# QED : Momentum of the photon-field

1. May 16, 2005

### 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: May 16, 2005
2. May 16, 2005

### dextercioby

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: May 16, 2005
3. May 16, 2005

### dextercioby

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.

4. May 16, 2005

### dextercioby

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.

5. May 16, 2005

### dextercioby

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: May 16, 2005
6. May 17, 2005

### Kalimaa23

Hmm, I like the second approach. Its alot 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...

7. May 17, 2005

### dextercioby

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

Daniel.