User:EditingPencil/sandbox/node theorem

Existence criteria
The existence of a bound state is guaranteed if any of the existence criteria are met.

By variational Method
By variational method of finding ground state, if there exists a trial wavefunction whose energy expectation value is negative, there exists ground state solution which is bounded.

$$\langle H\rangle\equiv{\frac{\int\psi H\psi d x}{\int\psi^{2}\,d x}} < 0$$

Bound states of Attractive potential
It can be shown that attractive potentials with negative spacial integral have at least one bound states in the dimensions $$d \leq 2$$. Since this means that the potential has to go to zero to give a non-divergent integral, the potential is offset by it's values at infinity and the condition expressed as:

$$\int_{-\infty}^{\infty}\left[V(x)-V(\pm\infty)\right] \, dx < 0$$

Node theorem
Node theorem states that n-th bound wavefunction ordered according to increasing energy has exactly n-1 nodes, ie. points $$x=a$$ where $$\psi(a)=0 \neq \psi'(a)$$. Rigorous derivation of the condition is treated in analysis of second order homogenous differential equations.

Due to the form of Schrodinger's time independent equations, it is not possible for a physical wavefunction to have $$\psi(a) = 0 = \psi'(a)$$ since it corresponds to $$\psi(x)=0$$ solution by the uniqueness of second order ordinary differential solutions. Considering a rigid wall around a point with $$x \pm a$$, if the walls expand to infinity, $$a\rightarrow \infty$$, the eigenfunctions of the trapped system morphs into eigenfunctions of the Hamiltonian. If the new nodes are formed during this, then at some value of $$a$$ and point $$c$$ within the walls, either the nodes begin to form at the middle with $$\psi'(c)=0 $$ or at the boundaries with $$\psi'(x+a)=0 $$ or $$\psi'(x - a) = 0 $$. Since wavefunction always remains zero at rigid walls, $$\psi'(x\pm a) \neq 0 = \psi(x\pm a) $$ condition ensures that nodes are not formed and that the sign of these derivatives remain the same. At the instant the node begins to form at $$c$$, it will be a local minima and hence have $$\psi(c) = 0 = \psi'(c) $$ which is not possible.

As a corollary, the no-node theorem states that the ground state of bound particles do not have any nodes.