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-iħ(∂/∂x)

Now lets say that I enact this operator on the famous solution to the 1-D particle in a box example:

Ψ= squrt(2/L) sin(πnx/L)

If the momentum operator operates on the above wave function, it yields:

-iħ * squrt(2/L) * (πn/L) * cos(πnx/L)

Now, it is true that squrt(2/L) * cos(πnx/L) is actually a solution to the 1D time independent Schrodinger equation that yields the energy eigenvalue E when you enact the Hamiltonian operator on it. However, it is still different from the original wave function (with sine in it). That is why I question whether or not the momentum operator operates on the same eigenstates as the Hamiltonian operator (since by definition of eigenstate, the result of enacting the operator on the wave function must yield an eigenvalue multiplied by the original wave function). So tell me:

Do the Hamiltonian and momentum operators work on the same eigenstates?

If not, what are the eigenstates of the momentum operator?