# Quantum mechanics ground state

1. Apr 20, 2010

### jasony

Why must the ground state not have a node？ And the first excited state must have 1 node.

2. Apr 20, 2010

### SpectraCat

Excellent questions. The first thing to understand is that the energy of a wavefunction is inversely proportional to its curvature ... more sharply curved --> higher energy. Second, if a wavefunction crosses the abscissa (i.e. goes through zero) exactly once, then it must necessarily have higher curvature than a wavefunction that never crosses the abscissa (assuming all the other requirements for QM wavefunctions are satisfied). So, for a given Hamiltonian, the energy of a one-node wavefunction must always be higher than a no-node wavefunction.

Now, that doesn't really answer your first question. To understand that, you need to know that the Hamiltonian is a Hermitian operator, and the eigenfunctions of a Hermitian operator form a complete, orthonormal set. Note also that the ground and first-excited state we are talking about here are eigenfunctions of the Hamiltonian as well. So, in most analytically solvable cases, the mathematical expressions for the eigenfunctions are actually in the form of orthogonal polynomials ... that is:

$$\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}$$

where n is a quantum number indexing the states, Nn is a normalization constant, gn is a node-less, non-polynomial function such as an exponential or a gaussian, and the cn,i are the expansion coefficients of a polynomial, which are chosen to ensure that polynomials of different order are orthogonal to each other. Clearly, there should be a polynomial of order zero, which will be node-less.

So, given the above points, I hope it is clear that, the node-less quality of the ground state is a fundamental property arising from the physical and mathematical aspects of quantum mechanics.

Note that I have made some simplifying assumptions in the above discussion, for example, if your system has symmetry, then it may not be true that the lowest energy state within a given irreducible representation is node-less, however it should be true that wavefunctions with more modes will have higher energies. I am not actually sure that it holds in *all* cases, although it certainly holds for intro level QM, and I cannot think of any cases where the ground state is *not* node-less.

3. Apr 22, 2010

### jasony

Can we always write the wavefunction as $$\psi_n(x)=N_ng_n(x)\sum_i{c_{n,i}x^i}$$?

Why?

Is there other simple proof for the nodeless property of ground state wavefunction?

4. Apr 23, 2010

### ansgar

that is just the most general polynomial function and the argument is actually the most simplest explanation I have seen. It is just theory of eigenfunctions and eigenvalues, go study functional analysis :)

5. Apr 23, 2010

### espen180

Since $$\sum_i{c_{n,i}x^i}$$ is a polynomial and the only requirement of $$g_n(x)$$ is that it is a non-polynomial, it follows that $$\psi_n(x)$$ can be any function which satisfies the normalization requirement. Therefore, it can also be any wavefunction.

6. Apr 24, 2010

### genneth

7. Apr 14, 2011