# Electric and Magnetic Fields due to Photon

1. Jun 14, 2006

### TriTertButoxy

Hi; I am curious about the correspondence between the E&M fields and photons in Quantum Electrodynamics. I want to calculate the electric and magnetic fields produced by a single photon. This may look like a rather long post containing an elaborate question designed to stump everyone, but it isn't. It is long simply because I am defining my notation here.

The way I approached this is by realizing that the best I can do is find the expected values for each of the four components of the electromagnetic vector potential, $A_\mu$. From there, I differentiate appropriately to arrive at the expected value of the electric and magnetic fields.

If my interpretation is right, the vector field operator, when sandwiched between two kets in Dirac's notation, should yield the expected value of the vector field for the system described by the ket, just as how one would expect ordinary operators in quantum mechanics to behave. So, in my case, I would like to calculate

$$\langle A(x)\rangle_\mu=\langle\Psi|\hat{A}_\mu(x)|\Psi\rangle$$,​

where $|\Psi\rangle$ represents the state of the system, which is the one-photon wavepacket,

$$|\Psi\rangle=\int \frac{d^3k}{(2\pi)^3}f(k)\hat{a}^{r\dagger}_k|0\rangle$$.​

Here, $f(k)$ is the fourier decomposition of the modes, which allows me to construct my wavepacket, and $\hat{a}^{r\dagger}_k$ is the creation operator which creates a photon with momentum, $k$ and with a polarization vector $r$, when it acts on the vacuum state, $|0\rangle$. The planewave solution to differential equation that describes free photon propagation is

$$\hat{A}_\mu=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}\sum_r\big[\hat{a}^r_p\epsilon_\mu^r(p)e^{-ip\cdot x}+\hat{a}^{r\dagger}_p\epsilon^{r*}_\mu(p)e^{ip\cdot x}\big][/itex].​ The integration, here, runs over all possible momentum, and $\omega_p$ is the frequency of mode $p$ to be interpreted as the energy of the particle. The summation runs over all polarization states, and $\epsilon^r_\mu$ is the polarization vector. So, all I have to do is substitute these relations into the first centered equation, right? Unfortunately, the bra and the ket states each contribute a single ladder operator and the field operator contributes one for each term. Thus, the expression contains many terms with an odd number of ladder operators sandwiched between the vacuum states, which we all know vanishes. So I find myself predicting a 0 expectation value for the electromagnetic field, which is clearly wrong. If I try to the extend this argument for many photon states, I arrive at the same dilema since the bra and the ket together will always contribute an even number of ladder operators. PLEASE, does anyone have any ideas? My intuition tells me this should be a straightforward calculation, so please help!!! :grumpy: Last edited: Jun 14, 2006 2. Jun 14, 2006 ### masudr What happens to EM waves when you sum over all polarisations? 3. Jun 15, 2006 ### TriTertButoxy Oh, I haven't properly explained the state of the system -- my fault. The state I intend to set up is a localized one-photon state centered at $x^\mu=(0,\,0,\,0,\,0)$ with a momentum $k^\mu=(\omega_{k_0},\,0,\,0,\,k_0)$, so that it is travelling up the z-axis with a right-handed polarization. So the state of the system is [tex]|\Psi\rangle=\int\frac{d^3k}{(2\pi)^3}f(k)\hat{a}^{\text{right}\dag}_k|0\rangle$$.​

The functional form of $f(k)$ takes care of the localization and momentum of the wavepacket. Since the photon now has a particular polarization (and not all polarizations), I would expect to have non-zero average electric and magnetic fields.

I would like to use this space here to type up my work. So the expected value of the electromagnetic vector potential is

$$\langle A(x)\rangle_\mu=\langle\Psi|\hat{A}_\mu(x)|\Psi\rangle$$.​

Subsituting everything in yields

$$=\left(\int\frac{d^3k_1}{(2\pi)^3}\langle 0|\hat{a}^{\text{right}}_{k_1}f^*(k_1)\right)\left(\int\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}\sum_\text{plr'ns}\big[\hat{a}^r_p\epsilon_\mu^r(p)e^{-ip\cdot x}+\hat{a}^{r\dagger}_p\epsilon^{r*}_\mu(p)e^{ip\cdot x}\big]\right)\left(\int\frac{d^3k_2}{(2\pi)^3}f(k_2)\hat{a}^{\text{right}\dag}_{k_2}|0\rangle\right)$$.​

A little rearrangements yields

$$=\iiint\frac{d^3k_1}{(2\pi)^3}\frac{d^3k_2}{(2\pi)^3}\frac{d^3p}{(2\pi)^3}\frac{1}{2\omega_p}f^*(k_1)f(k_2)\langle 0|\hat{a}^\text{right}_{k_1}\sum_\text{plr'ns}\left(\hat{a}^r_p\epsilon_\mu^r(p)e^{-ip\cdot x}+\hat{a}^{r\dagger}_p\epsilon^{r*}_\mu(p)e^{ip\cdot x}\right)a^{\text{right}\dagger}_{k_2}|0\rangle$$.​

At this stage, please pay attention to the second half of the expression, enclosed by the bra and the ket. The summation runs over the two available polarization states (left handed and right handed). Recall the commutation relations of the ladder operators:

$$[\hat{a}^r_p,\,\hat{a}^{r'\dagger}_{p'}]=(2\pi)^3\delta^{(3)}(p-p')\delta_{rr'}$$.​

The terms associated with left-handed polarizations will vanish because the ladder operators will commute through the left-most and right-most operators annhilating the vacuum.

All that I am left with is

$$\Big(\text{Integration Part}\Big)\langle 0|\hat{a}_{k_1} \hat{a}_{p} \hat{a}^{\dagger}_{k_2}\epsilon_\mu(p)e^{-ip\cdot x}+\hat{a}_{k_1}\hat{a}^{\dagger}_p\hat{a}^{\dagger}_{k_2}\epsilon^{*}_\mu(p)e^{ip\cdot x}|0\rangle$$.​

Since each term here has three (an odd number of) ladder operators sandwiched between the vacuum bra and the kets states, both terms will vanish.

So, you see, I've set up the problem incorrectly. I would really love to know how to calculate the expected electric and magnetic fields that represent a single photon.

Last edited: Jun 15, 2006
4. Jun 15, 2006

### nrqed

It is not clearly wrong since your calculation *is* correct!! The expectation values of the E and B quantum field operators in any fixed number photon state is zero.

The correspondence to the classical E and B fields is quite subtle. The quantum states that most closely ressemble classical fiels are coherent states which are *not* states with a fixed number of photons but are instead eigenstates of the annihilation operator.

I am always amazed that this is not covered in every quantum field theory book since it seems to me that the first thing (or almost) to do when introducing quantum fields should be to show clearly the correspondence with classical fields. For some reason, this is never shown!:grumpy: :grumpy:

A nice reference is http://people.deas.harvard.edu/~jones/ap216/lectures/ls_3/ls3_u3/ls3_unit_3.html [Broken]

Last edited by a moderator: May 2, 2017
5. Jun 15, 2006

### TriTertButoxy

Oooooooh! Thank you SO very much, nrqed. That problem was bugging the heck out of me. So if I understand correctly, there is an uncertainty in the number of photons in the system relative to that of the value of the field.

Oh, and that is true because the number operator, $\hat{a}^\dagger \hat{a}$ does not commute with the field operator, $A_\mu(x)\approx(a+a^\dagger)$! Everything now falls in place!

Thanks so much.

6. Jun 20, 2006

### TriTertButoxy

nrqed, your reference does not link to an active site; is there a typo?

Also, I'm having another problem. If I want to create a state that is a superposition of the 0-photon state (vacuum state) and a 1-photon state, the units don't match up right. Take a look:

Peskin and Schroeder's QFT book employs the normalization where $\langle 0|0\rangle=1$, for the vacuum state, and where $\langle\mathbf{p}|\mathbf{q}\rangle=2E_{\mathbf{p}} (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q})$ for the one particle state.

I understand the need for the factor of $E_\mathbf{p}$ in the normalization to make the expression Lorentz invariant. However, I run into problems where if I try to create the state,

$$|\Psi\rangle=|0\rangle + |\mathbf{p}\rangle$$​

which I believe is to be interpreted as the superposition of the vacuum state and the 1-particle state. If I assume states with different number of particles to be orthogonal, the inner product of such a state given by

$$\langle\Psi|\Psi\rangle=1+2E_{\mathbf{p}} (2\pi)^3\delta^{(3)}(0)$$​

doesn't quite seem to conform to the rules of dimensional analysis properly. The first term is unitless, while the second term has units [-2 energy] (Note that the three-dimensional delta function contributes the cube of the inverse of its argument).

I can't seem to find a fix to this problem? Any suggestions?

7. Jun 20, 2006

### masudr

What is the basis of your space that you have used? Are there any implicit tensor products employed?

8. Jun 20, 2006

### TriTertButoxy

For the problem I am doing; the basis I am using is the set of eigenkets of the hamiltonian of the free electromagnetic field.

However, to illustrate the problem I am having, the states I used in my previous post are eigenstates of the free scalar field.

I'm not sure how I may use tensor products here...
...could you show it to me?

PS: Since when did we make the change where the LaTeX displays no longer have a transparent background and where the antialiasing sucks? It was so nice a few months ago...

Last edited: Jun 21, 2006
9. Jun 24, 2006

### photonic

the basic conceptual problem here is that the photon is being treated as a particle possessing charge.its nothing but the change in E&M fields.So no question arises about findin the field due to a single photon

10. Jun 24, 2006

### TriTertButoxy

Thanks, but I'm not looking for the field due to a single photon; I'm looking for the correspondence between the photon and the electromagnetic fields.

My calculations are outlined above, and I'm having problems with units right now.

11. Jul 6, 2006