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A 'particle' in a given situation will be in a state [tex]|\psi>[/tex], which is determined by the Schrodinger Equation. After measurement, the particle will then go to a state [tex]|\omega>[/tex], where [tex]|\omega>[/tex] is an eigenvector of the operator corresponding to the measurement. The measurement you'll read will be the corresponding eigenvalue.

A 'particle' in a given situation will be in a state [tex]|\psi>[/tex], which is determined by the Schrodinger Equation. After measurement, the particle will then go to a state [tex]|\omega>[/tex], where [tex]|\omega>[/tex] is an eigenvector of the operator corresponding to the measurement. The measurement you'll read will be the corresponding eigenvalue.

Now this is where the confusion comes in. If I make any sort of errors here, feel free to correct me. I know there's a huge error here somewhere, but I just don't know where it is.

Let's say we have the particle in a box scenario, with the box going from 0 to a. In this case, the (normalized) solutions of the time independent Schrodinger Equation are of the form [tex]\psi_n=\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})[/tex] or [tex]\frac{i}{\sqrt{2a}}(e^{\frac{-in\pi x}{a}}+e^{\frac{in\pi x}{a}})[/tex].

Let's say we were to measure the momentum of the particle. We know that in the X basis, the momentum operator is [tex]i\hbar\frac{d}{dx}[/tex], with eigenvectors being (constant multiples of) [tex]e^{-\frac{ipx}{\hbar}}[/tex]. Since [tex]p=\pm\sqrt{2mE}=\pm\hbar k=\hbar \frac{n\pi}{a}[/tex], we can write the momentum eigenvectors as [tex]e^{\frac{-in\pi x}{a}}[/tex]. [tex]e^{\frac{in\pi x}{a}} [/tex] also happens to be a solution if you consider the negative value of momentum.

Since the general solution of the Schrodinger equation in this well are always linear combinations of [tex]e^{ikx}[/tex] and [tex]e^{-ikx}[/tex], we know that eigenstates of the momentum operator are also solutions to the Schrodinger equation. Everything is fine so far. The problem comes when you note that in order for a linear combination of the two forms of eigenfunctions for the momentum operator isn't a solution itself (unless one of the weights are zero). In other words, [tex]C_1e^{ikx} + C_2e^{-ikx}[/tex] is only an eigenfunction of the momentum operator if [tex]C_1[/tex] or [tex]C_2[/tex] is zero. However, the boundary conditions on the Schrodinger equation requires that the state vector [tex]C_1+C_2=0[/tex] (the boundary condition at the left point). This means that the only state that is a solution to the Schrodinger equation satisfying the boundary conditions

*and*happens to be an eigenfunction of momentum operator is the trival solution.

What I'm wondering is what happens after you measure the momentum of a particle. What state does it go to? It would seem that it can't collapse to a momentum eigenstate because the momentum eigenstate doesn't satisfy the Schrodinger equation after applying the boundary conditions.