### D.66 The tri­an­gle in­equal­ity

The nor­mal tri­an­gle in­equal­ity con­tin­ues to ap­ply for ex­pec­ta­tion val­ues in quan­tum me­chan­ics.

The way to show that is, like other tri­an­gle in­equal­ity proofs, rather cu­ri­ous: ex­am­ine the com­bi­na­tion of , not with , but with an ar­bi­trary mul­ti­ple of :

For 1 this pro­duces the ex­pec­ta­tion value of , for 1, the one for . In ad­di­tion, it is pos­i­tive for all val­ues of , since it con­sists of ex­pec­ta­tion val­ues of square Her­mit­ian op­er­a­tors. (Just ex­am­ine each term in terms of its own eigen­states.)

If you mul­ti­ply out, you get

where , , and rep­re­sents mixed terms that do not need to be writ­ten out. In or­der for this qua­dratic form in to al­ways be pos­i­tive, the dis­crim­i­nant must be neg­a­tive:

which means, tak­ing square roots,

and so

or

and tak­ing square roots gives the tri­an­gle in­equal­ity.

Note that this de­riva­tion does not use any prop­er­ties spe­cific to an­gu­lar mo­men­tum and does not re­quire the si­mul­ta­ne­ous ex­is­tence of the com­po­nents. With a bit of mess­ing around, the az­imuthal quan­tum num­ber re­la­tion can be de­rived from it if a unique value for ex­ists; the key is to rec­og­nize that where is an in­creas­ing func­tion of that stays be­low , and the val­ues must be half in­te­gers. This de­riva­tion is not as el­e­gant as us­ing the lad­der op­er­a­tors, but the re­sult is the same.