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I QFT for the Gifted Amateur Question?

  1. Sep 25, 2016 #1
    In chapter 9 "Quantum Mechanical Transformations", example 9.3, can anyone explain how $$\hat{p}$$ in the exponential converts itself to q? Apologies in advance if this is very basic, but thanks for looking.

    $$\hat{u}(a)|q>=e^{-i\hat{p}\cdot a} |q> $$

    $$\hat{u}(a)|q>=e^{-iq\cdot a} |q> $$
     
  2. jcsd
  3. Sep 25, 2016 #2
    I assume ##|q\rangle## is a momentum eigenstate? Then its just a matter of using ##\hat{p}|q\rangle = q |q\rangle##
     
  4. Sep 25, 2016 #3

    dextercioby

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    If that would be the case, then it's highly unfortunate to use q for labeling momentum in nonrelativistic QM. QM stays somewhere at the middle between Hamiltonian CM and QFT. In Hamiltonian CM "q" stands for generalized coordinate, while in QFT q goes along with p,r,s in Perturbation theory and Feynman diagrams to label 4-momenta. In "ordinary" QM, the momentum operator is p with a hat or without, its eigenket is |p> and its eigenvalue in the space of eigendistributions is also p. Why use q at all ?
     
  5. Sep 25, 2016 #4
    Wow, thanks for the quick replies. Fightfish is correct: |q> is a momentum state. That was stated in the example; sorry, I should have included it. Unfortunately, I am still a bit confused. Why is it OK to write $$\hat{p} \quad for \quad e^{-i\hat{p}\cdot a}$$ in the footnote 4 of chapter 8, Lancaster writes in general:
    $$e^{\hat{A}}=1+\hat{A}+\frac{1}{2!}\hat{A}\hat{A}+...$$
    That seems a bit messier.
     
  6. Sep 25, 2016 #5
    That is the definition of the exponential of an operator.
    It's not - ##\hat{p}## is not the same as ##e^{-i\hat{p}\cdot a}##. When we act ##e^{-i\hat{p}\cdot a}## on the momentum eigenstate, what we really are doing is to use that expansion, which is in powers of ##\hat{p}##, act it on the momentum eigenstate, and then collect the terms back together to obtain another exponential, this time in terms of the eigenvalue.
     
  7. Sep 26, 2016 #6
    One has:
    $$e^{-i\hat{\textbf{p}}\textbf{a}}|\textbf{q}\rangle=[1+(-i\hat{\textbf{p}}\textbf{a})+\frac{1}{2!}(-i\hat{\textbf{p}}\textbf{a})^2+\cdots]|\textbf{q}\rangle$$
    ##\hat{\textbf{p}}=-i\vec{\nabla}## so it commutes with ##\textbf{a}## then, with ##\hat{\textbf{p}}|\textbf{q}\rangle=\textbf{q}|\textbf{q}\rangle##, one has:
    $$(-i\hat{\textbf{p}}\textbf{a})|\textbf{q}\rangle=-i\textbf{qa}|\textbf{q}\rangle$$
    $$(-i\hat{\textbf{p}}\textbf{a})^2|\textbf{q}\rangle=(-i\textbf{qa})^2|\textbf{q}\rangle$$
    $$e^{-i\hat{\textbf{p}}\textbf{a}}|\textbf{q}\rangle=[1+(-i\textbf{qa})+\frac{1}{2!}(-i\textbf{qa})^2+\cdots]|\textbf{q}\rangle=e^{-i\textbf{qa}}|\textbf{q}\rangle$$
     
  8. Sep 26, 2016 #7
    Fightfish and mbond, thank you so much for your explanations; they perfectly filled in the steps I was missing.
     
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