# QED : Momentum of the photon-field

• Kalimaa23
In summary, the conversation discusses how to show that two expressions for momentum, denoted as \vec{P} in both cases, are equivalent. One expression involves a normal product, denoted as N, while the other involves a summation over a vector, denoted as \vec{k}, and a usual number operator, denoted as N(\vec{k}). The conversation goes on to explain the steps for calculating the normal ordering and the resulting terms, with one term causing trouble. The conversation then shifts to another method for finding the momentum, using the Lorenz-Lorentz gauge and the Hamiltonian. The resulting expression is compared to the previous expression and the normal ordering product is introduced to avoid an infinite vacuum energy and momentum. The
Kalimaa23
Greetings,

I have to show that

$$\vec{P}=\frac{1}{c^2} \int d^3 x N \left(\dot{A}^{\mu} \nabla A_{\mu}\right)$$

is equivalent to

$$\vec{P}= \sum_{\vec{k}} \hbar \vec{k} N(\vec{k})$$

N in the first expression denotes the normal product, and $$N(\vec{k})$$ is the usual number operator.

Now taking the normal ordering one gets
$$\vec{P}=\frac{1}{c^2} \int d^3 x \left\{\dot{A}^{\mu,+} \nabla A_{\mu}^{+} + \nabla A_{\mu}^{-} \dot{A}^{\mu,+} + \dot{A}^{\mu,-} \nabla A_{\mu}^{+} + \dot{A}^{\mu,-} \nabla A_{\mu}^{-}\right\}$$

The middle terms give the sougth expression, the exponentials in the expansion of the $$A_{\mu}$$ nicely cancelling. The first and the last term however is giving me trouble. I basically end up with an unwanted term of the form

$$\sum_{\vec{k},r} \frac{\hbar \vec{k}}{2} \int \frac{d^3x}{V} \left\{\epsilon^{\mu}_{r} (\vec{k}) \epsilon_{\mu,r} (\vec{k}) \left(a(\vec{k}) a(\vec{k})e^{-2ik.x} + a^{+} (\vec{k}) a^{+} (\vec{k}) e^{2ik.x}\right)\right\}$$

How to get rid of it?

Last edited:
I don't like that box normalization.Here's how i do it.

$$P_{\mu}=\int d^{3}x \ \left(-F^{0\rho}\partial_{\mu}A_{\rho}-\delta_{\mu}^{0}\mathcal{L}_{0}\right)$$ (1)

$$P_{j}=\int d^{3}x \ \left(-F^{0\rho}\partial_{j}A_{\rho}\right) =\int d^{3}x \ \left(-F^{0k}\partial_{j}A_{k}\right)$$ (2)

$$A^{\mu}=\int \frac{d^{3}p}{(2\pi)^{3}2p_{0}(\vec{p})}\left[ a^{\mu}(p)e^{-ipx}+a^{*\mu}(p)e^{ipx}\right]$$ (3)

$$\partial_{j}A_{k}(x) =\int \frac{d^{3}p}{(2\pi)^{3}2p_{0}(\vec{p})} \left(-ip_{j}\right) \left[a_{k}(p) e^{-ipx}-a_{k}^{*}(p) e^{ipx} \right]$$ (4)

$$F^{0k}(x)=\int \frac{d^{3}q}{(2\pi)^{3}2q_{0}(\vec{q})}\left(-i\right) \left\{\left[q_{0}(\vec{q})a^{k}(q)-q^{k}a^{0}(q)\right]e^{-iqx}-\left[q_{0}(\vec{q})a^{k}(q)-q^{k}a^{0}(q)\right]^{*}e^{iqx}\right\}$$ (5)

Therefore

$$P_{j}=\int \frac{d^{3}x \ d^{3}p \ d^{3}q}{(2\pi)^{6}2p_{0}(\vec{p}) \ 2q_{0}(\vec{q})} p_{j}\left[ a_{k}(p)e^{-ipx}-a^{*}_{k}(p)e^{ipx}\right]$$
$$\times \left\{\left[q_{0}(\vec{q})a^{k}(q)-q^{k}a^{0}(q)\right]e^{-iqx}-\left[q_{0}(\vec{q})a^{k}(q)-q^{k}a^{0}(q)\right]^{*}e^{iqx}\right\}$$ (6)

Can u do all calculations...?U'll have to break into 4 parts.2 of them give the desired result,while 2 cancel,as they are integrations of odd functions on even domains wrt the origin.

Daniel.

Last edited:
And of course,"p" is shorthand from the four-vector:

$$p^{\mu}=\left(p_{0}(\vec{p}),\vec{p}\right) \ ;p_{0}\left(\vec{p}\right)=\left|\vec{p}\right|$$

Daniel.

Here's another way of looking at it (other lecture notes,other conventions,blah,blah,blah).

$$A_{\mu}(x)=\frac{1}{(2\pi)^{3/2}}\int \frac{d^{3}k}{\sqrt{2k_{0}}} \sum_{\lambda=0}^{3} \epsilon_{\mu}^{\lambda} \left[a_{\lambda}(k)e^{-ikx}+a_{\lambda}^{*}(k) e^{ikx}\right]$$ (1)

Imposing the Lorenz-Lorentz gauge

$$a_{3}(k)=a_{0}(k)$$ (2)

The calculation is done for the hamiltonian and the momentum is found by covariance.

$$H=-\frac{1}{2}\int d^{3}x \left[(\partial_{0}A_{\mu})(\partial_{0}A^{\mu})+(\nabla A_{\mu})\cdot(\nabla A^{\mu})\right]$$

$$=\frac{1}{2(2\pi)^{3}}\int \frac{d^{3}x \ d^{3}k \ d^{3}q}{\sqrt{2k_{0}q_{0}}} \sum_{\lambda,\lambda'=0}^{3} \left[\epsilon^{\lambda}(k)\epsilon^{\lambda'}(q)\right] \left [\left(k_{0}q_{0}+\vec{k}\cdot\vec{q}\right)a_{\lambda}(k)a_{\lambda'}(q) e^{-i(k+q)x}$$

$$\left -\left(k_{0}q_{0}+\vec{k}\cdot\vec{q}\right)a_{\lambda}(k)a_{\lambda'}^{*}(q) e^{-i(k-q)x}-\left(k_{0}q_{0}+\vec{k}\cdot\vec{q}\right)a_{\lambda}^{*}(k)a_{\lambda'}(q) e^{i(k-q)x}+\left(k_{0}q_{0}+\vec{k}\cdot\vec{q}\right)a_{\lambda}(k)a_{\lambda'}^{*}(q) e^{-i(k+q)x} \right]$$

$$=\frac{1}{2}\int d^{3}k \frac{1}{2k_{0}}\sum_{\lambda,\lambda'=0}^{3} \left[\epsilon^{\lambda}(k)\cdot\epsilon^{\lambda'}(k)\right] \left[-2k_{0}^{2}a_{\lambda}(k)a_{\lambda'}^{*}(k)-2k_{0}^{2}a_{\lambda}^{*}(k)a_{\lambda'}(k)\right]$$

$$=-\frac{1}{2}\int d^{3}k \ k_{0}\sum_{\lambda,\lambda'} g^{\lambda\lambda'} \left[a_{\lambda}(k)a_{\lambda'}^{*}(k)+a_{\lambda}^{*}(k)a_{\lambda'}(k) \right]$$

$$=-\int d^{3}k \ k_{0} \sum_{\lambda,\lambda'} g^{\lambda\lambda'}a^{*}_{\lambda}(k)a_{\lambda'}(k)$$

$$=-\int d^{3}k \ k_{0}\left[a_{0}^{*}(k)a_{0}(k)-a_{1}^{*}(k)a_{1}(k)-a_{2}^{*}(k)a_{2}(k)-a_{3}^{*}(k)a_{3}(k) \right]$$ (3)

To be continued.

Daniel.

Using (3) & (2),one gets the desired result

$$H=\int d^{3}k \ k_{0} \left[a_{1}^{*}(k)a_{1}(k)+a_{2}^{*}(k)a_{2}(k)\right]$$ (4)

Now,through Lorentz covariance,u find easily

$$\vec{P}=\int d^{3}k \ \vec{k} \left[a_{1}^{*}(k)a_{1}(k)+a_{2}^{*}(k)a_{2}(k)\right]$$ (5)

So this is the classical theory.(5) is the momentum of the field.

Now realize the canonical quantization.The complex amplitudes thus far become operators of annihilation and creation.To avoid an infinit vacuum energy & momentum of the quantum field,one introduces the normal ordering product.

The "sought expression" is

$$:\hat{\vec{P}}: \ =\int d^{3}k \ \vec{k}\left[\hat{a}_{1}^{\dagger}(k)\hat{a}_{1}(k)+\hat{a}_{2}^{\dagger}(k)\hat{a}_{2}(k)\right]$$ (6)

Define the Number operator and reconsider hbar and you've got it.

Daniel.

Last edited:
Hmm, I like the second approach. Its a lot more insightful than the I'm working, but then again I'm stuck using the Mandl & Shaw conventions. Putting stuff in a box and then imposing boundary conditions has always struck me as weird...

It doesn't look too logical to me.Sides,there are better books than Mandl & Shaw.I've been taught after Bailin & Love.

Daniel.

## 1. What is QED?

QED stands for Quantum Electrodynamics, which is a branch of physics that studies the interactions between electrically charged particles and photons.

## 2. What is the momentum of the photon-field in QED?

In QED, the momentum of the photon-field is described by the energy-momentum tensor, which is a mathematical quantity that represents the flow of energy and momentum in a system.

## 3. How is the momentum of the photon-field calculated in QED?

The momentum of the photon-field in QED is calculated using the equations of motion, which are derived from the principles of quantum mechanics and special relativity.

## 4. What is the significance of the momentum of the photon-field in QED?

The momentum of the photon-field plays a crucial role in understanding the behavior of particles and their interactions in the quantum world. It helps explain phenomena such as the photoelectric effect and the emission of light from atoms.

## 5. Are there any applications of QED related to the momentum of the photon-field?

Yes, QED has numerous applications in fields such as electronics, photonics, and quantum computing. Understanding the momentum of the photon-field is essential in developing technologies that rely on the principles of quantum mechanics.

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