# Is single photon perfectly monochromatic?

Now, we do have equipment to generate single photon at a time, and LASERs are nearly monochromatic.

While typing the question, am realizing that successive photons in case of single photon generation may have slightly different wavelengths, but are its single photons essentially monochromatic? Because we may be able to use the phenomena to make highly coherent lasers.

• Are you asking if one photon as one wavelength? If so- yes. But that doesn't make the last sentence true. – jhobbie Jul 2 '14 at 13:50

Any measurement of the photon's energy (i.e. frequency, or free-space wavelength though making a direct identification of particle properties to wave properties is a little sketchy) will return a single value. Every time.

But ... you can't fool Heisenberg and if you have confined the position of the photons---say by insisting that it hit the detector---then

1. You can not predict exactly what the measurement will return
2. Measurements of many photons from a single source will show non-zero a width
• The corresponding uncertainty relation to position-momentum is energy-time. If you measure the wavelength (energy) perfectly, you must have no information on when you measured it. – Ross Millikan Jul 2 '14 at 15:21

Yes - but not really.

A single photon has a single value for it's related wavelength - so it's perfectly monochromatic, in a formal sense.

But, in a physical sense, the concept of a spectrum (like: monochromatic) basically does not apply to a single photon.

In practice, you do not have these "single photons, with some specific wavelength" to actively work with, as implied in the question - that would mean to ignore Heisenberg.
But you may measure you had one - even with a specific wavelength.

• @harshfi6 We may be arguing about language. My point of view comes from what I glean from books (Loudon, and Mandel and Wolf). (I may misrepresent them.) My photon: a single excitation of an energy eigenstate / number eigenstate. It has a single frequency, well-defined occupation number, and completely indeterminate phase. Anything else is a state of the field, but not a photon. Note that the states are degenerate. In free space, $\vec{k}$ can point in any direction, or be a superposition thereof, so with boundaries the field can take on non-space-filling shapes (e.g. cavities). – garyp Jul 3 '14 at 0:25