Free particle in quantum mechanics, Dirac formalism

[MManuel Abad]

The problem is very easy, maybe just something about eigenvectors that I'm missing. Go to the first two pages of the 5th chapter of ''Principles of Quantum Mechanics'', by Shankar, 2nd edition.

Homework Statement

Shankar wants to find the solution for a free particle in Quantum Mechanics. I've got a problem with what he calls ''trial solution'', and the change from an eigenvector to another.

Hamilton operator: H=\frac{P^{2}}{2m} for a free particle

P, the momentum operator, m the particle's mass

State vector (ket): \left|\psi\right\rangle

Homework Equations

The Schroedinger equation:

i\hbar\\left|\dot{\psi}\right\rangle = H\left|\psi\right\rangle = \frac{P^{2}}{2m}\left|\psi\right\rangle

The time independent Schrodinger equation:

H\left|E\right\rangle = \frac{P^{2}}{2m}\left|E\right\rangle = E\left|E\right\rangle

The Attempt at a Solution

Shankar then says that ''any eigenstate of P is also an eigenstate of P^{2}'', which is very clear. Then he says: ''feeding the trial solution \left|p\right\rangle '' into the time independent Schrodinger eq., he gets:

\frac{P^{2}}{2m}\left|p\right\rangle = E\left|p\right\rangle

But I don't get it. Why did he change from the \left|E\right\rangle eigenbasis to the \left|p\right\rangle eigenbasis, and still maintained the eigenvalue E? Is there atheorem of eigenbasis I'm missing? I mean, \left|p\right\rangle is an eigenvector of P, by definition, and then, of P^{2}. Then, from the time independent Schr. eq. we see that \left|E\right\rangle is an eigenvector of P^{2} too. But that doesn't mean it's the same eigenbasis, does it? Or what does ''trial'' mean in this context (forgive me, my mother language is spanish).

After that ''change of basis'' from \left|E\right\rangle to \left|p\right\rangle, everything is fine. So please, explain it to me.

Please, forgive me, I know this is a very stupid question, but I just don't get it.

Thankyou very much for your patience.

 

[vela]

The ket \vert p \rangle is an eigenstate of both the momentum operator \hat{p} and the Hamiltonian \hat{H}. If you apply Hamiltonian to it, you will get the corresponding eigenvalue of the Hamiltonian, namely E.

The \vert E \rangle and \vert p \rangle bases aren't the same. While a state \vert p \rangle is an eigenstate of \hat{H}, a state \vert E \rangle isn't necessarily an eigenstate of \hat{p}. For example, you can show that a state with the wave function \psi(x) = e^{ikx} + e^{-ikx} is an eigenstate of \hat{H} with eigenvalue E=\hbar^2k^2/2m but it's not an eigenstate of \hat{p}.

 

[MManuel Abad]

Dear Lord, how could I be so blind??! That's crystal clear! Thank you so much, vela, for taking the time for answering such a silly question! THANK YOU!