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

In summary, the question asks for the expression A=∫|x><x|dx to be written in terms of the K basis. The attempt at a solution involves introducing the identity twice and simplifying to get an answer of 1/2pi ∫∫dk dx. However, the correct solution should involve a triple integration and the use of the identity property e^{i(k-k')x}∝δ(k-k'). The final answer should be similar in form to |x><x|, but in terms of the K basis.
  • #1
rsaad
77
0

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!
 
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  • #2
Why is it a problem that the integrals have different limits?

More relevant question to help your forward: what is <k|x> ?
 
  • #3
<k|x>= exp(-ikx)/(2*pi)^0.5
 
  • #4
I am getting a very weird answer.
 
  • #5
I introduced the identity twice and on simplifying, I get 1/2pi ∫∫dk dx ??
 
  • #6
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
Yes, I know that. I simplified things and I got that answer.
 
  • #8
Could you please solve the solve question and suggest the steps?
 
  • #9
OK, so I was thinking

[tex]\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'[/tex]

Is that where you got to as well?

And then you go on to use
[tex]\int e^{i(k - k')x} \, dx \propto \delta(k - k')[/tex]
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|.
 
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FAQ: How Do You Convert Position to Momentum Space in Quantum Mechanics?

What is the k basis in Quantum Mechanics?

The k basis in Quantum Mechanics refers to a set of vectors that are used to represent the state of a quantum system. These vectors are called kets and are denoted by |k>. They are used to describe the position and momentum of a particle in a given quantum state.

How is the k basis used in Quantum Mechanics?

The k basis is used to represent the state of a quantum system and to perform calculations in Quantum Mechanics. The k basis allows for the description of the position and momentum of a particle, which are essential quantities in Quantum Mechanics.

What is the relationship between the k basis and the position and momentum operators?

The k basis is closely related to the position and momentum operators in Quantum Mechanics. The position operator, denoted by x, acts on a ket in the k basis to give the position of a particle in a given state. The momentum operator, denoted by p, acts on a ket in the k basis to give the momentum of a particle in a given state.

How does the k basis relate to the Schrödinger equation?

The Schrödinger equation, which is the fundamental equation of Quantum Mechanics, is written in terms of the k basis. The equation describes the time evolution of a quantum system and allows for the prediction of the future state of the system based on its current state in the k basis.

What is the significance of the k basis in Quantum Mechanics?

The k basis is essential in the understanding and application of Quantum Mechanics. It allows for the representation and description of quantum states, as well as the performance of calculations and predictions in this branch of physics. The k basis is fundamental to many concepts and equations in Quantum Mechanics and is crucial for studying and understanding the behavior of particles at the quantum level.

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