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 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.

Algebra

For one bosonic mode, the algebra reduces to

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

Equivalently,

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

This identity is the algebraic source of the familiar one-step offset between creating first and annihilating first. If one creates a quantum and then removes it, there is one extra way to return to the original state compared with removing first and then creating.

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

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

Since the number states form the standard basis of the single-mode bosonic Fock space, this is the identity operator on that space.

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.

Relation to Symmetric Fock Space

The bosonic Fock space over a one-particle Hilbert space $\mathcal h$ is

$$\mathcal F_B(\mathcal h) = \bigoplus_{N=0}^{\infty} \operatorname{Sym}^N\mathcal h.$$

The commutation relations are not an extra rule pasted on top of this space. They are the efficient operator representation of the symmetric tensor powers.

The same algebra also determines the commutator of the position-space field operators once the field is expanded in a one-particle basis; see Mode Expansions.

Given an orthonormal mode basis, a normalized occupation state is

$$\lvert n_1,n_2,\ldots\rangle_B = \prod_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle,$$

where only finitely many $n_i$ are nonzero in the finite-particle sector. The product ordering is irrelevant for bosonic creation operators because

$$[a_i^\dagger,a_j^\dagger]=0.$$

For two distinct one-particle modes $\lvert u\rangle$ and $\lvert v\rangle$,

$$a_u^\dagger a_v^\dagger\lvert0\rangle \longleftrightarrow \frac{1}{\sqrt2} \bigl( \lvert u\rangle_1\lvert v\rangle_2 + \lvert v\rangle_1\lvert u\rangle_2 \bigr).$$

For two bosons in the same mode,

$$\frac{(a_u^\dagger)^2}{\sqrt{2!}} \lvert0\rangle \longleftrightarrow \lvert u\rangle_1\lvert u\rangle_2.$$

The factorial in the occupation-state formula exactly compensates for the repeated symmetric copies of the same mode.

Examples

For one oscillator-like bosonic mode,

$$N=a^\dagger a, \qquad N\lvert n\rangle = n\lvert n\rangle.$$

The Hamiltonian of a harmonic oscillator can be written

$$H = \hbar\omega \left( N+\frac12 \right),$$

so the number basis is also the energy basis for that simple system.

For two modes, the normalized state with three bosons in mode $1$ and one boson in mode $2$ is

$$\lvert3,1\rangle_B = \frac{(a_1^\dagger)^3a_2^\dagger}{\sqrt{3!}} \lvert0\rangle.$$

Applying $a_1$ gives

$$a_1\lvert3,1\rangle_B = \sqrt3\,\lvert2,1\rangle_B.$$

Applying $a_2^\dagger$ gives

$$a_2^\dagger\lvert3,1\rangle_B = \sqrt2\,\lvert3,2\rangle_B.$$

These square roots are often where arithmetic mistakes enter many-body calculations.

Common Mistakes

• Writing $[a_i,a_j^\dagger]=1$ for all modes instead of $\delta_{ij}I$.
• Omitting the identity operator conceptually on the right-hand side of $[a,a^\dagger]=I$.
• Replacing commutators by anticommutators for bosonic modes.
• Dropping the factors $\sqrt n$ and $\sqrt{n+1}$ in actions on normalized number states.
• Assuming that bosonic occupation numbers stop at $1$.
• Confusing a harmonic oscillator ladder operator for one particle in a potential with a Fock-space operator that creates or removes a mode excitation.

Cross-Links

• Creation and Annihilation Operators
• Number Operators
• Mode Expansions
• Field Operators
• Normal Ordering
• Bosonic Fock Space
• Occupation-Number Basis
• Fermionic Anticommutation Relations
• Fock Space Exercises
• Ladder-Operator Solution
• Reference Bridge: Second Quantization
• Harmonic Oscillator to Fields

References

• 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.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.

Exercises

1. Verify $[a,a^\dagger]\lvert n\rangle=\lvert n\rangle$ from the action of $a$ and $a^\dagger$ on number states.

Solution

First,

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

Second,

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

Therefore

$$[a,a^\dagger]\lvert n\rangle = \bigl((n+1)-n\bigr)\lvert n\rangle = \lvert n\rangle.$$

2. Show that $a_1^\dagger a_2^\dagger\lvert0\rangle=a_2^\dagger a_1^\dagger\lvert0\rangle$ for two bosonic modes.

Solution

For bosons,

$$[a_1^\dagger,a_2^\dagger]=0.$$

Thus

$$a_1^\dagger a_2^\dagger = a_2^\dagger a_1^\dagger,$$

as operators, so they give the same result on the vacuum:

$$a_1^\dagger a_2^\dagger\lvert0\rangle = a_2^\dagger a_1^\dagger\lvert0\rangle.$$

3. 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

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

4. 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

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