What does amplitude in wavelength of light physically mean?What oscilates with time in photon?

Question

Like amplitude in wavelength of water waves signify the displacement of water particles about their mean position.

Answer 1

Although you can (as you obviously know) think of electromagnetic radiation as either a particle or a wave, it's easier in this case to think of it as a wave.

As a thought experiment, if you wave a magnet near a piece of wire, an electric potential will be induced in the wire. Likewise, if you pass current through a wire, a magnetic field will be produced around the wire. You can restate those two observations as: "a changing magnetic field produces an electric field; and a changing electric field produces a magnetic field."

If you work out the math to describe those interactions, you essentially get Maxwell's field equations. In a nutshell, if you want to produce an electromagnetic wave, you start by creating a time-changing electric field (say, by running electrons back and forth in an antenna). The changing electric field produces a changing magnetic field. The changing magnetic field in turn produces a changing electric field, et cetera. So an EM wave is just an electric field and a magnetic field leapfrogging their way through space.

To finally answer you question, the "amplitude" of the wave is the strength of the electric and magnetic fields involved. They aren't in units of distance, as are sound waves or water waves.

Answer 2

Before the photon is absorbed, i.e. whilst the photon quantum field is in a pure, one photon state, you can define the following probability amplitudes that uniquely specify the one-photon state:

$$\begin{array}{lcl}\vec{\phi}_E(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{E}}^+(\vec{r},t)\left|\left.\psi\right>\right.\\ \vec{\phi}_B(\vec{r},\,t)&=&\left<\left.0\right.\right| \mathbf{\hat{B}}^+(\vec{r},t)\left|\left.\psi\right>\right. \end{array}\tag{1}$$

and these probability amplitudes oscillate in space and time just like a classical light field. They have the physical meaning that the probability density to destructively detect the photon, i.e. absorb it with an ideal detector, when the state is properly normalised, is the analogue of the classical energy density (normalisation makes the classical energy density into a probability denstity), i.e.

$$p(\vec{r},\,t) = \frac{1}{2}\,\epsilon_0\,|\vec{\phi}_E|^2 + \frac{1}{2\,\mu_0}\,|\vec{\phi}_B|^2\tag{2}$$

In the above, $\psi$ is the photon field quantum state, $\mathbf{\hat{B}}^+,\,\mathbf{\hat{E}}^+$ are the positive frequency parts of the (vector valued) electric and magnetic field observables and, of course, $\left<\left.0\right.\right|$ is the unique ground state of the quantum photon field. Given the knowledge that the field is in a purely one photon state, the vector functions of space and time in (1) uniquely define the photon field's quantum state and contrariwise, so that they can be taken as being the quantum state. Moreover, the functions in (1) always fulfill Maxwell's equations and every classical solution of Maxwell's equations uniquely defines a one-photon state. So every solution to Maxwell's equations can indeed stand for three separate and different things (1) a classical EM field, (2) a one photon quantum state or (3) a coherent state of the the quantum light field (this state has a Poisson-distributed number of photons). So there is a one to one correspondence between the elements of all these three classes.

See my answer here for more information and references.

Now, you cannot physically measure an overall phase of the light field, but you can certainly, in principle, see the diffraction and other wave effects by destructively detecting successive photons passing through an experimental apparatus one photon at a time.

Answer 3

Your question is pretty vague, but does this help--

The E-M wave amplitude is a complex oscillating function, as you can see at the wikipedia page . Now, the power in any wave is the square of the amplitude, or more precisely, the product of the amplitude and its complex conjugate.
It turns out that E-M energy is quantized, so we can assign a specific energy to a photon based on its frequency.

If the wave diagrams at that page don't answer all your questions about wavelength, frequency, and orientation, please try to repost a more specific question.