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I Notation in QM

  1. Jun 14, 2017 #1
    Hello! I have a proof in my QM book that: ##\left<r|e^{-iHt}|r'\right> = \sum_j e^{-iHt} u_j(r)u_j^*(r')##, where, for a wavefunction ##\psi(r,t)##, ##u_j## 's are the orthonormal eigenfunctions of the Hamiltonian and ##|r>## is the coordinate space representation of ##\psi##. I am not sure I understand this. Like in general ##\psi = \sum_j a_j u_j## and I don't know where the ##a_j##'s disappeared. Also I am not sure what is the relation between ##u_j(r)## and ##u_j(r')##. Is there any formula linking them? Thank you!
     
  2. jcsd
  3. Jun 14, 2017 #2

    kith

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    Science Advisor

    Have a look at the following and check if you understand it. Note what happens with the exponent - you got this wrong in your post.
    ##
    \begin{eqnarray*}
    \langle r|e^{-iHt}|r'\rangle &=& \langle r|e^{-iHt} \left( \sum_j |E_j\rangle \langle E_j| \right) |r'\rangle\\
    &=& \langle r| \sum_j e^{-iHt} |E_j\rangle \langle E_j |r'\rangle\\
    &=& \langle r| \sum_j e^{-iE_jt} |E_j\rangle \langle E_j |r'\rangle\\
    &=& \sum_j e^{-iE_jt} \langle r|E_j\rangle \langle E_j|r' \rangle\\
    &=& \sum_j e^{-iE_jt} u_j(r) u_j^{*}(r')
    \end{eqnarray*}
    ##
     
  4. Jun 14, 2017 #3
    Thank you for your reply. I understand the logic of it. However I am a bit confused. From what I see ##\|E_j>## is an eigenfunction of the hamiltonian, but ##u_j## is that, too. What is the difference between them?
     
  5. Jun 14, 2017 #4

    kith

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    [itex]|E_j \rangle[/itex] is an eigenstate of the Hamiltonian, i.e. it is a vector in Hilbert space.

    [itex]u_j(r) = \langle r | E_j \rangle[/itex] is a representation of the state [itex]|E_j \rangle[/itex] which uses the position basis. You can use many different bases to represent the same state vector. [itex]\tilde u_j(p) = \langle p | E_j \rangle[/itex] for example uses the momentum basis and refers to the same state [itex]|E_j \rangle[/itex].
     
    Last edited: Jun 15, 2017
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