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Braket notation help

  1. Apr 14, 2012 #1
    If x,y,z are the position operators.

    Is it true that:

    <φ|x|φ> + <φ|y|φ> + <φ|z|φ> = <φ | x+y+z| φ> ?

    So that if, for example, one wanted to compute <φ|r|φ> (where r =x+y+z), then they would just have to sum the parts.

    I know that for scalars, a and b, we have the following:

    (a+b)|φ> = a|φ> + b|φ>
    But I don't know for sure if this is related at all to the case for operators (especially when they are sandwiched between the bra and the ket.
     
  2. jcsd
  3. Apr 14, 2012 #2
    Yes, it's true that [itex]\langle \psi | A + B | \psi \rangle = \langle \psi | A | \psi \rangle + \langle \psi | B | \psi \rangle[/itex]. But I'm not sure where you're going with the substitution [itex]r = x + y + z[/itex]. If you're trying to convert to spherical coordinates, that's not right--you need to do [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex] instead.
     
  4. Apr 14, 2012 #3
    I am given the set of overlaps <φ_i|φ_j> and <φ_i|x|φ_j> for all i and j (as well as y and z sets). So that's four sets (the overlaps, and the set with x y and z).

    So if I'm given those sets, how do I combine the sets {<φ_i|x|φ_j>, <φ_i|y|φ_j>, <φ_i|z|φ_j>} to get the single set {<φ_i|r|φ_j>}
     
  5. Apr 14, 2012 #4
    And the definition of [itex]r[/itex] is [itex]x + y + z[/itex]? Or something else?
     
  6. Apr 14, 2012 #5
    Could the following be a solution?


    <φ|r|φ> = <φ|x|φ>i-hat + <φ|y|φ>j-hat + <φ|z|φ>k-hat

    where i-hat is the unit vector of the x-axis, j-hat is the unit vector of the y-axis, and k-hat is the unit vector of the z-axis.
     
  7. Apr 14, 2012 #6
    I double checked my notes:

    r = x*i-hat +y*j-hat + z*k-hat *(by definition)
     
  8. Apr 14, 2012 #7
    Ah, I see, they're defining [itex]r[/itex] as a vector-valued operator. In that case, there's actually no new math here, it's just a notation trick. Specifically, it's a way of unifiying three separate operators together into one object that's easy to keep track of. So we have three operators [itex]\hat{x}, \hat{y}, \hat{z}[/itex], and we're defining the new object [itex]\hat{\textbf{r}} = (\hat{x}, \hat{y}, \hat{z})[/itex].

    In this case, applying the operator to a state is really applying three separate operators to the state, and using them to make an ordered triple. [itex]\langle \psi | \hat{\textbf{r}} | \psi \rangle = \langle \psi | (\hat{x}, \hat{y}, \hat{z}) | \psi \rangle = (\langle \psi | \hat{x} | \psi \rangle, \langle \psi | \hat{y} | \psi \rangle, \langle \psi | \hat{z} | \psi \rangle)[/itex].
     
  9. Apr 14, 2012 #8
    Final question:

    What does it mean to do the following:

    (<φ|r|φ> - <ψ|r|ψ>)^2

    Obviously if <ψ|r|ψ> is a vector, then the difference of two vectors is straightforward. What does it mean to take their square?
     
  10. Apr 14, 2012 #9
    Well, think about what it means to take the square of a normal vector. Since [itex]x^2 = x\cdot x[/itex] (where [itex]x[/itex] is a normal number), then the square of a vector must be [itex]\textbf{r}^2 = \textbf{r} \cdot \textbf{r}[/itex]. By the definition of the dot product, [itex]\textbf{r} \cdot \textbf{r} = r_x^2 + r_y^2 + r_z^2[/itex]. The generalization to the quantum case should be pretty straightforward.
     
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