# Are all wave functions energy-eigenstates?

1. May 25, 2014

### Nikitin

So I was reading this http://www.oberlin.edu/physics/dstyer/TeachQM/misconnzz.pdf, a list of common misconceptions students have after an intro course in QM.

I'm aware that energy eigenstates are the wave functions at "time = zero" and thus do not completely describe the system. However, it is correct that any wave function is a solution to the time independent Schroedinger equation (also known as the energy-operator).. Right?

Last edited: May 25, 2014
2. May 25, 2014

### hilbert2

Any continuously differentiable and square-integrable function is an allowed wavefunction of a quantum system. The energy eigenstates (solutions of the time-independent SE) are states that have a definite value of the total energy and that have time-independent square modulus. Not all allowed wavefunctions are energy eigenstates, but it is possible to write any allowed state as a linear combination of energy eigenstates (or a linear combination of the eigenstates of any hermitian operator).

3. May 25, 2014

### Nikitin

Ah yes, silly me. Obviously a wave function must be a linear combo of energy eigenstates, due to the Schroedinger equation being a linear PDF. But even though a linear combo of energy eigenstates $\Psi = \Sigma \psi_i$ is a solutions to the indepedent Schroedinger equation, $\Psi$ itself is not an eigenstate.

OK, thanks.

4. May 25, 2014

### Matterwave

Breaking the problem down to energy eigenstates is simply a way of solving for the time evolution of the wave function. This is because we know how an eigenstate evolves in time: $\psi_n (x,t)=\psi_n(x)e^{-iE_n t/\hbar}$.

But we need not solve the problem in this fashion, there are other ways of solving this problem. Therefore, the eigenstates of the Hamiltonian are just a tool. One could, for example, always express the wave function as linear combinations of the quantum harmonic oscillator energy eigenfunctions, the Hermite polynomials, even if the Hamiltonian is not for the harmonic oscillator. This is because those sets of polynomials are complete. It's just, for the case where the Hamiltonian is not the harmonic oscillator, the time evolution of your basis states are no longer trivial.

Last edited: May 25, 2014
5. May 25, 2014

### bhobba

Or position eigenstates, or momentum eigenstates, or spin eigenstates etc etc, and all at the same time - the principle of superposition at work for pure states ie they form a vector space.

But then we have non pure states that are different again - they are linear operators.

But to return to the original question a position eigenstate of a free particle is not in an energy eigenstate - its unstable and spreads - but is not in an energy eigenstate.

Thanks
Bill

Last edited: May 25, 2014
6. May 27, 2014

### moontiger

Rather, the wave function at t = 0 can be expressed as some linear superposition of eigenstates.

(And just a note that if you consider potentials that allow for scattering states, non-square-integrable states are allowable too.)

Last edited: May 27, 2014