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 UK university physics student.

So, the question basically was:"The wavefunction

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

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. This is where I got to:

Now, I am sure that that is incorrect somehow as it just doesn't seem to add up - however, my reasoning is that integrating the probability distribution function between -infinity and +infinity will give you an answer of 1; as the particle has to be somewhere in this one dimensional space. But then I'm not too sure about integrating the function with respect to x then t and multiplying them to give the answer 1 - as when I do this, the equations get into a horrible mess and I end up with a value of A = 0 for normalising the wavefunction; which is nonsensical.

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)