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

B Why is momentum the fourier transform of the wavefunction ?

  1. Jul 12, 2016 #1
    I think this is probably a very basic question: why does the Fourier transform of a wavefunction describing position probabilities gives us a function describing momentum probabilities ?

    Is there a fairly simple explanation for this ? What leads us to this relation ?
     
  2. jcsd
  3. Jul 12, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Everything follows from the Heisenberg algebra. For one-dimensional motion it reads
    $$[\hat{x},\hat{p}]=\mathrm{i},$$
    where I set ##\hbar=1## for convenience. From that you can derive that in the position representation
    $$\hat{p} \psi(x)=\langle x|\hat{p} \psi \rangle=-\mathrm{i} \partial_x \psi(x)=-\mathrm{i} \langle x|\psi \rangle.$$
    The generalized momentum eigenstates are definied by
    $$\hat{p} u_{p}(x)=p u_p(x).$$
    Solving this equations for the eigenstates leads to
    $$u_{p}(x)=N \exp(\mathrm{i} p x).$$
    The normalization is most conveniently chosen such that
    $$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x u_p^*(x) u_{p'}(x) = |N|^2 \int_{\mathbb{R}} \mathrm{d} p \exp[\mathrm{i}(p'-p)x] = (2 \pi)^3 |N|^2 \delta(p-p') \stackrel{!}{=} \delta(p-p') \; \Rightarrow \; N=\frac{1}{\sqrt{2 \pi}}.$$
    Thus we have
    $$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x).$$
    From this you finally get for the wave function in momentum representation
    $$\tilde{\psi}(p)=\langle p|\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x \langle p|x \rangle \langle x |\psi \rangle = \int_{\mathbb{R}} \mathrm{d} x u_p^*(x) \psi(x) = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi(x),$$
    which is nothing else than the Fourier transformation of the wave function in position representation. In the same way you immediately get the inverse transformation
    $$\psi(x)=\int_{\mathbb{R}} \mathrm{d} p u_p(x) \tilde{\psi}(p) = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} p x) \tilde{\psi}(p).$$
     
  4. Jul 14, 2016 #3
    Thank you for answering !

    This provided some insight. However my knowledge of QM and the associated maths is very incomplete and quite shaky. I think I'll come back to your answer later, it'll probably be much clearer then.

    There's just one part I wonder about. This :

    Here is what I read in plain english : "There exist a momentum operator and an associated eigenfunction. The eigenvalue is the momentum." If that's a correct reading, then why do we know there must exist a momentum operator, and its eigenfunction ? Hope the question makes sense. I wonder why someone would come with that idea.

    Also in the mean time I have read the first chapter of Griffiths's textbook "Introduction to Quantum Mechanics (2nd Edition)". To introduce momentum, he takes the time derivative of the expected value of position, and using Schrödinger equation, he ends up with an expression for the expected value of momentum. Which does look like a Fourier transform. He seems to say this isn't really good "theoretical foundation" though, I suppose it serves more as an intro.
     
  5. Jul 15, 2016 #4

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Yes, that's a very heuristic "derivation". Anyway, as it comes to the foundations and its subtleties, from what I read in this forum, Griffiths's textbook doesn't get the foundations very clear, and this causes a lot of misunderstandings and troubles for the uninitiated reader, which unfortunately is the adressee of this textbook ;-)). I recommend to read a book which is more careful in explaining the foundations. My first QM textbook at the university has been J. J. Sakurai, Modern Quantum Mechanics (2nd edition). I think that's a very good textbook. To get the math right, I strongly recommend Ballentine, Quantum Mechanics - a modern development, because it introduces the rigged-Hilbert space formalism without dwelling too much on a rigorous mathematical treatment.
     
  6. Jul 16, 2016 #5
    Thank you for the recommandations, I'll have a look at those !
     
  7. Jul 16, 2016 #6
    In plain English, the Fourier transform of the wavefunction in position gives the momentum because position and momentum are conjugate under uncertainty. The Fourier transform of the wavefunction of one therefore gives the wavefunction of the other. This is the underlying meaning of the math @vanhees71 explained above.

    The underlying physical reason for this is the properties of the vacuum (of spacetime, specifically) and the complementarity of conservation of momentum and continuous symmetry of physics over position under Noether's Theorem. Position and momentum are intimately related by the nature of spacetime itself, and this leads to complementary Heisenberg uncertainty of position and momentum as well as the fact about the Fourier transforms of their wavefunctions that you are trying to understand.
     
  8. Jul 17, 2016 #7
    Thanks, this adds some insight. At least it does illustrate a profound link between position and momentum in general. I'll come back to this question after I get more practice with QM and hopefully this will be clearer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted