# I Obtain simultaneous eigenfunctions?

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1. Nov 10, 2018

### John Greger

Let's consider two observables, H (hamiltonian) and P (momentum).

These operators are compatible since [H,P] = 0.

Let's look at the easy to prove rule:
1: "If the observables F and G are compatible, that is, if there exists a simultaneous set of eigenfunctions of the operators F and G, then these operators must commute; [F , G] = 0."

This can be turned around to yield,
2: " If the operators F and G commute, then it is possible to find a simultaneous set of eigenfunctions."

How would the simultaneous eigenfunctions look like for H and P? Would it be something on matrix form?

Or is the following functions simply the "simultaneous eigenfunctions":

$$\hat{H} u_{n,m}(x) = E_n u_{n,m}(x)$$
$$\hat{P} u_{n,m}(x) = P_m u_{n,m}(x)$$

where $\psi(x) = \Sigma_{n,m} C_{n,m} u_{n,m}(x)$ ?

And I think the spectrum cannot be continuous right...

EDIT: How would I express a generic wave function as a superposition of the above eigenstates if they are "equivalent"?

Last edited: Nov 10, 2018
2. Nov 10, 2018

### hilbert2

If you have a 1d free particle, the eigenfunctions of $\hat{p}$ are like $Ae^{ikx}$ with $A$ and $k$ some real-valued constants. These are all eigenfunctions of $\hat{H}$. The set of eigenfunctions of $\hat{H}$ also contains functions like $Ae^{ikx} + Be^{-ikx}$ which are not eigenfunctions of $\hat{p}$ because the sign of the 1d momentum is uncertain.

If you have a Hamiltonian operator that also depends on position $\hat{x}$ , then the $\hat{H}$ and $\hat{p}$ will not commute.