# Exp(i k.r) ~ A travelling wave ?

1. Aug 24, 2010

### mtrl

exp(i k.r) ~ A travelling wave ??

I am studying electrons in solids and the wavefunctions of free electrons, which have the form exp(i k.r), are said to be representing travelling waves. Isn't there supposed to be a term involving time in the exponential such as exp (i (k.r -wt)) so that it is a travelling wave?
If exp(i k.r) does not represent a standing wave, then what represents a standing wave?

2. Aug 24, 2010

### Dr Lots-o'watts

Re: exp(i k.r) ~ A travelling wave ??

r = r(x, y, z, t)

3. Aug 24, 2010

### mtrl

Re: exp(i k.r) ~ A travelling wave ??

There is a further explanation in the book which i think makes r=r(x,y,z,t) an invalid answer.

The book says that exp(+i k.r) and exp(-i k.r) are travelling waves. And a superposition of these two functions, exp(+i k.r) + exp(-i k.r) = 2 cos(k.r) is a standing wave. If r=r(x,y,z,t), then the superposition is a traveling wave too, isn't it?

4. Aug 24, 2010

### nasu

Re: exp(i k.r) ~ A travelling wave ??

As a general observation, in some books the time factor is left out for convenience but the complete form is assumed to be as you show above.

In quantum mechanics, if you plug in the complete (time included) solution in Schrodinger equation, the temporal part can be separated and then you only look at the time independent Schrodinger equation. The solutions of this last one have only position dependence but in the end you should (if necessary) add the time factor.

5. Aug 24, 2010

### mtrl

Re: exp(i k.r) ~ A travelling wave ??

Thank you nasu. I have understood it now.
The book on solid state physics by Kittel is really driving me mad. There are almost no explanations during formula derivations and lots of steps are omitted.

6. Aug 25, 2010

### Dr Lots-o'watts

Re: exp(i k.r) ~ A travelling wave ??

nasu's answer is better. Indeed, r usually doesn't include the time coordinate. My answer needs an unconventional definition for r.

7. Aug 25, 2010

### nasu

Re: exp(i k.r) ~ A travelling wave ??

I agree with you. It's a good reference if you already know the stuff but not the best introduction.