1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculation of probabilities in QM

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\psi(x,y,z)=\psi(\vec{r})[/tex] be the normalized wave function of a particle.Express in terms of [itex]\psi(\vec{r})[/itex] the probability for a simultaneous measurements o X y P_z to yield :
    [tex]x_1 \leq x \leq x_2[/tex]
    [tex] p_z \geq 0[/tex]

    2. Relevant equations
    [tex]<\vec{p}|\vec{r}>=\frac{1}{(2\pi\hbar)^{3/2}}e^{-i\vec{p}.\vec{r}/\hbar} [/tex]
    [tex]<\vec{p}|\psi>=\frac{1}{(2\pi\hbar)^{3/2}}\int \psi(\vec{r}) e^{-i\vec{p}.\vec{r}/\hbar} dr^3[/tex]

    3. The attempt at a solution
    I have reached the following 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]
    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?
    Last edited: Sep 20, 2009
  2. jcsd
  3. Sep 20, 2009 #2
    Sorry, I am not answering your question. But could you explain how you arrived at the expression [tex] <\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}>
  4. Sep 20, 2009 #3
    I evalueted de expression [itex]<\psi|P_2P_1|\psi>[/itex] where P_1 and P_2 are the proyectors:
    But I'm sure whether what I'm doing is correct
    Last edited: Sep 20, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook