(a) Define \(V(\lambda, x) = \lambda V_2 (x) +(1 - \lambda )V_1(x).\) Obviously V(0, x) = V1(x), V(1, x) = V2(x), \(\partial V/\partial \lambda = V_2(x) - V_1(x) \ge 0.\)
The Hamiltonian is then
we have \(E_{1n} = E_n (0) \le E_n(1)\), and the theorem is proved. Note that we have used \((n \lambda|n \lambda ) = 1\).
(b) Let V(x) = kx2/2. Then V(x) ≥ U(x). If E, is an energy level for the potential U(x), then \(E_n\le (n + 1/2)h \omega\), where \(\omega = \sqrt{k /m}\). For a bound state, En ≤ ka2/2. Solving (N + 1/2)h\(\omega\) ≤ ka2/2, we find
\(N \le \frac{m \omega ^2}{2h} - \frac 12 = \left[\frac{m\omega a}{2h}\right]^2\)
where [A] indicates the maximum integer that is less than A.
We now choose for V(x) a square well of finite depth,
V(x) = ka2/2, |x| > a,
V(x) = 0, |x| ≤ a
The number of bound states of U(x) is less than that of V(x), which for the latter is [2m\(\omega\)a2/7rh] + 1. We can take the upper bound to the number of bound states of U(x) as [2m\(\omega\)a2/πh] as for N >> 1 the term 1 can be neglected. Taken together, we get that the number of bound states is between [m\(\omega\)a2/2h] and [2m\(\omega\)a2/πh].