# 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.

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.)