Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 35B / b

The condition $2m={\hslash }^2$ makesForthe squared norm is the Gaussian integralSince ${\psi }_a^{\prime }=-ax{\psi }_a$, integration by parts gives the kinetic quotientThe potential quotient isThe variational principle therefore yields the Gaussian variational bound for an attractive Gaussian well

The positive zero of $E(a)$ satisfiesSquaring and using $a>0$ givesso

Furthermore,At an extremum this vanishes, and henceSquaring gives the required equation

As $a\downarrow 0$,so $E(a)<0$ for all sufficiently small positive $a$. On the other hand,The derivative satisfiesThe extremum equationhas exactly one positive solution because its left side is strictly increasing. Thus $E$ has one global minimum, and that minimum is negative.

The potential tends to zero at spatial infinity, so zero is the threshold of the continuous spectrum. A trial state with negative energy expectation forces the spectrum below zero; the corresponding lowest spectral value is a normalizable bound state. Hence every $V_0>0$ produces at least one bound state.

Sinceexcept at the single point $x=0$, and the kinetic-energy expectation is nonnegative, every normalized wavefunction satisfiesTherefore

For the upper bound, use the admissible choiceThenThe Rayleigh-Ritz variational principle now givesIndeed, the displayed calculation provides the sharper remainder $O(V_0^4)$.