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Calculation of probabilities in QM

  1. Sep 21, 2009 #1
    I posted this question in the homework page but I got no answers, so will try here.
    Given the wave function of a (spinless) particle I need to expres in terms of [itex]\psi(\vec{r})[/itex] the probability for simultaneous measurements of X y P_z to yield :

    [tex]x_1 \leq x \leq x_2[/tex]
    [tex] p_z \geq 0[/tex]

    I got the result:
    [tex]\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx \int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z <\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}> [/tex]

    To get this I evaluated the expression [itex]<\psi|P_2P_1|\psi>[/itex] where P_1 and P_2 are the proyectors:
    [tex]P_1=\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx|x,y,z><x,y,z|[/tex]
    [tex]P_2=\int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z|p_x,p_y,p_z><p_x,p_y,p_z|[/tex]

    I need to know two things: 1) is my result correct? 2) in case it is correct, is there any other more simple or concrete answer?
     
  2. jcsd
  3. Sep 21, 2009 #2

    Avodyne

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    1) Yes.
    2) Yes.
     
  4. Sep 21, 2009 #3
    Thank you Avodyne, could you be so kind as to explain a little how to get an easier or more concrete answer?
     
  5. Sep 21, 2009 #4

    Avodyne

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    Well, notice that if it wasn't for the pz question, you could do everything with the position-space wave function. So you really only need to Fourier transform on z. Your version has a lot of extra Fourier-transforming and untransforming in x and y.
     
  6. Sep 21, 2009 #5
    You can go further by writing <p|r> and <Psi|p> in a function form like you did with <x|Psi>. Generally, you want to put things in a single representation, like the position space, so you might ask yourself how far you can go to do that.

    If you have a free particle, e.g., then Psi(r)=Psi_x Psi_y Psi_z, and similarly for the momentum space wavefunction. Can you see how doing the above steps would make things much more simplified?
     
  7. Sep 21, 2009 #6
    Thank you both. I think what javier says is simple to write explictly the expresions for <p|r> and <psi|p>
    in space representation but that would only complicate even more the final expression(sorry if I misunderstood you)
    So will try Avodyne's approach and I will very much appretiate any comments on this :
    Since we need to express the final result in terms of [itex]\psi(\vec{r})[/itex] only, we will change momentarily from X,Y,Z to the C.S.C.O X,Y,P_z. The desired probabilty is [itex]<\psi|P_n|\psi>[/itex]

    where
    [tex]P_n=\int_{-\infty}^\infty dy\int_{x_1}^{x_2}dx\int_0^\infty dp_z |x,y,p_z><x,y,p_z|[/tex]
    [tex]<\psi|P_n|\psi>=\int_{-\infty}^\infty dy \int_{x_1}^{x_2}dx \int_0^\infty dp_z |<x,y,p_z|\psi>|^2[/tex]
    where
    [tex]<x,y,p_z|\psi>=\int<x,y,p_z|\vec{r'}><\vec{r'}|\psi>d^3r'=\int<x,y,p_z|x',y',z'><x',y',z'|\psi>d^3r' [/tex]
    [tex] =\int\delta(x-x')\delta(y-y')<p_z|z'>\psi(\vec{r'})d^3r'=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty dz' \psi(x,y,z')e^{-ip_z z'/\hbar}[/tex]
    is this what you meant Avodyne?
     
  8. Sep 21, 2009 #7

    Avodyne

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  9. Sep 24, 2009 #8
    Thank you Avodyne! I really appreate your help because a have no other means to deal with this matters.
     
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