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.

  • $\begingroup$ 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

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