Is the Wave Function for a Free Particle with a Rigid Wall at x=0 Correct?

[Ed Quanta]

Ok, so suppose there is a free particle of mass m that moves in a one-dimensional space in the interval 0<=x, with energy E. There is a rigid wall at x=0. Write down a time independent wave function G(x) which satisfies these conditions, in terms of x and k, where k is the wave vector of motion. State the relation between k and E.


So what do I do? The solution G(x)=xAexp(ikx) seems to obey the one boundary condition we have here, but can this be correct?And could this possibly be an eigenfunction of the Hamiltonian for this system?

 

[nrqed]

Ok, so suppose there is a free particle of mass m that moves in a one-dimensional space in the interval 0<=x, with energy E. There is a rigid wall at x=0. Write down a time independent wave function G(x) which satisfies these conditions, in terms of x and k, where k is the wave vector of motion. State the relation between k and E.


So what do I do? The solution G(x)=xAexp(ikx) seems to obey the one boundary condition we have here, but can this be correct?And could this possibly be an eigenfunction of the Hamiltonian for this system?

Did you check if your wavefunction satisfies Schroedinger's equation?

There are two issues: the wavefunction must obey Schroed's equation and it must obey the boundary conditions. But you should *first* solve the equation and *then* impose the boundary conditions. If you try the other way around (writing functions which satisfy the bc's and then plugging them in the equation to see if it is satisfied) is terribly terribly inefficient! Go the other way around. Solving the equation for a free particle (V=0) is easy. Then your general solution will be a linear combination of the solutions found in the previous step. And then you impose the boundary conditions. (But your solution will not be normalizable which I hope is ok with you).

Pat

 

[reilly]

The solution for any k is sinkx for x>0 and 0 for x<0, up to a normalization constant.
Regards,
Reilly Atkinson

(X exp(ikx) dos not satisfy the SE)

 

[Ed Quanta]

Hey reilly, what about when x =0? Here is my problem.

Aexp(ikx) is the solution to the space component of the Schrodinger Differential equation. But if I impose the boundary condition that the wave function =0 when x=0, then A must equal zero and thus we have a trivial solution.

I recognize that the the step function when x>=0, sinkx
when x<0, 0 works here

But did you just guess at this solution? Or is there a way to get the general solution Aexp(ikx) and eliminate the cos part of the solution. Sorry if I am being slow.

And nrqed, I am fine with the solution not being normalizable, I just haven't seen any wave functions with discontinuities before in my limited studies.

Thanks to both of you.

 

[nrqed]

Hey reilly, what about when x =0? Here is my problem.

Aexp(ikx) is the solution to the space component of the Schrodinger Differential equation. But if I impose the boundary condition that the wave function =0 when x=0, then A must equal zero and thus we have a trivial solution.

I recognize that the the step function when x>=0, sinkx
when x<0, 0 works here

But did you just guess at this solution? Or is there a way to get the general solution Aexp(ikx) and eliminate the cos part of the solution. Sorry if I am being slow.

And nrqed, I am fine with the solution not being normalizable, I just haven't seen any wave functions with discontinuities before in my limited studies.

Thanks to both of you.

It is not a guess. If you consider the SE in the region where V = infinity, the only solution that makes sense is psi=0. In the region where V=0, the general solution (it's a second order ordinary DE (we are considering the time independent SE), so the general solution contains two arbitrary coefficients)
can be written in two equivalent forms

\psi(x) = A sin(kx) + B cos (kx)

or

\psi(x) = C e^{i kx} + D e^{-ikx}

where, of course, one could relate A, B to C, D.

Do you see that these are the most general solutions to the SE when V=0? That's the key point. It's not a guess. For V=0 the SE is pretty much the simplest nontrivial ordinary DE that one can write down. If you have some backgorund on differential eqs, this will be obvious to you. If you have no background in DE's, you should look carefully at the equation and let us know if you don't see that this is the solution.


Then you impose the boundary condition, \psi(0) = 0. With the first form, the solution is obvious, B=0. In the second form, you must impose C=-D which gives you back something proportional to a sine function.

Regards

Pat