Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How is this equation for <x|P|x'> derived?

  1. Jan 20, 2016 #1
    Image: http://imgur.com/pynKr8q

    How is the boxed equation arrived at when looking at the step before?

    I think I need clarification on a few key ideas:

    in words, does < x | p | ψ > mean, "The momentum operator acting on the state vector, then projected onto the position basis"? If so, wouldn't that mean there would always be a factor of iћ since [x,p] = iћ?

    Furthermore, is the reason the integral of |x'><x'| there just so you get <x'|ψ> and project ψ to the x' basis. I know it equals and identity matrix but can you do this?

    Thank you
  2. jcsd
  3. Jan 20, 2016 #2


    User Avatar

    Staff: Mentor

    I don't think you can get it from the equation above, but rather from the equation below. For eq. (75) to hold, which it should given the definition of ##\hat{P}## as a differential operator and ##\psi(x) = \langle x | \psi \rangle##, then you need eq. (74) to be true.

    Yes, and the factor of iћ is there.

    The author is inserting the identity operator between ##\hat{P}_x## and ##\psi##, which you always do and the result should be unchanged. This is how the value of ##\langle x | \hat{P}_x | x' \rangle## can be figured out.
  4. Jan 20, 2016 #3
    Thanks for the reply!

    but the whole reason we need to figure out < x | Px | x' > in the first place is because the identity operator is inserted. What's the point?

    Also how to do you do this derivation step by step? I'm very confused I think my prof. is skipping steps
    : http://imgur.com/tl07k5g
  5. Jan 22, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper

    It seems like the sentence above is more logical if it were written as
    "but the whole reason we can figure out < x | Px | x' > in the first place is because the identity operator is inserted".
    The reason why the author needs to figure out ##\langle x|\hat{P}_x|x'\rangle## is because he wants to make the reader know how this quantity looks like, and also possibly because its equivalent form being sought will be needed in the other derivations to come, it is not because the identity operator is inserted.

    To me, the presented steps look sufficiently complete. Which particular part of this derivation you can't understand?
  6. Jan 22, 2016 #5
    Thanks for replying. The 2nd line to the 3rd line is where I get lost.
  7. Jan 22, 2016 #6


    User Avatar
    Science Advisor
    Homework Helper

    The second line reads as
    \langle x | \left[ \sum_{n=0}^\infty \frac{x'^n}{n!} \frac{\partial^n V(X)}{\partial X^n} \right] |x'\rangle
    The factor in the big square bracket is now just a number/scalar, it's not an operator anymore like it was in the first line. Therefore, you can bring ##|x'\rangle## out to the left till it meets ##\langle x|##, leaving you with the expression
    \langle x | x'\rangle \left[ \sum_{n=0}^\infty \frac{x'^n}{n!} \frac{\partial^n V(X)}{\partial X^n} \right]
    Note that the factor in the square bracket is the Taylor expansion of a scalar ##V(x')##. So you obtain the third line.
  8. Jan 22, 2016 #7
    but the 3rd line has <x'|x> not <x|x'>. Am I missing something?
  9. Jan 22, 2016 #8


    User Avatar
    Science Advisor
    Homework Helper

    Ah you are right, I must have missed that part. Fortunately, that's not a big trouble because ##\langle x|x' \rangle = \delta(x-x') = \delta(x'-x) = \langle x'|x \rangle##.
  10. Jan 27, 2016 #9


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I've not followed the entire discussion, but perhaps, the following helps (I'm setting ##\hbar=1## for simplicity)
    $$\langle x|\hat{p}|x' \rangle=\int \frac{\mathrm{d} p}{(2 \pi)} \langle{x}|\hat{p}|p \rangle \langle p|x' \rangle=\int \frac{\mathrm{d} p}{(2 \pi)} \exp[\mathrm{i} p (x-x')] p = -\mathrm{i} \partial_x \int \frac{\mathrm{d} p}{(2 \pi)} \exp[\mathrm{i} p(x-x')]=-\mathrm{i} \partial_x \delta(x-x')=+\mathrm{i} \partial_{x'} \delta(x-x').$$
    Now you can use it to determine the momentum operator in position representation on an arbitrary Hilbert-space vector (in the domain of the momentum operator!):
    $$\hat{p} \psi(x):=\langle x|\hat{p} \psi \rangle=\int \mathrm{d} x' \langle x|\hat{p}|x' \rangle \psi(x')=\int \mathrm{d} x' \psi(x') \mathrm{i} \partial_{x'} \delta(x-x') = \int \mathrm{d} x' (-\mathrm{i} \partial_{x'}) \psi(x') \cdot \delta(x-x')=-\mathrm{i} \partial_x \psi(x).$$
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: How is this equation for <x|P|x'> derived?
  1. Question about <x|P> (Replies: 6)

  2. Commutator of x and p (Replies: 3)

  3. Why x and p (Replies: 1)