[UG]

I read somewhere '...a wavefunction of the form e^ikx describes a particle with linear momentum...'
What does that little 'i' mean?

[renge ishyo]

It's the imaginary number: i² = -1 or i = sqroot(-1)

If you want me to save you the frustration and the headaches UG, I spent a VERY large amount of effort a few years ago trying to figure out why an imaginary number appeared in the formulas in quantum mechanics, and so I will just spill out the reason why it is there for you below if you would like  

Does the use of imaginary numbers have some physical meaning, or is it just a mathematical trick? It is actually a mathematical trick. It turns out that when you raise an exponential to an imaginary power (e^ix) you can use the expression as an algebraic substitute for writing the trigonometric functions via the relationships:

e^ix = cos x + i sin x

or its complex conjugate:

e^-ix = cos x - i sinx

You will also see places in quantum mechanics where you often have to multiply the wave function by the complex conjugate in places to get real solutions (i.e. this takes place when you "square" the wavefunctions to get the probability of finding an electron in a certain place). With the trig displayed, the operation looks like this.

(e^ix)(e^-ix) = cos² x + sin² x = 1

That trig function you see there should look familiar, and the "i" disappears The mathematics is much more of a hassle if you do wavefunctions with straight trig, so the physicists mostly use the exponentials with the imaginaries instead of the trig for the wavefunctions to make the calculus easier (but at the same time, most physics books that I have read don't tell you that this is what they are doing because they assumed you have been taught all this in your math classes...well, not in MY math classes at least

[UG]

Righto, thanks renge ishyo, now I'll have to get my head around this beast