# Quantum Mechanics: k basis

1. Mar 4, 2013

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

write the following in K basis:

A=∫|x><x|dx where the integral limits are from -a to a

2. Relevant equations

3. The attempt at a solution

I tried solving it by inserting the identity
I=∫|k><k|dk where the integral limits are from -∞ to +∞

but then I do not know how to proceed from there. What to do about the two integrals with varying limits!

2. Mar 4, 2013

### CompuChip

Why is it a problem that the integrals have different limits?

More relevant question to help your forward: what is <k|x> ?

3. Mar 4, 2013

<k|x>= exp(-ikx)/(2*pi)^0.5

4. Mar 4, 2013

I am getting a very weird answer.

5. Mar 4, 2013

I introduced the identity twice and on simplifying, I get 1/2pi ∫∫dk dx ??

6. Mar 4, 2013

### CompuChip

If you introduce the identity twice, you should use two different integration variables. So I expect a triple integration, e.g. over x, k and k'.

7. Mar 4, 2013

Yes, I know that. I simplified things and I got that answer.

8. Mar 4, 2013

Could you please solve the solve question and suggest the steps?

9. Mar 4, 2013

### CompuChip

OK, so I was thinking

$$\int |x\rangle \langle x | \, dx = \iiint |k\rangle \langle k | x\rangle \langle x | |k'\rangle \langle k' | \, dx \, dk \, dk' \propto \iiint e^{-i(k - k')x}|k\rangle \langle k' | \, dx \, dk \, dk'$$

Is that where you got to as well?

And then you go on to use
$$\int e^{i(k - k')x} \, dx \propto \delta(k - k')$$
but I don't see how the |k> <k'| disappeared from your suggested answer... after all, what you should get is similar in form to |x> <x|.

Last edited: Mar 4, 2013