Wavefunction Normalisation

[x(x-y)]

So, I was just messing around with some practice questions in my university physics textbook and I came across a particular question which was fairly easy to answer but which I would like to take further - however, it may be too advanced for me at the moment being a first year university physics student.

 

So, the question basically was: "The wavefunction

 

[latex]\psi (x,t) = Ae^{i(k_1 x - \omega_1 t)} + Ae^{i(k_2 x - \omega_2 t)}[/latex]

 

is a superposition of 2 free-particle wavefunctions, both k1 and k2 are positive.

 

a) Show that this wavefunction satisfies the Schrödinger equation for a free particle of mass m.

b) Find the probability distribution function for ψ(x,t)."

 

So, the first question (a) was simple of course - just taking partial derivatives then a bit of algebra. And the next part (b) gave me an answer of

 

[latex]|\psi (x,t)|^2 = 2A^2(1 + cos[(k_2 - k_1)x + (\omega_1 - \omega_2)t])[/latex]

 

using the Born interpretation procedure (complex multiplied by complex conjugate gives mod. complex squared which is the probability density - or "distribution function").

 

So, I was wondering how one could use this result for the probability distribution function to normalise the wavefunction - as I have done for simpler complex wavefunctions before. Now I know that this result should be true:

 

[latex]\displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = 1[/latex]

 

as the particles must be somewhere in this one dimensional space. However, when working through this I found that A = 0; which is nonsensical and obviously incorrect.

 

I have also tried to compute the problem using the identity:

 

[latex]\frac{d}{dt} \displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = 0[/latex]

 

and then using Schrödinger's 1D time dependent equation, one arrives at the result:

 

[latex]\frac{d}{dt} \displaystyle \int \limits_{-\infty}^{\infty} |\psi (x,t)|^2 \, dx = \frac{i\hbar}{2m}\left[\psi (x,t)^{\ast}\frac{\partial \psi (x,t)}{\partial x} - \psi (x,t)\frac{\partial \psi (x,t)^{\ast}}{\partial x}\right]_{-\infty}^{\infty} = 0[/latex]

 

However, yet again when working through this I found that A = 0 again! After reviewing page 38 of "Quantum Mechanics by E.Merzbacher, 2nd Edition, 1970" I found that the above mechanism only works if the expression below is true:

 

[latex]\displaystyle \lim_{x \to \infty} \psi (x,t) = 0[/latex]

 

which the wavefunction in my case doesn't obey.

 

So, I was wondering how one may go about computing such a situation - it may be too advanced for me at the moment, but I love to learn! By the way, no this is not a "homework" question - I am not the kind of person who cheats in such a way for homework answers; however, evidently I cannot prove this so believe this is a homework question if you want to, it makes no difference to me as I know it isn't!

 

Thanks,

 

x(x-y)

 

P.S:

 

Here is a more complete review of what I've done for this problem so far on another forum

[mathematic]

Your description has the same problem for a single particle. |wave function|^2 = 1 for all x and t.

[ajb]

The plane wave is technically not square integrable, so some care needs to be taken. One has to extend the Hilbert space of square integrable functions to include things like the plane wave, which if you think about it have a continuous spectrum. You need to think about Rigged Hilbert spaces.

[x(x-y)]

The plane wave is technically not square integrable, so some care needs to be taken. One has to extend the Hilbert space of square integrable functions to include things like the plane wave, which if you think about it have a continuous spectrum. You need to think about Rigged Hilbert spaces.

 

Ah, thank you - I suspected that this wavefunction can't be normalised in a way that I know of (as shown by the link in the OP). Anyway, I'm afraid that link are your explanation seem a bit too advanced for my current level; however, I love to learn and shall take a closer look later on. Thanks!

[ajb]

Ah, thank you - I suspected that this wavefunction can't be normalised in a way that I know of (as shown by the link in the OP). Anyway, I'm afraid that link are your explanation seem a bit too advanced for my current level; however, I love to learn and shall take a closer look later on. Thanks!

 

No problem.

 

Loosely, the space of wave functions is supposed to be the Hilbert space of square integrable functions. This is okay if we have a discrete spectra, like you have for bound states, so say for the Harmonic oscillator or the bound states of the hydrogen atom. The problem comes when we have free states. The basic idea is to extend the Hilbert space to include such things, and in fact the more general distributions. You should also note that these things come up in the mathematical framework of C*-algebras.

 

This is all a little outside my area of expertise and most physicists would not worry about this too much.

[x(x-y)]

Hmmm... Yes, this is a bit beyond my current level of understanding - I suspect that I will have to wait until the Quantum Mechanics 2 course of the second year of my Physics degree to begin to gain an understanding of these more advanced processes. I very much look forward to it - quantum mechanics is one of my favourite topic of physics, it's so profound and weird; yet incredibly interesting and exciting to study!

[ajb]

Hmmm... Yes, this is a bit beyond my current level of understanding - I suspect that I will have to wait until the Quantum Mechanics 2 course of the second year of my Physics degree to begin to gain an understanding of these more advanced processes.

I would imagine that the notion of a rigged Hilbert space will be glossed over and not discussed in any detail.