1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    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!

Scalar Field Theory-Vacuum Expectation Value

  1. Dec 24, 2011 #1
    1. The problem statement, all variables and given/known data
    I am given an equation for a quantized, neutral scalar field expanded in creation and destruction operators, and need to find the vacuum expectation value of a defined average field operator, squared. See attached pdf.


    2. Relevant equations
    Everything is attached, but I can include more.


    3. The attempt at a solution
    I've solved part (a) (I think- if someone could check my work it would be appreciated), so I need some idea on how to do part (b). The question clearly asks for the expectation of the squared value, but I don't have a clue how to solve the integral in the first place, which must then be squared to find the expectation value.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 24, 2011 #2
    switch to spherical coordinates... put your z axis toward the direction of k i.e. measure your \theta angle from k. dont forget to use the correct integral measure...
     
  4. Dec 24, 2011 #3
    So use [tex] k_{\mu}=\bigg(\mathbf{k},\frac{i\omega}{c}\bigg),~~ x_{\mu}=(\mathbf{x},ict),~~
    \mathbf{k}\cdot\mathbf{x}=k_{\mu}x_{\mu}-\omega t=k_1x_1+k_2x_2+k_3x_3-\omega t
    =kr(\sin\theta \cos\phi +\sin\theta \sin\phi +\cos\theta)-\omega t [/tex]? Regardless of the θ or ∅ dependence, [tex]\int^{\infty}_0 dr~r^2~ e^{r^2/2b}~e^{ikr}[/tex] gives a horrible result, even when squared.
     
  5. Dec 25, 2011 #4
    nono that is not correct.
    The kx in the power of the exponential is already only the 3-part of them. the expression is not in a "covariant" form, but the usual separate spatial and time parts.

    So If you put your z axis towards the direction of k, and measure the angle \theta from there your integral will be:

    [tex]\int_0^{2\pi}d\phi\int_0^{\infty}dr \, r^2 e^{-r^2/2b^2} \int_{-1}^{1}d\cos{\theta} e^{ikr\cos\theta}[/tex]


    The theta part can be integrated easily.

    U will get the kind of integral wt you wrote:

    [tex]\int_0^{\infty}dr\, r^2 e^{-r^2/2b^2}e^{\pm ikr}[/tex]

    now try bringing the integrand into the form:[itex]r^2e^{-ax^2} [/itex] i.e. complete the square on the exponentials. this kind of integral can be simply expressed with gamma functions.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook