# A QFT and transitions between momentum states

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1. Jul 8, 2016

### hilbert2

Hi,

I'm trying to learn some QFT at the moment, and I'm trying to understand how interactions/nonlinearities are handled with perturbation theory. I started by constructing a classical mechanical analogue, where I have a set of three coupled oscillators with a small nonlinearity added. The eqns. of motion of that system are:

$m\frac{d^2 x_1(t)}{dt^2}=-kx_1(t)+k'(x_2(t)-x_1(t))-\lambda [x_1(t)]^3$
$m\frac{d^2 x_2(t)}{dt^2}=-kx_2(t)+k'(x_1(t)-x_2(t))+k'(x_3(t)-x_2(t))-\lambda [x_2(t)]^3$
$m\frac{d^2 x_3(t)}{dt^2}=-kx_3(t)+k'(x_2(t)-x_3(t))-\lambda [x_3(t)]^3$

Now, if $\lambda = 0$, this system is easy to solve by finding out the linear combinations of $x_1,x_2,x_3$ that are the normal modes of this system and behave like independent oscillators.

In the perturbed case, where $\lambda > 0$, I can still plot how the system oscillates in the normal modes of the unperturbed system, but now because of the nonlinear term, energy (kinetic+potential) is transferred between those normal modes (some transitions are symmetry-disallowed, though).

The QFT equivalent of this would be a one-dimensional Klein-Gordan field with the phi-fourth perturbation, which is obviously a system of infinite number of oscillators (normal modes of the KG field), unless I arbitrarily assign a cut-off at some value of momentum and discretize the k-space.

Now, suppose I have an initial state that describes a particle with definite value of momentum:

$\left|p\right>=a_p^\dagger\left|0\right>$

Now I'd like to calculate the temporary rate at which there happen transitions to states with some other values of momentum. From the Schrödinger equation, I have:

$\frac{\partial\left|\psi (t)\right>}{\partial t} = -iH\left|\psi (t)\right>$ ,

where H is the hamiltonian operator. The initial transition amplitude from momentum state $p_1$ to momentum state $p_2$ is:

$-i< 0 | a_{p_2} H a_{p_1}^{\dagger} |0>$ .

Now, the question I run into, is how do I simplify the term $-i< 0 | a_{p_2} [\hat{\psi} (x)]^4 a_{p_1}^{\dagger} |0>$, where the psi-hat is the field operator? I should somehow be able to write that fourth power of $\hat{\psi}(x)$ in terms of the creation and annihilation operators. All textbooks I've read seem to do these things in the Heisenberg or interaction pictures, but I'd personally understand this easier if I saw it done in the Schrödinger picture.

EDIT: I understand that a particle of one momentum can't simply turn into a particle with some other momentum, because momentum must be conserved... It seems, however, that an excitation with some value of p could be converted to two excitations with different momenta.

Last edited: Jul 8, 2016
2. Jul 9, 2016

### hilbert2

If I write the phi-fourth interaction term with creation and annihilation operators, I get a sum like this:

$e^{i(p_{1}+p_{2}+p_{3}+p_{4})x}a_{p_1} a_{p_2} a_{p_3} a_{p_4} +e^{i(p_{1}+p_{2}+p_{3}-p_{4})x}a_{p_1} a_{p_2} a_{p_3} a_{p_3}^\dagger +e^{i(p_{1}+p_{2}-p_{3}+p_{4})x}a_{p_1} a_{p_2} a_{p_3}^\dagger a_{p_4}$

$+e^{i(p_{1}+p_{2}-p_{3}-p_{4})x}a_{p_1} a_{p_2} a_{p_3}^\dagger a_{p_3}^\dagger +e^{i(p_{1}-p_{2}+p_{3}+p_{4})x}a_{p_1} a_{p_2}^\dagger a_{p_3} a_{p_4} +e^{i(p_{1}-p_{2}+p_{3}-p_{4})x}a_{p_1} a_{p_2}^\dagger a_{p_3} a_{p_3}^\dagger$

$+e^{i(p_{1}-p_{2}-p_{3}+p_{4})x}a_{p_1} a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4} +e^{i(p_{1}-p_{2}-p_{3}-p_{4})x}a_{p_1} a_{p_2}^\dagger a_{p_3}^\dagger a_{p_3}^\dagger+e^{i(-p_{1}+p_{2}+p_{3}+p_{4})x}a_{p_1}^\dagger a_{p_2} a_{p_3} a_{p_4}$
$+e^{i(-p_{1}+p_{2}+p_{3}-p_{4})x}a_{p_1}^\dagger a_{p_2} a_{p_3} a_{p_3}^\dagger +e^{i(-p_{1}+p_{2}-p_{3}+p_{4})x}a_{p_1}^\dagger a_{p_2} a_{p_3}^\dagger a_{p_4}$

$+e^{i(-p_{1}+p_{2}-p_{3}-p_{4})x}a_{p_1}^\dagger a_{p_2} a_{p_3}^\dagger a_{p_3}^\dagger +e^{i(-p_{1}-p_{2}+p_{3}+p_{4})x}a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3} a_{p_4}$

$+e^{i(-p_{1}-p_{2}+p_{3}-p_{4})x}a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3} a_{p_3}^\dagger +e^{i(-p_{1}-p_{2}-p_{3}+p_{4})x}a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4} +e^{i(-p_{1}-p_{2}-p_{3}-p_{4})x}a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_3}^\dagger$

and it is multiplied with a factor that is proportional to $\frac{1}{\sqrt{E_1 E_2 E_3 E_4}}$ and integrated over all possible values of $x$ and the momenta $p_1 , p_2 , p_3 , p_4$. The exponential factors in the sum above are probably meant to create a momentum-conserving delta function, if I've understood correctly(?)

If I'm trying to calculate probabilities for 2->2 processes where particles with momenta $p_1 , p_2$ collide and become particles with momenta $p_3, p_4$, am I supposed to ignore all terms that don't contain an equal number of a and a-dagger operators?