Solve Srednicki Problem 3.3: Show U(\Lambda) Invariance

[WannabeNewton]

Homework Statement

Use $U(\Lambda)^{-1}\varphi(x)U(\Lambda) = \varphi(\Lambda^{-1}x)$ to show that $U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) = a(\Lambda^{-1}\mathbf{k})$ and $U(\Lambda)^{-1}a^{\dagger}(\mathbf{k})U(\Lambda) = a^{\dagger}(\Lambda^{-1}\mathbf{k})$ and hence that $U(\Lambda)|k_1...k_n\rangle = |\Lambda k_1...\Lambda k_n \rangle$ where $|k_1...k_n\rangle$ is a multi-particle state of $n$ particles with momenta $k_1,...,k_n$.

Homework Equations

The real free scalar field: $\varphi(x) = \int d\tilde{k} [a(\mathbf{k})e^{ikx} + a^{\dagger}(\mathbf{k})e^{-ikx}]$ where $d\tilde{k}$ is the Lorentz-invariant measure and $kx = k_{\mu}x^{\mu}$.

The Attempt at a Solution

I can't even start the problem because of what Srednicki writes for the transformation of the annihilation and creation operators under the unitary representation $U(\Lambda)$ of the Lorentz transformation $\Lambda$. In particular, he writes $a(\Lambda^{-1}\mathbf{k})$ and $a^{\dagger}(\Lambda^{-1}\mathbf{k})$ but $\mathbf{k}$ is a 3-vector (in particular the 3-momentum) so what does $\Lambda^{-1}\mathbf{k}$ even mean? Matrix representations of Lorentz transformations have Lorentz indices and hence must be contracted with other Lorentz indices so as to result in Lorentz indices after the contraction e.g. $k^{\mu'} = (\Lambda^{-1})^{\mu'}{}{}_{\nu}k^{\nu} = \Lambda_{\nu}{}{}^{\mu'}k^{\nu}$. But a 3-vector like $\mathbf{k}$ doesn't have Lorentz indices so I can't make any sense of $\Lambda^{-1}\mathbf{k}$. Thanks in advance!

 

[vanhees71]

This is very sloppy notation. Of course, you must use the energy-momentum four vector to define the Lorentz transformation of momentum. For free particles, the four-momentum is onshell, i.e.,
$$E(\vec{p})=\sqrt{m^2+\vec{p}^2} \; \Leftrightarrow \; p_{\mu} p^{\mu}=m^2.$$
Of course also the Lorentz transformed four vector,
$$p'^{\mu} = {\Lambda^{\mu}}_{\nu} p^{\nu}$$
obeys the on-shell condition.

Just use the "local" transformation property of the field operators and the mode decomposition of the field operator with respect to annihilation and creation operators with respect to the momentum eigenbasis, as well as the Lorentz invariance of the invariant integration measure over on-shell momenta,
$$\frac{\mathrm{d}^3 \vec{p}}{2 E(\vec{p})}=\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2).$$

 

[WannabeNewton]

Thanks vanhees! Should I Fourier transform $\varphi(x)$ to its momentum representation first? In chapter 3, Srednicki only gives the mode decomposition of $\varphi(x)$ in the coordinate representation: $\varphi(x) = \int d\tilde{k} [a(\mathbf{k})e^{ikx} + a^{\dagger}(\mathbf{k})e^{-ikx}]$; will this problem be substantially easier if I work with $\tilde{\varphi}(k) = \int d^4 x e^{-ikx}\varphi(x)$ and derive the mode decomposition of $\tilde{\varphi}(k)$ in terms of $a(\mathbf{k})$ and $a^{\dagger}(\mathbf{k})$ using the mode decomposition of $\varphi(x)$?

Also he defines the Lorentz invariant measure as $d\tilde{k} = \frac{d^3 k}{(2\pi)^3 2\omega}$ so I'm not sure how to go from this to the expression you wrote with the delta function, theta function, and $d^4 k$.

 

[strangerep]

Should I Fourier transform $\varphi(x)$ to its momentum representation first?

Don't over-think this. I reckon this exercise can be done in just a few lines once you recognize some "cut-through" techniques.

Hmm, now... how can I hint at those techniques without actually giving you the answer?

Let's see...

1) What else do you know about $\Lambda$? I.e., what matrix properties does it satisfy?

2) What can you say (if anything) about linear independence of $e^{ikx}$ and $e^{-ikx}$ ?

3) Write $kx$ in more explicitly vectorial notation. E.g., similar to how we can write $u \cdot v$ as $u^T v$. If we apply an inner-product preserving transformation to $v$, how can we re-express the inner product with the transformation acting only $u$ ?

Also he defines the Lorentz invariant measure as $d\tilde{k} = \frac{d^3 k}{(2\pi)^3 2\omega}$ so I'm not sure how to go from this to the expression [Hendrik] wrote with the delta function, theta function, and $d^4 k$.

That's a different issue, and (I think?) not essential for this exercise. Anyway, Srednicki mentions this in (3.16) and (3.17).

BTW, keep in mind that the expressions for $\phi$ should perhaps be more explicitly written like this: $$
\phi(x) ~=~ \int \Big[ \cdots \Big]_{\omega^2 - k^2 = m^2}
$$ which emphasizes that $\phi$ has support only the mass hyperboloid in momentum space.

 

[WannabeNewton]

Hi Strangerep thanks for replying. I was on a 6 hour bus ride from my university back to my home for Christmas break so I didn't see your post until now; however during my bus ride I had already started working on the problem using the method I mentioned earlier. I'll post what I have thus far and it would be awesome if you could help me figure out the last few steps because I'm stuck. In the meanwhile I'll use the prompts you gave in your post to see if I can do the problem using the faster/more efficient method that you had in mind. Thanks again!

So the Fourier transform of the free scalar field will be given by $\tilde{\varphi}(k) = \int d\tilde{k}'d^{4}x[a(\mathbf{k}')e^{i(k' - k)x} + a^{\dagger}(\mathbf{k}')e^{-i(k' + k)x}] \\= 2\pi \int dk'^{0}d^{3}k' \delta(k'^2 + m'^2)\theta(k'^0)[a(\mathbf{k}')\delta^{4}(k - k') +a^{\dagger}(\mathbf{k}')\delta^{4}(k + k') ] \\= 2\pi \int dk'^{0} \delta(k^2 + m^2) \theta(k'^0) [a(\mathbf{k}) \delta(k^0 - k'^0) +a^{\dagger}(\mathbf{-k}) \delta(k^0 + k'^0)] \\= 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)]$

where I have used (3.16).

Furthermore, $U^{-1}\tilde{\varphi}(k)U = \int d^{4}xe^{-ikx}U^{-1}\varphi(x)U = \int d^{4}xe^{-ikx}\varphi(\Lambda^{-1}x)$.

Define $x' = \Lambda^{-1}x$ so that $\int d^{4}xe^{-ikx}\varphi(\Lambda^{-1}x)=\int d^{4}x'e^{-ik(\Lambda x')}\varphi(x') = \int d^{4}x' e^{-i(\Lambda^{-1}k)x'}\varphi(x') = \tilde{\varphi}(\Lambda^{-1}k)$

where I have used the fact that $d^4 x = d^4 x'$ under a Lorentz transformation since the space-time measure is Lorentz-invariant and $\Lambda^{\mu}{}{}_{\nu}k_{\mu}x'^{\nu} = \Lambda_{\mu}{}{}^{\nu}k^{\mu}x'_{\nu} = (\Lambda^{-1})^{\nu}{}{}_{\mu}k^{\mu}x'_{\nu}$.

So collecting everything together we have
$\tilde{\varphi}(k) = 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)]$ and $U^{-1}\tilde{\varphi}(k)U = \tilde{\varphi}(\Lambda^{-1}k)$.

This gives us $\delta(k^2 + m^2)[\theta(k^0) U^{-1}a(\mathbf{k})U+\theta(-k^0)U^{-1}a^{\dagger}(\mathbf{-k})U] \\= \delta(k^2 + m^2)[\theta(\Lambda^{-1}k^0)a(\Lambda^{-1}\mathbf{k}) +\theta(-\Lambda^{-1}k^0)a^{\dagger}(-\Lambda^{-1}\mathbf{k})]$

This is where I'm stuck. I'm not sure how to go from the above equation to the desired result.

 

[strangerep]

You could invoke something$^*$ to restrict to orthochronous $\Lambda$, and that would simplify your $\theta$ expressions. But that doesn't get you to the final result.

TBH, I think that by Fourier transforming as your first step, you lost the ability to apply my hint (2).
[Edit: Or maybe not. After fixing up your $\theta$ functions, you can note that for any given $k_0$, only 1 of the terms is non-zero. That gets you a little closer, I think.]

But to re-start with the very first eqn in the exercise, i.e.,
$$
U(\Lambda)^{-1} \phi(x) U(\Lambda) ~=~ \phi(\Lambda^{-1} x) ~.
$$
Note that the 2nd eqn in (3.34) is just the adjoint of the 1st, so you only need to prove the 1st. So try to get equations involving $a$ and $a^\dagger$ separately.

[$^*$] [Edit: The "something" I mentioned above is on Srednicki p17. He excludes parity and time reversal from his treatment at this early stage -- to be revisited later.]

 

[vanhees71]

Despite the fact that you use the east-coast instead of the west-coast metric (which I'm used to; what is Srednicky using?), you already used the relation for the invariant integration measure over momenta on the mass shell I gave. It's, however better to simply use the local transformation properties of the scalar field operator (it's constructed such that it fulfills that relation, because you aim at a local QFT from the very beginning!).

Using also the east-cost convention for the metric this reads
$$U^{-1}(\Lambda) \phi(x) U(\Lambda)=\phi(\Lambda^{-1} x).$$
This you translate into the mode decomposition for the uncharged scalar field
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [U^{-1}(\Lambda) a(\vec{p}) U(\Lambda) \exp(\mathrm{i} x \cdot p) + U^{-1}(\Lambda) a^{\dagger}(\vec{p}) U(\Lambda) \exp(-\mathrm{i} x \cdot p) \right]_{p^0=E(\vec{p})} \\ \stackrel{!}{=} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [a(\vec{p}) \exp(\mathrm{i} (\Lambda^{-1} x) \cdot p) + a^{\dagger}(\vec{p}) \exp(-\mathrm{i} (\Lambda^{-1}x) \cdot p) \right]_{p^0=E(\vec{p})}.$$
Now after some playing with the Lorentz-transformation properties of four-vectors on the right-hand side, doing an appropriate substitution in the momentum integral and then comparing the Fourier coefficients you should get the desired transformation properties of the annihilation and creation operators (note that $U(\lambda)$ is a unitary transformation!).

 

[andrien]

you have $\tilde{\varphi}(k) = 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)]$,So for positive k₀ you have
$2\pi \delta(k^2 + m^2)a(\mathbf{k})=\tilde{\varphi}(k)$.Now make the orthochronous Lorentz transformation on it and use it once again while doing to get the result.

 

[WannabeNewton]

Thanks for the help guys! Strangerep thanks I'll redo the problem starting from your hint (2) right after I get one last thing cleared up based on what andrien said directly above: does an orthochronous Lorentz transformation necessarily preserve the positivity of $k^0$? We have $k^{0'} = \Lambda^{0'}_{0}k^{0} + \Lambda^{0'}_{i}k^{i}$ and we know that $\Lambda^{0'}_{0}k^{0} >0$ but what about $\Lambda^{0'}_{i}k^{i}$?

Also, there is no loss of generality in restricting ourselves to positive $k^0$ right?

Thanks again everyone.

 

[George Jones]

Despite the fact that you use the east-coast instead of the west-coast metric (which I'm used to; what is Srednicky using?)

Srednicky uses east-coast, which, as you note, is quite unusual for qft/high energy.

does an orthochronous Lorentz transformation necessarily preserve the positivity of $k^0$?

Yes, hence the name "orthochronous". See theorem 1.3.3 of Naber.

restricting ourselves to positive $k^0$

In the rest frame of the particle, what does this mean?

 

[WannabeNewton]

Yes, hence the name "orthochronous". See theorem 1.3.3 of Naber.

Awesome thanks!

In the rest frame of the particle, what does this mean?

Ok so in the rest frame of the particle, $k^0 = m > 0$ since rest mass is positive. Then using the theorem you referenced in Naber, restricting ourselves to orthochronous Lorentz transformations we have $k^0 > 0$ in all reference frames so we basically can only use $k^0 > 0$ right? It wouldn't even make sense to consider $k^0 < 0$?

Thanks George!

 

[strangerep]

It wouldn't even make sense to consider $k^0 < 0$?

For this exercise you could also consider $k^0 < 0$.
Think about what the last line of your earlier post #5 simplifies to in this case...

 

[WannabeNewton]

For this exercise you could also consider $k^0 < 0$.
Think about what the last line of your earlier post #5 simplifies to in this case...

Right so if we happened to restrict ourselves to $k^0 < 0$ then we would get $U^{-1} a^{\dagger}(-\mathbf{k})U = a(-\Lambda^{-1}\mathbf{k})$ but if we restrict ourselves to $k^0 < 0$ will $U^{-1} a^{\dagger}(-\mathbf{k})U = a(-\Lambda^{-1}\mathbf{k})$ hold for all $\mathbf{k}$? That's what we want right? We want it to hold for all possible 3-momenta $\mathbf{k}$.

Thanks!

 

[strangerep]

Right so if we happened to restrict ourselves to $k^0 < 0$ then we would get $U^{-1} a^{\dagger}(-\mathbf{k})U = a(-\Lambda^{-1}\mathbf{k})$

I presume that's a typo on the rhs, and you meant: $\dots = a^\dagger(-\Lambda^{-1}\mathbf{k})$ ?

[...] That's what we want right? We want it to hold for all possible 3-momenta $\mathbf{k}$.

Yes.

 

[WannabeNewton]

Sorry, yes that was a typo. Also I think I phrased my question ambiguously. I don't get why we don't lose any generality by restricting ourselves to $k^0 < 0$ and subsequently deriving $U^{-1}a^{\dagger}(-\mathbf{k})U = a^{\dagger}(-\Lambda^{-1} \mathbf{k})$. Why does this necessarily hold for all $\mathbf{k}$ even after restricting to $k^0 < 0$? I only ask because $\mathbf{k}$ isn't independent of $k^0$. Thanks!

EDIT: also for the very last part of the problem i.e. to show that $U|k_1...k_n \rangle = |\Lambda k_1...\Lambda k_n \rangle$, can I simply say that since $U^{-1}a^{\dagger}(\mathbf{k})U = a^{\dagger}(\Lambda^{-1} \mathbf{k})$ we have as a result $U^{-1}a^{\dagger}(\Lambda \mathbf{k})U = a^{\dagger}(\mathbf{k})$ hence
$U|k_1...k_n \rangle \\= U a^{\dagger}(\mathbf{k}_1)...a^{\dagger}(\mathbf{k}_n)|0\rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)Ua^{\dagger}(\mathbf{k}_2)...a^{\dagger}(\mathbf{k}_n)|0 \rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)...a^{\dagger}(\Lambda \mathbf{k}_n)|0\rangle \\= |\Lambda k_1...\Lambda k_n \rangle$
?

 

[strangerep]

[...] I don't get why we don't lose any generality by restricting ourselves to $k^0 < 0$ and subsequently deriving $U^{-1}a^{\dagger}(-\mathbf{k})U = a^{\dagger}(-\Lambda^{-1} \mathbf{k})$. Why does this necessarily hold for all $\mathbf{k}$ even after restricting to $k^0 < 0$? I only ask because $\mathbf{k}$ isn't independent of $k^0$.

Note that I didn't say "restrict" to $k^0 < 0$. I said "consider".

But ok, let's go back to the full equation at the end of your post #5. This holds for all 4-vectors $k$ satisfying the mass-shell constraint $k^2 = m^2$. We can rewrite that constraint as $k^0 = \pm\sqrt{m^2 + |\mathbf{k}|^2}$ . Therefore it is sufficient to consider your equation for these 2 cases separately. For each sign of $k^0$, only one of $\theta$'s is nonzero. So you get the transformation rule for $a$ and $a^\dagger$ separately. In each case, the rules apply to all 3-vector $\mathbf{k}$ -- only the sign of $k^0$ is different. And (just as well!), the rules turn out to be consistent with each other, as can be seen by taking adjoints.

also for the very last part of the problem i.e. to show that $U|k_1...k_n \rangle = |\Lambda k_1...\Lambda k_n \rangle$, can I simply say that since $U^{-1}a^{\dagger}(\mathbf{k})U = a^{\dagger}(\Lambda^{-1} \mathbf{k})$ we have as a result $U^{-1}a^{\dagger}(\Lambda \mathbf{k})U = a^{\dagger}(\mathbf{k})$ hence
$U|k_1...k_n \rangle \\= U a^{\dagger}(\mathbf{k}_1)...a^{\dagger}(\mathbf{k}_n)|0\rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)Ua^{\dagger}(\mathbf{k}_2)...a^{\dagger}(\mathbf{k}_n)|0 \rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)...a^{\dagger}(\Lambda \mathbf{k}_n)|0\rangle \\= |\Lambda k_1...\Lambda k_n \rangle$
?

Yes, that's the basic idea, since the vacuum state is Lorentz-invariant.

 

[George Jones]

Also, there is no loss of generality in restricting ourselves to positive $k^0$ right?

Another way of looking at this, related to strangerep's comment

And (just as well!), the rules turn out to be consistent with each other, as can be seen by taking adjoints

is through

So the Fourier transform of the free scalar field will be given by $\tilde{\varphi}(k) = 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)]$

What is $\tilde{\varphi}(-k)$?

Ok so in the rest frame of the particle, $k^0 = m > 0$ since rest mass is positive. Then using the theorem you referenced in Naber, restricting ourselves to orthochronous Lorentz transformations we have $k^0 > 0$ in all reference frames so we basically can only use $k^0 > 0$ right? It wouldn't even make sense to consider $k^0 < 0$?

These negative mass solutions are the infamous negative energy solutions, and are extremely important. Interpreted correctly, they give *positive* mass antiparticles. If you want, I can expand substantially on this.

 

[WannabeNewton]

In each case, the rules apply to all 3-vector $\mathbf{k}$ -- only the sign of $k^0$ is different. And (just as well!), the rules turn out to be consistent with each other, as can be seen by taking adjoints.

Ok so to make sure I understand, say we take an arbitrary $\mathbf{k}$ then there are two valid choices for $k^0$ that differ by sign. The positive choice gives us the equation for $a$ and the negative choice gives us the equation for $a^{\dagger}$ but regardless the equations will hold for all $\mathbf{k}$ since $\mathbf{k}$ was taken to be arbitrary and we can branch off into freedom to take both negative and positive $k^0$ in the on-shell relation.

So, in other words, the set of all possible $\mathbf{k}$ along with the set of all associated positive $k^0$ gives us the equation for $a$ whereas the set of all possible $\mathbf{k}$ along with the set of all associated negative $k^0$ gives us the equation for $a^{\dagger}$.

Does that sound about right? And thanks a million for the help!

What is $\tilde{\varphi}(-k)$?

Ah right since $\tilde{\varphi}^{\dagger}(k) = \tilde{\varphi}(-k)$. Thanks!

If you want, I can expand substantially on this.

I would absolutely love it if you could! Srednicki's explanation of the issue was not at all clear to me (he does it very briefly in chapter 1). Thanks!

 

[strangerep]

Does that sound about right?

Yes.

 

[George Jones]

Ah right since $\tilde{\varphi}^{\dagger}(k) = \tilde{\varphi}(-k)$.

So, in some sense, the negative mass hyperboloid doesn't give any information not already given by the positive mass hyperboloid. This is because this thread deals with a real scalar field, and, for real scalar fields, particles are their own antiparticles.

I would absolutely love it if you could! Srednicki's explanation of the issue was not at all clear to me (he does it very briefly in chapter 1). Thanks!

This may take some time, but I (with the help of Folland and others) definitely will do this. It might be better if I put the material in this thread

https://www.physicsforums.com/showthread.php?t=727571.

 

[vanhees71]

Despite the fact that you use the east-coast instead of the west-coast metric (which I'm used to; what is Srednicky using?), you already used the relation for the invariant integration measure over momenta on the mass shell I gave. It's, however better to simply use the local transformation properties of the scalar field operator (it's constructed such that it fulfills that relation, because you aim at a local QFT from the very beginning!).

Using also the east-cost convention for the metric this reads
$$U^{-1}(\Lambda) \phi(x) U(\Lambda)=\phi(\Lambda^{-1} x).$$
This you translate into the mode decomposition for the uncharged scalar field
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [U^{-1}(\Lambda) a(\vec{p}) U(\Lambda) \exp(\mathrm{i} x \cdot p) + U^{-1}(\Lambda) a^{\dagger}(\vec{p}) U(\Lambda) \exp(-\mathrm{i} x \cdot p) \right]_{p^0=E(\vec{p})} \\ \stackrel{!}{=} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [a(\vec{p}) \exp(\mathrm{i} (\Lambda^{-1} x) \cdot p) + a^{\dagger}(\vec{p}) \exp(-\mathrm{i} (\Lambda^{-1}x) \cdot p) \right]_{p^0=E(\vec{p})}.$$
Now after some playing with the Lorentz-transformation properties of four-vectors on the right-hand side, doing an appropriate substitution in the momentum integral and then comparing the Fourier coefficients you should get the desired transformation properties of the annihilation and creation operators (note that $U(\lambda)$ is a unitary transformation!).

For completeness, let's solve the problem with this method, which I think is simpler.

To get the desired transformation law for the annihilation and creation operators (which for the neutral scalar field are just conjugates to each other), you simply substitute $$p=\Lambda^{-1} p'$$ for the on-shell four-momentum $p$, using the fact that the Lorentz transformed four-momentum of course also obeys the on-shell condition, i.e., $p^2=p'^2=m^2$, the invariance of $\mathrm{d}^3 \vec{p}/E(\vec{p})$, and that for orthochronous Lorentz transformations the sign of $p^0$ doesn't change. This gives for the right-hand side of my previous equation in the above quote
$$\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}'}{(2 \pi)^3 E(\vec{p}')} \left [a(\overrightarrow{\Lambda^{-1} p'}) \exp(\mathrm{i} p' \cdot x') +\text{c.c.} \right ]_{p'^0=E(\vec{p}')}.$$
Comparing this with the left-hand side of the last equation of my quote, gives the transformation rule
$$U^{-1}(\Lambda)a(\vec{p}) U(\Lambda) = a(\overrightarrow{\Lambda^{-1} p}).$$
Taking the Hermitean conjugate of this equation together with the unitarity of $U(\Lambda)$ shows that the same rule holds for $a^{\dagger}$, which makes the equation consistent.

 

[strangerep]

For completeness, let's solve the problem with this method, which I think is simpler. [...]

Aargh! I was trying to get WBN to work that method through for himself...

 

[WannabeNewton]

Haha actually I attempted to do the problem in the exact same way vanhees did it above when I first looked at the problem so it's not all in vain!

I had $\int \frac{d^{3}k}{(2\pi)^3 2\omega}[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\= \int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\mathbf{k})e^{ik(\Lambda^{-1}x)} + a^{\dagger}(\mathbf{k})e^{-ik(\Lambda^{-1}x)}] \\= \int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\mathbf{k})e^{i(\Lambda k)x} + a^{\dagger}(\mathbf{k})e^{-i(\Lambda k)x}]$

Then I let $k' = \Lambda k$ to get $[a(\mathbf{k})e^{i(\Lambda k)x} + a^{\dagger}(\mathbf{k})e^{-i(\Lambda k)x}] = [a(\Lambda^{-1}\mathbf{k}')e^{ik'x} + a^{\dagger}(\Lambda^{-1}\mathbf{k}')e^{-ik'x}]$ but this is where I got stuck because the integral is over momentum 3-space not 4-space so I thought that the measure $\frac{d^{3}k}{(2\pi)^3 2\omega}$ transformed in a non-trivial way under the transformation $k' = \Lambda k$ and switched over to Fourier transforms so that I could work with 4-space quantities wherein I knew for a fact that the space-time volume element was Lorentz-invariant.

I totally forgot that the measure $\frac{d^{3}k}{(2\pi)^3 2\omega}$ was Lorentz-invariant, despite me repeatedly calling it the "Lorentz-invariant" measure in this and other threads! So of course under $k' = \Lambda k$ we would get $\frac{d^{3}k}{(2\pi)^3 2\omega} = \frac{d^{3}k'}{(2\pi)^3 2\omega'}$ and hence $\int \frac{d^{3}k}{(2\pi)^3 2\omega}[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\= \int \frac{d^{3}k'}{(2\pi)^3 2\omega'}[a(\Lambda^{-1}\mathbf{k}')e^{ik'x} + a^{\dagger}(\Lambda^{-1}\mathbf{k}')e^{-ik'x}] \\ =\int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\Lambda^{-1}\mathbf{k})e^{ikx} + a^{\dagger}(\Lambda^{-1}\mathbf{k})e^{-ikx}]$

where I relabeled the dummy index $k'$ to $k$ since it's just an integration variable and I have the freedom to do that.

How would I formally go about getting the desired results for the creation and annihilation operators from the final equation above?

Also where exactly did we need that $\Lambda$ is orthochronous and that $k'^2 = k^2 = -m^2$ in the above method? All I can see being needed is that the momentum 3-space measure given by $\frac{d^{3}k}{(2\pi)^3 2\omega}$ is Lorentz-invariant.

Thanks!

 

[vanhees71]

You need the "orthochronizity" of $\Lambda$ just that you are allowed to conclude from $p^0=+E(\vec{p})$ that $p'^0=+E(\vec{p})$. For a general Lorentz transformation you can only use the on-shell condition, which yields only $p'^0=\pm E(\vec{p})$.

It is physically important to keep in mind that "Lorentz invariance/covariance" refers only to invariance under proper orthochronous Lorentz transformations. The analysis of the unitary ray representations of the Poincare group (leading uniquely to those representations that can be deduced from unitary representations of the covering group, which substitutes $\mathrm{SO}(1,3)^{\uparrow}$ with its covering group $\mathrm{SL}(2,\mathbb{C})$. That leads to the various types of fields, characterized by two spin-like quantum numbers $(j,j')$ with $j,j' \in \{0,1/2,1,\ldots \}$. A local QFT with stable ground state then is also automatically invariant under the CPT transformation (C=charge conjugation=exchange of particles and antiparticles, P=parity=spatial reflections, T=time reversal=(time reflection)).

If a certain model is in addition also invariant/covariant under larger Lorentz groups, this implies that it is invariant under some (or all) of the "discrete" transformations that can be built from parity (space-reflection), time-reversal, and charge-conjugation transformations. E.g., the electromagnetic force obeys all three of these transformations (and thus also all combinations). This implies that for the electrons you must use the Dirac field, i.e., the direct sum of the (1/2,0) and the (0,1/2) representation of the $\mathrm{SL}(2,\mathbb{C})$. The same holds true for the strong interaction and QCD (where it is still a puzzle what mechanism implies CP invariance of the strong interaction). The weak force violates P, CP (which both is experimentally proven), and thus also T (which follows from the CPT theorem and the violation of P and CP, but is not directly independently proven by experiment as far as I know).

 

[WannabeNewton]

Thanks! So if I understood correctly, the necessity of $\Lambda$ being orthochronous is implicit in the Lorentz-invariance of the measure?

And thanks for all of the additional information, I appreciate it :)

 

[andrien]

The way the invariance of measure is shown goes like d³p/2E=∫d⁴p[δ(p₀-√(|p|²+m²))]/2√(|p|²+m²)
=∫d⁴p[δ(p₀-√(|p|²+m²))+δ(p₀+√(|p|²+m²))]/2√(|p|²+m²),where we restrict now p₀>0 which equals
∫d⁴pδ(p²+m²).So you see that this quantity to be lorentz invariant,you have to restrict to one of the hyperboloids.

 

[WannabeNewton]

Gotcha, thanks andrien!

However I still need one last thing clarified. Going back to that final equation:

$\int \frac{d^{3}k}{(2\pi)^3 2\omega}[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\=\int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\Lambda^{-1}\mathbf{k})e^{ikx} + a^{\dagger}(\Lambda^{-1}\mathbf{k})e^{-ikx}]$

What's the quickest explicit way of getting from this final equation to the desired relations:

$U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) = a(\Lambda^{-1}\mathbf{k})$ and $U(\Lambda)^{-1}a^{\dagger}(\mathbf{k})U(\Lambda) = a^{\dagger}(\Lambda^{-1}\mathbf{k})$?

Because the only thing I could think of was to rewrite the equation as

$\int \frac{d^{4}k}{(2\pi)^3}\delta(k^2 + m^2)\theta(k^0)[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\=\int \frac{d^{4}k}{(2\pi)^3}\delta(k^2 + m^2)\theta(k^0)[a(\Lambda^{-1}\mathbf{k})e^{ikx} + a^{\dagger}(\Lambda^{-1}\mathbf{k})e^{-ikx}]$

and then apply $\int d^{4}x e^{-ik'x}$ to both sides of the equation to get

$\int d^{4}k'\delta(k'^2 + m^2)\theta(k'^0)[U^{-1}(\Lambda)a(\mathbf{k'})U(\Lambda)\delta^{4}(k - k') +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k'})U(\Lambda)\delta^{4}(k + k') ] \\=\int d^{4}k'\delta(k'^2 + m^2)\theta(k'^0)[a(\Lambda^{-1}\mathbf{k'})\delta^{4}(k - k') + a^{\dagger}(\Lambda^{-1}\mathbf{k'})\delta^{4}(k + k')]$

which reduces to

$U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)\theta(k^0) +U^{-1}(\Lambda)a^{\dagger}(-\mathbf{k})U(\Lambda)\theta(-k^0) \\=a(\Lambda^{-1}\mathbf{k})\theta(k^0) + a^{\dagger}(-\Lambda^{-1}\mathbf{k'})\theta(-k^0)$

which is no different from the Fourier transform method I employed earlier in the thread.

Thanks in advance.

 

[strangerep]

Going back to that final equation:

$\int \frac{d^{3}k}{(2\pi)^3 2\omega}[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\=\int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\Lambda^{-1}\mathbf{k})e^{ikx} + a^{\dagger}(\Lambda^{-1}\mathbf{k})e^{-ikx}]$

What's the quickest explicit way of getting from this final equation to the desired relations:

$U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) = a(\Lambda^{-1}\mathbf{k})$ and $U(\Lambda)^{-1}a^{\dagger}(\mathbf{k})U(\Lambda) = a^{\dagger}(\Lambda^{-1}\mathbf{k})$?

Go back to the classical case for a moment, and look at Srednicki's paragraph following his eq(3.12).
The $a(\mathbf{k})$ are arbitrary functions of the wave vector $\mathbf{k}$.

So take your first equation above and move everything under the same integral, so that you have something like $\int \cdots = 0$. Then observe that the integrand is essentially an arbitrary function of $\mathbf{k}$. At the "level of rigor of theoretical physics"$^*$, one can usually go direct from that to a conclusion that the integrand itself is 0. (To justify this more rigorously involves functional analysis: extension of Cauchy integration theory and Fourier transforms to general Banach spaces, and something called the "Gel'fand transform", iiuc.)

Then you just... oh, wait... I should let you figure out the rest from my original 2nd hint in post #4.
[Edit 1: Hmm, my hint might not be enough. Also think about functional (in)dependence of an arbitrary complex variable $z$ and its conjugate $\bar z$.]

[Edit 2: On reflection, I suppose I must concede that this doesn't really qualify as an "explicit" method.][*] Arnold Neumaier is the copyright holder for the quote used above.

 

[vanhees71]

May be my conclusion was indeed a bit too quick ;-) at the final step. To make the argument strict you need to use some "orthogonality properties" of the plane-wave solutions of the KG equation. Let's define the mode functions as
$$u_p(x)=\frac{1}{(2 \pi)^3 2 E(\vec{p})}\exp(-\mathrm{i} E(\vec{p}) t+\mathrm{i} \vec{p} \cdot \vec{x}).$$
Using the common notation
$$A(x) \overleftrightarrow{\partial}_t B(x)=A \partial_t B-(\partial_t A) B$$
for any fields $A$ and $B$, you find
$$\mathrm{i} u_q^*(x) \overleftrightarrow{\partial}_t u_p(x)=\frac{1}{(2 \pi)^6} [E(\vec{q})+E(\vec{p})]\exp(-\mathrm{i} [E(\vec{p})-E(\vec{q}) t + \mathrm{i} (\vec{p}-\vec{q}) \cdot \vec{x})].$$
Integrating this wrt. $\vec{x}$, you get
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathrm{i} u_q^*(x) \overleftrightarrow{\partial}_t u_p(x) = \frac{2 E(\vec{p})}{(2 \pi)^3} \delta^{(3)}(\vec{p}-\vec{q}).$$
These plain waves are thus "orthogonal" in the sense of the (indefinite!) functional bilinear form defined by the integral
$$\langle A | B \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} A^*(x) \overleftrightarrow{\partial}_t B(x).$$
In the same way you prove that
$$\langle u_q^*|u_p \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_q(x) \overleftrightarrow{\partial}_t u_p(x)=0.$$
Now you just need to use these "orthogonality relations" on both sides of the equation
$$\int \frac{d^{3}k}{(2\pi)^3 2\omega}[U^{-1}(\Lambda)a(\mathbf{k})U(\Lambda)e^{ikx} +U^{-1}(\Lambda)a^{\dagger}(\mathbf{k})U(\Lambda)e^{-ikx} ] \\=\int \frac{d^{3}k}{(2\pi)^3 2\omega}[a(\Lambda^{-1}\mathbf{k})e^{ikx} + a^{\dagger}(\Lambda^{-1}\mathbf{k})e^{-ikx}],$$
and you are done.