What is the relation between a Fock state and a qudit? Is the Fock state $|n\rangle, n=0, 1, 2, ...$ a qudit having $d=\infty$?

  • $\begingroup$ Some approaches to QC approximate a qudit by taking the first d Fock states and ignoring the rest. Mathematically that's not very clean, since the other states are still there and the system can in principle drift into them, but experimentally that can still be used to do QC. $\endgroup$
    – unknown
    Aug 25 at 17:49

I would say it is not exactly the same.

A qudit of size $d$ has a finite number of states. A Fock state can have discretely infinite in principle.

The goal of having multiple qudits, let's say $N$, is to make a larger space of size $d^N$. Rarely you consider product of Fock spaces (but it can be done). Usually, you define a Fock space that includes all configurations.

Fock spaces can be fermionic or bosonic (the ladder operators have defined commutation relations). The qudits are usually neither fermions nor bosons (check how weird is the commutation algebra of qutrits).

If you have a chain of qudits, there are mappings to converting them to Fock states and viceversa. But without a mapping I would not force them to be equivalent.

You can also have continuous Fock spaces, but it would be weird to have continuous qudits?

Any way if you want to think about $d=\infty$ qudits, consider that you can always write a Fock space to describe not only your $\infty$-qdit but also a chain of $\infty$ number of $\infty$-qdit. The Fock space is just more general.

  • $\begingroup$ The commutation algebra you link to is that of $su(3)$; I guess you can call it the algebra of qutrits but there are probably subtleties. If they're exactly the same, then "quarks" might be a better name for "qutrit". Do you see $su(2)$ as the algebra of qubits? $\endgroup$
    – unknown
    Aug 25 at 17:53
  • $\begingroup$ @unknown is it not? Why are the subtleties? $\endgroup$
    – Mauricio
    Aug 25 at 18:32
  • $\begingroup$ For one, $su(3)$ is 8 dimensional (the link defines 8 matrices that can serve as basis for it). Exponentiating these gives you $SU(3)$. QC uses $U(3)$... $\endgroup$
    – unknown
    Aug 25 at 19:26
  • $\begingroup$ @unknown sure but working with $U(n)$ or $SU(n)$ is equivalent for quantum computation (same projective unitary group). $\endgroup$
    – Mauricio
    Aug 25 at 21:21
  • 1
    $\begingroup$ @unknown exactly, check this quantumcomputing.stackexchange.com/a/17763/15775 $\endgroup$
    – Mauricio
    Aug 25 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.