Do different bases of Fock space commute?

Question

$\newcommand\dag\dagger$ Suppose we have a Fock space $\mathcal{F}$ with two different bases of creation and annihilation operators $\{a_\lambda, a^\dag_\lambda\}$ and $\{a_{\tilde \lambda}, a^\dag_{\tilde \lambda}\}$. The change of basis formula for these operators is given by:

$$ a^\dag_{\tilde \lambda} = \sum_\lambda \langle \lambda | \tilde{\lambda} \rangle a^\dag_\lambda $$

and similar equations for $a_{\tilde \lambda}$, etc. However, this seems to imply that $a^\dag_{\tilde \lambda}$ and $a^\dag_{\lambda}$ commute or anti-commute (depending on if the space describes bosonic or fermionic particles, respectively). This seems surprising to me, so I had the following two questions:

1. Is this actually true?
2. If not, why not? If it is, why should we expect this to be true and is there a physical interpretation/intuition for this?

Answer 1

For a given decomposition of the Fock space into $n$-particle subspaces, creation operator commute. For a $1$ particle state $|\psi\rangle$, there is an associated annihilation operator $a(\psi)$. If $|\psi\rangle,|\psi'\rangle$ are any two $1$-particle states, then an elementary calculation shows that $a(\psi)$ and $a(\psi')$ (anti)-commute (depending on the statistics). An other way to see this, as OP pointed out, is to see that the change of basis is of the form $$b_\mu = \sum_{\lambda} \alpha_{\mu,\lambda} a_{\lambda}$$ i.e. it is linear in annihilation operators.

There is a more general kind of linear transformation possible on a Fock space, called a Bogoliubov transformation. This is a transformation linear in both creation and annihilation operators : $$b_\mu = \sum_{\lambda} \alpha_{\mu,\lambda} a_{\lambda}+\beta_{\mu,\lambda}a^\dagger_\lambda$$ (with some constraints on the coefficients, to ensure that the commutation relations are preserved).

Then, the vacuum (and more generally $n$-particle subspaces) are not preserved. This has physical implications in various contexts, from Hawking radiation to superfluidity and quantum optics.

Answer 2

It is evident linear algebra. Intuition? Try a space of just two bosonic oscillators, i=1,2, $$ [a_i,a^\dagger_j]=\delta_{ij}, \qquad [a_i,a_j]=0, $$ and ditto for the conjugates.

Rotate them to $$ \tilde a_1=\cos\theta ~ a_1+ \sin\theta ~ a_2=\sum_k\langle k|\tilde 1\rangle a_k,\\ \tilde a_2=\cos\theta ~ a_2- \sin\theta ~ a_1=\sum_k\langle k|\tilde 2\rangle a_k, $$ and ditto for the conjugates.

It is then evident that $$ [\tilde a_i,\tilde a_j]=0, \qquad [\tilde a_i, a_j]=0, $$ and ditto for the conjugates, as well as $$ [\tilde a_i,\tilde a_j^\dagger ]=\delta_{ij}, \qquad [ a_i, \tilde a_j^\dagger]=\langle i|\tilde j\rangle. $$