Quantization of vector field in the Coulomb gauge

[diracologia]

I have a technical question and at the time being I can't ask it to a professor. So, I'm here:

If I try to quantize the vector field in the Coulomb gauge (radiation gauge)

$$A_0(x)=0,\quad \vec\nabla\cdot\vec A=0.$$

by imposing the equal-time commutation relation

$$[A_i(x),E_j(y)]=-i\delta_{ij}\delta(\vec x-\vec y)$$

then I should find

$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]=0,$$
since $\vec\nabla\cdot\vec A=0,$ which is inconsistent with $\partial_i\delta_{ij}\delta(\vec x-\vec y)\neq 0$.

My question is simply how to take this divergence

$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]$$

I'm getting
$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]+A_i\partial_i E_j-(\partial_i E_j)A_i .$$
I must be missing something in the math here. Can anyone help me?

 

[Fredrik]

I don't remember much of this, but if you can write $\vec E=-\nabla\phi$, then the last two terms cancel each other out.

 

[samalkhaiat]

I have a technical question and at the time being I can't ask it to a professor. So, I'm here:

If I try to quantize the vector field in the Coulomb gauge (radiation gauge)

$$A_0(x)=0,\quad \vec\nabla\cdot\vec A=0.$$

by imposing the equal-time commutation relation

$$[A_i(x),E_j(y)]=-i\delta_{ij}\delta(\vec x-\vec y)$$

then I should find

$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]=0,$$
since $\vec\nabla\cdot\vec A=0,$ which is inconsistent with $\partial_i\delta_{ij}\delta(\vec x-\vec y)\neq 0$.

My question is simply how to take this divergence

$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]$$

I'm getting
$$\partial_i[A_i,E_j]=[\vec\nabla\cdot\vec A,E_j]+A_i\partial_i E_j-(\partial_i E_j)A_i .$$
I must be missing something in the math here. Can anyone help me?

$$\partial^{x}_i[A_i(x),E_j(y)]=[\vec\nabla\cdot\vec A,E_j(y)]$$

you are not differentiating with respect to y. If you want to avoid confussion just set y = 0.

Sam

 

[Bill_K]

Kaku's QFT p.110 seems to be addressing your question:

"If we impose canonical commutation relations, we find a further complication.

[A_i(x,t), E_j(y,t)] = −iδ_ijδ(x⃗ − y⃗)

However, this cannot be correct because we can take the divergence of both sides of the equation. The divergence of A_i is zero, so the left-hand side is zero, but the right hand side is not. As a result, we must modify the canonical commutation relations as follows:

[A_i(x,t), E_j(y,t)] = −iδ_ijδ(x⃗ − y⃗)

where the right-hand side must be transverse; that is:

δ_ij = ∫d³k/(2π)³ exp(ik·(x-x') (δ_ij - k_ik_j/k²)

[In other words, in Coulomb gauge only the transverse part is quantized, so only the transverse part appears in the commutator.]

EDIT: In other books they make this even more explicit by putting a "transverse part" operator on both A and E on the left hand side.

 

[diracologia]

Thank you all,

Sam, you solve my puzzle. I just puted $\partial_i$ and forgot that this is a differentiation only over $x$. Shame on me!