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I How to prove Schrodinger's equation in momentum space?

  1. Mar 5, 2016 #1
    Specifically, i do not know hot to express the potential in momentum space. If someone would provide me with a link of source that has the proof in it, it would be appreciated.
     
  2. jcsd
  3. Mar 5, 2016 #2

    jtbell

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    Staff: Mentor

    Hint: find the position operator in momentum space, analogous to the momentum operator in position space.
     
  4. Mar 5, 2016 #3
    Hello and thanks for the reply. What about the potential? How will we convert that to momentum space.
     
  5. Mar 5, 2016 #4
  6. Mar 5, 2016 #5

    blue_leaf77

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    In momentum space, you cannot consider the potential in a separate mathematical expression with the state it acts on. In position space, you have ##V(x)\psi(x)##, then performing Fourier transform of this term will give you a convolution, ##\int_{-\infty}^\infty \tilde{V}(p-p')\tilde{\psi}(p')dp'## where ##\tilde{V}(p)## and ##\tilde{\psi}(p)## are the Fourier transforms of the potential and the state, respectively.
     
  7. Mar 5, 2016 #6

    jtbell

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    The potential is a function of position, right? Substitute the position operator for x. Consider how you do something similar for functions of momentum, e.g. p2, in position space.
     
  8. Mar 5, 2016 #7

    samalkhaiat

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    1) If you know Dirac’s notation
    Start with [tex]i\partial_{t}|\Psi (t) \rangle = H (X , P) | \Psi (t) \rangle = \frac{P^{2}}{2m} |\Psi (t)\rangle + V(X) |\Psi (t) \rangle ,[/tex] where [itex]X[/itex] and [itex]P[/itex] are operators but [itex]x[/itex] and [itex]p[/itex] will be used to mean ordinary numbers. In fact, we will work in 3D, so the operators [itex]X = (X_{1},X_{2},X_{3})[/itex], [itex]P = (P_{1},P_{2},P_{3})[/itex], while [itex]x = (x_{1},x_{2},x_{3})[/itex] and [itex]p=(p_{1},p_{2},p_{3})[/itex]. Multiply from the left by the bra [itex]\langle p |[/itex] and use [tex]\langle p | \frac{P^{2}}{2m} = \frac{p^{2}}{2m}\langle p | .[/tex]
    Introduce the wave function notation [itex]\langle p | \Psi (t) \rangle = \Psi (p ,t)[/itex]
    [tex]i\partial_{t}\Psi (p,t) = \frac{p^{2}}{2m}\Psi (p,t) + \langle p | V(X) | \Psi (t)\rangle .[/tex] Now, let us work on the potential term. Insert the completeness relation [itex]\int d^{3}x \ |x\rangle \langle x | = 1[/itex] between [itex]\langle p|[/itex] and [itex]V(X)[/itex], then use
    [tex]\langle x | V(X) | \Psi (t)\rangle = V(x) \langle x | \Psi (t)\rangle [/tex]
    and
    [tex]\langle p | x \rangle = e^{i x \cdot p } .[/tex]
    So the potential term becomes
    [tex]\langle p | V(X) | \Psi (t)\rangle = \int d^{3}x \ V(x) \ e^{i x \cdot p } \langle x | \Psi (t) \rangle .[/tex]
    Now, insert the completeness relation [itex]\int d^{3}\bar{p} \ | \bar{p}\rangle \langle \bar{p} | = 1[/itex] between [itex]\langle x |[/itex] and [itex]| \Psi (t)\rangle[/itex] and use
    [tex]\langle x | \bar{p}\rangle = e^{- i x \cdot \bar{p}}, \ \ \langle \bar{p}| \Psi (t)\rangle = \Psi ( \bar{p},t)[/tex]
    So, the potential term becomes
    [tex]\langle p | V(X) | \Psi (t)\rangle = \int d^{3}\bar{p} \left( \int d^{3}x \ V(x) \ e^{i x \cdot (p - \bar{p})} \right) \ \Psi (\bar{p},t) .[/tex]
    The integral in the bracket is just the Fourier transform of [itex]V(x)[/itex]
    [tex]V(p - \bar{p}) = \int d^{3}x \ V(x) \ e^{i x \cdot (p - \bar{p})} .[/tex]
    So, we rewrite the potential term as
    [tex]\langle p | V(X) | \Psi (t)\rangle = \int d^{3}\bar{p} \ V(p - \bar{p}) \Psi (p,t) ,[/tex]
    and the whole Schrodinger’s equation becomes just an ordinary integral equation
    [tex]i\partial_{t}\Psi (p,t) = \frac{p^{2}}{2m} \Psi(p,t) + \int d^{3}\bar{p} \ V(p - \bar{p}) \Psi (p,t) .[/tex]
    2) If you do not know the Dirac notation
    Start with
    [tex]i\frac{\partial}{\partial t}\Psi (x,t) = - \frac{1}{2m} \nabla^{2} \Psi (x,t) + V(x) \Psi (x,t) .[/tex]
    Multiply by [itex]\exp (i x \cdot p )[/itex], integrate over [itex]x[/itex] and use
    [tex] \Phi (p,t) = \int d^{3}x \ e^{i x \cdot p} \ \Psi (x,t) .[/tex]
    In the differential term on the right, integrate by parts twice and neglect surface term:
    [tex]\int d^{3}x \ e^{i x \cdot p} \ \nabla^{2}\Psi (x,t) = \int d^{3}x \ \Psi (x,t) \nabla^{2}\left( e^{i x \cdot p} \right) = - p^{2} \Phi (p,t) .[/tex]
    Okay, now for the potential term, use
    [tex]\Psi(x,t) = \int d^{3}y \ \Psi(y,t) \delta ( x - y) ,[/tex] and for the delta function, use the integral representation
    [tex]\delta (x-y) = \int d^{3}\bar{p} \ e^{- i (x-y) \cdot \bar{p}} .[/tex]
    So, after changing the order of integrations, the potential term can be written as
    [tex]\int d^{3}x \ V(x) \Psi(x,t) e^{i x \cdot p} = \int d^{3}\bar{p} \left( \int d^{3}x \ V(x) e^{i x \cdot (p - \bar{p}) } \right) \left( \int d^{3}y \ e^{ i y \cdot \bar{p}} \Psi(x,t) \right) .[/tex]
    Well, the integrals in the brackets on right hand of this are just the Fourier transforms of [itex]V(x)[/itex] and [itex]\Psi (x,t)[/itex]. So we can rewrite the potential term as
    [tex]\int d^{3}x \ V(x) \Psi(x,t) e^{i x \cdot p} = \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .[/tex]
    And, the whole equation becomes
    [tex]i \partial_{t} \Phi (p,t) = \frac{p^{2}}{2m} \Phi (p,t) + \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .[/tex]
     
    Last edited: Mar 5, 2016
  9. Mar 10, 2016 #8
    Thanks, but i should point a mistake. In the third line from the end, in the third bracket, it must be Ψ(y,t) rather than Ψ(x,t). Also, in the last line, it must be [tex]\Phi(\bar{p},t) [/tex]
    Oh, and thanks again for taking the time to derive it for me!
     
    Last edited: Mar 10, 2016
  10. Mar 10, 2016 #9

    samalkhaiat

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    Yes, clearly those were typos. Thanks, at least I know that you have read it all.
     
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