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I Finding components in momentum representation

  1. Jul 26, 2016 #1
    Dear all,

    I am trying to understand how they get they get the following components:
    $$c_\textbf{p} = \langle \textbf{p} | \psi\rangle = \int \frac{d\textbf{r}}{\sqrt{V}}e^{-\frac{i}{\hbar}\textbf{p}\cdot\textbf{r}}\psi(\textbf{r})$$
    Where ##|\textbf{p}\rangle## are the plane waves
    $$
    |\textbf{p}\rangle = \frac{1}{\sqrt{V}}e^{-iEt/\hbar}e^{i\textbf{p}\cdot\textbf{r}/\hbar}$$
    I understand the step
    $$\langle \textbf{p} | \psi\rangle = \int d\textbf{r}\langle\textbf{p}|\textbf{r}\rangle\langle\textbf{r}|\psi\rangle = \int d\textbf{r} \langle\textbf{p}|\textbf{r}\rangle \psi(\textbf{r})$$

    But not how
    $$ \langle\textbf{p}|\textbf{r}\rangle = \frac{1}{\sqrt{V}}e^{-\frac{i}{\hbar}\textbf{p}\cdot\textbf{r}}$$
     
  2. jcsd
  3. Jul 26, 2016 #2

    vanhees71

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    You should be much more careful in your notation. An abstract ket is not a wave function to begin with. Using the correct Dirac notation, which is utmost mnemonic, helps a lot in getting this things straight. If you have a normalized Hilbert-space vector, the wave function in position representation is given by
    $$\psi(\vec{r})=\langle \vec{r}|\psi \rangle.$$
    The generalized momentum eigenstates in a finite volume (most simply a cube with wave functions defined with periodic boundary conditions, because otherwise momentum doesn't make any sense as an observable) are given in position representation by (setting ##\hbar=1##)
    $$\langle \vec{x}|\vec{p} \rangle =\frac{1}{\sqrt{V}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}),$$
    as can be shown by the fact that ##\hat{\vec{p}}## is the operator of spatial translations. With the boundary conditions the momenta are discretized as
    $$\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3, \quad L^3=V.$$
    Now you have the completeness relations
    $$\sum_{\vec{p}} |\vec{p} \rangle \langle \vec{p}|=\mathbb{1}, \quad \int_{Q} \mathrm{d}^3 \vec{x} |\vec{x} \rangle \langle \vec{x}|=\mathbb{1}.$$
    Via these completeness relations you get indeed the correct transformation from the wave function in position representation to the wave function in momentum representation and vice versa as you've written in your posting #1. In this case it's nothing else than the usual Fourier-series theory for periodic functions. In the limit ##L \rightarrow \infty## momentum becomes continuous with the spectrum ##\vec{p} \in \mathbb{R}^3##, and the Fourier-series transformations go into Fourier-integral transformations. The orthonormalization conditions ##\langle \vec{p}|\vec{p}' \rangle=\delta_{\vec{p},\vec{p}'}## for the finite volume become ##\delta##-distributions in the infinite-volume limit ##\langle \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}')##.
     
  4. Jul 27, 2016 #3
    Thank you for the explanation, you have made it a lot more clear. The problem was indeed that I described ##|\textbf{p}\rangle## as a wave function instead of a momentum eigenstate.
     
  5. Jul 27, 2016 #4
    What happens when the completeness relations are unequal to 1? Is it not possible then to find momentum eigenstates in coordinate representation?
     
  6. Jul 27, 2016 #5

    vanhees71

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    Well, then your orthonormal system of (generalized) vectors is incomplete, and you must find the missing vectors to get a complete basis.
     
  7. Jul 27, 2016 #6
    That makes sense, thank you.
     
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