# How fast is wave propagation in interference?

When someone performs Young's Double Slit experiment, the person sees an interference pattern on the screen. What is the time taken to for the pattern to appear on the screen? Is it distance between slit and the screen divided by speed of light? Another way to put the question is when photons are converted to waves is wave propagation speed = speed of light ?

This question is all about the signal to noise ratio you achieve in your experimental setup, so the details are highly dependent on the latter. Here are the physical principles you would use to calculate how long it takes a fringe pattern to form.

Assuming the source sends unentangled photons, each photon propagates following Maxwell's equations. So the propability density of absorption as a function of time can be calculated as a classical intensity as a function of time. Simply put, this means that the time taken for the pattern to reach the screen is simply the propagation delay: the propagation distance $\ell$ divided by $c$.

However, most of the time for the interference pattern to form is the time taken for each detector - each pixel, if you like - in the interference pattern to register enough photons that it can report, with the appropriate level of statistical confidence, that the number of photons it has registered is lower or higher than that of the neighbouring detectors such that the data gathered from the whole detector array bespeak what we would call "fringes".

This is probably more easily explained by a simple calculation. Suppose we have an array of CCD detectors lined up along the detection screen. The fringe pattern will form fastest when the detector spacing is exactly the fringe spacing. If the fringe visibility is $\mathscr{V}$, then the ration of light intensity in the troughs to that in the peaks is:

$$I_{min} = I_{\max}\frac{1-\mathscr{V}}{1+\mathscr{V}}\tag{1}$$

If each detector's area is $A$ then the mean number of photons arriving each second is

$$\mu(I)=\frac{I\,A\,\lambda}{h\,c}\tag{2}$$

where $\lambda$ is the light's wavelength. Photon arrivals from most CW sources like lasers follow Poisson statistics, so if a detector's light gathering time is $\delta t$, then the number of photons actually gathered in that time will be Poisson-distributed with mean:

$$\mu(I,\,\delta t) = \frac{I\,A\,\lambda\,\delta t}{h\,c}\tag{3}$$

so what you're looking for is a $\delta t$ such that there is "overwhelming" probability that the number of photons detected in each photon field peak will be greater than the number detected in each trough. Here is where statistical confidence levels enter the calculation. In symbols, we want the probability that:

$$N_p\left(\left(\frac{\eta\,I_{\max}\frac{1-\mathscr{V}}{1+\mathscr{V}}\,A\,\lambda}{h\,c} + \sigma_D\right)\,\delta t\right) < N_p\left(\left(\frac{\eta\,I_{max}\,A\,\lambda}{h\,c} + \sigma_D\right)\,\delta t\right)\tag{4}$$

to be greater than some "reasonable" confidence level; say $0.9$ for a rough calculation. Here $N_p(\mu)$ is the number of photon measurement events that Poisson variable with mean $\mu$ actually assumes in a given observation. I have also added a residual constant detector noise $\sigma_D$ which will always present owing to noise in the detector electronics and so forth. It is the number of "false positive" detections per unit time and most often, with photodetectors, represents the "dark current". I have also added a quantum efficiency $\eta$ for the detectors; this is a probability of a "false nagative" detection and for modern detectors $\eta \approx 0.8$ is reasonable. The above is quite an involved calculation to do properly, because the difference between two Poisson RVs is not a Poisson RV (Poisson distributions do not have the nice self replication property under summations that normal or Chi-squared distributions have), so we make a normal approximation for a back of the envelope calculation. Re-arranging (4) in this case shows that the peak detector photon number is greater than the trough detector photon number by an approximately normal random variable whose mean $\mu$ and standard deviation $\sigma$ are:

$$\mu = \frac{2\,\eta\,I\,\mathscr{V}\,A\,\lambda}{h\,c}\,\delta t;\;\sigma^2 = 2\,\left(\frac{\eta\,I\,A\,\lambda}{h\,c}+\sigma_D\right)\delta t\tag{5}$$

(here I've simply added the two Poisson variable variances, given that a Poisson RV's variance equals its mean) and we want to choose $\delta t$ such that the probability of this random variable's being positive is equal to our confidence level. Here I've used $I = I_{max} / (1+\mathscr{V})$ to rewrite my equation in terms of the mean fringe pattern intensity $I$ instead of the peak $I_{max}$. Let's put some numbers for an experiment in. Suppose we do our experiment with a light intensity of $10^3\mathrm{W m^{-1}}$; this is a reasonable laboratory intensity if a $100mW$ laser (beware: this is at least a class 3B laser if you're doing this) lights an interference pattern that is a centimetre across, and suppose that our pixel area is $10^{-10} m^2$, this corresponds to very big CCD cells. Then, at $\lambda = 5\times 10^{-7}m$, a quantum efficiency of $\eta=0.8$, a fringe visibility of 0.5 and a perfectly clean measurement ($\sigma_D=0$), we get:

$$\mu \approx 2\times 10^{11}\,\delta t;\;\sigma^2 \approx 4\times 10^{11}\,\delta t\tag{6}$$

(here the $2\,\sigma^2$ comes from the fact that we subtract and with a critical value of the normal distribution for $\alpha=0.9=90\%$ confidence of $\sqrt{2}\,\mathrm{erf}^{-1}(\alpha) \,\sigma \approx 1.64 \sigma$, we need $\mu-1.64\,\sigma\geq0$ or

$$2\times 10^{11}\,\delta t \geq 1.64\times \sqrt{4\times 10^{11}\,\delta t}\,\Rightarrow\,\delta t \geq \frac{1.64^2\times 4\times 10^{11}}{2^2\times 10^{22}}\tag{7}$$

$$3\times 10^{11} \times10^{-10} \times 5\times 10^{-7}/(h\,c)\approx 15$$

photons per pixel. Your electronics is not likely to be this fast, so that the electronic delay is the dominant factor. More typically, light is spread much more widely and you need to gather light for much longer to get high quality fringes: an interferometer I have used to test small lenses takes at least several microseconds to gather a fringe pattern and I have calculated it to be very near to achieving the ideal situation studied above. Quantum noise (photon arrival variation with the Poisson statistics described above) limits high speed interferometry and microscopy much more often than you might think.

Another, qualitative way to answer your question is to get Mathematica (or other numerical simulation) to calculate simulated fringe patterns by assigning random photon positions according to an intensity pattern, and then increasing the total number of photons until your fringe pattern looks clear. Then you need to calculate what signal acquistion time you need given the calculated intensities in your experimental setup. But you will find about 15 photons per pixel to be a pretty representative figure. In the instruments I have designed, a common "design" standard is to aim for 100 photons per pixel; this gives you about a 10dB signal to noise ratio, given the variance is then $\sqrt{100}=10$ photons per pixel.

• thanks for nice and practical answer but i was looking at propagation speed which I think is 'c'. – user43794 Apr 4 '14 at 9:18
• @user43794 Yes it is $c$ but the light will not all take one path. So suppose you have pulsed source. You then solve Maxwell's equations for each frequency, and then add the superposition together to match the pulsed source. You'll find for a general setup the pulse is spread out in time at the detector owing to multipathing. So there will not be one simple $c/\ell$ delay. Lone photons propagate exactly following Maxwell's equations, so once you've got your solution, you find the classical intensity as a function of time at your detector and that will be proportional to probability density ... – Selene Routley Apr 4 '14 at 9:56
• @user43794 ...for the photon event registration. So the answer you seek is my second paragraph in my answer. See my answer here for more on interpretting Maxwell equation solutions as photon absorption probability waves. – Selene Routley Apr 4 '14 at 9:57
• Accepted your answers. I understand photon arrival has some probabilistic distribution. It is akin to saying that photons do not jump the barrier at once. Therefore it might take even few seconds for a screen that is far off. Fair enough but the speed remain c and its the intensity which varies with time. Thanks for the answers. It enlightened me a little bit more about mysterious world of QM. – user43794 Apr 7 '14 at 4:12
• @user43794 Again, you've got the right idea, particularly since a lens's purpose is to equalise the transit time for all light paths. However, it does this by delaying the central (on-axis) field relative to the field at the lens edges. So your transit time will be something like $(f+n\,t)/c$, where $t$ is the lens thickness at its centre and $n$ its refractive index. Also, for many photons, its a question of source intensity. You need to calculate how often a particular source is adding photons to the electromagnetic field, and so a particular source may add significantly to the photon ... – Selene Routley Apr 7 '14 at 5:39

Light is a wave, it does not change its nature from particle to wave. Light, and this is true for any particle, is a an excitation is some field, and we model its propagation/time evolution using waves, but also quantise some of that wave's parameters, and in a sense, modelling it as a particle when convenient.

Now answering your question, light during the experiment always propagates at the speed of light $c$.