1. Dec 29, 2008

### KFC

If I know the explicit form of potential, the energy and a specific eigenstate, but I don't know the general form of eigenstates and eigenvalues, can I tell if the state is ground state or not?

2. Dec 29, 2008

### malawi_glenn

No i dont think so, why are you asking?

3. Dec 29, 2008

### clem

A ground state wave function usually has no nodes.

4. Dec 29, 2008

### tim_lou

I think you can prove this statement by noting that the ground state wave function minimizes
$$\left < \psi \right| H \left | \psi \right>$$

5. Dec 30, 2008

### intervoxel

The Variational Principle guarantees that

$$E_g \leq \langle \psi|H|\psi \rangle \equiv \langle H \rangle$$

Last edited: Dec 30, 2008
6. Dec 30, 2008

### msumm21

In doing some QM problems a few weeks ago I noticed that, for the PARTICULAR potential I was working with, the product of the standard deviation of the position (sx) and the standard deviation of the momentum (sp) was exactly hbar/2 in the ground state (sx*sp=hbar/2), and the product grew larger for higher energy states. So the uncertaintly principle was closest to getting violated in the lowest energy state. I assume that's NOT true for other potentials, but maybe something you could look into depending upon what exactly you are trying to do.