A. So the distinct solutions areknwith n = 1, 2, 3, ....[2.22]aCuriously, the boundary condition at x = a does not determine the constant A,but rather the constant k, and hence the possible values of E:h°k; n?n?h?EnGriffithsIntroductioquantummechanics[2.23]2m2ma?In sharp contrast to the classical case, a quantum particle in the infinite square wellcannot have just any old energy-only these special allowed values. Well, how dowe fix the constant A? Answer: We normalize :14? sin'(kx)dx =|A 1,аso |4? == -.aThis only determines the magnitude of A, but it is simplest to pick the positive realroot: A = /2/a (the phase of A carries no physical significance anyway). Inside thewell, then, the solutions are2020You moved an iteNTGriffitheVn(x) = .- sin[2.24]aaAs promised, the time-independent Schrödinger equation has delivered an infi-nite set of solutions, one for each integer n. The first few of these are plotted in Fig-ure 2.2; they look just like the standing waves on a string of length a. 1, which car-ries the lowest energy, is called the ground state; the others, whose energies increasePepeojdn noAGiffithsroun the functions lk (r) have conditions.)Problem 2.4 Solve the time-independent Schrödinger equation with appropriateboundary conditions for an infinite square well centered at the origin [V (x) = 0, for-a/2 < x < +a/2; V (x) = ∞ otherwise]. Check that your allowed energies areconsistent with mine (Equation 2.23), and confirm that your y's can be obtained frommine (Equation 2.24) by the substitution x x - a/2.Droblo m 25 Celaulnte lu).2

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Answered: Problem 2.4 Solve the time-independent…

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