Mathematical Methods for Physics and Engineering - Matematica.NET

QUANTUM OPERATORShence formally an eigenstate of L 2 with l = 0, all three components of angularmomentum could be (and are) known to be zero.◮Working in one dimension, show that the minimum value of the product ∆p x × ∆x foraparticleis 1 . Find the form of the wavefunction that attains this minimum value for a2particle whose expectation values for position and momentum are ¯x and ¯p, respectively.We have already seen, in (19.23) that the commutator of p x and x is −i, aconstant.Therefore, irrespective of the actual form of |ψ〉, the RHS of (19.38) is 1 4 2 (see observation(ii) above). Thus, since all quantities are positive, taking the square roots of both sides ofthe equation shows directly that∆p x × ∆x ≥ 1 . 2Returning to the derivation of the Uncertainty Principle, we see that the inequality becomesan equality only when(〈u | v〉 + 〈v | u〉) 2 =4〈u | u〉〈v | v〉.The RHS of this equality has the value 4||u|| 2 ||v|| 2 and so, by virtue of Schwarz’s inequality,we have4‖u‖ 2 ‖v‖ 2 =(〈u | v〉 + 〈v | u〉) 2≤ (|〈u | v〉| + |〈v | u〉|) 2≤ (‖u‖‖v|| + ‖v‖‖u‖) 2=4‖u‖ 2 ‖v‖ 2 .Since the LHS is less than or equal to something that has the same value as itself, all ofthe inequalities are, in fact, equalities. Thus 〈u|v〉 = ‖u‖‖v‖, showing that |u〉 and |v〉 areparallel vectors, i.e. |u〉 = µ|v〉 for some scalar µ.We now transform this condition into a constraint that the wavefunction ψ = ψ(x)must satisfy. Recalling the definitions (19.37) of |u〉 and |v〉 in terms of |ψ〉, we have(−i d )dx − ¯p ψ = µi(x − ¯x)ψ,dψ[ µ(x − ¯x) − i¯p ]ψ =0.dx + 1 [ µ(x − ¯x)2The IF for this equation is exp− i¯px ], giving2 { [d µ(x − ¯x)2ψ expdx2− i¯px ]}=0,which, in turn, leads to[ ] ( )µ(x − ¯x)2 i¯pxψ(x) =A exp − exp .2From this it is apparent that the minimum uncertainty product ∆p x × ∆x is obtained whenthe probability density |ψ(x)| 2 has the form of a Gaussian distribution centred on ¯x. Thevalue of µ is not fixed by this consideration and it could be anything (positive); a largevalue for µ would yield a small value for ∆x but a correspondingly large one for ∆p x . ◭666