Mathematical Quantum Field Theory - Lagrangians - Comments

[Urs Schreiber]

Greg Bernhardt submitted a new PF Insights post

Mathematical Quantum Field Theory - Lagrangians

Continue reading the Original PF Insights Post.

 

[strangerep]

In Example 5.4,

For this to be physical unit-free in the sense of remark 5.2 the physical unit of the parameter $m$ must be that of the inverse metric, hence must be an inverse length

I would have thought this is because of the 2 derivatives of $\phi$ by the coordinate $x^\mu$. Normally the metric $\eta_{\mu\nu}$ would be dimensionless.

And btw, what are your dimensions of $\phi$? I'm a bit confused about how you're apparently dispensing with physical units here. In ordinary Lagrangian mechanics, we'd have $$S ~=~ \int L \,dt$$where $S$ has dimensions of action ($ML^2/T$), hence $L$ must have dimensions $ML^2/T^2$, i.e., energy. If we express $L$ in terms of a density ${\mathcal L}$ via $$L ~=~ \int {\mathcal L} \; d^3x$$then ${\mathcal L}$ must have dimensions $M/LT^2$.

But perhaps you mean to make $S$ dimensionless by replacing it by $S/\hbar$ ?

 

[Urs Schreiber]

In Example 5.4,
I would have thought this is because of the 2 derivatives of $\phi$ by the coordinate $x^\mu$. Normally the metric $\eta_{\mu\nu}$ would be dimensionless.

Right, these are two different ways to see the dimension. These two ways lead to equivalent dimensions over Minkowski spacetime, but only the other one, which i use in the series, generalizes well to curved spacetime: On curved spacetime it makes no sense to "rescale coordinates", but it always makes sense to rescale the metric. Even though I don't discuss this generalization in the series, my aim is to present everything in a form such that this generalization will be straightforward.

The alternative perspective that you have in mind I have spelled out on the nLab here

 

[strangerep]

On curved spacetime it makes no sense to "rescale coordinates", [...]

Pardon my ignorance, but,... why not? I would have thought one could at least perform a position-dependent rescaling of the (co-)tangent space at a point, hence rescaling $dx^\mu$ and $\partial/\partial x^\mu$.

 

[strangerep]

Re: "shell": I'm glad you're using this term in a unified way in both the classical and quantum cases. I once used the term "on-shell" in a classical context to mean "in the space of trajectories", but was scolded for doing so.

 

[strangerep]

Prop 5.8, 1st para: Is ${\mathbf {LL}}$ a typo ?

Also, in defn 5.18: typo: mathfb

 

[strangerep]

In the examples from 5.20 onwards: it wouldn't hurt to illustrate the pre-symplectic current explicitly in each case. (I must say I had great difficulty getting my head around $\Theta_{BFV}$ in the abstract.)

 

[Urs Schreiber]

Pardon my ignorance, but,... why not? I would have thought one could at least perform a position-dependent rescaling of the (co-)tangent space at a point, hence rescaling $dx^\mu$ and $\partial/\partial x^\mu$.

If you carry that thought to the end, you arrive at th rescaling of the metric: Because what does it mean to "rescale the (co)tangent bundle"? This can only mean to 1) have a scale on it and then 2) change that. But a "scale" on the (co)tangent bundle, that's precisely a (pseudo-)Riemannian metric.

 

[Urs Schreiber]

In the examples from 5.20 onwards: it wouldn't hurt to illustrate the pre-symplectic current explicitly in each case. (I must say I had great difficulty getting my head around $\Theta_{BFV}$ in the abstract.)

That's spelled out earlier, in examples 5.11 and 5.13!

 

[Urs Schreiber]

Prop 5.8, 1st para: Is ${\mathbf {LL}}$ a typo ?

Also, in defn 5.18: typo: mathfb

Thanks! Fixed now.

 

[Urs Schreiber]

Re: "shell": I'm glad you're using this term in a unified way in both the classical and quantum cases. I once used the term "on-shell" in a classical context to mean "in the space of trajectories", but was scolded for doing so.

Hm, saying "on-shell" for "when the equations of motion hold" is widely adopted standard. You should feel relaxed about saying it!:-)

 

[vanhees71]

In Example 5.4,
I would have thought this is because of the 2 derivatives of $\phi$ by the coordinate $x^\mu$. Normally the metric $\eta_{\mu\nu}$ would be dimensionless.

And btw, what are your dimensions of $\phi$? I'm a bit confused about how you're apparently dispensing with physical units here. In ordinary Lagrangian mechanics, we'd have $$S ~=~ \int L \,dt$$where $S$ has dimensions of action ($ML^2/T$), hence $L$ must have dimensions $ML^2/T^2$, i.e., energy. If we express $L$ in terms of a density ${\mathcal L}$ via $$L ~=~ \int {\mathcal L} \; d^3x$$then ${\mathcal L}$ must have dimensions $M/LT^2$.

But perhaps you mean to make $S$ dimensionless by replacing it by $S/\hbar$ ?

Usually we use natural units in HEP physics. Then $\hbar=c=1$. So actions are dimensionless. In (1+3)-Minkowski space (physical case) thus the Lagrangian (to be precise the Lagrangian density) must be of dimension $\text{Length}^{-4}$. Of course the fundamental form $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ (or the opposite sign, depending on your convention) is dimensionless. The kinetic term for a KG field is
$$\mathcal{L}=(\partial_{\mu} \phi)^*(\partial^{\mu} \phi).$$
Each derivative has dimension $1/\text{Length}$. Consequently the KG field must have also dimension $1/\text{length}$ such that $\mathcal{L}$ has dimension $1/\text{Length}^4$, such that the action is dimensionless.

In natural units energies, momenta, and masses have of course the dimension $1/\text{Length}$ or the other way around length has dimension $1/\text{Energy}$. Usually one counts in terms of energ-momenta-mass dimensions. Then the counting rules for the superficial degree of divergence in renormalization theory becomes more natural, and you can simply state that a Lagrangian leads to a superficially renormalizable theory (in (1+3) spacetime dimensions) if the energy-mopmetnum-mass dimensions of the coefficients in the lagrangian (derivatives, masses, and coupling constants) are not negative.

 

[Urs Schreiber]

Of course the fundamental form $\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ (or the opposite sign, depending on your convention) is dimensionless.

A detailed study of QFT with respect to "physical scaling" of the metric tensor is in https://arxiv.org/abs/1710.01937, following https://arxiv.org/abs/1411.1302. This is well adapted to QFT on curved spacetimes. Maybe I find time to come back to this later.

 

[strangerep]

[$\Theta_{BFV}$] spelled out earlier, in examples 5.11 and 5.13!

I did read those examples, of course, but failed to get a grip on what the presymplectic current means physically. I've been able to (more-or-less) follow your other stuff, since I could relate it to ordinary classical mechanics and pedestrian Lagrangian field theory. But the presymplectic current didn't ring a bell with anything I'd learned before. Probably I'm just missing something important.

 

[strangerep]

Usually we use natural units in HEP physics. Then $\hbar=c=1$. So actions are dimensionless.

I don't think "dimensionless" is the right word here. Just because you give something a value of 1 doesn't cause it to lose its physical type. E.g., if both $\hbar$ and $c$ were dimensionless, I should be able to add them together. But that could never make sense because they're different types. You can't sensibly add a velocity to an action.

 

[strangerep]

If you carry that thought to the end, you arrive at th rescaling of the metric: Because what does it mean to "rescale the (co)tangent bundle"? This can only mean to 1) have a scale on it and then 2) change that. But a "scale" on the (co)tangent bundle, that's precisely a (pseudo-)Riemannian metric.

Actually, when I carry that thought further, I think of $$ds^2 ~=~ g_{\mu\nu} dx^\mu dx^\nu ~,$$and I want $ds$ to have the same dimensions as $dx$. Also, if I re-scale $dx^\mu \to k dx^\mu$ I want that to induce $ds \to kds$. That seems to force $g$ to be dimensionless.

OTOH,... I suppose that if one wanted $ds$ to be invariant under $dx^\mu \to k dx^\mu$, then $g$ must scale (inversely) also.

 

[vanhees71]

I don't think "dimensionless" is the right word here. Just because you give something a value of 1 doesn't cause it to lose its physical type. E.g., if both $\hbar$ and $c$ were dimensionless, I should be able to add them together. But that could never make sense because they're different types. You can't sensibly add a velocity to an action.

I don't know, how you call a quantity which is just a number like, e.g., the Sommerfeld fine-structure constant $\alpha=\frac{e^2}{4 \pi} \simeq 1/137$ (where $e$ is again a pure number in the Heaviside-Lorentz units with $\hbar=c=1$), which conventionally is defined as ($\alpha=\frac{e^2}{4 \pi \epsilon_0 \hbar c}$ in SI units) or radians for angles. Of course, it doesn't make physical sense to add $\hbar$ and $c$. Also in conventional units you have examples for this: E.g., it doesn't make sense to add an energy (dimension Force times Distance) to a torque (also dimension Force times Distance).

The usual and most convenient convention in HEP is to set $\hbar=c=1$. These are just conversion constants depending on the used system of units. It is expected that next year the entire SI will be redefined by giving these constants a fixed value (as is already the case for the speed of light, coupling the definition of the metre to that of the second, which is defined by a hyperfine transition of Cs).

The only thing you need to know to convert from these natural units of the theoreticians to SI units is that $\hbar c \simeq 0.197 \text{GeV}\,\text{fm}$. In the natural system you have the choice of only one unit left. Usually one works with GeV for masses, energies, momenta (in the conventional system multiplied with the appropriate powers of $c$) and fm for times (conventionally $\text{fm}/c$) and distances. This, together with the value of $\hbar c$, given in these units, is all you need in HEP, and the natural units make dimensional analysis, as exemplified in the previous posting, easier.

 

[Urs Schreiber]

I did read those examples, of course, but failed to get a grip on what the presymplectic current means physically. I've been able to (more-or-less) follow your other stuff, since I could relate it to ordinary classical mechanics and pedestrian Lagrangian field theory. But the presymplectic current didn't ring a bell with anything I'd learned before. Probably I'm just missing something important.

Oh, I see. i should maybe add some outlook explanation in that section. The role of the presymplectic current will be elucidated in the chapter "Phase Space": it is the jet bundle avatar of the symplectic form on the phase space. In other words, it is the current whose conserved charge with respect to some spacelike Cauchy surface is the symplectic form for the phase space corresponding to that Cauchy surface.

You can see this well in the example of the scalar field: if you write "q" for the field "phi" and "p" for the field derivative, then the presymplectic current has the form $\delta p \wedge \delta q \wedge volume$. After transgression this becomes the familiar $d P \wedge d Q$ on the phase space of the scalar field.

I don't have time today to add an outlook remark on this in the present chapter. But please remind me to do so.

 

[strangerep]

I don't know, how you call a quantity which is just a number like, e.g., the Sommerfeld fine-structure constant $\alpha=\frac{e^2}{4 \pi} \simeq 1/137$ (where $e$ is again a pure number in the Heaviside-Lorentz units with $\hbar=c=1$), which conventionally is defined as ($\alpha=\frac{e^2}{4 \pi \epsilon_0 \hbar c}$ in SI units) or radians for angles. [...]

I call $\alpha$ dimensionless because, in full gory detail, it is: $$ \alpha ~:=~ \frac{1}{4\pi\varepsilon_0} \, \frac{e^2}{\hbar c} ~.$$

 

[vanhees71]

Yes, so what's wrong with what I wrote before? In natural units velocities are measured with dimensionless numbers, because you set $c=1$. In the SI it's just $\beta=v/c$ (with $v$ and $c$ measured in metre over second, m/s).

 

[strangerep]

$\beta$ is dimensionless (regardless of which system of units you choose to use).

I might have misunderstood you, but the only thing I thought was "wrong" is the statement that a physical quantity ("$S$", say) with dimensions of "action" can be made "dimensionless" by setting $\hbar=1$. However, if one works instead with $S/\hbar$, that quantity is now indeed dimensionless, and by choosing units s.t. $\hbar=1$ one can reduce some of the mess.

 

[vanhees71]

Of course, you can make quantities dimensionless by choosing appropriate systems of units. In Planck units all quantities are dimensionless! Of course, sometimes it's good to keep dimensions, i.e., for the WKB approximation in QT (aka loop expansion in Feynman-diagram language) you keep $\hbar$ and make an expansion in powers of $\hbar$. The same holds for the non-relativistic limit of relativistic models, where you keep $c$ and expand in powers of $1/c$ etc. etc. The choice of units is just driven by convenience. Correct physics is always independent of this choice.

 

[Urs Schreiber]

failed to get a grip on what the presymplectic current means physically

But please remind me to do so.

Did so now: There is now a remark 5.12 that, hopefully, elucidates what the presymplectic current is about, by way of providing a preview of what will become of it in Chapter 8. Phase space.

Please let me know if this helps.

 

[strangerep]

There is now a remark 5.12 [...]
Please let me know if this helps.

Yes, a little.

But I suspect I'm at, or beyond, the limit of what I can properly understand in this abstract treatment.

 

[Urs Schreiber]

Yes, a little. But I suspect I'm at, or beyond, the limit of what I can properly understand in this abstract treatment.

Maybe it helps to re-consider this in terms of variation of the action: The usual prescription for deriving the Euler-Lagrange equations asks you to keep the variations at the boundaries fixed. Now re-do the computation of the EL-equations without that assumption. As a result there is a non-trivial boundary term being picked up. If you compute this, you find that it is the integral over the boundary of what is called $\Theta_{BFV}$ here. Hence the variation of that boundary term, that's the integral over $\Omega_{BFV}$. The important point is that this boundary term is what induces the phase space structure, hence the Poisson bracket, hence the quantum commutator on the field theory.

 

[strangerep]

Maybe it helps to re-consider this in terms of variation of the action: The usual prescription for deriving the Euler-Lagrange equations asks you to keep the variations at the boundaries fixed. Now re-do the computation of the EL-equations without that assumption. As a result there is a non-trivial boundary term being picked up. If you compute this, you find that it is the integral over the boundary of what is called $\Theta_{BFV}$ here.

Hmm, ok. Indeed, I had wondered about correct handling of boundary terms at temporal infinity earlier when you mentioned integration by parts when deriving the E-L eqns. If the variations are not fixed to be $0$ at $t=\pm\infty$, then either one has a nontrivial term remaining, as you say, or else one must rely on the term being a function of conserved quantities (so that what comes in at $t=-\infty$ necessarily goes out at $t=+\infty$).

Hence the variation of that boundary term, that's the integral over $\Omega_{BFV}$. The important point is that this boundary term is what induces the phase space structure, hence the Poisson bracket, hence the quantum commutator on the field theory.

Now you've lost me again. I don't follow at all how $\Omega_{BFV}$ "induces" a Poisson bracket.

 

[Urs Schreiber]

Now you've lost me again. I don't follow at all how $\Omega_{BFV}$ "induces" a Poisson bracket.

Do you know how a symplectic form induces a Poisson bracket? The $\Omega_{BFV}$ becomes the (pre-)symplectic form of the field theory after integration over a Cauchy surface.

 

[strangerep]

Do you know how a symplectic form induces a Poisson bracket?

Well,... I'll take the simplest case in classical mechanics where the basic dynamical variables are a single configuration variable $q$ and corresponding canonical momentum $p$. Then, suppose we have 2 functions $f,g$ on the corresponding phase space, i.e., $f\leftrightarrow f(q,p)$ and $g\leftrightarrow g(q,p)$, then their total differentials are phase space vectors given by $$df ~=~ f,_q dq + f,_p dp ~;~~~~~~ (\mbox{and similarly for}~ dg).$$A symplectic form $\omega$ on this space is (by definition) a non-degenerate skewsymmetric bilinear 2-form, so$$\omega(df,dg) ~=~ f,_q g,_p - f,_p g,_q ~=:~ \{f,g\}_{PB} ~,$$

The $\Omega_{BFV}$ becomes the (pre-)symplectic form of the field theory after integration over a Cauchy surface.

Hmm. I need to work that through some more to see it properly...

If we have a simple action $I$ as follows: $$I ~=~ \int\!dt\; L(q,\dot q)$$ then, proceeding with the usual variation computation...
$$0 ~=~ \delta I ~= \int\!dt\;\Big[L(q+\delta q,\, \dot q + \delta\dot q) - L(q,\dot q)\Big] ~= \int\!dt\;\Big(L_q \delta q + L_{\dot q} \delta\dot q \Big) ~.$$Performing an integration-by-parts on the 2nd term gives: $$\Big[L_{\dot q} \delta q \Big]^{t=+\infty}_{t=-\infty} ~-~ \int\!dt\; L_{\dot q} \delta q ~.$$Normally, we'd assume that the variations $\delta q$ vanish at $t\to\pm\infty$, so the term in square brackets vanishes, and the remaining terms in the integrand yield the usual E-L equations of motion.

For the simple case of a free particle, with $L=m\dot q^2/2$ we have $p := L_{\dot q} = m\dot q$, so the square bracket term would be $$\Big[p \, \delta q \Big]^{t=+\infty}_{t=-\infty} ~~,$$but how do we get from here to a "(pre-)symplectic form" ?

 

[Urs Schreiber]

but how do we get from here to a "(pre-)symplectic form" ?

You did rederive the presymplectic potential $\Theta_{BFV}$, which in the case you considered is $p \delta q$. The presymplectic form is the variational derivative of that $\Omega_{BFV} := \delta \Theta_{BFV}$. In the case you considered this yields $\delta p \wedge \delta q$, which is the standard symplectic form that you expect to see.

 

[strangerep]

You did rederive the presymplectic potential $\Theta_{BFV}$, which in the case you considered is $p \delta q$. The presymplectic form is the variational derivative of that $\Omega_{BFV} := \delta \Theta_{BFV}$. In the case you considered this yields $\delta p \wedge \delta q$, which is the standard symplectic form that you expect to see.

If it had been $p\,dq$, where $d$ is the exterior derivative, then $d(p\,dq) = dp \wedge dq$ (since $d^2q=0$).
But we're using variational (a.k.a. functional) derivatives here, aren't we? What am I missing?