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.

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

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.

Proof

Accumulation of nodes is not possible for non trivial wavefunction. If there exists a limit point of nodes, then, the value of $\psi '(p)=0$ for the limit point $p$ but this corresponds to trivial solution having initial values $\psi (p)=\psi '(p)=0$. From Schrodinger's equations: $(E_{i}-V(x,t)){\Psi _{i}(x,t)}=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\Psi _{i}(x,t)}{\partial x^{2}}}$ Multiplying with $\Psi _{j}(x,t)$ and subtracting equations of $i=1\,,j=2$ and $i=2\,,j=1$ $(\psi _{1}''\psi _{2}-\psi _{2}''\psi _{1})=(\psi _{1}^{\prime }\psi _{2}-\psi _{2}^{\prime }\psi _{1})^{\prime }=(E_{2}-E_{1})\psi _{1}\psi _{2}$ If $x_{1}$ and $x_{2}$ are consecutive nodes of $\psi _{1}$, assuming it to be positive in the region, implies $\psi _{1}^{\prime }(x_{1})>0$ and $\psi _{1}^{\prime }(x_{2})<0$. Computing the following integration: $\psi _{1}^{\prime }(x_{2})\psi _{2}(x_{2})-\psi _{1}^{\prime }(x_{1})\psi _{2}(x_{1})=(E_{2}-E_{1}){\int _{x_{1}}^{x_{2}}\!dx\,\psi _{1}(x)\psi _{2}(x)}$ If $E_{2}>E_{1}$, $\psi _{2}$ cannot be entirely positive or entirely negative in the region and must therefore have node within the consecutive nodes of $\psi _{1}$. Note that this is independent on assuming sign of $\psi _{1}$ in this region. Thus wavefunctions with different energy levels always have different number of nodes or conversely wavefunctions with same number of nodes cannot have different energies. Thus in a 1D bound system where degeneracy is not allowed, all energy eigenfunctions differ by the number of nodes and when arranged in increasing order of energies, give increasing number of node.

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

References

Methods of Mathematical Physics Vol. I - Courant and Hilbert pp 451-455

https://arxiv.org/pdf/quant-ph/0702260.pdf

See Also

Sturm–Picone comparison theorem

Sturm separation theorem