Green's Function for a harmonic oscillator

[TheBigDig]

Homework Statement

Express the equation of motion for a classical harmonic oscillator in 1-dimension in the general form above. Indicate how the harmonic oscillator can be treated by the Green function method. Obtain the corresponding Green function (in frequency domain) for this case

Relevant Equations

$$\ddot{x}+\omega_0 ^2 x = A(t)$$
$$x(t)=\int dt' g(t-t')A(t')$$
$$\ddot{g}+\omega_0 ^2 g = \delta(t-t')$$
$$G(\omega) = \frac{1}{\omega_0 ^2-\omega^2}$$
$$g(t-t') = \frac{1}{2\pi}\int d\omega e^{-i\omega(t-t')}G(\omega) = -\frac{1}{2\pi} \int d\omega \frac{e^{-i\omega(t-t')}}{\omega^2-\omega_0^2}$$
$$\oint dz f(z) = 2\pi i R$$

I know that due to causality g(t-t')=0 for t<t' and I also know that for t>t', we should get
g(t-t&#039;)=\frac{sin(\omega_0(t-t&#039;))}{\omega_0}
But I can't seem to get that to work out.
Using the Cauchy integral formula above, I take one pole at -w_0 and get
\frac{ie^{i\omega_0(t-t&#039;)}}{2\omega_0}
and the other pole at +w_0
\frac{-ie^{i\omega_0(t-t&#039;)}}{2\omega_0}
Summing these together I get
g(t-t&#039;) = \frac{i(e^{i\omega_0(t-t&#039;)}-e^{-i\omega_0(t-t&#039;)})}{2\omega_0}
I'm sure I'm just missing a factor of -1 or something that could get this in the correct form but I've been through it twice and can't find my slip up

 

[vela]

Perhaps check whether you should be closing the contour in the upper-half plane or lower-half plane.

 

[TheBigDig]

How does closing the contour in the upper half differ from closing in the lower half? Apologies, I'm a physicist and have had a crash course in complex analysis so this isn't really my strong suit

 

[vela]

Given the sign of $t-t'$, the integral will converge in one of the half planes and diverge in the other, so you have to choose the one in which it will converge. A simple way to see which one is correct is to consider $e^{-iz(t-t')}$ with $z=iy$. Do you need $y>0$ or $y<0$ as $y \to \pm\infty$ to get the exponential to go to 0? Depending on which contour you end up using, you'll either enclose the poles with a contour that goes clockwise or counterclockwise.

 

[Cryo]

The trick I like to use it to introduce some loss into the system, and then take it to zero. The trouble with the poles arises because your equations are symmetric with respect to time reversal, but to get a causal solution you have to select the direction of time flow (to get cause and effect). One intuitive way is to add loss.

So let your equation be

$\ddot{x}+\gamma \dot{x} + \omega_0^2 x = A(t)$

Clearly taking the limit $\gamma\to 0$ corresponds to what you want. In the abscence of driver, $A=0$, and in the limit of very small loss $\left|\gamma\right|\ll \omega_0$, the solutions are:

$x_{hom}\approx \exp\left(-\gamma t/2\right)\left(const \cdot \cos\omega_0 t + const\cdot\sin\omega_0 t\right)$

So if you want your system to loose energy as the time goes forward, $\gamma\ge 0$. Now do your contour integration, keep using the fact that $\gamma \ll \omega_0$ to keep expressions simple. You will find that this $\gamma$ displaces your poles off the real axis, which will allow you to carry out the integration without any problems. Finally you will be able to set $\gamma\to 0^+$ to arrive at the end result.

In my notation (assuming that $G(t-t')=\frac{1}{2\pi}\int d\omega \exp\left(-i\omega\left(t-t'\right)\right)\tilde{G}\left(\omega\right)$)

$\tilde{G}\left(\omega\right)\approx\frac{1}{\left(\left(\omega_0 + i\gamma/2\right)-\omega\right)\left(\left(\omega_0-i\gamma/2\right) +\omega\right)}$

So you will have poles at $\omega=\pm \omega_0 + i\gamma/2$, so your contour will be enclosing the poles counter-clock-wise.

 

[Orodruin]

The trick I like to use it to introduce some loss into the system, and then take it to zero.

You do not need to do this if you accept that the solution is to be seen as a distribution rather than a normal function and don't insist on having to compute the Green's function by Fourier transform. You can just assert $G(t-t_0) = \theta(t-t_0) g(t)$, where $\theta$ is the Heaviside function. Insertion into the differential equation for the Green's function directly gives $g(t)$ as a linear combination of a sine and cosine as well as the initial conditions at $t = t_0$ to fix the constants.

If one insists on using Fourier transforms, one needs to take great care in terms of which way you go around the poles because the poles are on the real axis. This prescription (together with the different ways of closing the contours at infinity as mentioned in #2) will lead you to either the retarded Green's function, the advanced Green's function, or (if passing the poles on opposite sides) the Feynman Green's function.

 

[Cryo]

You do not need to do this if you accept that the solution is to be seen as a distribution rather than a normal function and don't insist on having to compute the Green's function by Fourier transform.

I agree, but the OP was asking about contours. Also, proper handling of generalized functions requires some familiarity with them which is not always included in physics courses, the contour integration methods on the other hand are bread-and-butter.

 

[TheBigDig]

Given the sign of $t-t'$, the integral will converge in one of the half planes and diverge in the other, so you have to choose the one in which it will converge. A simple way to see which one is correct is to consider $e^{-iz(t-t')}$ with $z=iy$. Do you need $y>0$ or $y<0$ as $y \to \pm\infty$ to get the exponential to go to 0? Depending on which contour you end up using, you'll either enclose the poles with a contour that goes clockwise or counterclockwise.

I see. So I'm taking the lower half contour which should account for a missing factor of -1 in my solution.

 

[Cryo]

poles are on the real axis

Hence, why I like my trick :-)

 

[Orodruin]

Also, proper handling of generalized functions requires some familiarity with them which is not always included in physics courses

Proper handling in mathematical terms, yes. However, the only property you really need to argue for is $\theta’ = \delta$. If you are going to talk about Green’s functions you need at least some familiarity with $\delta$ and it is not difficult to argue for this being reasonable. (Technically you also need $\delta’$, but you can argue that it is a different beast altogether and all you really need is that.) That amount of rigor is about the same as you would get from a ”physics” FT course.

I see. So I'm taking the lower half contour which should account for a missing factor of -1 in my solution.

Note that you also need to worry about the pole prescription!