Basic Photon Question single, understand, electric, everyone

amy

Hi everyone,

I understand that a photon is interlinked packet of energy with oscillating electric and magnetic fields. My question is what is the electric field strength associated with a single photon? For a capacitor we of course use V/d and could use this to calculate the e-field of an atom (using the separation of the proton and electron for d), but what about the e-field of a photon so one can compare between the two?

Thanks,

L  

    The expectation value of the E field for a single photon (or, more generally, for ANY state with a well-defined number of photons; i.e  number state) is zero.
However, the variance of the E field is non-zero.  

    Photons are packets of energy and momentum, it doesn't really make sense to say that a photon has an intrinsic field.

Claude.  

    The energy density of an EM wave =  in Gaussian notation.
Multiplying this by a reasonable length of the wave packet times a reasonable area for the wave front gives the energy in the EM wave.  This energy equals hf for a single photon.
Using this, E for a single photon can be found.  

                                   Originally Posted by pam                   The energy density of an EM wave =  in Gaussian notation.
Multiplying this by a reasonable length of the wave packet times a reasonable area for the wave front gives the energy in the EM wave.  This energy equals hf for a single photon.
Using this, E for a single photon can be found.                  
No it can not. E (or, more correctly since it is QM, lt;Egt;) for a single photon is zero. It is NOT quot;undefinedquot; or anything like that; it is just exactly zero; lt;n|E|ngt;=0
Any text about basic quantum optics will cover this.

Or, see e.g.
~jone...s3_unit_3.html

Eq III-3A is the E-field operator; if you take the expectation value of this with respect to a number state (any state with a specfic number of photons, e.g. one) you always get zero.

I know this isn't exactly a pedagogical explanation; but unfortunately it is impossible to avoid math when dealing with photons; I don't think there is an quot;intuitivequot; explanation.  

    I think Pam is referring to the amplitude of E.  Since E is oscillating its expectation value is zero.  

    There is a flaw in this method.                              Originally Posted by pam                   The energy density of an EM wave =  in Gaussian notation.
Multiplying this by a reasonable length of the wave packet times a reasonable area for the wave front gives the energy in the EM wave.  This energy equals hf for a single photon.
Using this, E for a single photon can be found.                  
Having a wave-packet of finite length implies that the packet contains a spectrum of frequencies. It is therefore wrong to imply that you can define a specific energy for the photon. At best all you can do is define the width of the frequency spectrum.

Claude.  

                                   Originally Posted by Claude Bile                   There is a flaw in this method.

Having a wave-packet of finite length implies that the packet contains a spectrum of frequencies. It is therefore wrong to imply that you can define a specific energy for the photon. At best all you can do is define the width of the frequency spectrum.

Claude.                  
That is the importance of the phrase quot;reasonable lengthquot;.
For instance, if a ruby laser of a given power is pulsed on for  1 nanosecond,
the average amplitude of E per photon can be calculated.  

    That still doesn't make sense. An ordinary laser will not generate a Fock state (more likely a coherent state), so the number of photons is stricly speaking not fixed. Hence, there can't be an quot;E field per photonquot;.
Of course it is always possible to use a semi-classical calculation if the power is high enough, e.g. just divide the energy with . That will give you a handle on the average number of photons in the system and is useful for e.g. looking at the thermal occupancy of states.  However, for a single photon system this expectation value will be very,very small (much smaller than the fluctations) meaning it still doesn't make sense to talk about this in terms of the field per photon.