Time Dependent Perturbation Theory - Klein Gordon Equation

1. Oct 30, 2012

Sekonda

Hey,

I'm struggling to understand a number of things to do with this derivation of the scattering amplitude using time dependent perturbation theory for spinless particles.

We assume we have some perturbation 'V' such that :

$$\left ( \frac{\partial^2 }{\partial t^2}-\triangledown ^2 + m^2 \right )\psi = \delta V\psi$$

We also assume plane wave solutions of the wavefunction such that the input wavefunction is:

$$\psi _{in}=\psi _{i}(x)e^{-iE_{i}t}$$

A single eigenwavefunction of the wavefunction psi. This input interacts and we get an output wavefunction which can be expanded like so:

$$\psi _{out}=\sum_{n}a_{n}(t)\psi _{n}(x)e^{-iE_{n}t}$$

We substitute this output wavefunction into the perturbed Klein Gordon equation above and attain, (by assuming the second derivative of a(t) with respects to time is small):

$$\frac{\partial^2 }{\partial t^2}\sum_{n}a_{n}(t)\psi _{n}(x)e^{-iE_{n}t}=\delta V\sum_{n}a_{n}(t)\psi _{n}e^{-iE_{n}t}$$

Upon assuming the second derivative of a(t) is small we obtain the simplified equation:

$$-2i\sum_{n}\dot{a}_{n}(t)\psi _{n}(x)e^{-iE_{n}t}=\delta V\sum_{n}a_{n}(t)\psi _{n}e^{-iE_{n}t}$$

Though I'm not exactly sure why all these terms cancel... Nonetheless, to specify a value within the sum we use the orthogonality of wavefunctions - we want to attain the 'final' wavefunction and amplitude (denoted by subscript 'f') and so we multiply both sides of the above equation by:

$$\int_{-\infty }^{\infty }d^{3}x\psi_{f}^{*}$$

We then attain this equation upon use of orthogonality:

$$-2iE_{f}\dot{a}_{f}e^{-iE_{f}t}=\int_{-\infty }^{\infty }d^{3}x\psi _{f}^{*}\delta V\sum_{n}a_{n}(t)\psi _{n}e^{-iE_{n}t}$$

We then simplify by saying at t=0, all a(t)=0 apart from the initial a(0)=1 (so essentially we have one eigenwavefunction coming in) - this holds true for small 't'. The equation then becomes:

$$-2iE_{f}\dot{a}_{f}e^{-iE_{f}t}=\int_{-\infty }^{\infty }d^{3}x\psi _{f}^{*}\delta V \psi _{i}e^{-iE_{i}t}$$

to

$$\dot{a}_{f}(t)=\frac{i}{2E_{f}}\int_{-\infty }^{\infty }d^{3}x\psi _{f}^{*}\delta V \psi _{i}$$

and finally attaining solution:

$$a_{f}(t)=\frac{i}{2E_{f}}\int_{-\infty }^{\infty }d^{4}x\psi _{f}^{*}\delta V \psi _{i}$$

(the d4x including the time differential)

Now I'm unsure of a number of things including the output wavefunction form - I think it's just a sum of wavefunctions related to the input but the input is just a single wavefunction?

I'm unsure on why terms cancel in the assumption that the second derivative of 'a' with respects to time is small, though I will try doing the differentiation now and see if I can do it.

Basically, I'd be grateful if someone could check that this derivation follows through and if someone could explain why the assumptions have been made that'd be great.

Thanks guys,
SK

2. Oct 31, 2012

Sekonda

The jump from the 4th to the 5th equation is confusing me, upon applying the time derivative and laplacian operator we attain a number of expressions that disappear but I'm not sure why. Can someone explain why these terms cancel or =0?

Thanks

3. Nov 1, 2012

andrien

They just disappear because apart from an,the Ïˆ satisfies the homogeneous part,which will be equal to zero in absence of any potential.

4. Nov 5, 2012

Sekonda

So am i correct in thinking that the perturbation is instantaneous and so the outgoing wavefunction can be treated as a free particle and so solves the free Klein-Gordon equation?

Thanks

5. Nov 6, 2012

andrien

of course,that is the lowest order approximation to treat the outgoing wavefunction as a free particle,that is what is done in general theory.But I do think that perturbation must be treated adiabatic in character.

6. Nov 6, 2012

Sekonda

Thanks, that's essentially what my professor said today - I was incorrect in describing the perturbation as instantaneous.

Cheers!

7. Jul 28, 2013

plasmon

Dear i wish to know what is the validity of assumption that the second order derivative of a(t) is neglected. Kindly clarify the issue.