also writing x=xzpf(a+a†)x=x_{\mathrm{zpf}}(a+a^\dagger).
  • Expanding (a+a†)r(a+a^\dagger)^r as though aa and a†a^\dagger commute.
  • Concluding that an odd perturbation has no effect because its first-order energy vanishes.
  • Forgetting the lower-state terms with positive denominators in an excited-state second-order sum.
  • Calling the pure real cubic oscillator a stable bound-state problem.
  • Treating the exact quadratic frequency change as proof that every perturbation series converges.
  • Assuming a weak coupling for n=0n=0 remains weak at large nn.
  • Duplicating the quartic model calculation instead of linking its canonical worked and multi-method pages.
  • Trusting a truncated oscillator basis without varying its cutoff or reference frequency.

Show that ⟨m∣xr∣n⟩\langle m\vert x^r\vert n\rangle can be nonzero only when ∣m−n∣≤r\lvert m-n\rvert\le r and m−nm-n has the same parity as rr.

Solution

Since

xr=xzpfr(a+a†)r,x^r = x_{\mathrm{zpf}}^r (a+a^\dagger)^r,

every ordered term contains rr ladder operations. If a term contains uu creation operators and r−ur-u annihilation operators, its net number change is

Δn=u−(r−u)=2u−r.\Delta n = u-(r-u) = 2u-r.

Therefore

∣Δn∣≤r\lvert\Delta n\rvert\le r

and

Δn≡r(mod2).\Delta n\equiv r\pmod2.

Commuting operators into normal order removes pairs through [a,a†]=1[a,a^\dagger]=1, changing the degree by two and preserving the same parity condition.

Starting from the sum-over-states formula, compute the second-order energy for V=FxV=Fx and show that it equals the exact displaced-oscillator shift.

Solution

Only m=n±1m=n\pm1 contribute:

En(2)=F2xzpf2nℏω−F2xzpf2(n+1)ℏω.\begin{aligned} E_n^{(2)} ={}& \frac{ F^2x_{\mathrm{zpf}}^2 n }{ \hbar\omega } \\ &- \frac{ F^2x_{\mathrm{zpf}}^2(n+1) }{ \hbar\omega }. \end{aligned}

Thus

En(2)=−F2xzpf2ℏω=−F22mω2.E_n^{(2)} = -\frac{ F^2x_{\mathrm{zpf}}^2 }{ \hbar\omega } = -\frac{F^2}{2m\omega^2}.

Completing the square gives an exact constant offset

−λ2F22mω2,-\frac{\lambda^2F^2}{2m\omega^2},

so the perturbative result is exact and independent of nn.

Expand the exact frequency

Ω(λ)=ω2+2λκm\Omega(\lambda) = \sqrt{ \omega^2+\frac{2\lambda\kappa}{m} }

through second order and recover En(1)E_n^{(1)} and En(2)E_n^{(2)}.

Solution

Factor out ω\omega:

Ω(λ)=ω1+2λκmω2.\Omega(\lambda) = \omega \sqrt{ 1+\frac{2\lambda\kappa}{m\omega^2} }.

Using 1+z=1+z/2−z2/8+O(z3)\sqrt{1+z}=1+z/2-z^2/8+O(z^3),

Ω(λ)=ω+λκmω−λ2κ22m2ω3+O(λ3).\begin{aligned} \Omega(\lambda) ={}& \omega + \lambda\frac{\kappa}{m\omega} \\ &- \lambda^2 \frac{\kappa^2}{2m^2\omega^3} + O(\lambda^3). \end{aligned}

Multiplication by ℏ(n+1/2)\hbar(n+1/2) gives

En(1)=κℏmω(n+12)E_n^{(1)} = \frac{\kappa\hbar}{m\omega} \left(n+\frac{1}{2}\right)

and

En(2)=−κ2ℏ2m2ω3(n+12).E_n^{(2)} = -\frac{\kappa^2\hbar}{2m^2\omega^3} \left(n+\frac{1}{2}\right).

4. Compute the cubic ground-state coefficient

Section titled “4. Compute the cubic ground-state coefficient”

For V=γx3V=\gamma x^3, use q3∣0⟩q^3\lvert0\rangle to compute E0(2)E_0^{(2)}. Then state why the result requires a stability caveat.

Solution

The cubic action on the ground state is

q3∣0⟩=3∣1⟩+6 ∣3⟩.q^3\lvert0\rangle = 3\lvert1\rangle + \sqrt6\,\lvert3\rangle.

Therefore

E0(2)=γ2xzpf6[9−ℏω+6−3ℏω]=−11γ2xzpf6ℏω.\begin{aligned} E_0^{(2)} ={}& \gamma^2x_{\mathrm{zpf}}^6 \left[ \frac{9}{-\hbar\omega} + \frac{6}{-3\hbar\omega} \right] \\ ={}& -\frac{ 11\gamma^2x_{\mathrm{zpf}}^6 }{ \hbar\omega }. \end{aligned}

The algebra is correct as a formal local coefficient, but a real x3x^3 term makes the potential unbounded below on one side. A stable global model needs additional stabilizing structure.

Use

q4=:q4:+6:q2:+3q^4=:q^4:+6:q^2:+3

to derive ⟨n∣q4∣n⟩\langle n\vert q^4\vert n\rangle.

Solution

In :q4::q^4:, only the term with two creation and two annihilation operators is diagonal:

⟨n∣:q4:∣n⟩=6⟨n∣(a†)2a2∣n⟩=6n(n−1).\begin{aligned} \langle n\vert:q^4:\vert n\rangle &= 6 \langle n\vert (a^\dagger)^2a^2 \vert n\rangle \\ &= 6n(n-1). \end{aligned}

In :q2::q^2:, the only diagonal term is 2a†a2a^\dagger a, so

6⟨n∣:q2:∣n⟩=12n.6\langle n\vert:q^2:\vert n\rangle = 12n.

Adding the contraction constant gives

⟨n∣q4∣n⟩=6n(n−1)+12n+3=6n2+6n+3.\begin{aligned} \langle n\vert q^4\vert n\rangle &= 6n(n-1)+12n+3 \\ &= 6n^2+6n+3. \end{aligned}

6. Design a parity-resolved numerical matrix

Section titled “6. Design a parity-resolved numerical matrix”

For H=H0+gx4H=H_0+gx^4, explain how parity and bandwidth reduce a matrix calculation in the number basis. What convergence checks remain necessary?

Solution

x4x^4 connects only

Δn=0,±2,±4.\Delta n=0,\pm2,\pm4.

The Hamiltonian therefore separates into an even block built from

∣0⟩,∣2⟩,∣4⟩,…\lvert0\rangle,\lvert2\rangle,\lvert4\rangle,\ldots

and an odd block built from

∣1⟩,∣3⟩,∣5⟩,….\lvert1\rangle,\lvert3\rangle,\lvert5\rangle,\ldots.

Within either block the matrix is banded, so forbidden entries need not be stored or computed. One must still increase the maximum number state, monitor eigenpair residuals, check target eigenvalue stability, and test whether a different reference frequency improves convergence. Parity reduces cost but does not control truncation error by itself.

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