# Use standing wave to understand mode locked laser

Recently when I was trying to explain the principle of kerr-lens mode-locked laser, however I find that I don't even understand it myself.

People around me always say the longtitude modes of the cavity can be understood as standing wave, and when the amplified modes have "fixed" phase relationship the laser is mode locked.

In reality when I add different standing wave together mathematically I get two pulses propogate at opposite direction in the cavity. It makes sense since the standing wave is the superposition of two counter-propogating waves. If it's true then the repetition rate of the laser is $$f=c/L$$ if c is the speed of light and L is the cavity length, instead of $$f=c/2L$$.

I don't know which part of my understanding is wrong even this sounds ridiculous.

Now I begin to doubt my understanding of the term "standing wave". I am thinking if it means the averaged effect over long time compared with the traveling time used in the cavity. If you track the wave which is shorter than cavity length at light speed, you don't see standing wave at this high time resolution.

Update

Here is the small script I use to add up 100 different standing waves:

import numpy as np
from matplotlib import pyplot as plt
from scipy.constants import speed_of_light as lc
print(lc)

L=5000*1e-9
x=np.arange(0,L,L/300000)

omega=lambda n: 2*np.pi*lc/L*n
wave = lambda n,x,t: np.sin(2*np.pi*x/L*n)*np.cos(omega(n)*t)#assume same phase at t=0
ax = plt.subplot(111)

t = lambda n: 2*np.pi/omega(n)
stack = lambda n,t: sum([wave(i,x,t) for i in range(n-50,n+50)],0)

ax.plot(x,stack(10000,t(10000)*3000),linewidth=3.0)

ax.spines['right'].set_visible(False)
ax.spines['top'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['bottom'].set_visible(False)
#plt.xlabel('t/fs')
plt.yticks([])
plt.xticks([])
plt.savefig('pulse',transparent = True)
plt.show()


The output is:

if you change the time then you can see these two pulses travel to different direction.

You can find the cavity modes $$\nu_n$$ by using the stationarity of the electric field.

Suppose $$E_0$$ is your field at $$t=0$$, then when it returns to its starting point, it has travelled a distance $$2L$$. The stationarity states that :

$$$$E_0=E_0e^{jk2L} \rightarrow k2L = 2n\pi$$$$

Using $$k=\frac{2\pi\nu}{c}$$, you finally get the longitude modes of the cavity $$\nu_n=\frac{nc}{2L}$$.

Concerning Kerr-lens mode-locking, I just know it's about non-linear effects in your cristal so the index of refraction depends on the strength of the incoming electric field. The index will be greater at the center so it's gonna act as a lens to focalise light so you have less losses.

Edit : (2 years later...)

The cavity modes i.e. the frequencies that can oscilate in the cavity are $$\nu_n=\frac{nc}{2L}$$. The total field in the cavity is the sum of all the fields of frequencies $$\nu_{n}$$ i.e. $$$$E(t)=\sum_n E_n\exp{[-i(\omega_n t+\phi_n)]}$$$$ where $$\omega_n=\omega_0 +\omega_{n}$$, with $$\omega_n=2\pi \nu_n$$ and $$\phi_{n}$$ the phase associated to the mode $$n$$.

The idea of mode locking is to put in phase every fields i.e. $$\phi_{n}=\phi_{0}$$ (phase matching). If we suppose that $$E_{n}=E_0$$, then we can rewrite the sum like $$$$E(t)=E_0e^{-i(\phi_0+\omega_0 t)}\sum_n \exp{(-i\omega_n t)}$$$$

The intensity, $$I(t) \propto E(t)E^*(t)=E_0^2\frac{sin^2 (N\omega t/2)}{sin^2(\omega t /2)}$$ where $$\omega=\pi c/L$$ and $$N$$ the number of standing waves considered.

The time $$\tau$$ between two pulses i.e. two maximums of the intenity is for $$\omega \tau/2=\pi$$, so $$\tau=\frac{2\pi}{\omega}=\frac{1}{\nu}$$. So the repetition rate $$f=1/\tau$$ is equal to the frequency of the first resonant mode $$\frac{c}{2L}$$.

• Thanks for the answer. I update the question and I am confused by the two pulses instead of one if we assume different modes as standing wave. Commented Apr 25, 2022 at 11:24