Bosonic Commutation Relations

Bosonic commutation relations are the operator algebra that makes bosonic occupation-number notation work. They encode two facts at once: a bosonic mode may be occupied by any nonnegative integer number of particles or excitations, and creation operators for different modes commute because exchanging identical bosons does not introduce a minus sign.

For orthonormal modes labeled by $i,j$, the canonical bosonic algebra is

$$[a_i,a_j^\dagger] = \delta_{ij}I, \qquad [a_i,a_j] =0, \qquad [a_i^\dagger,a_j^\dagger] =0.$$

Here

$$[A,B] = AB-BA$$

is the commutator, and $I$ is the identity operator on the relevant Fock space. Many texts suppress $I$ and write $[a_i,a_j^\dagger]=\delta_{ij}$, but the right-hand side is still an operator identity.

Helpful background. Creation and Annihilation Operators gives the ladder action; Bosonic Fock Space defines the underlying Hilbert space.

Canonical bosonic commutators

For one bosonic mode, the algebra reduces to

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

Equivalently,

$$aa^\dagger = a^\dagger a+I.$$

On a normalized state with occupation $n$, creating then removing multiplies the state by $n+1$, while removing then creating multiplies it by $n$. Their difference is one. This is an amplitude identity, not a comparison of normalized classical paths.

For many modes,

$$[a_i,a_j^\dagger] =0 \qquad \text{when }i\ne j.$$

Thus operations on distinct bosonic modes commute. Creating a boson in mode $i$ and then one in mode $j$ gives the same state as doing those operations in the opposite order:

$$a_i^\dagger a_j^\dagger\lvert0\rangle = a_j^\dagger a_i^\dagger\lvert0\rangle.$$

This is the operator form of bosonic exchange symmetry.

Action on Number States

Let $\lvert n\rangle$ be the normalized occupation-number state of one bosonic mode. The creation and annihilation operators act by

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

with

$$a\lvert0\rangle =0.$$

These formulas imply the commutation relation on every number state:

$$aa^\dagger\lvert n\rangle = (n+1)\lvert n\rangle, \qquad a^\dagger a\lvert n\rangle = n\lvert n\rangle.$$

Subtracting gives

$$[a,a^\dagger]\lvert n\rangle = \lvert n\rangle.$$

The identity therefore holds on finite linear combinations of number states, a common invariant dense domain for these unbounded operators. Applying arbitrary products to a general infinite superposition also requires that the state lie in their domains.

The square-root factors are forced by this algebra together with normalization. If one tried to use $a^\dagger\lvert n\rangle=\lvert n+1\rangle$ for every $n$, the commutator would not equal the identity on the normalized number basis.

Number Operators

The number operator for mode $i$ is

$$N_i = a_i^\dagger a_i.$$

On the occupation-number basis,

$$N_i \lvert n_1,n_2,\ldots\rangle_B = n_i \lvert n_1,n_2,\ldots\rangle_B.$$

The total number operator is

$$N = \sum_i N_i.$$

Using the commutation relations, one finds

$$[N_i,a_j^\dagger] = \delta_{ij}a_j^\dagger, \qquad [N_i,a_j] = -\delta_{ij}a_j.$$

These identities say exactly what the words say: $a_j^\dagger$ raises the occupation of mode $j$ by one, while $a_j$ lowers it by one. For $i\ne j$, the occupation of mode $i$ is unchanged.

Multiple Modes

A bosonic occupation state is written

$$\lvert n_1,n_2,\ldots,n_i,\ldots\rangle_B.$$

The mode-$i$ operators act only on the $i$th occupation number:

$$a_i^\dagger \lvert\ldots,n_i,\ldots\rangle_B = \sqrt{n_i+1}\, \lvert\ldots,n_i+1,\ldots\rangle_B,$$

and

$$a_i \lvert\ldots,n_i,\ldots\rangle_B = \sqrt{n_i}\, \lvert\ldots,n_i-1,\ldots\rangle_B.$$

For two distinct modes $1$ and $2$,

$$a_1a_2^\dagger \lvert n_1,n_2\rangle_B = \sqrt{n_1}\sqrt{n_2+1}\, \lvert n_1-1,n_2+1\rangle_B.$$

Because $[a_1,a_2^\dagger]=0$, the same result is obtained if one applies $a_2^\dagger a_1$ instead.

Repeated creation in the same mode is allowed. Starting from the vacuum,

$$(a_i^\dagger)^n\lvert0\rangle \ne 0 \qquad \text{for every }n=1,2,\ldots .$$

This is the sharp operator contrast with fermions, where the corresponding creation operator squares to zero.

Symmetric Fock space and finite cutoffs

The normalized bosonic Fock representation realizes these relations by adding and removing modes in symmetric tensor powers. It allows every nonnegative occupation. A finite occupation cutoff changes the algebra at its top state.

Let $P_K$ retain $|0\rangle,\ldots,|K\rangle$ and $a_K=P_KaP_K$. Direct multiplication gives

$$[a_K,a_K^\dagger] =I_{K+1}-(K+1)|K\rangle\langle K|.$$

For $n<K$ the usual result survives; on $|K\rangle$, raising is blocked and the commutator gives $-K|K\rangle$. No finite matrices can obey $[a,a^\dagger]=I$ exactly, because the trace of any finite commutator is zero. In numerical work, increasing the cutoff and checking the weight near its boundary is therefore essential. Replacing a truncated product by its untruncated commutator can silently change the model.

This occupation cutoff is different from retaining a finite set of spatial modes while leaving each bosonic occupation unbounded. A spatial-mode projection changes the field completeness kernel; see Mode Expansions.

Exercises

1. Compute $[N_i,a_i^\dagger]\lvert n_i\rangle$ for one mode $i$.

Solution

Using $N_i\lvert n_i\rangle=n_i\lvert n_i\rangle$,

$$N_i a_i^\dagger\lvert n_i\rangle = N_i\sqrt{n_i+1}\,\lvert n_i+1\rangle = (n_i+1)\sqrt{n_i+1}\,\lvert n_i+1\rangle.$$

Also,

$$a_i^\dagger N_i\lvert n_i\rangle = n_i\sqrt{n_i+1}\,\lvert n_i+1\rangle.$$

Subtracting gives

$$[N_i,a_i^\dagger]\lvert n_i\rangle = \sqrt{n_i+1}\,\lvert n_i+1\rangle = a_i^\dagger\lvert n_i\rangle.$$

2. Normalize $(a^\dagger)^3\lvert0\rangle$.

Solution

By repeated use of the creation formula,

$$(a^\dagger)^3\lvert0\rangle = \sqrt{1}\sqrt{2}\sqrt{3}\,\lvert3\rangle = \sqrt{3!}\,\lvert3\rangle.$$

Therefore the normalized state is

$$\lvert3\rangle = \frac{(a^\dagger)^3}{\sqrt{3!}}\lvert0\rangle.$$

3. **Cutoff error.** For a normalized truncated state with top-level probability $p_K$, find $\langle[a_K,a_K^\dagger]\rangle$. Is small $p_K$ alone a cutoff-independent criterion?

Solution

The expectation is $1-(K+1)p_K$. The deviation from the canonical value is $(K+1)p_K$, so the weighted boundary probability must be small. Other observables can demand stronger convergence checks, especially if they contain high powers of ladder operators.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.