I am reading Berkeley Physics Course vol. 4 (Quantum Mechanics) , chapter 4 (photons).

(1) Section 46: book says: consider a typical photon emitted by the source. It can be regarded as a a wave train of finite duration, spreading out in all directions in space and carrying a total energy $\hbar\omega$.

If this is true, a photon emitted by a star which is few light years away, would have spread over a sphere of radius of few light years. Then it is not possible to "detect" that photon using a detector.

Then why does the book say like this?

In section 47, the book gives same example of photon spreading over light years. But in this section, it says that how the people can use this as a paradox; and clarifies that paradox by saying that classical expression for "energy density" refers to a large number of photons, and not for single photon.

So does photon really spread in all directions?

(2) In Section 38 book says: almost monochromatic photons can not be split into 2 photons of the same frequency which carry only a fraction of the energy of the original photon. While in section 48, it says (with reference to double-slit diffraction experiment) photon came through BOTH the slits.

Does a photon REALLY goes through both the slits?

Doesn't it indicate that a single photon can be split into two photons of same frequency ?

Any help will be appreciated. Thanks


2 Answers 2


Your confusion arises because the light is not a photon and it's not a wave. It's a quantum field, specifically the photon field, and this quantum field can interact with other matter in particle like or wave like ways. My preferred way of thinking about this is that the photon is the unit of interaction of the quantum field with something else, so the photon can't be said to exist until the field interacts.

So your light source interacts with the photon field and transfers energy to it. In your example this energy propagates isotropically outwards. When the energy reaches e.g. our CCD detector the photon field transfers the energy to our CCD in a chunk of energy $h\nu$, and this is when we see the photon. The interaction with the CCD localises it at some point in space and time.

The next question is how the energy spread over a several light year sphere gets to our CCD in an instant when it can't move faster than light. The answer is that it can move faster than light. You can think of the light at the instant before interaction as being in a superposition of all possible locations of the photon. Viewed this way the interaction with the CCD collapses the wavefunction, and wavefunction collapse happens at arbitrarily high speeds. The superluminal collapse doesn't violate special relativity because it cannot transmit any information.

This also explains the Young's slits result. A localised particle cannot pass through two different slits at the same time, but when the light passes through the slits it is delocalised and does not have a position. We can only assign a position when the light interacts with the screen at the far side of the slits. Just as in the example above it's the interaction with the screen that localises the light to produce what we recognise as a photon. So the light really does pass through both slits.

  • 1
    $\begingroup$ According to your (@John) view, if photon field exists already, then electron in the atom interacts with this photon field and emits "energy packet" (which we call as a photon!) whenever electron "jumps" to lower/higher energy state. Can you elaborate what this photon-field exactly is? First I want to know whether it is "discrete" field or "continuous" one? $\endgroup$
    – physics
    Nov 19, 2013 at 11:40
  • $\begingroup$ @user34185: you're asking me to explain quantum field theory, and that would take a large book not a few comments! The photon field, like all quantum fields, is a continuous field that exists everywhere. It's discrete in the sense that energy is transferred to or from it in chunks (i.e. photons) but of course a photon can have any energy so the chunk can be any size. There is a fairly detailed discussion on Wikipedia, or Matt Strassler's blog has lots of articles on QFT. $\endgroup$ Nov 19, 2013 at 11:46
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    $\begingroup$ This answer claims that light moves faster than the speed of light. That's very confusing. $\endgroup$ Nov 26, 2013 at 23:17
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    $\begingroup$ @BlueRaja-DannyPflughoeft: no, only the collapse of the wavefunction happens faster than light. But you're right that it is confusing, and it has confused generations of physicists starting with Einstein! $\endgroup$ Nov 27, 2013 at 6:52
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    $\begingroup$ If a quantum field is not made of photons what is it? $\endgroup$ Sep 3, 2016 at 20:34

(1) If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called photon or wave packet. If you link the appropriate formulas from QM and E&M waves, the result is the diameter of the wave packet (about lambda/2). The radius and the direction of propagation do not change. If this wave packet hits your detector, you can measure the energy hf. Some say: The photon has arrived. Other say: The E&M wave has induced a small current in the antenna. This depends on the type of the receiver.

Nothing is spread over a sphere, the energy is not diluted! Read about Pulsars: Those rotating neutron stars emit very strong signals into a single direction. If this beam impinges on earth, astronomers receive a signal. Most of the time the beam points to different directions.

(2) Wrong: In "Spontaneous parametric down-conversion" a single photon is converted into two photons of lower energy. Nobody knows, why.

The double-slit diffraction experiment can not be explained with photons. Nonsense to claim, that a bullet came through BOTH the slits. Nonsens.

  • $\begingroup$ Part (2) is quoted as saying that a monochromatic photon cannot be split into two photons of the exact same frequency which each carry only a fraction of the energy of the original photon. Whereas parametric down-conversion converts a photon into multiple photons of lower frequency, so is not a counter-example. $\endgroup$
    – Timaeus
    Jan 24, 2015 at 20:20

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