This is the way I understand it. Correct me if I'm wrong.(adsbygoogle = window.adsbygoogle || []).push({});

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 conditionsandhappens 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.

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# Confusion about eigenstates and wavefunctions.

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