5 Quantization of energy

We have seen how Bohr’s pos­tu­late accounted for the quan­ti­za­tion of energy. Let us now take a look at how Schrödinger’s theory does it.

We begin by observing that if the poten­tial V is inde­pen­dent of time, the Schrödinger equa­tion (2.3.7) has solu­tions of the form ψ(x,y,z,t) = Ψ(x,y,z)[1:−ωt], where Ψ(x,y,z) is (obvi­ously) inde­pen­dent of time. (You may want to pay atten­tion to the dif­fer­ence between ψ and Ψ.) These solu­tions are sta­tionary in the sense that the prob­a­bil­i­ties defined by ψ(x,y,z,t) are inde­pen­dent of time. Inserting such a solu­tion into Eq. (2.3.7) yields the time-​​independent Schrödinger equation:

(2.5.1)   EΨ = −(ℏ2/​2m) [(∂x)2 + (∂y)2 + (∂z)2]Ψ + VΨ.

Returning to one spa­tial dimen­sion, we cast Eq. (2.5.1) into the fol­lowing form:

(2.5.2) (∂x)2Ψ = AΨ   with   A = (2m/​ℏ2)(V − E).

Because Eq. (2.5.2) does not con­tain any com­plex num­bers (apart from, pos­sibly, Ψ itself), it has real-​​valued solu­tions. So let us assume that Ψ is real.

The first thing we notice is that if V is greater than E then Ψ and (∂x)2Ψ have the same sign, and if E is greater than V then Ψ and (∂x)2Ψ have oppo­site signs. To see what this means, we need to know a bit more about the oper­ator ∂x. If we plug in the func­tion Ψ(x), out pops another func­tion Ψ’(x):

Ψ’(x) = ∂xΨ(x).

Its value at any par­tic­ular place x equals the slope of Ψ(x) at that place, and this equals the slope of the tan­gent on the graph of Ψ at that place. (The slope of a straight line such as this tan­gent tells us how steeply it ascends or, if neg­a­tive, descends from left to right.)

The oper­ator (∂x)2Ψ = ∂x (∂xΨ) = ∂xΨ’, accord­ingly, yields the slope of the slope of Ψ. What does this mean? Simply, if the slope of the slope of Ψ is pos­i­tive, the slope of Ψ increases (from left to right), and the graph of Ψ curves upward as a result. Sim­i­larly, if the the slope of the slope of Ψ is neg­a­tive, the slope of Ψ decreases (from left to right), and the graph of Ψ curves down­ward as a result.

Thus if Ψ and (∂x)2Ψ have the same sign, the graph of Ψ curves upward wher­ever Ψ is pos­i­tive (that is, where its graph lies above the x-​​axis), and it curves down­ward wher­ever Ψ is neg­a­tive (that is, where its graph lies below the x-​​axis). In either case it bends away from the x-​​axis. On the other hand, if Ψ and (∂x)2Ψ have oppo­site signs, the graph of Ψ curves down­ward wher­ever Ψ is pos­i­tive, and it curves upward wher­ever Ψ is neg­a­tive. In either case it bends toward the x-​​axis.

So if V is greater than E, the graph of Ψ bends away from the x-​​axis, and if E is greater than V, the graph of Ψ bends toward the x-​​axis. In the first case it crosses the x-​​axis at most once; in the second case it keeps crossing and re-​​crossing the x-​​axis exactly like a wave.

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potential well

Figure 2.5.1 A poten­tial well.

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We are now ready to sketch solu­tions of Eq. (2.5.2) that describe a par­ticle trapped inside a poten­tial well like that in Fig. 2.5.1. Between x1 and x2, the particle’s total energy E exceeds the poten­tial energy V, and Ψ exhibits wave­like behavior. To the left of x1 and to the right of x2, V exceeds E, and the graph of Ψ bends away from the x-​​axis. (A clas­sical par­ticle would oscil­late back and forth between these two points.)

To obtain a nor­mal­ized solu­tion (for which the prob­a­bility of finding the par­ticle any­where equals unity), we must make sure that the graph of Ψ approaches the x-​​axis asymp­tot­i­cally in both the limits x→−∞ and x→+∞. Together with the par­tic­ular form of V(x), these con­di­tions deter­mine both the value and the slope of Ψ at x1 and at x2, and these must be exactly matched by the graph of Ψ between these points. For such a match to occur, the value of E must be exactly right. Once again the pos­sible (“allowed”) ener­gies form a sequence En, starting with n=0, n counting the number of nodes. (Nodes are points at which Ψ crosses the x-​​axis.)

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the first six solutions

Figure 2.5.2 The first six solu­tions of Eq. (2.5.2).

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Let us remind our­selves in con­cluding of the sense in which the solu­tions of Fig. 2.5.2 describe a par­ticle trapped inside a poten­tial well. Like every prob­a­bility algo­rithm in quantum mechanics, Ψ serves one and only one pur­pose: to cal­cu­late the prob­a­bil­i­ties of pos­sible mea­sure­ment out­comes on the basis of actual out­comes (as well as, gen­er­ally, some clas­sical boundary con­di­tions). Given the poten­tial V (a clas­sical boundary con­di­tion), and given a par­ticle whose energy has been found to equal Em, the prob­a­bility of finding the par­ticle in the interval I between any two points x=a and x=b (pro­vided that the appro­priate mea­sure­ment is made) is

p(I) = ∫(a,b)dx |Ψ(x)|2.

Here the symbol ∫(a,b)dx indi­cates a sum­ma­tion over the points con­tained I: each con­tributes a com­plex number |ψ(x)|2. The Fourier trans­form of Ψ(x), on the other hand, allows us to cal­cu­late the prob­a­bility of finding the particle’s momentum in any given interval of the p– or k-​​axis.

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