Ground state has no nodes in one-dimension
In one-dimension, the ground state of the Schrödinger equation can be proven to have no nodes.
Consider the average energy of a state with a node at x = 0; i.e., ψ(0)=0. The average energy in this state would be:
where V(x) is the potential.
Now, consider a small interval around ; i.e., . Take a new (deformed) wave function ψ' (x) to be defined as , for ; and , for ; and constant for . If is small enough, this is always possible to do, so that ψ' (x) is continuous.
Assuming around , one may write
where is the norm.
Note that the kinetic energy density everywhere because of the normalization. More significantly, the average kinetic energy is lowered by by the deformation to ψ' .
Now, consider the potential energy. For definiteness, let us choose . Then it is clear that, outside the interval , the potential energy density is smaller for the ψ' because there.
On the other hand, in the interval we have
which is holds to order .
However, the contribution to the potential energy from this region for the state with a node, ψ, is
lower, but still of the same lower order as for the deformed state ψ' , and subdominant to the lowering of the average kinetic energy. Therefore, the potential energy is unchanged up to order , if we deform the state with a node into a state without a node ψ' , and the change can be ignored.
We can therefore remove all nodes and reduce the energy by , which implies that ψ' cannot be the ground state: the ground-state wave function cannot have a node. This completes the proof. (The average energy may then be further lowered by eliminating undulations, to the variational absolute minimum.)
Initial wave functions for the first four states of a one-dimensional particle in a box
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