# I Any EM-field in terms of photon

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1. May 13, 2016

### olgerm

I know formula $p=\frac{h}{λ}$
• p is photon momentum
• h is plankc constant
• λ is EM-wave wavelenght
but it is only valid for one wave.

How to describe most general EM- field in terms of photons? Is there always discrete number of photons? If EM field is given in terms of
a)EM-vectorfield:
Ex(T;x;y;z) ; Ey(T;x;y;z) ; Ez(T;x;y;z) ; Bx(T;x;y;z) ; By(T;x;y;z) ; Bz(T;x;y;z)
for any x,y,z and one T.

b) potentsials and loren(t?)z gauge

then what are momentums of photons?

2. May 13, 2016

### vanhees71

The only way to describe "photons" right is quantum electrodynamics, a relativistic quantum field theory. The free electromagnetic field isn't so easily quantized, because of gauge invariance, but one way to do that for the free photon field is to start from classical (Maxwell) electrodynamics and fix the gauge completely. A convenient choice is the socalled radiation gauge, which uses the gauge constraints
$$A^0=\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
The vector potential is then decomposed into transverse plane-wave solutions via [corrected in view of #7]
$$\vec{A}(t,\vec{x})=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{k}}{(2 \pi)^{3/2} \sqrt{2 \omega_{\vec{k}}}} \left [A(\vec{k},\lambda) \vec{\epsilon}(\vec{k},\lambda) \exp(-\mathrm{i} \omega_{\vec{k}} t+\mathrm{i} \vec{k} \cdot \vec{x}) + \text{c.c.} \right ].$$
Here $\vec{\epsilon}(\vec{k},\vec{\lambda})$ are the two transverse, i.e., $\vec{k} \cdot \vec{\epsilon}(\vec{k},\vec{\lambda})=0$, left- and right-circular polarization vectors (leading to helicity states for the photons) and $\omega_{\vec{k}}=|\vec{k}|$ (setting $\hbar=c=1$ for convenience).

Then you can quantize the theory by assuming the canonical commutation relations for the field, i.e., substituting $A(\vec{k},\lambda) \rightarrow \hat{a}(\vec{k},\lambda)$. Then the operators fulfill
$$[\hat{a}(\vec{k},\lambda),\hat{a}(\vec{k}',\lambda')]=0, \quad [\hat{a}(\vec{k},\lambda),\hat{a}^{\dagger}(\vec{k}',\lambda')]=\delta^{(3)}(\vec{k}-\vec{k}'),$$
and are thus are annihilation and creation operators for massless particles with momentum $\vec{k}$ and helicity $\lambda$, because the commutation relations are those of an continuous number of idependent harmonic oscillators.

A convenient complete set of orthonormal basis vectors are the simultaneous eigenvectors of the number operators
$$\hat{N}(\vec{k},\lambda)=\hat{a}^{\dagger}(\vec{k},\lambda) \hat{a}(\vec{k},\lambda).$$
The ground state (aka "the vacuum") is given by the state, where all eigenvalues of these operators $N(\vec{k},\vec{\lambda})=0$, fulfilling
$$\hat{a}(\vec{k},\lambda)|\Omega \rangle=0.$$
Then any basis state is given by
$$|\{N(\vec{k},\lambda) \} \rangle =\prod_{\vec{k},\lambda} \frac{1}{\sqrt{N(\vec{k},\lambda)!}} \hat{a}^{\dagger N(\vec{k},\lambda)}|\Omega \rangle,$$
where the product runs over any finite set of momenta and helicities and $N(\vec{k},\lambda) \in \{0,1,2,\ldots \}$. These are then states with a definite photon number, the socalled Fock states.

It's also clear that due to the use of commutation relations for the field operators, photons are bosons, i.e., interchanging the annihilation and creation operators within the product to build the Fock states, doesn't change the state in any way, i.e., the occupation number basis consists of totally symmetrized products of single-photon states.

A general state in this Hilbert space consists of arbitrary superpositions of such Fock states. It can be a state with a definite photon number or some superposition of such states. E.g., the states most close to a classical electromagnetic field (like the field produced by a laser) are the socalled coherent states, where the photon number is not determined but follows a Poisson distribution.

For a very nice introduction to quantum optics, see

Scully, M. O., Zubairy, M. S.: Quantum Optics, Cambridge University Press, 1997

Last edited: May 14, 2016
3. May 13, 2016

### olgerm

Thank you for thorough answer. I am quite good in math, but do not understand many of notations, you have used.

Are both of these equations coulomb gauge constraints?
$\Phi=0$ is true only at time T?
$\vec{\nabla} \cdot \vec{A}=0$ is true any time?
$\sum_{\lambda=\pm 1}(f(λ))=f(λ)+f(-λ)$?
What does c.c. mean?
What do bracets ( [] ) mean?
What does † mean?
What does a^ mean?

What does that mean?

Sorry I do not know a lot about QM. Hopefully you can explain these basic things to me.

4. May 13, 2016

### f95toli

Then you need to start by learning a bit more about QM. Photons are -unfortunately- very complicated animals, and the full formalism/theory is so complicated that it is not really covered until you get the graduate level QM and even then you will only cover the basics. You are unlikely to ever learn this properly unless you work in quantum optics (and even then only if you do theory).
(for the record: I do NOT understand all of the maths, only the bits and pieces necessary for my own research )

5. May 13, 2016

### vanhees71

Ok, the quantum theory, we cannot treat in terms of forum discussions. You should use quantum theory textbooks to study this. My favorite to start with is

J.J. Sakurai, Modern Quantum Mechanics

Then you need some QFT to understand photons. For that, I recommend the above mentioned book by Scully rather than a usual relativistic QFT book for particle physicists (although QED of course comes originally from that field), because there other aspects are more emphasized than the nature of light itself.

What we can discuss, however is the basics of classical electrodynamics of free fields, which is crucial to understand before you turn to quantum theory anyway.

You start from Maxwell's equations. In the following I'm using Heaviside-Lorentz units with $c=1$. First we consider only the homogeneous equations,
$$\vec{\nabla} \times \vec{E}+\partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0.$$
The 2nd equation tells you, using, Helmholtz's decomposition theorem that $\vec{B}$ is a pure solenoidal field, i.e., it is the curl of another vector field, $\vec{A}$, the vector potential:
$$\vec{B}=\nabla \times \vec{A}.$$
Plugging this into the first equation (Faraday's Law) you get
$$\vec{\nabla} \times (\vec{E}+\partial_t \vec{A})=0,$$
and again from Helmholtz's decomposition theorem (or Poincare's lemma for that matter) you see that there must be a scalar potential such that
$$\vec{E}+\partial_t \vec{A}=-\vec{\nabla} \Phi.$$
Thus we have up to now
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
Now you can forget about the homogeneous Maxwell equations for a moment, because writing the fields in terms of the scalar and vector potential, they are automatically fulfilled.

Now let's turn to the inhomogeneous Maxwell equations, i.e.,
$$\vec{\nabla} \times \vec{B}-\partial_t \vec{E}=\vec{j} , \quad \vec{\nabla} \cdot \vec{E}=\rho,$$
where $\rho$ and $\vec{j}$ are the electric-charge and electric-current density. For free fields no charges or currents are present, and we'll set them to 0 in a moment, but first let's plug in the potentials into these equations.
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})-\partial_t(-\partial_t \vec{A}-\vec{\nabla} \Phi)=\vec{j}.$$
Using Cartesian coordinates you can simplify them by using the identity $\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}$, where the Laplacian is defined as $\Delta=\vec{\nabla} \cdot \vec{\nabla}$. Using this, the above equation becomes
$$\partial_t^2 \vec{A}-\Delta \vec{A} +\vec{\nabla}(\vec{\nabla} \cdot \vec{A}+\partial_t \Phi)=\vec{j}.$$
Now we can simplify the solution of these equations tremendously by recognizing that the potentials are not uniquely defined by the physically observable fields, $\vec{E}$ and $\vec{B}$. Indeed you can add the gradient of an arbitrary scalar field, $\chi$, to the vector potential, without changing $\vec{B}$, i.e.,
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi$$
is as good as $\vec{A}$ to get $\vec{B}=\vec{\nabla} \times \vec{A}=\vec{\nabla} \times \vec{A}'$. At the same time we have to make sure that
$$\vec{E}=-\partial_t \vec{A}-\vec{\nabla} \Phi=-\partial_t (\vec{A}'+\vec{\nabla} \chi)-\vec{\nabla \Phi} = -\partial_t \vec{A}' - \vec{\nabla}(\Phi+\partial_t).$$
Thus we set
$$\Phi'=Phi+\partial_t \chi$$
so that we have the electromagnic field in the same way given by the primed as by the unprimed potentials. The transformation
$$\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\partial_t \chi$$
is called gauge transformation, and what we just derived tells us that electrodynamics is invariant under this gauge transformation for the potential.

Now because of this arbitrariness, we can put an extra condition on the potentials, and it's convenient to use the Lorenz-gauge condition
$$\partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0,$$
because then our above equation for the potential from the one inhomogeneous Maxwell equation simplifies to
$$(\partial_t^2-\Delta) \vec{A}=\vec{j},$$
i.e., the three Cartesian components decouple from each other, and each other fulfills just an inhomogeneous wave equation with the current density as source.

Now we turn to the last equation, Gauss's Law. Plugging in the electric field in terms of the potentials we get
$$\vec{\nabla} \cdot \vec{E} = -\vec{\nabla} \cdot (\partial_t \vec{A}+\vec{\nabla} \Phi)=\rho,$$
but using the Lorenz gauge condition we can as well eliminate $\vec{A}$ from this equation, because we can put
$$\vec{\nabla} \cdot \vec{A}=-\partial_t \Phi,$$
which finally also gives a wave equation for $\Phi$, decoupled from all components of the vector potential:
$$(\partial_t^2-\Delta)\Phi=\rho.$$
Now we specialize to free electromagnetic fields, i.e., we assume that there are no charges and currents present anywhere, i.e., $\rho=0$ and $\vec{j}=0$. Then we have
$$(\partial_t^2-\Delta) \Phi=0, \quad (\partial_t^2-\Delta) \vec{A}=0.$$
Now the Lorenz gauge condition does not completely fix the potentials together with these equations, because we can still take a gauge transformation, for which the gauge field $\chi$ itself also satisfies the homogeneous wave equation, i.e., we set
$$\Phi'=\Phi+\partial_t \chi, \quad \vec{A}'=\vec{A}-\vec{\nabla} \chi$$
for a function $\chi$ satisfying
$$(\partial_t^2-\Delta \chi)=0.$$
Then also the new potentials fulfill the Lorenz gauge condition as do the old ones, because we have
$$\partial_t \Phi'+\vec{\nabla} \cdot \vec{A}'=\partial_t \Phi+\vec{\nabla} \cdot \vec{A} + (\partial_t^2-\Delta)\chi=0.$$
Now we can choose this remaining restricted gauge freedom to envoke the additional constraint that
$$\Phi'=\Phi+\partial_t \chi=0.$$
Since we have $(\partial_t^2-\Delta \Phi)=0$, this is consistent with $(\partial_t^2-\Delta) \chi=0$, and that's only the case for free fields, i.e., for $\rho=0$. Thus now we have the simplified equations
$$(\partial_t^2-\Delta) \vec{A}'=0, \quad \vec{\nabla} \cdot \vec{A}'=0,$$
where the latter equation follows from the Lorenz gauge condition for $(\Phi',\vec{A}')$ and our additional condition $\Phi'=0$. This is called the radiation gauge, and as stressed above, it can be used only for free fields (or for fields with $\rho=0$). The above derivation shows that the gauge constraints $\Phi'=0$, $\vec{\nabla} \cdot \vec{A}=0$ must hold for all times.

You are right concerning the sum over $\lambda=\pm 1$; c.c. means "complex conjugate" and thus $f+\text{c.c.}=f+f^*$ for any complex valued function $f$. The brackets are commutators of operators, $\dagger$ stands for the hermitean conjugate of a linear operator in Hilbert space, and $\hat{a}$ is such an operator in Hilbert space (see Quantum Mechanics textbooks).

6. May 13, 2016

### phyzguy

These gauge constraints are true at all times.
Yes.
c.c. means 'complex conjugate'. Basically A has to be real, so by adding the expression in the brackets to its complex conjugate, you guarantee the result is real.
The square brackets denote commutator. So, for two operators A and B, [A,B] = AB - BA
The dagger denotes the Hermitian conjugate of an operator, also called the adjoint or Hermitian adjoint. You can look up its definition. If the operator is written as a square matrix, then the Hermitian conjugate is the conjugate transpose of the matrix.
It denotes that a is an operator.

7. May 13, 2016

### olgerm

Do I understand it correctly: You decompose magnetic potential fiels into fictional plane-waves with different frequencies and sum over the plane-waves parameters $\omega_{\vec{p}}$, $\omega_{\vec{k}}$, $A(\vec{k},\lambda)$, $\vec{\epsilon}(\vec{k},\lambda)$, $\vec{k}$ and $\vec{x}$ ?

8. May 13, 2016

### Staff: Mentor

Shouldn't $\sum_{\lambda=\pm 1}(f(λ))=f(+1)+f(-1)$?

9. May 13, 2016

### phyzguy

Yes, of course you're right. Sorry for the confusion.

10. May 14, 2016

### vanhees71

Sorry, there's a typo. It must be $\omega_{\vec{k}}=|\vec{k}|$ everywhere. I don't know, what you mean bei "fictional". It's the usual Fourier decomposition of the electromagnetic field in plane-wave modes, i.e., momentum-helicity eigenstates. It's not fictional but just a mathematical tool to describe the electromagnetic field. An alternative basis is to use energy-angular-momentum eigenstates (partial waves, multipole expansion).

11. May 16, 2016

### Demystifier

It's Lorenz gauge, but most books erroneously write Lorentz gauge.