# Wave packet - Integration by parts

1. Sep 8, 2012

### ibysaiyan

1. The problem statement, all variables and given/known data
Hi ,

I am reading a little on introductory QM , initial chapters on waves.
They have given an integral for a wavepacket , assuming at t= 0.
Which is: ψ(x,0) = $\int A cosk'x dk'$ (I don't know how to define limits to the integral in Latex upper = k+Δk , lower limit = k-Δk)
which gives ψ(x,0) = S(x)coskx , where S (x) = 2AΔK sin(Δkx)/(Δkx).

2. Relevant equations

Integration by parts ?

3. The attempt at a solution
Now I am totally brain stuck , it surely can't be anything complex so I should be able to do it by parts.
Let u =cosk'x and dv = dk'

Or am I overlooking something ?

2. Sep 8, 2012

### vela

Staff Emeritus
Yeah, it's just a straightforward integration. Remember x acts like a constant here because you're integrating with respect to k'. When you integrate cos k'x, you get sin k'x/x.

3. Sep 8, 2012

### vela

Staff Emeritus
You can look at the code in this post to see how to write the integral with limits.
$$\int_{k-\Delta k}^{k+\Delta k} A\cos k'x\,dk'$$ You also want a backslash before cos to get the function name to typeset correctly, and a little space (\,) before dk' makes it look better.

4. Sep 8, 2012

### voko

It really is simple. x is constant during integration.

5. Sep 16, 2012

### ibysaiyan

Thanks for the responses everyone but I still don't get it.
$$\int_{k-\Delta k}^{k+\Delta k} A\cos k'x\,dk'$$
As it has been mentioned that X and A are constant, so I end up with :

Asink'x/x but when I plug in the limits I get:

2Asin(Δkx)/x... how do they end up with the cos. identity ?

6. Sep 16, 2012

### vela

Staff Emeritus
After you integrate, you have
$$\left.\frac{A}{x}\sin k'x\right|_{k-\Delta k}^{k+\Delta k} = \frac{A}{x}[\sin ((k+\Delta k)x) - \sin((k-\Delta k)x)].$$ How did you manage to go from this last expression to what you got?

Last edited: Sep 16, 2012
7. Sep 16, 2012

### ibysaiyan

Yes. I did get the above step after which I expanded the brackets , which resulted in the cancelling of -kx +kx terms , right ?

8. Sep 16, 2012

### vela

Staff Emeritus
Are you claiming that sin(a+b)-sin(a-b) = 2 sin b by simply subracting the arguments of the sines?

Last edited: Sep 16, 2012
9. Sep 16, 2012

### ibysaiyan

Oh no, I was greatly mistaken !
I just realized sum formula is all I need....

I feel so stupid :

Just how the hell am I going to become a physicist if I keep doing such silly mistakes...

10. Sep 16, 2012

### ibysaiyan

Ok.. so I have just used the identified that I previously mentioned.. I still don't end up with the form in the OP.

$$\left.\frac{A}{x}\sin k'x\right|_{k-\Delta k}^{k+\Delta k} = \frac{A}{x}[\sin ((k+\Delta k)x) - \sin((k-\Delta k)x)].$$

Now sin (a+b) = sinAcosB+sinBcosA , sin (a-b) = sinAcosB-sinBcosA

I used the above to only end up with

$2A sin(Δkx) coskx / x$

They seem to have two extra Δk for some reason..