Bosonic Occupation Bases

Bosonic numerical bases require either a fixed total particle number or an explicit occupation cutoff. The cutoff is part of the approximation: it changes the oscillator commutator at the boundary and can contaminate low-energy results when population reaches that boundary.

Helpful background. Bosonic Fock space explains occupation states; tensor products in code explains their array ordering.

Occupation tuples and ladder factors

For $M$ modes use tuples $\boldsymbol n=(n_0,\ldots,n_{M-1})$ and

$$a_j|\boldsymbol n\rangle=\sqrt{n_j}\,|\boldsymbol n-\boldsymbol e_j\rangle, \qquad a_j^\dagger|\boldsymbol n\rangle=\sqrt{n_j+1}\, |\boldsymbol n+\boldsymbol e_j\rangle.$$

There is no parity sign from other modes. A local cutoff $0\leq n_j\leq n_{\max}$ gives $(n_{\max}+1)^M$ states. A fixed total number $\sum_jn_j=N$, with no additional local cutoff, gives $\binom{N+M-1}{N}$ states by the stars-and-bars count.

Enumerate tuples recursively by assigning one mode and distributing the remaining particle number. Store a tuple-to-index map; tuple ordering should be deterministic so that saved vectors remain interpretable.

For hopping $a_i^\dagger a_j$ with $i\ne j$, the amplitude is $\sqrt{(n_i+1)n_j}$. Apply the annihilation before creation. For $i=j$, use $n_j$, not the offsite formula.

Truncated commutator

For one mode represented on $|0\rangle,\ldots,|n_{\max}\rangle$, set the creation action on the top state to zero. Then

$$[a,a^\dagger] =I-(n_{\max}+1)|n_{\max}\rangle\langle n_{\max}|.$$

This follows because the ordinary commutator is one below the top level, while at the top $aa^\dagger=0$ and $a^\dagger a=n_{\max}$. No finite matrix pair can satisfy $[a,a^\dagger]=I$, since the trace of a commutator is zero.

The defect is therefore expected, not a coding failure. Its expectation is small only if the top-level occupation is small enough for the requested accuracy. Small population alone is a diagnostic, not a universal error bound for every unbounded observable.

Projection also fails to preserve arbitrary operator products. With $P$ projecting onto the retained occupations and $Q=I-P$,

$$PABP=(PAP)(PBP)+PAQBP.$$

For the dimensionless oscillator coordinate $X=a+a^\dagger$, the missing excursion to $|n_{\max}+1\rangle$ gives

$$PX^2P-(PXP)^2 =(n_{\max}+1)|n_{\max}\rangle\langle n_{\max}|.$$

Thus squaring the truncated coordinate matrix differs from projecting the continuum $X^2$ operator. Define which finite Hamiltonian is intended; when approximating a specified continuum polynomial, construct its matrix elements before truncation or include the boundary correction. Both constructions must converge in the target low-energy regime, but their finite-cutoff spectra need not agree. This issue is distinct from the absence of a fermionic parity sign.

Bose–Hubbard dimer benchmark

For two sites with two particles,

$$H=-J(a_0^\dagger a_1+a_1^\dagger a_0) +\frac U2\sum_{j=0}^1n_j(n_j-1).$$

In basis $(|20\rangle,|11\rangle,|02\rangle)$,

$$H=\begin{pmatrix} U&-\sqrt2J&0\\ -\sqrt2J&0&-\sqrt2J\\ 0&-\sqrt2J&U \end{pmatrix}.$$

The antisymmetric state $(|20\rangle-|02\rangle)/\sqrt2$ has energy $U$. The other energies are $(U\pm\sqrt{U^2+16J^2})/2$. At $U=0$ they are $-2J,0,2J$, matching two bosons occupying single-particle energies $\pm J$.

This example detects missing square-root factors. A cutoff $n_{\max}=1$ removes both doubly occupied states and changes the model to hard-core bosons; it does not approximate this dimer at arbitrary $U/J$.

Cutoff convergence

For a driven oscillator or non-number-conserving model, increase $n_{\max}$ while holding propagation settings fixed. Compare energy, occupation, and the target observable, and inspect the occupation tail. A drive can push probability toward the cutoff only at later times, so inspect the full time interval.

For a number-conserving model with $N$ particles, a local cutoff at least $N$ includes the entire fixed-$N$ space. Further local-cutoff increases add no states. In contrast, grand-canonical thermal calculations require controlling both the particle-number range and Boltzmann tail.

Exercise

Find the expectation of the truncated commutator for a state with top-level probability $p_{\rm top}$.

Solution

It is $1-(n_{\max}+1)p_{\rm top}$. Thus a top population of $10^{-4}$ at $n_{\max}=99$ already changes this expectation by $10^{-2}$. A fixed probability tolerance independent of cutoff can be misleading.

References

• QuTiP developers, “Manipulating States and Operators”, QuTiP documentation — finite oscillator matrices.
• Weinberg, P., and Bukov, M., “QuSpin. Part II: bosons, fermions and higher spins,” SciPost Physics 7, 020 (2019), arXiv:1804.06782.