How Do You Convert Position to Momentum Space in Quantum Mechanics?

[rsaad]

Homework Statement

write the following in K basis:

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

Homework Equations

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!

 

[CompuChip]

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

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

 

[rsaad]

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

 

[rsaad]

I am getting a very weird answer.

 

[rsaad]

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

 

[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'.

 

[rsaad]

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

 

[rsaad]

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

 

[CompuChip]

OK, so I was thinking

\int |x\rangle \langle x | \, dx = <br /> \iiint |k\rangle \langle k | x\rangle \langle x | |k&#039;\rangle \langle k&#039; | \, dx \, dk \, dk&#039; <br /> \propto \iiint e^{-i(k - k&#039;)x}|k\rangle \langle k&#039; | \, dx \, dk \, dk&#039;

Is that where you got to as well?

And then you go on to use
\int e^{i(k - k&#039;)x} \, dx \propto \delta(k - k&#039;)
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|.