More information on the momentum eigenstate in the position basis

  • Context: Undergrad 
  • Thread starter Thread starter hmparticle9
  • Start date Start date
  • Tags Tags
    Quantum mechanics
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
hmparticle9
Messages
157
Reaction score
26
The following sentence appears in my book (Introduction to Quantum Mechanics by Griffiths (3rd Edition)) :

"## \langle x | p \rangle ## is the momentum eigenstate (with eigenvalue p) in the position basis." Page 122.

I am not sure why this is the case. I have done some digging... on a previous page (114) it says

"The momentum space wave function ##\Phi(p ,t)## is the ##p## component in the expansion of ## | S(t) \rangle ## in the basis of momentum eigenfunctions:
$$\Phi(p ,t) = \langle p | S(t) \rangle$$
(with ## | p \rangle ## standing for the eigenfunction of ##\hat{p}## with eigenvalue ##p##)"

So is it convention that ## | p \rangle ## is the eigenfunction/eigenstate of ##\hat{p}##? If that is the case then that is one problem solved.

The second problem is, why does application of ## \langle x|## to ## | p \rangle ## mean that we are taking ##| p \rangle## with respect to the position basis?

From doing some further reading it seems as though that my second question could have the same answer as the first.
 
Last edited:
Physics news on Phys.org
The text inside a ket is merely a label, so in general we need to know what convention is being used to assign labels to know what is meant by ##|whatever\rangle##.

There are some conventions that are so commonly used and generally accepted that authors often won't bother stating them explicitly. You've just figured two of them out: unless the context tells us otherwise, ##|p\rangle## is usually the eigenket of ##\hat{p}## with eigenvalue ##p## and likewise for ##|x\rangle## and ##\hat{x}##.

The expression ##\langle x|\psi\rangle## is a number, so for a fixed ##|\psi\rangle## and variable ##x## it maps values of ##x## to numbers - which makes it a function of ##x##, the position-space representation of ##|\psi\rangle##.
 
Reply
  • Like
Likes   Reactions: hmparticle9 and PeroK