Regular interference pattern can be observed if the two wave sources are coherent.
Two sources are coherent if they have the same frequency and constant phase difference.
Could someone please explain to me what constant phase difference implies?
[Physics] Coherent sources for interference
coherenceinterferencewaves
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I remember working this out in the opposite direction, hoping to get a paradox: that the amplitude of the sound produced by a choir of 100 singers is only 10 times the amplitude of the sounds produced by one of them.
Each individual sound can be represented as a vector in 2-dimensional space, with length 1 (since all are equally loud) and random phase. The resultant sound heard will be the sum of all those vectors.
Adding the vectors one at a time is thus effectively undertaking a random walk in 2 dimensions. The following statements are therefore equivalent:
The expected distance travelled in an $n$-step random walk is $\sqrt{n}$.
The expected amplitude of $n$ coherent equally loud sound sources is $\sqrt{n}$ times the amplitude of one of them.
It is entirely up to you to decide which one of them you prove from which, but in this case you can assume the truth of 1 (it is a standard result) and deduce the truth of 2, which, squared, is the result you asked for in your question.
(And my paradox wasn't a paradox because I had forgotten that loudness is proportional to the square of amplitude, so 100 choristers are 100 times as loud as one of them).
Incoherent sound sources yield essentially the same answer because you can think of them as coherent sources with varying phase - so again you end up with a total of random phases.
Coherent states are eigenvectors for the (bosonic) annihilator,$$\hat a ~|\alpha\rangle = \alpha~|\alpha\rangle,$$and if we define the position and momentum quadratures as $\hat x = \hat a^\dagger + \hat a,$ $\hat p = i \hat a^\dagger - i \hat a,$ we have $[\hat x, \hat p] = 2i$ and the dimensionless Hamiltonian $\hbar\omega ~ \hat a^\dagger \hat a = \frac12 \hbar\omega~x^2 + \frac12 \hbar\omega~p^2 + \text{const.}$ to guide us. We can immediately see that in the coherent state we have $\langle x \rangle = \alpha^* + \alpha = 2 ~\Re~{\alpha}$ whereas $\langle p \rangle = i~\alpha^* - i ~ \alpha = 2 ~\Im~\alpha, $ so the position and momentum plane is basically just the complex plane $\mathbb C$ that $\alpha$ lives on.
Now this Hamiltonian of course has an eigenbasis $\hat a^\dagger \hat a ~ |n\rangle = n ~|n\rangle$ and in terms of that basis we see a recurrence that if $|\alpha\rangle = \sum_n c_n |n\rangle$ then we can work out that $\alpha |\alpha\rangle = \hat a |\alpha\rangle$ implies $$c_n \sqrt{n} = \alpha c_{n-1},\text { so } c_n = c_0 \frac{\alpha^n}{\sqrt{n!}}.$$Then working out $1 = \langle \alpha|\alpha\rangle = c_0\sum_n \big(|\alpha|^2\big)^n/n! = c_0 \exp\big(|\alpha|^2\big)$ gives the normalization constant $c_0.$
However note that under this Hamiltonian, $|n\rangle \mapsto e^{-in\omega t} |n\rangle$ and therefore, $$|\alpha\rangle \mapsto \exp\left(-|\alpha|^2\right) \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} ~ e^{-i~\omega t~n} |n\rangle,$$which we see on the right hand side combines by normal exponent rules to form $(\alpha e^{-i\omega t})^n.$ In other words the time evolution is that $|\alpha(t)\rangle = |\alpha_0 e^{-i\omega t}\rangle,$ and our coherent state simply makes a circle on the complex plane as it evolves.
It is in this precise sense that I understand the word "coherent," it is the meaning "it stays perfectly together as it goes along its journey." It's the same way I would say "lasers are a coherent phenomenon; light by its nature wants to spread out in a $1/r^2$ law but in a laser, the different wave packets are all arranged with just the right phase differences so that they destructively interfere for the waves that are trying to get out of the beam proper, and constructively interfere in the next position of the beam."
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I agree with @Floris that the statement doesn't make sense at face value, but since I know what he is trying to say, I might be able to translate.
No signal has a single frequency. There is always a very small spread in a signal's frequency content. (Pure single frequency implies a signal that started in the infinite past and will continue into the infinite future.) But a nearly-single-frequency wave will be indistiguishable from a true-single-frequency wave if you look at it for a finite time interval.
So take two waves, and look at them during the same time interval. They both will look like waves with some frequency $f$, with some phase shift $\phi$ between them.
Now wait a while and look at the same two waves again. Again, they both will look like waves of frequency $f$. And they will have a phase shift between them, call it $\phi_\mathrm{new}$.
If you wait a short enough time between measurements, the two phase values will be the same. If you wait long enough, they will have been seen to have drifted apart.
Over time, $\phi_\mathrm{new}$ drifts away from $\phi$ because of those tiny fluctuations in the frequency. If $\phi_\mathrm{new} = \phi$ we could say that there is a constant phase shift; if not, then we can say that the phase shift is not constant.
Optical detectors average intensity over a long time, often on the order of seconds. If the phase shift is not constant during that interval (an incandescent source for example), interference patters will be washed out, and not visible. The light is incoherent over that interval. If the phase shift is constant over the (e.g. a laser), interference patterns are visible. The light is said to be coherent over the interval.