- #1
Sekonda
- 201
- 0
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
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