# Effect of phase difference on interference pattern

1. Aug 1, 2012

### San K

When two wave functions are in perfect phase = constructive interference

when the phase difference is half (the wavelength) = destructive interference

What happens when the phase difference is more than one wavelength in case of:

Case 1: single particle interference
Case 2: two-photon (i.e. a pair of entangled photons) interference

i.e say twice the wavelength
How about phase difference thrice the wavelength?

in short: does single particle interference happen if the phase difference is more than one wavelength?

Last edited: Aug 1, 2012
2. Aug 1, 2012

### Staff: Mentor

As long as it is well below the coherence length, take the phase difference mod 2pi (edit: thanks). If the phase difference gets comparable to the coherence length, interference will begin to vanish. This is true for both cases.

If you look at double slit patterns, you can usually see several maxima, they correspond to wavelength differences of (edit: thanks) 0, 1, 2, 3, ...

Last edited: Aug 1, 2012
3. Aug 1, 2012

### Staff: Mentor

I would say "path [length] difference" instead of "phase difference" in the paragraph above, to make the units consistent. Path [length] and wavelength are distances, whereas phase is an angle.

Correspondingly, "wavelength difference" should be "path [length] difference" above. (Remember, lambda is the wavelength! )

Last edited: Aug 1, 2012
4. Aug 1, 2012

### Staff: Mentor

To (hopefully) address San K's original question:

Consider two-slit interference using a normal-intensity coherent light source (e.g. a laser) that produces bazillions of photons.

http://fuff.org/interference/two-slit-experiment_light.png

(In this diagram, I would label the maxima below the central max as n = -1, -2, -3, etc.)

At the central maximum on the screen, the path difference and the phase difference are both zero. (This point is equidistant from both slits.)

At the n = 1 maximum, the path difference (path from lower slit minus path from upper slit) equals one wavelength, and the phase difference is one cycle (or 360° or 2π radians, whichever units you like to use for angles).

At the n = 2 maximum, the path difference equals two wavelengths, and the phase difference is two cycles (or 720° or 4π radians).

At the n = -1 maximum the path difference (again, path from lower slit minus path from upper slit) equals -1 wavelength (because now the path from the upper slit is longer) and the phase difference is -1 cycle or -360° or -2π radians.

Etc. for other values of n.

Now, if we switch to a single-photon source and send one photon per hour through the apparatus, we get a new small spot on the screen every hour. If the screen can "save" the position of each spot (e.g. a piece of photographic film) and we let this go on long enough, we eventually build up exactly the original interference pattern, except for statistical fluctuations in intensity ("graininess" in the image on the screen), which decrease (in percentage terms) as more and more photons arrive at the screen.

5. Aug 3, 2012

### San K

thanks jtbell and mfb

at what (approx) point, per the Schrodinger wave equation, does the interference happen? specially in cases where the phase length difference is a multiple of wavelength....for example when the difference is, say, 8 pie radians (= four wavelengths ?)

Also
Do the wavefunctions need to arrive at the same point in time-space to interfere?

Last edited: Aug 4, 2012