Is there a difference between a three-level atomic system and a qutrit? If yes, please tell me what it is and how to relate both quantum systems.

  • 1
    $\begingroup$ For me the 3-level atomic system can be used to represent a qutrit, or a qutrit can be realized by a 3-level atomic system. $\endgroup$ – XXDD Jul 4 '17 at 11:57
  • $\begingroup$ Thanks. But I noticed something concerning potential transitions between the levels. Do you think it has something to do ? $\endgroup$ – T. Arthur Jul 4 '17 at 12:02
  • $\begingroup$ I understand the transition between different levels as a way to achieve logical operation on the qutrit. $\endgroup$ – XXDD Jul 4 '17 at 12:04
  • $\begingroup$ It looks like all transitions are available for a three-level atomic system, while only some can be within a qutrit. What is your opinion. Thanks $\endgroup$ – T. Arthur Jul 4 '17 at 12:05

In a qutrit, every transition between pairs of levels is independent, in the sense that $\vert 1\rangle \leftrightarrow \vert 2\rangle$ is independent from $\vert 1\rangle \leftrightarrow \vert 3\rangle $ which is independent from $\vert 2\rangle\leftrightarrow \vert 3\rangle$.

Thus, not every three-level system is a qutrit. For instance, the angular momentum states with $\ell=1$ do not form a qutrit if we restrict the observables to angular momentum operators, since, for instance the matrix elements $\langle 1,-1\vert L_- \vert 1,0\rangle = \langle 1,0\vert L_-\vert 1,1\rangle=\sqrt{2}$, i.e. the transitions between adjacent levels are not independent. In addition, a pulse resonant for the $m=-1\to m=0$ transition is also resonant for the $m=0\to m=1$ transition: in the case of $\ell=1$ states, not very transition can be accessed independently so it's NOT a true qutrit.

Unfortunately too many people do confuse a spin triplet for a true qutrit. This is because the states $\{\vert 1,m\rangle\}$ do span the space, but the difficulty is with the operators: when restricted to angular momentum operators, not every transition can be accessed separately. You need to use the full set $su(3)$ operators acting between basis states to properly described a qutrit.

An important and useful way to see the difference between angular momentum and $su(3)$ operators is that the standard ladder operators in $su(3)$ connect only 1 pair of states, whereas angular momentum operators connect two pairs of states when $\ell=1$.

  • $\begingroup$ "When restricted to..." ─ so don't. The spin triplet isn't a useful qutrit because come of the transitions are hard to implement using the initial toolbox, but that doesn't mean it's not a qutrit. $\endgroup$ – Emilio Pisanty Jul 4 '17 at 17:13

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.