Modifying photon states to get a gauge symmetry

[JD_PM]

Homework Statement

Let $|\Psi_T \rangle$ be a state which contains transverse photons only and let

$$|\Psi'_T \rangle = \Big\{1+ a \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] \Big\} |\Psi_T \rangle \tag{1}$$

Where $a$ is a constant. Show that replacing $|\Psi_T \rangle$ by $|\Psi'_T \rangle$ corresponds to a gauge transformation, so that we get

$$\langle \Psi'_T | A^{\mu} (x) | \Psi'_T \rangle = \langle \Psi_T | A^{\mu} (x) + \partial^{\mu} \Lambda (x) | \Psi_T \rangle \tag{2}$$

Where

$$\Lambda (x) = \Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \text{Re}(ia e^{-ik \cdot x}) \tag{3}$$

This is exercise 5.3 of Mandl & Shaw's beautiful book on QFT.

Relevant Equations

$$\langle \Psi'_T | A^{\mu} (x) | \Psi'_T \rangle = \langle \Psi_T | A^{\mu} (x) + \partial^{\mu} \Lambda (x) | \Psi_T \rangle \tag{2}$$

I see that this procedure helps to get rid of the two extra degrees of freedom (due to the scalar and longitudinal photons) one firstly encounters while writing the electromagnetic field theory in a Lorentz-covariant way; it indeed shows that modifying the allowed admixtures of longitudinal and scalar photons is equivalent to apply a gauge transformation between two potentials. In other words: one of these $2$ extra degrees of freedom is shown to correspond to the arbitrary choice of a Lorentz gauge.

The other extra-degree of freedom is removed thanks to the Gupta-Bleuler condition

$$\partial_{\mu} A^{\mu +} (x) | \Psi \rangle = 0$$

Which in momentum space becomes

$$[a_3(\vec k) - a_0(\vec k)] | \Psi \rangle=0$$

But let's focus on the exercise. I proceeded as follows:

We know that the Fourier-expansion of the free-electromagnetic field is $A^{\mu} (x) = A^{\mu +} (x) + A^{\mu -} (x)$, where

$$A^{\mu +} (x) = \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \epsilon_r^{\mu} (\vec k) a_r (\vec k) e^{-i k \cdot x} \tag{4}$$

$$A^{\mu -} (x) = \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \epsilon_r^{\mu} (\vec k) a_r^{\dagger} (\vec k) e^{i k \cdot x} \tag{5}$$

The polarization vectors are defined as follows

$$\epsilon_0^{\mu} (\vec k) = n^{\mu} := (1, 0, 0, 0)$$

$$\epsilon_r^{\mu} (\vec k) = (0, \vec \epsilon_r (\vec k)), \ r=1,2,3$$

Out of $(1)$ we get

$$ \langle \Psi'_T | = \langle \Psi_T | \Big\{1+ a \Big[a_3(\vec k) - a_0(\vec k)\Big] \Big\}$$

Thus, based on $(2)$ we get

$$\langle \Psi'_T | A^{\mu} (x) | \Psi'_T \rangle = \langle \Psi_T | \Big\{1+ a \Big[a_3(\vec k) - a_0(\vec k)\Big] \Big\} A^{\mu} (x) \Big\{1+ a \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] \Big\} |\Psi_T \rangle$$ $$=\langle \Psi_T | A^{\mu} (x) | \Psi_T \rangle + a\langle \Psi_T | A^{\mu} (x) \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] | \Psi_T \rangle + a\langle \Psi_T | \Big[a_3(\vec k) - a_0(\vec k)\Big] A^{\mu} (x) | \Psi_T \rangle$$

Where we noted that the term containing $\Big[a_3(\vec k) - a_0(\vec k)\Big] A^{\mu} (x) \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big]$ vanishes.

We only have to evaluate the terms $a\langle \Psi_T | A^{\mu} (x) \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] | \Psi_T \rangle, a\langle \Psi_T | \Big[a_3(\vec k) - a_0(\vec k)\Big] A^{\mu} (x) | \Psi_T \rangle$

$$a\langle \Psi_T | A^{\mu} (x) \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] | \Psi_T \rangle = a\langle \Psi_T | \Big( A^{\mu +} (x) + A^{\mu -} (x) \Big) \Big[a_3^{\dagger}(\vec k) - a_0^{\dagger}(\vec k)\Big] | \Psi_T \rangle $$ $$=a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k)) e^{-i k \cdot x} | \Psi_T \rangle$$

Where I've used the commutation relation $[a_r (\vec k), a_s^{\dagger} (\vec k')] = \rho_r \delta_{rs} \delta_{\vec k \vec k'}$ (in this specific problem $[a_3 (\vec k), a_3^{\dagger} (\vec k)] = \rho_3 \delta_{33} \delta_{\vec k \vec k} = 1$ and $[a_0 (\vec k), a_0^{\dagger} (\vec k)] = \rho_0 \delta_{00} \delta_{\vec k \vec k} = -1$)

Analogously we get

$$a\langle \Psi_T | \Big[a_3(\vec k) - a_0(\vec k)\Big] A^{\mu} (x) | \Psi_T \rangle = a\langle \Psi_T |
\Big[a_3(\vec k) - a_0(\vec k)\Big] \Big( A^{\mu +} (x) + A^{\mu -} (x) \Big) | \Psi_T \rangle $$ $$=a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k)) e^{i k \cdot x} | \Psi_T \rangle$$

Where I've used the same commutation relation as above.

My issue now is how to show that

$$a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(e^{i k \cdot x} + e^{-i k \cdot x})| \Psi_T \rangle = \langle \Psi_T | \partial^{\mu} \Lambda (x) | \Psi_T \rangle$$

Thank you!

 

[TSny]

In equations (4) and (5), should the numerical factor $\large \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big)$ be raised to the $1/2$ power?

Overall, your work looks good to me.

My issue now is how to show that

$$a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(e^{i k \cdot x} + e^{-i k \cdot x})| \Psi_T \rangle = \langle \Psi_T | \partial^{\mu} \Lambda (x) | \Psi_T \rangle$$

Have you tried using the given expression for $\Lambda (x)$ to see if you can get $\langle \Psi_T | \partial^{\mu} \Lambda (x) | \Psi_T \rangle$ to reduce to $$a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(e^{i k \cdot x} + e^{-i k \cdot x})| \Psi_T \rangle \rm \,\,\,?$$

I'm not knowledgeable in this area. I've only done a little self-study. But I think I was able to get the expressions to match (except for a factor of $c$).

The constant $a$ in equation (1) of your post is not specified to be real. I think things should work out with a complex value of $a$. Some of your expressions would then contain both $a$ and $a^*$.

 

[JD_PM]

Hi TSny, it is always a pleasure to discuss with you!

In equations (4) and (5), should the numerical factor $\large \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big)$ be raised to the $1/2$ power?

You are right, that was a typo.

Have you tried using the given expression for $\Lambda (x)$ to see if you can get $\langle \Psi_T | \partial^{\mu} \Lambda (x) | \Psi_T \rangle$ to reduce to $$a \sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(e^{i k \cdot x} + e^{-i k \cdot x})| \Psi_T \rangle \rm \,\,\,?$$

I'm not knowledgeable in this area. I've only done a little self-study. But I think I was able to get the expressions to match (except for a factor of $c$).

The constant $a$ in equation (1) of your post is not specified to be real. I think things should work out with a complex value of $a$. Some of your expressions would then contain both $a$ and $a^*$.

Oh I did not try it! Let's see

As we know, $\partial^{\mu} := \partial / \partial x_{\mu}$. So if I am not mistaken we get (let me use $\Re$ notation from now on)

$$\partial^{\mu} \Lambda (x) = \Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \Re(ake^{-ik \cdot x})$$

Where $k=k_0+ \vec k=\omega_{\vec k}/c+\vec k$

Let's assume $a \in \Bbb C$. Then out of $(1)$ we get

$$\langle \Psi'_T | = \langle \Psi_T | \Big\{1+ a^* \Big[a_3(\vec k) - a_0(\vec k)\Big] \Big\}$$

Following the above procedure we end up with

$$\sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(a^* e^{i k \cdot x} + a e^{-i k \cdot x})| \Psi_T \rangle$$

Thus what we want to show is that

$$\sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big) \langle \Psi_T | (\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(a^* e^{i k \cdot x} + ae^{-i k \cdot x})| \Psi_T \rangle = \langle \Psi_T | \Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \Re (ake^{-ik \cdot x}) | \Psi_T \rangle$$

Or, alternatively, show that

$$\sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big)(\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))(a^* e^{i k \cdot x} + ae^{-i k \cdot x})=\Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \Re (ake^{-ik \cdot x}) \tag{6}$$

Yikes! But let's not panic. We know that

$$\Re (z)=\frac{z+z^*}{2}$$

Thus

$$\Re (ake^{-ik \cdot x})=k\Big(\frac{ae^{-ik \cdot x}+a^*e^{ik \cdot x}}{2}\Big)$$

And the RHS of $(6)$ becomes

$$\Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \Re (ake^{-ik \cdot x})=\Big( \frac{\hbar c^2}{2 V \omega_k^3} \Big)^{1/2} k(a^* e^{i k \cdot x} + ae^{-i k \cdot x})=\Big( \frac{\hbar c^2}{2 V \omega_k^3} \Big)^{1/2}(\omega_{\vec k}/c+\vec k)(a^* e^{i k \cdot x} + ae^{-i k \cdot x})$$

OK this looks (to me) like a step in the right direction! OK so the issue now is to show that

$$\sum_{r \vec k} \Big( \frac{\hbar c^2}{2 V \omega_{\vec k}} \Big)(\epsilon_3^{\mu} (\vec k) + \epsilon_0^{\mu} (\vec k))=\Big( \frac{\hbar c^2}{2 V \omega_k^3} \Big)^{1/2}(\omega_{\vec k}/c+\vec k) \tag{7}$$

I am much closer now! I think the trick may be on using properties of the polarization vectors. What I've studied is that

$$\vec k \cdot \vec \epsilon_r( \vec k) =0, \ \ r=1, 2 \tag{*}$$

$$\epsilon_r( \vec k) \cdot \epsilon_s( \vec k) = \delta_{rs}, \ \ r,s=1,2,3 \tag{**}$$

I do not think we need this to solve the exercise but anyway: the longitudinal polarization vector can be written in the following covariant form (and honestly, I do not know how to prove it; that goes beyond the exercise of course)

$$\epsilon_3^{\mu} ( \vec k) = \frac{k^{\mu}-(kn)n^{\mu}}{((kn)^2-k^2)^{1/2}} \tag{***}$$

Or I may be over-complicating things

 

[TSny]

Hi TSny, it is always a pleasure to discuss with you!

Thank you.

Oh I did not try it! Let's see

As we know, $\partial^{\mu} := \partial / \partial x_{\mu}$. So if I am not mistaken we get (let me use $\Re$ notation from now on)

$$\partial^{\mu} \Lambda (x) = \Big( \frac{2 \hbar c^2}{V \omega_k^3} \Big)^{1/2} \Re(ake^{-ik \cdot x})$$

The right hand side is not quite correct. Note the appearance of the index $\mu$ on the left side. So, this index should also appear on the right side.

$\frac{\partial }{\partial x_{\mu}} e^{-ik \cdot x} = -ik^{\mu} e^{-ik \cdot x}$

Can you show that $\large k^\mu = \frac{\omega_{\vec k}}{c }\left( \epsilon_0^\mu + \epsilon_3^\mu \right)$?

 

[JD_PM]

Can you show that $\large k^\mu = \frac{\omega_{\vec k}}{c }\left( \epsilon_0^\mu + \epsilon_3^\mu \right)$?

Oh I see that showing so would basically solve the problem!

I think we have

$$k^{\mu}=k_0^{\mu}+k_3^{\mu} = \frac{\omega_{\vec k}}{c }n^{\mu}+k_3^{\mu}= \frac{\omega_{\vec k}}{c }\epsilon_0^\mu+k_3^{\mu}$$

Naive question: why $k_3^{\mu}= \frac{\omega_{\vec k}}{c }\epsilon_3^\mu$?

 

[TSny]

$$k^{\mu}=k_0^{\mu}+k_3^{\mu} = \frac{\omega_{\vec k}}{c }n^{\mu}+k_3^{\mu}= \frac{\omega_{\vec k}}{c }\epsilon_0^\mu+k_3^{\mu}$$

The four-vector $k$ has components $k^\mu$, where $\mu = 0, 1, 2, 3$. There is not a four-vector $k_0$ or $k_3$. So, I don't believe the notation $k_0^\mu$ and $k_3^\mu$ is appropriate.

In order to verify $\large k^\mu = \frac{\omega_{\vec k}}{c }\left( \epsilon_0^\mu + \epsilon_3^\mu \right)$, you can just check that the left side equals the right side for all four choices $\mu = 0, 1, 2, 3$.

Start by writing out explicitly the 4 components $k^\mu$ in terms of $|\vec k|$ or $\frac{\omega_{\vec k}}{c }$.

 

[JD_PM]

Start by writing out explicitly the 4 components $k^\mu$ in terms of $|\vec k|$ or $\frac{\omega_{\vec k}}{c }$.

Alright, I think I got it. We know that

$$k^\mu=(\frac{\omega}{c }, \vec k)$$

When $k^0$ we have

$$k^0= \frac{\omega}{c}= \frac{\omega}{c}(1+0)= \frac{\omega}{c}(\epsilon_0^0+\epsilon_3^0) \tag{8}$$

When $\vec k$ we have

$$\vec k= \frac{\omega}{c} \hat k= \frac{\omega}{c} (0+\hat k)= \frac{\omega}{c}(\epsilon_0^r+\epsilon_3^r) \tag{9}$$

Adding up $(8)$ and $(9)$ yields

$$\large k^\mu = \frac{\omega_k}{c }\left( \epsilon_0^\mu + \epsilon_3^\mu \right)$$

I used $\frac{\omega_k}{c }$ instead of $\frac{\omega_{\vec k}}{c }$ to emphasize k is a fourth vector. Maybe I am missing something and it is indeed a 3 vector.

 

[TSny]

Alright, I think I got it. We know that

$$k^\mu=(\frac{\omega}{c }, \vec k)$$

When $k^0$ we have

$$k^0= \frac{\omega}{c}= \frac{\omega}{c}(1+0)= \frac{\omega}{c}(\epsilon_0^0+\epsilon_3^0) \tag{8}$$

Yes. Here $\omega$ depends on $\vec k$: $\, \omega = c |\vec k|$. So, we often write $\omega$ as $\large \omega_{\vec k}$ .

When $\vec k$ we have

$$\vec k= \frac{\omega}{c} \hat k= \frac{\omega}{c} (0+\hat k)= \frac{\omega}{c}(\epsilon_0^r+\epsilon_3^r) \tag{9}$$

Yes, this is the right idea. You could quibble over the notation. Your last expression has a specific index $r$ to denote one of the spatial components of $\vec{\epsilon}_0$ or $\vec{\epsilon}_3$; whereas, the other expressions in the equation do not have the index $r$. So, this is a little inconsistent. Instead, you could write $$\large \vec k= \frac{\omega_{\vec k}}{c} \hat k= \frac{\omega_{\vec k}}{c} (0+\hat k)= \frac{\omega_{\vec k}}{c} \left(\vec{\epsilon}_0+\vec{\epsilon_3} \right) $$ or you could write

$$ \large k^r= \frac{\omega_{\vec k}}{c} \left(\epsilon_0^r+\epsilon_3^r \right), \,\,\,\, \normalsize r = 1, 2, 3$$

As you said, combining the results of (8) and (9) yields
$$\large k^\mu = \frac{\omega_{\vec k}}{c }\left( \epsilon_0^\mu + \epsilon_3^\mu \right)$$

I used $\frac{\omega_k}{c }$ instead of $\frac{\omega_{\vec k}}{c }$ to emphasize k is a fourth vector. Maybe I am missing something and it is indeed a 3 vector.

Yes, k is a 4-vector and $\vec k$ is the 3-vector part of k. It's probably better to write $\large \omega_{\vec k}$ rather than $\large \omega_k$ in equations (4) and (5) since you are summing over the 3-vectors $\vec k$.

Note that in (4) and (5) you are also summing over the subscript $r$ that denotes the different polarization directions. This $r$ is not to be confused with the superscript $r$ we used above to denote the components of a 3-vector.

Anyway, this is how I see it. As I said, I'm still wet behind the ears on this stuff. So, I could be mistaken in some of what I wrote.