# What are some molecular systems which have quantum states that are measurably incompatible?

What I'm looking for is a molecular system, which we can produce in a lab, that is,

1. A discrete variable system (ie, a finite dimensional hilbert space)
2. It has observables that are incompatible with eachother (ie, two observables who's commutators equal $$i\hbar$$, or equivalently that their corresponding states are fourier transform dials of eachother).

Are there actually molecular systems out there that are relatively inexpensive to produce and measure in a lab?

• No such system exists, your condition 1 and 2 are mutually exclusive, cf. physics.stackexchange.com/q/149786/50583. But why do you equate being "incompatible" with being Fourier conjugate variables? Usually, one would call compatible observables those that commute with each other, but the negation of that is simply...not commuting, which is much weaker a condition than having a constant commutator of $\mathrm{i}\hbar$. – ACuriousMind Dec 26 '18 at 13:35
• Sorry, I got the terminology mixed up, Ignore the part about the incompatiable observables, what I meant was that I'm looking for a molecular system which has states that are simultaneously unmeasurable (like measuring position and momentum states), except for a discrete variable system. – wigglywinks Dec 26 '18 at 13:49
• I'm not sure what you're asking for then - any system has states that are eigenstates of one observable but not of another. Is anything about an atom with spin - as used e.g. in Stern-Gerlach experiments - and measuring spin along two different axes insufficient for you? Please edit your question so answerers will know better what you are looking for – ACuriousMind Dec 26 '18 at 13:53
• Are you thinking of something like benzene, where the central ring is often shown as alternating single and double bonds. But in fact the bonds are sp2 hybrid orbitals. chemguide.co.uk/basicorg/bonding/benzene2.html – mmesser314 Dec 26 '18 at 14:00
• As pointed out by ACuriousMind, you've asked for something that doesn't exist. Please clarify the question by using the "edit" button. If you just want a case with two observables that don't commute, then get rid of the bit about the commutator being equal to $i \hbar$. – DanielSank Dec 26 '18 at 15:40