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Zee QFT Derivation Question

  1. Jul 1, 2015 #1
    On page 12 of Zee's QFT in a nutshell, he rewrites an integral from dirac notation to a mixture of schrodinger/dirac

    My question is, since NumberedEquation2.gif , what happened to the integrals, and shouldn't there be four wave functions instead of two?

    I tried writing out the integrals explicitly, thinking maybe theyre integrated out with a delta function or something, but with no luck - I don't see how to get to psi(q_f).

    Any help would be appreciated.
     
    Last edited: Jul 1, 2015
  2. jcsd
  3. Jul 1, 2015 #2

    samalkhaiat

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    Let me rewrite the matrix element as [tex]\langle \Psi_{F}| e^{- i H T} | \Psi_{I} \rangle .[/tex] Now insert the completeness relations [tex]\int d X_{a} \ | X_{a} \rangle \langle X_{a} | = 1 , \ \ \mbox{for} \ \ a = I , F .[/tex] You find [tex] \langle \Psi_{F}| e^{- i H T} | \Psi_{I} \rangle = \int d X_{F} d X_{I} \langle \Psi_{F}| X_{F} \rangle \langle X_{F}| e^{ - i H T} | X_{I} \rangle \langle X_{I} | \Psi_{I} \rangle .[/tex] Okay, now use the definitions of the wavefunctions [tex]\Psi_{I} ( X_{I} ) = \langle X_{I} | \Psi_{I} \rangle , \ \ \Psi^{*}_{F}(X_{F}) = \langle \Psi_{F} | X_{F} \rangle .[/tex]
     
  4. Jul 1, 2015 #3
    I think I'm missing something extremely simple here... why is dotting ⟨X_I| into |psi_i⟩ the same as evaluating psi_I at X_I?
     
  5. Jul 2, 2015 #4

    samalkhaiat

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    And yes, you need to know the difference between state vector [itex]|\Psi \rangle[/itex] and its coordinate representation, i.e. wave-function [itex]\Psi ( x)[/itex]. Almost all textbooks explain the Bra-Ket notations and their connection to “wave-functions”.
    You seem to have no problem with [tex]\langle \Phi | \Psi \rangle = \int dx \ \Phi^{*} (x) \Psi (x) . \ \ \ \ (A)[/tex] Okay, let us work on the left-hand side by inserting the completeness relation [itex]1=\int dx |x\rangle \langle x|[/itex]:
    [tex]\langle \Phi | \Psi \rangle = \langle \Phi | ( \int dx |x \rangle \langle x | ) | \Psi \rangle = \int dx \ \langle \Phi | x \rangle \langle x | \Psi \rangle .[/tex]
    Now, if you compare the RHS of this equation with the RHS of Eq(A), what would you get?

    Similar state of affair exists in elementary vector algebra. You know how to expand a vector [itex]|\vec{V}\rangle[/itex] in terms of some orthogonal unit-vectors [itex]| e_{i}\rangle[/itex] [tex]|\vec{V}\rangle = \sum_{i} V_{i} \ | e_{i} \rangle .\ \ \ \ (1)[/tex] You also know the ortho-normality condition [tex]\langle e_{i} | e_{j} \rangle = \delta_{ij} . \ \ \ \ \ \ \ (2)[/tex] You should also know how to calculate the components of the vector from [tex]V_{i} = \langle e_{i} | \vec{V}\rangle . \ \ \ \ \ \ \ (3)[/tex] You can now substitute (3) in (1) to obtain another familiar equation in vector algebra [tex]| \vec{V} \rangle = \sum_{i} | e_{i}\rangle \langle e_{i} | \vec{V} \rangle . \ \ \ \ \ \ (4)[/tex] This leads to the completeness relation for the unit vectors [tex]\sum_{i} | e_{i} \rangle \langle e_{i}\rangle = 1 . \ \ \ \ \ \ (5)[/tex] And finally, you know the scalar product [tex]\langle \vec{V}| \vec{U} \rangle = \sum_{i} V^{t}_{i} U_{i} .\ \ \ \ \ (6)[/tex] Now, imagine the vector space to be an infinite-dimensional complex vector space spanned by un-countable infinity of ortho-normal functions (vectors) [itex]|e(x)\rangle \equiv | x \rangle[/itex], i.e. just pass to the continuous limits [tex]i \to x , \ \ \ | e_{i}\rangle \to | x \rangle , \ \ \sum_{i} \to \int dx ,[/tex] and [tex]V_{i} \to V(x), \ \ \ V^{t}_{i} \to V^{*}(x) .[/tex] So, we can translate Eq(1)-Eq(6) into the continuous language as follow [tex]\mbox{Eq(1)} \to \ \ \ | V \rangle = \int dx \ V(x) \ | x \rangle ,[/tex] [tex]\mbox{Eq(2)} \to \ \ \ \langle x | y \rangle = \delta (x - y) .[/tex] [tex]\mbox{Eq(3)} \to \ \ \ V(x) = \langle x | V \rangle ,[/tex] which is what you were asking about [itex]\Psi (x) = \langle x | \Psi \rangle[/itex]. [tex]\mbox{Eq(4)} \to \ \ \ | V \rangle = \int dx \ | x \rangle \langle x | V \rangle .[/tex] [tex]\mbox{Eq(5)} \to \ \ \ \int dx \ | x \rangle \langle x | = 1 .[/tex] And finally the equation you know [tex]\mbox{Eq(6)} \to \ \ \langle V | U \rangle = \int dx \ V^{*}(x) U(x) .[/tex]
     
    Last edited by a moderator: Jul 2, 2015
  6. Jul 2, 2015 #5
    Thank you for your help, I understand now.
     
    Last edited by a moderator: Jul 2, 2015
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