Show the invariance of the complex-scalar-field Lagrangian

[JD_PM]

Homework Statement

The charge conjugation for the complex Klein-Gordon scalar field $\phi(x)$ is defined by the following transformation



$$\phi \rightarrow U \phi U^{-1} = \eta_c \phi^{\dagger}$$



Where $U$ is a unitary operator which leaves the vacuum invariant (i.e. $U | \ 0 > = | \ 0>$) and $\eta_c$ is a phase factor.



a) Show that $\mathcal{L} = \partial_{\mu} \phi^{\dagger} \partial^{\mu} \phi -m^2 \phi^{\dagger} \phi$ is invariant under such a transformation.



b) Show that the charge current $s^{\alpha} = \frac{-iq}{\hbar} ((\partial^{\alpha} \phi^{\dagger}) \phi - (\partial^{\alpha} \phi) \phi^{\dagger})$ picks up a change of sign under such a transformation.

Source: (part of) Ex. 3.5 QFT book by Franz Mandl and Graham Shaw.

Relevant Equations

$$\phi \rightarrow U \phi U^{-1} = \eta_c \phi^{\dagger}$$

a)

Alright, I think that the trick here is to consider $\phi^{\dagger}$ and $\phi$ as independent scalar fields.

I've read that the unitary matrices read as follows

$$U = e^{i \epsilon}$$

Thus here we have to consider two separate transformations

$$\phi \rightarrow \phi' = e^{i \epsilon} \phi \sim (1 + i \epsilon) \phi$$

$$\phi^{\dagger} \rightarrow \phi'^{\dagger} = e^{i \epsilon} \phi^{\dagger} \sim (1 - i \epsilon) \phi^{\dagger}$$

Both are infinitesimal transformations

$$\phi \rightarrow \phi' = \phi + \delta \phi = \phi + i \epsilon \phi$$

$$\phi^{\dagger} \rightarrow \phi'^{\dagger} = \phi^{\dagger} + \delta \phi^{\dagger} = \phi -i \epsilon \phi^{\dagger}$$

OK this is all the theory I've learned but how can I actually prove invariance of the Lagrangian $\mathcal{L} = \partial_{\mu} \phi^{\dagger} \partial^{\mu} \phi -m^2 \phi^{\dagger} \phi$ under such transformations? Could you please give me a hint?

Thank you.

 

[Gaussian97]

Alright, I think that the trick here is to consider $\phi^{\dagger}$ and $\phi$ as independent scalar fields.

Yes, in general when you deal with complex fields you need to treat $\phi$ and $\phi^\dagger$ as independent fields (if you are not comfortable with that, you can always write $\phi = \frac{1}{\sqrt{2}}\left(\phi_1 + i \phi_2\right)$ with $\phi_1$ and $\phi_2$ independent real fields and you get the same answer)

I've read that the unitary matrices read as follows

$$U = e^{i \epsilon}$$

It's true that any unitary matrix can be written as $U=e^{iH}$ where $H$ is a Hermitian matrix ($H^\dagger = H$), if you mean that then ok, but $\epsilon$ is a strange notation for a Hermitian matrix if you think of $\epsilon$ as a number, then no, what you say is not true unless $U$ is a 1x1 matrix (which don't need to be the case).

Thus here we have to consider two separate transformations

$$\phi \rightarrow \phi' = e^{i \epsilon} \phi \sim (1 + i \epsilon) \phi$$

$$\phi^{\dagger} \rightarrow \phi'^{\dagger} = e^{i \epsilon} \phi^{\dagger} \sim (1 - i \epsilon) \phi^{\dagger}$$

Yes, you need to consider two transformations, how $\phi$ transform, and how $\phi^\dagger$ transform, i.e.
$$\phi\rightarrow \phi' = U\phi U^{-1} = \eta_c \phi^\dagger, \qquad \phi^\dagger \rightarrow (\phi')^\dagger=?$$

But, the exercise tells you to prove that $\mathcal{L}$ is invariant under finite transformations, not only under infinitesimal ones, so you cannot prove this using infinitesimal transformation.

So, before trying to prove that $\mathcal{L}$ is invariant you must find the transformation for $\phi^\dagger$ (the finite one)

 

[JD_PM]

It's true that any unitary matrix can be written as $U=e^{iH}$ where $H$ is a Hermitian matrix ($H^\dagger = H$), if you mean that then ok, but $\epsilon$ is a strange notation for a Hermitian matrix if you think of $\epsilon$ as a number, then no, what you say is not true unless $U$ is a 1x1 matrix (which don't need to be the case).

Oh I missed the Hermitian matrix; my book says

$$U=e^{i \epsilon H}$$

Where $H$ is a Hermitian matrix ($H^\dagger = H$) and $\epsilon$ is a small 'real continuously variable parameter'.

So, before trying to prove that $\mathcal{L}$ is invariant you must find the transformation for $\phi^\dagger$ (the finite one)

OK. the exercise doesn't explicitly show it, but I guess it has to be $\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1} = \eta_c \phi$

But what method should I go for, in order to show invariance?

 

[Gaussian97]

Oh I missed the Hermitian matrix; my book says

$$U=e^{i \epsilon H}$$

Where $H$ is a Hermitian matrix ($H^\dagger = H$) and $\epsilon$ is a small 'real continuously variable parameter'.

Well, I don't know what the book is trying to say, but for sure not all unitary matrices $U$ can be written as you said, for example
$$\begin{pmatrix}
0&1\\
1&0\\
\end{pmatrix}
$$
So I would better drop the $\epsilon$. Anyway, I think we don't need that to do the exercise.

OK. the exercise doesn't explicitly show it, but I guess it has to be $\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1} = \eta_c \phi$

But what method should I go for, in order to show invariance?

I agree with $\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1}$, but no with $\phi^\dagger \rightarrow \phi'^\dagger =\eta_c \phi$ which is the important one here. Notice that, in general, $\eta_c \in \mathbb{C}$

 

[JD_PM]

OK, so assuming that $\eta_c \in \mathbb{C}$ we have

$$\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1} = -\eta_c \phi$$

But what method should I go for, in order to show invariance? I just need a hint to get it I think...

 

[Gaussian97]

Not quite right yet :)
Show what is your reasoning to get this transformation.

Once you have the correct transformation is simply compute the transformed Lagrangian; you have
$$\mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger)=\partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi$$
So under the transformation you have
$$\mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger) \rightarrow \mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger)$$
And you should prove that
$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger)$$
in order to prove that the Lagrangian is invariant.

 

[JD_PM]

Alright so this was my logic; I was at this point:

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (?) \partial^{\mu} (\eta_c \phi^{\dagger}) - m^2(?)(\eta_c \phi^{\dagger})$$

This means that, if we want to get the same Lagrangian after the transformation we need

$$\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1} = \eta^c \phi^{\dagger}$$

Where

$$\eta^c \eta_c = I$$

Thus indeed we have solved a)

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (\eta^c \phi^{\dagger}) \partial^{\mu} (\eta_c \phi^{\dagger}) - m^2(\eta^c \phi^{\dagger})(\eta_c \phi^{\dagger}) = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi = \mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger)$$

This is OK, but it feels like cheating!

If you see it OK, I am going to attack b) next ;)

 

[Gaussian97]

Jajaja no, of course, it's not OK.
First of all, with this transformation $\phi^\dagger \rightarrow \phi'^\dagger= \eta^c \phi^{\dagger}$ doesn't keep the Lagrangian invariant! You have
$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger)
= \partial_\mu \phi^\dagger \partial^\mu \phi^\dagger - m^2 \phi^\dagger \phi^\dagger$$

Then, putting this aside, you cannot choose the transformation of $\phi^\dagger$ to be whatever you need to keep the Lagrangian invariant, once you know the transformation of $\phi$, the transformation of $\phi^\dagger$ is immediately fixed (I think you are thinking too much on this, it's much more easy than what you think).

But you're on the good track with $\eta^c$, although there's a better way to call a number that, multiplied by $\eta_c$ gives 1.

 

[JD_PM]

Damn, I messed it up! Let me correct #7 (the idea is exactly the same though).

Alright so this was my logic; I was at this point (nothing changes here):

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (?) \partial^{\mu} (\eta_c \phi^{\dagger}) - m^2(?)(\eta_c \phi^{\dagger})$$

This means that, if we want to get the same Lagrangian after the transformation we need (I think now is OK)

$$\phi^\dagger \rightarrow \phi'^\dagger = U\phi^\dagger U^{-1} = \eta^c \phi$$

Thus indeed we have solved a) (I think now is OK)

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (\eta^c \phi) \partial^{\mu} (\eta_c \phi^{\dagger}) - m^2(\eta^c \phi)(\eta_c \phi^{\dagger}) = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi = \mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger)$$

Now it should be OK, what do you think?

 

[Gaussian97]

Well, now the transformation for $\phi^\dagger$ is ok, notice that the number $\eta^c$ is just $\eta_c^{-1}$... But still not the correct way to get it, you haven't proved that the Lagrangian is invariant, you have seen that the field must transform in that way if we want the Lagrangian to be invariant, so now you need to actually prove that the field transforms this way. Is very easy, think about how can you prove that
$$\phi \rightarrow \phi' = \eta_c \phi^\dagger \Longrightarrow \phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1} \phi$$

After that, proving b) is equally easy than proving a), so you are almost done.

 

[JD_PM]

But still not the correct way to get it, you haven't proved that the Lagrangian is invariant

Mmm I am afraid I do not get your point. I have explicitly shown (via plug and chug) that the Lagrangian doesn't change under such transformations.

What do you think is left/wrong?

 

[Gaussian97]

You have proven that IF the field transform as $\phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1} \phi$ THEN your lagrangian is invariant.
But you need to prove that IF $\phi \rightarrow \phi' = \eta_c \phi^\dagger$ THEN your Lagrangian is invariant. Is not the same.

As I said, you still need to prove $\phi \rightarrow \phi' = \eta_c \phi^\dagger \Longrightarrow \phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1} \phi$

 

[JD_PM]

Thinking...

 

[JD_PM]

Ahhh I think I get your point now.

So now we have this scenario

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (\eta^c \phi ) \partial^{\mu} (?) - m^2(\eta^c \phi )(?)$$

If $\phi \rightarrow \phi' = \eta_c \phi^\dagger$ then we get that the Lagrangian is invariant; explicitly:

$$\mathcal{L}(\phi', \phi'^\dagger, \partial_\mu \phi', \partial_\mu \phi'^\dagger) = \partial_{\mu} (\eta^c \phi) \partial^{\mu} (\eta_c \phi^{\dagger}) - m^2(\eta^c \phi)(\eta_c \phi^{\dagger}) = \partial_\mu \phi^\dagger \partial^\mu \phi - m^2 \phi^\dagger \phi = \mathcal{L}(\phi, \phi^\dagger, \partial_\mu \phi, \partial_\mu \phi^\dagger)$$

I think you agree with me now! :D

 

[JD_PM]

For b) we use the exact same idea

Here we're dealing with

$$s^{\alpha} = \frac{-iq}{\hbar} ((\partial^{\alpha} \phi^{\dagger}) \phi - (\partial^{\alpha} \phi) \phi^{\dagger})$$

We have to prove both transformations separately.

1) Assuming $\phi \rightarrow \phi' = \eta_c \phi^\dagger$

$$s'^{\alpha} = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} ?) \eta_c \phi^\dagger - (\partial^{\alpha} \eta_c \phi^\dagger) ? \Big)$$

If $\phi^{\dagger} \rightarrow \phi'^{\dagger} = \eta^c \phi$ then we get that the charge picks up a change of sign; explicitly:

$$s'^{\alpha} = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} \eta^c \phi) \eta_c \phi^\dagger - (\partial^{\alpha} \eta_c \phi^\dagger) \eta^c \phi \Big) = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} \phi) \phi^\dagger - (\partial^{\alpha} \phi^\dagger) \phi \Big) = \frac{iq}{\hbar} ((\partial^{\alpha} \phi^{\dagger}) \phi - (\partial^{\alpha} \phi) \phi^{\dagger})$$

Thus $s^{\alpha} = -s'^{\alpha}$, as expected.

2) Assuming $\phi^{\dagger} \rightarrow \phi'^{\dagger} = \eta^c \phi$

$$s'^{\alpha} = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} \eta^c \phi) ? - (\partial^{\alpha} ?) \eta^c \phi \Big)$$

If $\phi \rightarrow \phi' = \eta_c \phi^\dagger$ then we get that the charge picks up a change of sign; explicitly:

$$s'^{\alpha} = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} \eta^c \phi) \eta_c \phi^\dagger - (\partial^{\alpha} \eta_c \phi^\dagger) \eta^c \phi \Big) = \frac{-iq}{\hbar} \Big( (\partial^{\alpha} \phi) \phi^\dagger - (\partial^{\alpha} \phi^\dagger) \phi \Big) = \frac{iq}{\hbar} ((\partial^{\alpha} \phi^{\dagger}) \phi - (\partial^{\alpha} \phi) \phi^{\dagger})$$

Thus $s^{\alpha} = -s'^{\alpha}$, as expected.

PS: I had so so much fun with this one!

 

[Gaussian97]

No, you're not getting my point, forget about the Lagrangian and the current, focus ONLY on the transformation $$\phi \rightarrow \phi'=\eta_c\phi^\dagger$$
Now, you need to prove that
$$\phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1}\phi$$
Tell me ideas, how you think you can prove such a thing?

 

[JD_PM]

No, you're not getting my point, forget about the Lagrangian and the current, focus ONLY on the transformation $$\phi \rightarrow \phi'=\eta_c\phi^\dagger$$
Now, you need to prove that
$$\phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1}\phi$$
Tell me ideas, how you think you can prove such a thing?

It is true that I did not prove $\phi^\dagger \rightarrow \phi'^\dagger = \eta^c \phi$ (that is why I said it felt like cheating).

Well, I see that we have to take the hermitian on both sides of the equation $\phi'=\eta_c\phi^\dagger$ to get $\phi'^{\dagger}=(\eta_c)^{\dagger}(\phi^\dagger)^{\dagger}$

So do you mean I have to prove the following equations:

$$(\phi^\dagger)^{\dagger} = \phi$$

$$(\eta_c)^{\dagger} = \eta^c$$

?

 

[Gaussian97]

Exact, if you prove these two things ($(\phi^\dagger)^\dagger=\phi$ and $\eta_c^\dagger = \eta_c^{-1}$, then you're done.

 

[JD_PM]

1) Prove $(\phi^\dagger)^{\dagger} = \phi$

We know that the conjugate transpose is defined as follows:

$$\phi^\dagger := (\bar \phi)^T$$

Where the overbar denotes the complex conjugate of $\phi$ and $T$ the transpose of $\phi$

$(\phi^\dagger)^{\dagger}$ implies we have to take the transpose of the transposed $\phi$ and the conjugate of the conjugated $\phi$.

But we already know that taking the transpose of the transpose yields $\phi$ itself and that taking the conjugate of the conjugate yields $\phi$ itself as well.

Thus

$$(\phi^\dagger)^{\dagger} = \phi$$

(I have used maybe too many words though).

2) Prove $\eta_c^{\dagger} = \eta^c$

Well, in this problem I knew this equality held due to the fact that we're dealing with unitary matrices, which have such a property.

I think we have nothing to prove here, it just a property.

 

[Gaussian97]

Well, $\eta_c$ is not a matrix, just a complex phase, which indeed has the property $\eta_c^\dagger \equiv \eta_c^* = \eta_c^{-1}$.
Notice that proving
$$\phi \rightarrow \phi' = \eta_c \phi^\dagger \Longrightarrow \phi^\dagger \rightarrow \phi'^\dagger = \eta_c^{-1} \phi $$
is a trivial thing and you didn't need any Lagrangian nor current, now you have proved that the Lagrangian is invariant, and that the current changes sign.

 

[JD_PM]

OK but my point is that $\eta_c^\dagger \equiv \eta_c^* = \eta_c^{-1}$ cannot be proven; it is just a property.

 

[Gaussian97]

Well... As everything is just a matter of definition, you can define a complex phase $\eta_c$ as a complex number with $|\eta_c|=1$, then $\eta_c^*=\eta_c^{-1}$ is a property, but of course can (and must) be proven, in this case, the inverse of a complex number is defined as
$$z^{-1}=\frac{z^*}{|z|^2}$$
So, applying this definition to $\eta_c$ with $|\eta_c|=1$ you have
$$\eta_c^{-1}=\frac{\eta_c^*}{|\eta_c|^2}=\eta_c^*$$

Of course, you don't need to prove this every time, and it's fine if you say "$\eta_c$ is a complex phase so $\eta_c^{-1}=\eta_c^*$", but this doesn't mean that can't be proved.

 

[samalkhaiat]

Oh I missed the Hermitian matrix; my book says
$$U=e^{i \epsilon H}$$
Where $H$ is a Hermitian matrix ($H^\dagger = H$) and $\epsilon$ is a small 'real continuously variable parameter'.

This true ONLY for continuous groups of transformations (Lie groups) like rotations, space-time translations and gauge transformations. For such groups, the concept of “near the identity” or “infinitesimal transformation” makes sense because these (continuous) groups have infinite number of elements. So you can write $U(\epsilon) \approx 1 + \epsilon \cdot J$.

Charge conjugation, like parity and time reversal, belong to discrete groups of transformations. These groups consist of two elements $G = \{ e , a \}$ with the property $a \cdot a = e$. So, there is no such thing as “infinitesimal” or “near the identity” transformation. The rule for these transformation is "You either do one thing (by $a$) or nothing ( by $e$)"

 

[JD_PM]

Charge conjugation, like parity and time reversal, belong to discrete groups of transformations. These groups consist of two elements $G = \{ e , a \}$ with the property $a \cdot a = e$. So, there is no such thing as “infinitesimal” or “near the identity” transformation. The rule for these transformation is "You either do one thing (by $a$) or nothing ( by $e$)"

Hi samalkhaiat, thank you for your reply.

What kind of elements are $e$ and $a$?

 

[samalkhaiat]

Hi samalkhaiat, thank you for your reply.

What kind of elements are $e$ and $a$?

$e$ is the identity transformation (it does nothing), $a$ depends on the type of transformations you are considering, it can be the Parity operator, time-reversal operator or the Charge Conjugation operator.