# Scalar Field Theory-Vacuum Expectation Value

1. Dec 24, 2011

### Mick83

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:

• ###### QFT_2.pdf
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2. Dec 24, 2011

### Thaakisfox

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...

3. Dec 24, 2011

### Mick83

So use $$k_{\mu}=\bigg(\mathbf{k},\frac{i\omega}{c}\bigg),~~ x_{\mu}=(\mathbf{x},ict),~~ \mathbf{k}\cdot\mathbf{x}=k_{\mu}x_{\mu}-\omega t=k_1x_1+k_2x_2+k_3x_3-\omega t =kr(\sin\theta \cos\phi +\sin\theta \sin\phi +\cos\theta)-\omega t$$? Regardless of the θ or ∅ dependence, $$\int^{\infty}_0 dr~r^2~ e^{r^2/2b}~e^{ikr}$$ gives a horrible result, even when squared.

4. Dec 25, 2011

### Thaakisfox

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:

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

The theta part can be integrated easily.

U will get the kind of integral wt you wrote:

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

now try bringing the integrand into the form:$r^2e^{-ax^2}$ i.e. complete the square on the exponentials. this kind of integral can be simply expressed with gamma functions.