Alternative definitions of energy?

[juanrga]

Do you actually know anyone who pays attention to the IUPAP guidelines?

Several journals (including tops from the APS) follow IUPAP recommendations.

I mean, they're nice folks and all, but in most fields of physics, it's more important to follow the conventions in the literature of the field than the recommendations of some folks who have spent a little too much time in the company of chemists, if you know what I mean.

I do not know what you mean. Please explain with detail.

For example, IUPAP recommends using the term "enplenthy" for the amount-of-substance-usually-just-called-number-of-moles. Do you know anyone who uses this terminology? Along these lines, IUPAP discourages the use of "molar volume" and instead prefers "enplenthic volume". There are a lot of idiosyncratic recommendations in the IUPAP guidelines. Do you know anyone who uses "L" for Avogadro's number?

The term enplenty was suggested by one IUPAC commision, but has not still been accepted and is not part of IUPAC official recommendations, less still of IUPAP.

It is not Avogadro number but Avogadro constant, and IUPAP, IUPAC, and ISO recommend both symbols L and N_A. I understand rationale beyond both symbols: L is used in honor of Josef Loschmidt.

I tend to follow the advice of Howard Georgi, which is not to get too hung up on notation. You have to know what the symbols mean, but getting fussy about whether the speed of light in vacuum is c or c₀ (the IUPAP recommendation) is a waste of time.

Both of you missed the point. The point is not to change A by B because the launch of a coin suggested it. The point is the development of a modern, consistent, and interdisciplinar notation, terminology, and units.

Regarding the ISO,IUPAP,IUPAC symbol for the speed of light you seem to miss the recommendation to omit the subscript denoting vacuum, «when there is no risk of ambiguity».

I've seen a lot of different ways to write the first law; I usually prefer the version with the slashed d's for heat and work inexact differentials, but that can sometimes be a typographic challenge. So sometimes I write

du = δq - δw

and let the δ stand in for slash-d. I'm okay with

ΔU = Q₁₂ - W₁₂

for a finite process between initial state 1 and final state 2, but the bare version ΔU = Q - W looks a little funny. But like I said, it's not worth it to get hung up on notation.

BBB

This is fine

du = δq - δw

and the finite counterpart

ΔU = Q - W

is fine as well. The subindices in your

ΔU = Q₁₂ - W₁₂

are redundant, but it is still acceptable.

What is not acceptable is the notation given here by Andrew Mason, ΔU = ΔQ - W, because this is an inconsistent (and misleading) notation.

 

[juanrga]

The horrendous nomenclature rears its ugly head again.

In thermodynamics, heat flux=heat flux density. Yet in electrodynamics, magnetic flux/area=magnetic flux density, which makes more sense.

Exactly, heat flux=heat flux density makes no sense, because you are equating a quantity with the density of that quantity.

Although in previous years the IUPAP also did the mistake heat flux=heat flux density, they have corrected its mistake now.

The IUPAP defines heat flux density=heat flux/area in its more recent recommendations.

http://www.google.com/search?q=heat+flux
http://search.yahoo.com/search?p=heat+flux
http://msxml.excite.com/info.xcite.psp/search/web?q=heat+flux
http://www.dogpile.com/info.dogpl.t2.7/search/web?fcoid=417&fcop=topnav&fpid=27&q=heat+flux&ql=
http://www.ask.com/web?q=heat+flux

You can find all kind of misconceptions and inconsistencies on the web. Not an argument

 

[juanrga]

The real problem is that the IUPAP, ISO, and IUPAC do not have a dictatorship that forces these definitions to be the only ones taught (and available to learn) via publications. If they had that, we wouldn't have multiple definitions for the word heat flux like what we already have for heat, which helps no one to have, and the wrong people wouldn't be blamed for not being aware of this by others like yourself. Problem is, they don't have that kind of control. Though, it would actually be helpful if they did.

It is good to be aware of last international standards and recommendations; but, of course, I do not encourage to follow anything said by them.

Personally, I try to understand the rationale behind their recommendations and I follow a lot of them, but not all.

 

[Andrew Mason]

Regarding the ISO,IUPAP,IUPAC symbol for the speed of light you seem to miss the recommendation to omit the subscript denoting vacuum, «when there is no risk of ambiguity».
...

What is not acceptable is the notation given here by Andrew Mason, ΔU = ΔQ - W, because this is an inconsistent (and misleading) notation.

The Δ in ΔQ means "difference" in heat flows in the process (|heat flow in| - |heat flow out|). Some times the first law is written ΔU = ΔQ - ΔW where ΔW refers to the difference in work done in the process ie.: |work done by the gas| - |work done on the gas|. ΔU means difference in internal energy. Since internal energy refers to a state rather than a process, it means difference in internal energy between initial and final states.

So long as everyone understands what it means and there is no risk of ambiguity, what is wrong with it?

AM

 

[bbbeard]

Yes, D is a function of the Hamiltonian, just as the non-dissipative part L_H is function as well; the difference is that D is a very very complicated function (as said before). Therefore I prefer not to write it in explicit form here.

As also explained in #56, the Lagrangian formulation (including the Euler Lagrange equations) can be derived after applying approximations on the general equation of motion; the approximations include the neglect of dissipation. You can try to add, by hand, some dissipative term to the final Euler Lagrange equations but the resulting theory will be totally inconsistent {**}.

Well, first, the microscopic Lagrangian is the Standard Model Lagrangian (with the exception of gravity -- we don't know the appropriate way to integrate gravity with either the microscopic Lagrangian or microscopic Hamiltonian in a way that works above tree-level). Most of the time that people do calculations in the Standard Model, they use the Lagrangian formulation, whether the calculation is perturbative or non-perturbative. But Hamiltonian approaches to the Standard Model are equivalent, just usually not as easy to use.

Macroscopic Lagrangians and Hamiltonians are always approximations to collective microscopic dynamics. I don't know of any exceptions to this.

At the macroscopic level, dissipation is incorporated into the Lagrangian approach with the Rayleigh dissipation function (usually denoted with F). F is a functional of ${x}$ and $\dot{x}$, just like the Lagrangian. However, F is not a "function" of the Lagrangian. Both Lagrangian and Hamiltonian functionals just involve energies, so I am unclear how dissipation can be modeled with something that is *just* a "function" of the Hamiltonian.

Let's consider the simplest example of a classical dissipative system: a block of mass m sliding on a horizontal plane with a frictional force that is proportional to velocity. The Lagrangian is

$L=\frac{1}{2}m\dot{x}^2$

and the Rayleigh function is

$F=\frac{1}{2}c\dot{x}^2$.

The Euler-Lagrange equation is

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}+\frac{\partial F}{\partial \dot{x}}=0$

which gives

$m\ddot{x}+c\dot{x}=0$

In the method you advocate, using the dissipator D, what is the corresponding formulation? And why do you think this model is "totally inconsistent"?

{**} Recall the same online reference that you introduced in a previous post. Read the part where says you that the Lagrangian formalism does not work for systems with dissipation.

I've ignored your note before because you're misinterpreting this Wikipedia entry, possibly because English is not your first language, not that there's anything wrong with that. But in the interests of clarity, the (recently edited) Wikipedia entry on Noether's theorem says

Noether's theorem has become a fundamental tool of modern theoretical physics and the calculus of variations. A generalization of the seminal formulations on constants of motion in Lagrangian and Hamiltonian mechanics (developed in 1788 and 1833, respectively), it does not apply to systems that cannot be modeled with a Lagrangian alone (e.g. systems with a Rayleigh dissipation function). In particular, dissipative systems with continuous symmetries need not have a corresponding conservation law.

What this means is that Noether's theorem does not apply to systems with a dissipation term, for example, there is no conserved current in the sliding block system I presented above. It does not say that systems with dissipation cannot be modeled in the Lagrangian formulation of mechanics. Dissipation is modeled with the Rayleigh dissipation function, which is different from the Lagrangian. Systems with dissipation do not conserve energy -- unless one proceeds to model the system at a lower level, e.g. with a Lagrangian that incorporates the degrees of freedom that absorb the dissipated energy. Ultimately, the microscopic Standard Model Lagrangian is reversible.

BBB

 

[juanrga]

The Δ in ΔQ means "difference" in heat flows in the process (|heat flow in| - |heat flow out|). Some times the first law is written ΔU = ΔQ - ΔW where ΔW refers to the difference in work done in the process ie.: |work done by the gas| - |work done on the gas|. ΔU means difference in internal energy. Since internal energy refers to a state rather than a process, it means difference in internal energy between initial and final states.

So long as everyone understands what it means and there is no risk of ambiguity, what is wrong with it?

AM

All of this was corrected before

 

[juanrga]

Well, first, the microscopic Lagrangian is the Standard Model Lagrangian (with the exception of gravity -- we don't know the appropriate way to integrate gravity with either the microscopic Lagrangian or microscopic Hamiltonian in a way that works above tree-level). Most of the time that people do calculations in the Standard Model, they use the Lagrangian formulation, whether the calculation is perturbative or non-perturbative. But Hamiltonian approaches to the Standard Model are equivalent, just usually not as easy to use.

This is plain wrong. As correctly emphasized by Weinberg in the volume 1 of his quantum theory of fields. It is the Hamiltonian, not the Lagrangian, which is used to obtain the S-matrix elements, after confronted with experiments at lab.

The only utility of the Lagrangian in the SM is that it is more easy to check certain symmetries (which do no need to be fundamental). The Lagrangian is only an easy tool to obtain a Hamiltonian (H=pv-L) with certain physical requirements, nothing more. Since the Lagrangian plays absolutely no central role in the dynamics of particle physics (the generator of time translations is the Hamiltonian), Weinberg even speculates about future Hamiltonian theories not obtained from some previous Lagrangian.

Macroscopic Lagrangians and Hamiltonians are always approximations to collective microscopic dynamics. I don't know of any exceptions to this.

At the macroscopic level, dissipation is incorporated into the Lagrangian approach with the Rayleigh dissipation function (usually denoted with F). F is a functional of ${x}$ and $\dot{x}$, just like the Lagrangian. However, F is not a "function" of the Lagrangian. Both Lagrangian and Hamiltonian functionals just involve energies, so I am unclear how dissipation can be modeled with something that is *just* a "function" of the Hamiltonian.

Let's consider the simplest example of a classical dissipative system: a block of mass m sliding on a horizontal plane with a frictional force that is proportional to velocity. The Lagrangian is

$L=\frac{1}{2}m\dot{x}^2$

and the Rayleigh function is

$F=\frac{1}{2}c\dot{x}^2$.

The Euler-Lagrange equation is

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}+\frac{\partial F}{\partial \dot{x}}=0$

which gives

$m\ddot{x}+c\dot{x}=0$

In the method you advocate, using the dissipator D, what is the corresponding formulation? And why do you think this model is "totally inconsistent"?

Why this kind of nonsense again? The Euler-Lagrange equations are only valid when one ignores dissipation and takes the pure state approximation.

The Euler Lagrange equation for $L$ is

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$

if you add some term $A=A(v)$ to the Lagrangian the equation for the new Lagrangian $L' = L+A$ would be

$\frac{d}{dt}\frac{\partial L+A}{\partial \dot{x}}-\frac{\partial L+A}{\partial x}=0$

But this is not what you are doing.

What you do first is to agree that the Euler-Lagrange equation for the Lagrangian $L$ is not valid

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} \neq 0$

and second to add a non Lagrangian term $\frac{\partial F}{\partial \dot{x}} \neq \frac{\partial A}{\partial \dot{x}}$

$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x} + \frac{\partial F}{\partial \dot{x}} = 0$

You are confirming that dissipation cannot be studied using the Lagrangian formalism, as you even agree-disagree-agree-disagree-agree-disagree...

Let us now to be serious about dissipation. Applying a number of approximations to the Dρ in the dissipative equation of motion given before, we can obtain the following approximate equation

$m\dot{v} = -\zeta v + \tilde{f}$

Applying the further approximation that the block of mass is placed in an environment with zero temperature $T=0 \mathrm{K}$, this is reduced to

$m\dot{v} = -\zeta v$

which cannot be derived from a Lagrangian, as has been known for centuries.

I've ignored your note before because you're misinterpreting this Wikipedia entry, possibly because English is not your first language, not that there's anything wrong with that. But in the interests of clarity, the (recently edited) Wikipedia entry on Noether's theorem saysWhat this means is that Noether's theorem does not apply to systems with a dissipation term, for example, there is no conserved current in the sliding block system I presented above. It does not say that systems with dissipation cannot be modeled in the Lagrangian formulation of mechanics. Dissipation is modeled with the Rayleigh dissipation function, which is different from the Lagrangian. Systems with dissipation do not conserve energy -- unless one proceeds to model the system at a lower level, e.g. with a Lagrangian that incorporates the degrees of freedom that absorb the dissipated energy. Ultimately, the microscopic Standard Model Lagrangian is reversible.

BBB

In the interest of clarity, Bbbeard edited the Wikipedia the day 27 November 2011 (i.e. yesterday)‎. The entry before your modification said

it does not apply to systems that cannot be modeled with a Lagrangian; for example, dissipative systems

Which is correct and just what I said. This is a beautiful example of why the Wikipedia has a bad fame among scholars. Everyone can go and put any nonsense that (s)he want.P.S: I solicited you more information about your 'insinuations' about the IUPAP and the chemists. What happened?

 

[bbbeard]

The Δ in ΔQ means "difference" in heat flows in the process (|heat flow in| - |heat flow out|). Some times the first law is written ΔU = ΔQ - ΔW where ΔW refers to the difference in work done in the process ie.: |work done by the gas| - |work done on the gas|. ΔU means difference in internal energy. Since internal energy refers to a state rather than a process, it means difference in internal energy between initial and final states.

So long as everyone understands what it means and there is no risk of ambiguity, what is wrong with it?

All of this was corrected before

I don't see the problem with what Andrew is saying. Even the IUPAP Red Book says it's fine to use Δ to indicate a finite change in a quantity (on p.48).

BBB

 

[juanrga]

I don't see the problem with what Andrew is saying. Even the IUPAP Red Book says it's fine to use Δ to indicate a finite change in a quantity (on p.48).

BBB

It is fine for me if both of you do not see any problem with your incorrect and ambiguous notation and terminology and it is also fine for me if both of you decide to re-write the first law as ΔU=ΔQ-ΔW, ignoring all that was said to both.

Fortunately, the rest of us (including ISO, IUPAP, and IUPAC, and standard textbooks) can sanely avoid such abhorrences and write nicely stuff as ΔU=Q-W

 

[Dadface]

Energy is a property of moving bodies and of anything which has the capacity to create movement.

 

[bbbeard]

This is plain wrong. As correctly emphasized by Weinberg in the volume 1 of his quantum theory of fields. It is the Hamiltonian, not the Lagrangian, which is used to obtain the S-matrix elements, after confronted with experiments at lab.

The only utility of the Lagrangian in the SM is that it is more easy to check certain symmetries (which do no need to be fundamental). The Lagrangian is only an easy tool to obtain a Hamiltonian (H=pv-L) with certain physical requirements, nothing more. Since the Lagrangian plays absolutely no central role in the dynamics of particle physics (the generator of time translations is the Hamiltonian), Weinberg even speculates about future Hamiltonian theories not obtained from some previous Lagrangian.

Well, you are welcome to prefer whichever approach you'd like, but the Hamiltonian and Lagrangian frameworks are two sides of the same coin. It's like arguing whether a periodic function is better represented in time domain or frequency domain; because the Fourier transform is invertible, both domains contain the same "essence" of the function. As you yourself have pointed out, the Legendre transformation connects the Hamiltonian and Lagrangian. Equivalently, you can go back and forth between the time-evolution operator (i.e. Hamiltonian) formulation and the path integral (Lagrangian) formulation. If you pick up any textbook, as I'm sure you often do, you will notice that we go back and forth between the two approaches nearly without comment.

However, the Lagrangian is more directly connected to physical symmetries, via the afore-mentioned Noether theorem. In general, you can't just inspect a Hamiltonian and deduce the symmetries that have conserved currents. More to the point, particle physics proceeds by formulating Lagrangians that have desired symmetries and then calculating the observational consequences. Weinberg says this (Vol I, p.298-9):

Having seen that various realistic theories may be cast in the canonical formalism, we must now face the question of how to choose the Hamiltonian. As we will see in the next section, the easiest way to enforce Lorentz invariance and other symmetries is to choose a suitable Lagrangian and use it to derive the Hamiltonian. There is not much loss of generality in this; given a realistic Hamiltonian, we can generally reconstruct a Lagrangian from which it could be derived, be reversing the process that we are going to describe here of deriving Hamiltonians from Lagrangians... But although we can go from Hamiltonians to Lagrangians or Lagrangians to Hamiltonians, it is easier to explore physically satisfactory theories by listing possible Lagrangians, rather than Hamiltonians.

Indeed, this is the basis of chiral perturbation theory -- you start with a Lagrangian that contains all the possible terms (to the desired order) that respect the symmetries of the parent theory, you work out the observables, and you match the undetermined coefficients with the data from other sources. A beautiful example of this is the 1996 paper by Beard and Wiese, which provides numerical estimates of the parameters of the CPT formulation of Hasenfratz and Niedermayer for the anti-ferromagnetic Heisenberg model.

A more pragmatic approach is advocated by Halzen and Martin (Quarks and Leptons, p.313):

What is the relation between the Lagrangian approach and the perturbative method based on Feynman rules obtained from single-particle wave equations? To each Lagrangian, there corresponds a set of Feynman rules; and so, once we identify these rules, the connection is made. We can then calculate physical quantities by just following the methods presented in Chapters 4 and 6.

The identification of the Feynman rules proceeds as follows: [rules are given]...

In the orthodox approach to quantum field theory, we would now proceed to formally derive these assertions. In order to do this, the classical Lagrangian is quantized... The end result of this lengthy and formal approach can always be translated into a set of Feynman rules which are exactly those we just described. So, we might as well take these rules and proceed to investigate the physical implications of a given Lagrangian using the methods with which we are already familiar. The canonical formalism was formerly regarded as more rigorous and can be found in many books... We do not present it, as we will never use it. We hereby subscribe to the growing belief that "the diagrams contain more truth than the underlying formalism" ['t Hooft and Veltman (1973)]

You know I could go on. Practically every book on quantum field theory uses the same approach. What I said is true: "Most of the time that people do calculations in the Standard Model, they use the Lagrangian formulation, whether the calculation is perturbative or non-perturbative." You lift the Feynman rules from the Lagrangian, not the Hamiltonian. To formulate a theory that respects a desired set of symmetries, you write down a Lagrangian, not a Hamiltonian.

And this is what you seem unwilling to face: the Lagrangian is not just a "check", and not just a way point on the way to a Hamiltonian, but it's how particle physics is done. We go straight from the Lagrangian to Feynman rules to calculating amplitudes, or, we go straight from the Lagrangian to numerical methods. Now again, I'm not claiming primacy for Lagrangians over Hamiltonians in some metaphysical sense -- as I said, they are two sides of the same coin. It's your eccentric view that the Hamiltonian approach is somehow more fundamental than Lagrangian approach.

Why this kind of nonsense again? The Euler-Lagrange equations are only valid when one ignores dissipation and takes the pure state approximation... if you add some term $A=A(v)$ to the Lagrangian... But this is not what you are doing... You are confirming that dissipation cannot be studied using the Lagrangian formalism, as you even agree-disagree-agree-disagree-agree-disagree...

Honest to gosh, Juan, you act like you've never studied Lagrangian mechanics. You don't add the Rayleigh dissipation function to the Lagrangian. And the Rayleigh function is just an elegant way to insert external forces, which are more often computed using the method of virtual work. And all this is part of Lagrangian mechanics -- go get a book.

Let us now to be serious about dissipation. Applying a number of approximations to the Dρ in the dissipative equation of motion given before, we can obtain the following approximate equation

$m\dot{v} = -\zeta v + \tilde{f}$

Applying the further approximation that the block of mass is placed in an environment with zero temperature $T=0 \mathrm{K}$, this is reduced to

$m\dot{v} = -\zeta v$

which cannot be derived from a Lagrangian, as has been known for centuries.

Nor can it be derived from the Hamiltonian. The term $-\zeta v$ is inserted "by hand" just like the Rayleigh function. It is not, as you have claimed, a "function" of the Hamiltonian.

In the interest of clarity, Bbbeard edited the Wikipedia the day 27 November 2011 (i.e. yesterday)‎. The entry before your modification said

it does not apply to systems that cannot be modeled with a Lagrangian; for example, dissipative systems

Which is correct and just what I said. This is a beautiful example of why the Wikipedia has a bad fame among scholars. Everyone can go and put any nonsense that (s)he want.

On the contrary, this is the strength of Wikipedia -- that knowledgeable contributors can correct and refine the information without an interminable review process. In fact all I did was to clarify what the previous text meant -- that some systems require more than just the Lagrangian to describe. It is not the case that the systems cannot be treated within the Lagrangian framework; what was meant was that these systems are not described by a Lagrangian alone. This is an elementary point.

However, the fact that Wikipedia is compiled by volunteers can mean that the quality is variable. I am willing to cite Wikipedia when I feel that the content is appropriate for pedagogy, but it is rarely, if ever, reasonable to cite it as authority.

BBB

 

[juanrga]

Well, you are welcome to prefer whichever approach you'd like, but the Hamiltonian and Lagrangian frameworks are two sides of the same coin. It's like arguing whether a periodic function is better represented in time domain or frequency domain; because the Fourier transform is invertible, both domains contain the same "essence" of the function. As you yourself have pointed out, the Legendre transformation connects the Hamiltonian and Lagrangian. Equivalently, you can go back and forth between the time-evolution operator (i.e. Hamiltonian) formulation and the path integral (Lagrangian) formulation. If you pick up any textbook, as I'm sure you often do, you will notice that we go back and forth between the two approaches nearly without comment.

It is the Hamiltonian, not the Lagrangian, which enters in the computation of the S-matrix. Standard textbooks will give you the definition of the S-matrix in terms of the Hamiltonian...

However, the Lagrangian is more directly connected to physical symmetries, via the afore-mentioned Noether theorem. In general, you can't just inspect a Hamiltonian and deduce the symmetries that have conserved currents. More to the point, particle physics proceeds by formulating Lagrangians that have desired symmetries and then calculating the observational consequences. Weinberg says this (Vol I, p.298-9):

What he says in that quote is essentially what I said and is still quoted at the start of your above message.

Indeed, this is the basis of chiral perturbation theory -- you start with a Lagrangian that contains all the possible terms (to the desired order) that respect the symmetries of the parent theory, you work out the observables, and you match the undetermined coefficients with the data from other sources. A beautiful example of this is the 1996 paper by Beard and Wiese, which provides numerical estimates of the parameters of the CPT formulation of Hasenfratz and Niedermayer for the anti-ferromagnetic Heisenberg model.

A more pragmatic approach is advocated by Halzen and Martin (Quarks and Leptons, p.313):

You know I could go on. Practically every book on quantum field theory uses the same approach. What I said is true: "Most of the time that people do calculations in the Standard Model, they use the Lagrangian formulation, whether the calculation is perturbative or non-perturbative." You lift the Feynman rules from the Lagrangian, not the Hamiltonian. To formulate a theory that respects a desired set of symmetries, you write down a Lagrangian, not a Hamiltonian.

And this is what you seem unwilling to face: the Lagrangian is not just a "check", and not just a way point on the way to a Hamiltonian, but it's how particle physics is done. We go straight from the Lagrangian to Feynman rules to calculating amplitudes, or, we go straight from the Lagrangian to numerical methods. Now again, I'm not claiming primacy for Lagrangians over Hamiltonians in some metaphysical sense -- as I said, they are two sides of the same coin. It's your eccentric view that the Hamiltonian approach is somehow more fundamental than Lagrangian approach.

This whole part is rather incorrect.

(i) Precisely the fact of that many 'old' textbooks on quantum field theory lack precision and rigor was the starting point why Weinberg decided to write his own book, without worrying about «historical precedent», and using a new and fresh perspective about QFT.

(ii) You cite partially Weinberg in your message:

the easiest way to enforce Lorentz invariance and other symmetries is to choose a suitable Lagrangian and use it to derive the Hamiltonian.

In that part he is confirming what I said before about the role of the Lagrangian as check for obtaining a suitable Hamiltonian. But Weinberg also says in his book:

It is the Hamiltonian formalism that is needed to calculate the S-matrix (whether by operator or path-integral methods)

Which is a kind of trivial statement: first, because the Hamiltonian is the generator of time-translations and, second, because the S-matrix is defined using this generator. However, I understand that he needs to emphasize this kind of trivial stuff, because the old literature on QFT is full of misconceptions and nonsense {*}.

(iii) Regarding the Feynman rules, precisely Weinberg offers an example of a kind of theories for the which

using the naive Feynman rules derived directly from the Lagrangian density would yield an S-matrix that is not only wrong but even non-unitary

. He then derives the path-integral formalism from the Hamiltonian formalism, supplementing the simplest version of the Feynman path-integral.

Honest to gosh, Juan, you act like you've never studied Lagrangian mechanics. You don't add the Rayleigh dissipation function to the Lagrangian. And the Rayleigh function is just an elegant way to insert external forces, which are more often computed using the method of virtual work. And all this is part of Lagrangian mechanics -- go get a book.

You are so confused about the Lagrangian formalism that you not even understand that derivation what you pasted in this forum.

Let me explain what you really did. You introduced a 'Rayleigh function' for accounting for dissipation, because dissipation cannot be studied using a Lagrangian and the Euler-Lagrange equations.

If dissipation could be studied using the Lagrangian formalism, one would simply write down some Lagrangian and then derive the dissipative equation of motion for the mass using the Euler-Lagrange equations. Since this is not possible, you have amended the Euler Lagrange equation with an ad hoc non-Lagrangian term.

Precisely, this is the reason, which the Wikipedia article correctly said that dissipative systems could not be described by a Lagrangian.

The Wikipedia article, before you edited it, said exactly:

it does not apply to systems that cannot be modeled with a Lagrangian; for example, dissipative systems

External forces (non-dissipative) can be introduced in the Lagrangian-formalism by adding a potential term to the Lagrangian (L → L - U) and then deriving the equation of motion using Euler-Lagrange. It is not needed to use a 'Rayleigh function' for introducing external force as you claim. You are mixing apples and oranges once again.

Nor can it be derived from the Hamiltonian. The term $-\zeta v$ is inserted "by hand" just like the Rayleigh function. It is not, as you have claimed, a "function" of the Hamiltonian.

If you read the part what you quoted, you can see me saying that the term $-\zeta v$ was derived from the dissipator (which itself is a function of the Hamiltonian).

But there is more, if you continue reading the part what you quote, you will discover how we go beyond your approach and obtain a correction term $\tilde{f}$. This term, absent in your ad hoc equation, is needed to correct your equation in well-defined situations.

When the particle subject to friction is relatively small and placed in a heat bath, the equation that I derived gives the correct asymptotic behavior, whereas yours ad hoc equation miserably fails. Your equation miserably fails because does not account correctly for dissipation and, as a consequence, predicts an incorrect final thermal state.

On the contrary, this is the strength of Wikipedia -- that knowledgeable contributors can correct and refine the information without an interminable review process. In fact all I did was to clarify what the previous text meant -- that some systems require more than just the Lagrangian to describe. It is not the case that the systems cannot be treated within the Lagrangian framework; what was meant was that these systems are not described by a Lagrangian alone. This is an elementary point.

However, the fact that Wikipedia is compiled by volunteers can mean that the quality is variable. I am willing to cite Wikipedia when I feel that the content is appropriate for pedagogy, but it is rarely, if ever, reasonable to cite it as authority.

BBB

Evidently, the problem is when those «knowledgeable contributors» have absolutely no idea of the topic. That is the reason for which scholarly alternatives to the Wikipedia only allow to «knowledgeable contributors» (i.e., scholars with credentials) to edit the articles.

{*} An author of a well-known handbook did the silly claim that QED is a Lagrangian theory and that no Hamiltonian exists for QED. Silliness of his claim increases when the same handbook contains a chapter (by another author) devoted to give the Hamiltonian of QED...

 

[Studiot]

This is a most interesting discussion, gentlemen.

Please do not let the rhetoric become overheated to the point of obscuring the physics.

 

[bbbeard]

This is a most interesting discussion, gentlemen.

Please do not let the rhetoric become overheated to the point of obscuring the physics.

Yes, well, apparently quantum field theory has not yet reached the point of "consensus science"...

BBB

 

[kith]

juanrga, I'm not sure if I understand your claim that the Dissipator is a function of the Hamiltonian correctly.

The full Hamiltonian for a dissipative system (S) and the relevant part of it's environment (E) reads H=H_S+H_E+H_int. The Dissipator of the system can't be expressed by the system Hamiltonian H_S alone, because it describes interactions of the system with it's environment. It has to include H_int.

So did you mean that the Dissipator is a function of the full Hamiltonian?

The difference between the Lagrangian formulation and the Hamiltonian formulation would then be the following: Under certain physically justifiable approximations, we can derive an effective description of the dissipative system from the full Hamiltonian. We can't do this using the full Lagrangian. Instead we have to start with the Lagrangian of the system without dissipation and include dissipation by hand.

 

[juanrga]

juanrga, I'm not sure if I understand your claim that the Dissipator is a function of the Hamiltonian correctly.

The full Hamiltonian for a dissipative system (S) and the relevant part of it's environment (E) reads H=H_S+H_E+H_int. The Dissipator of the system can't be expressed by the system Hamiltonian H_S alone, because it describes interactions of the system with it's environment. It has to include H_int.

So did you mean that the Dissipator is a function of the full Hamiltonian?

The difference between the Lagrangian formulation and the Hamiltonian formulation would then be the following: Under certain physically justifiable approximations, we can derive an effective description of the dissipative system from the full Hamiltonian. We can't do this using the full Lagrangian. Instead we have to start with the Lagrangian of the system without dissipation and include dissipation by hand.

As showed in #56 the Hamiltonian formalism is derived when dissipation is ignored and a pure approximation is applied. As also said in #56

Of course, we can derive the Lagrangian formalism from the Hamiltonian formalism, using the Legendre transformation L(q,v) = pv - H(q,p)

That is, the Lagrangian formalism is valid only when dissipation is ignored and a pure approximation is applied.

Of course, you can start with a «Lagrangian of the system without dissipation and include dissipation by hand». But there is not guarantie. Precisely even the trivial equation obtained by bbbeard fails in several well-known cases. Whereas the equation obtained from the dissipative theory gives the correct final thermal state.

The equation of motion given in #56 is exact and, so far as I know, describes any system either dissipative or not.

As also said in #56 both the Liouvillian and the dissipator are functions of the Hamiltonian.

The dissipator is a very complex object that includes several powers of the interaction Liouvillian L_int (which is itself a function of H_int), 'propagators' exp(L₀t), where the 'free' Liouvillian L₀ is itself a function of H_S and H_E, and more stuff.

If you know the Hamiltonian you can obtain both Liouvillian and dissipator. Another thing is if after obtaining both you find a powerful enough supercomputer to solve the resulting equation.

The Lagrangian formalism cannot account for dissipation, as the Wikipedia article cited by bbbeard correctly said before bbbeard corrupted it. Moreover, the Lagrangian formalism plays absolutely no role in the dynamics of quantum mechanics or in QFT scattering. As Weinberg correctly notices in his celebrated textboook in QFT:

It is the Hamiltonian formalism that is needed to calculate the S-matrix (whether by operator or path-integral methods)

In QFT, the Lagrangian is only a simple way to check that the resulting quantum Hamiltonian formalism satisfies certain symmetries. Again Weinberg is very clever:

the easiest way to enforce Lorentz invariance and other symmetries is to choose a suitable Lagrangian and use it to derive the Hamiltonian

In classical mechanics, the Lagrangian formalism is, in principle, equivalent to the Hamiltonian formalism (see #56). The two difficulties are (i) those cases in which the Legendre transformation cannot be solved exactly and (ii) the well-known computational difficulties associated to the Lagrangian formalism (second order equations). In practice, is much simpler to design computational algorithms for the Hamiltonian formalism (first order equations).

 

[kith]

As showed in #56 the Hamiltonian formalism is derived when dissipation is ignored and a pure approximation is applied.

Ah, I just realized that I assumed that the term "Hamiltonian formulation" would also apply to the Liouvillian / von Neumann equation, which is probably wrong. Now your posts make much more sense to me. ;-)

So the true advantage of the Hamiltonian formalism is that it generalizes easily to mixed states, while the Lagrangian formalism does not?

The equation of motion given in #56 is exact and, so far as I know, describes any system either dissipative or not.

I don't think it is the general form of dynamics for an arbitrary subsystem of a closed system. In deriving such structures from the full Liouville / von Neumann equation, one usually makes certain assumptions (Markov approximation, weak coupling) which may not be fulfilled in general.

If you know the Hamiltonian you can obtain both Liouvillian and dissipator. Another thing is if after obtaining both you find a powerful enough supercomputer to solve the resulting equation.

If I know the full Hamiltonian and don't care about computational power, I could also just solve the Liouville / von Neumann equation for the whole system and trace out the environmental degrees of freedom later.

 

[azaharak]

This is not generally true. It is only true if the applied force F is the only force acting on the body.
AM

I think you are not interpreting what was initially said or meant.

If you want to correct the statement, it would be that the NET WORK done on a body is equal to its change in kinetic energy. I think that was implied.

Where the NET work done is the sum of all the works done on the body. For the car moving up a hill at constant speed, the net work done is zero.

 

[Andrew Mason]

I think you are not interpreting what was initially said or meant.

I was responding to the statement that $W = \int F\cdot ds = mv^2/2$

This is not true unless the force, F, applied to the body is the only force (or is the net force) acting on the body. The work done by an applied force to a body against gravity does not result in kinetic energy of mv^2/2, for example.

That is all I was saying.

AM

 

[azaharak]

I was responding to the statement that $W = \int F\cdot ds = mv^2/2$

That is all I was saying.

AM

The 1st statement written was

"Work done on a system is defined as the change in Kinetic Energy (KE) of that system. While The total energy of a system is the potential energy (PE) plus the kinetic energy, E=PE+KE."

There isn't anything wrong with this, one can read that is that the NET work done on a body is equal to its change in kinetic energy.

Furthermore, if you read the responding post there is clearly a summation over (i)


W12=∑i∫21Fi⋅dsi=∑i∫21miv˙i⋅vidt=∑i∫21d(12miv2i)=T2−T1 where T=12∑imiv2i is the kinetic energy of the system.


The summation clearly takes into account all the forces and gives you the net work done on the object. So I don't know what your talking about only one force, it works for many forces.

For instance if a block slides down a plane at constant speed, two forces contribute non zero work, the work done by gravity and the work done by friction. The NET work done on the object is zero, hence its kinetic energy remains constant.

Both original posts are correct.

 

[juanrga]

Ah, I just realized that I assumed that the term "Hamiltonian formulation" would also apply to the Liouvillian / von Neumann equation, which is probably wrong. Now your posts make much more sense to me. ;-)

Sometimes the term «Liouville formulation» is applied to the Liouville von Neumann equation. I think that I never used the term «Hamiltonian formalism» for this equation.

In my posts I only said that the Liouvillian was a function of the Hamiltonian, and showed how the «Hamiltonian formulation» was derived under certain approximations/assumptions from a more general formulation that uses the Liouvillian and the dissipator.

So the true advantage of the Hamiltonian formalism is that it generalizes easily to mixed states, while the Lagrangian formalism does not?

In #86 I gave some advantages of the classical Hamiltonian formalism over the classical Lagrangian formalism, as the computational advantages. I also explain how the quantum Hamiltonian formalism is used in QFT to obtain the S-matrix (after utilized in the lab).

Since that the Hamiltonian (not the Lagrangian) is the generator of time translations for pure states. Generalizations to mixed states, unstable states, dissipative systems, etc. utilize formalism based in the existence of a Hamiltonian.

I don't think it is the general form of dynamics for an arbitrary subsystem of a closed system. In deriving such structures from the full Liouville / von Neumann equation, one usually makes certain assumptions (Markov approximation, weak coupling) which may not be fulfilled in general.

The equation given includes non-Markovian corrections and coupling to any order. That is the reason for which the form of the dissipator D is so complex that I apologized for not writing it in explicit form. Of course applying a Markov approximation and taking coupling only up to second order (weak coupling) the dissipator simplifies a lot of.

If I know the full Hamiltonian and don't care about computational power, I could also just solve the Liouville / von Neumann equation for the whole system and trace out the environmental degrees of freedom later.

No exactly. The equation for the whole system is not the ordinary Liouville / von Neumann equation (which is unitary and time-reversible) but a generalized equation, with emergent elements beyond the Hilbert space, which is non-unitary and time-irreversible. See for instance:

1997 "The Liouville Space Extension of Quantum Mechanics" T. Petrosky and I. Prigogine Advances in Chemical Physics Volume 99, 1-120

One application of this recent formalism to an extension of scattering theory

http://prola.aps.org/abstract/PRA/v53/p4075_1

an introduction to this paper is given here

http://www.ph.utexas.edu/~gonzalo/3bgraphs.html

See also other applications of the generalized Liouville von Neumann equation:

2001 "Quantum transitions and dressed unstable states" G. Ordonez, T. Petrosky and I. Prigogine Phys. Rev. A 63, 052106

2000 "Quantum transitions and nonlocality" T. Petrosky, G. Ordonez and I. Prigogine Phys. Rev. A 62 42106

 

[kith]

The equation given includes non-Markovian corrections and coupling to any order.

You are talking about corrections and power series expansion. To me, this sounds like there are some underlying assumptions. Can you please give a reference?

No exactly. The equation for the whole system is not the ordinary Liouville / von Neumann equation (which is unitary and time-reversible) but a generalized equation, with emergent elements beyond the Hilbert space, which is non-unitary and time-irreversible. See for instance: [...]

As far as I can see, your references are talking about the dynamics of systems approaching equilibrium. From the viewpoint of fundamental dynamics, this means there is again an environment involved. So this just adds a layer of complexity. Again the equations of motion can in principle be derived from the Hamiltonian of a larger system. Namely the combined system "whole" system + relevant part of it's environment. This system evolves accordingly to the Liouville / von Neumann equation.

But I can see where this is going. We're going to end up in another discussion about the question, if the time-evolution of closed systems is unitarian. ;-)

 

[Studiot]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

One thing puzzles me, juanrga.

What exactly is a 'pure state' - it is not a term I am familiar with.

Further since you make the distinction what alternatives are there ie what might non pure state be and what are they called?

Thanks.

 

[kith]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

I don't know much about QFT myself. The last few posts were about more fundamental questions, concerning the Hamiltonian formalism which can be considered the framework for QFT and many other branches of physics.

What exactly is a 'pure state' - it is not a term I am familiar with.

A pure state is a state of maximum knowledge. In statistical mechanics, you typically only know macroscopic variables like temperature. These macro variables do not specify your micro state completely, but lead to a variety of possible states. So the fundamental objects in statistical mechanics are not pure states, but so-called mixed states.

Classically, a mixed state ρ is a probability distribution on the space of states. The Hamiltonian equations of motion generalize to the Liouville equation. Quantum mechanically, a mixed state ρ is an an operator on the space of states. The Schrödinger equation generalizes to the von Neumann equation.

If you want to include dissipation, the situation gets more complicated and that's one of the things which have been discussed throughout this thread.

 

[Andrew Mason]

The 1st statement written was

"Work done on a system is defined as the change in Kinetic Energy (KE) of that system. While The total energy of a system is the potential energy (PE) plus the kinetic energy, E=PE+KE."

There isn't anything wrong with this, one can read that is that the NET work done on a body is equal to its change in kinetic energy.

So then, what is kinetic energy? You cannot then define kinetic energy the ability to do work (by virtue of its motion) because that is circular.

This is not so much a definition of net Work as it is a statement that net work results in a change of kinetic energy. Work is still defined as force applied through a distance and energy as the ability to do work.

Furthermore, if you read the responding post there is clearly a summation over (i)W12=∑i∫21Fi⋅dsi=∑i∫21miv˙i⋅vidt=∑i∫21d(12miv2i)=T2−T1 where T=12∑imiv2i is the kinetic energy of the system.The summation clearly takes into account all the forces and gives you the net work done on the object. So I don't know what your talking about only one force, it works for many forces.

For instance if a block slides down a plane at constant speed, two forces contribute non zero work, the work done by gravity and the work done by friction. The NET work done on the object is zero, hence its kinetic energy remains constant.

Both original posts are correct.

There is only one v. There are not a v_i for each F_i. So I am not sure what v_i means. Perhaps it should be:

$W = \int_1^2 (\sum \vec{F_i})\cdot d\vec{s} = \int_1^2 \vec{F_{net}}\cdot d\vec{s} = \int _1^2 m\dot vdv = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE$ where v is the speed of the centre of mass of the body.

AM

 

[cbetanco]

So then, what is kinetic energy? You cannot then define kinetic energy the ability to do work (by virtue of its motion) because that is circular.

The Kinetic energy is $KE=\frac{1}{2}mv^2$. There is no mention of work in that definition. The Work done on a system between the initial and final state is defined to be the change in kinetic energy of the system in the initial state to its final state. It is NOT circular to define a new quantity to be the change of another quantity from the initial to the final state.

This is not so much a definition of net Work as it is a statement that net work results in a change of kinetic energy. Work is still defined as force applied through a distance and energy as the ability to do work.

Yes, it is still defined as the force applied through a distance, but what I had showed you is that this is equivalent to the change in Kinetic Energy of the system

There is only one v. There are not a v_i for each F_i. So I am not sure what v_i means. Perhaps it should be:

$W = \int_1^2 (\sum \vec{F_i})\cdot d\vec{s} = \int_1^2 \vec{F_{net}}\cdot d\vec{s} = \int _1^2 m\dot vdv = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta KE$ where v is the speed of the centre of mass of the body.

AM

The $F_i$ is the NET external force acting on the ith mass, $m_i$. The i is not an index over individual forces, its an index running over all the masses of the system. These are point particles with mass $m_i$ in the strictly classical sense, but it also works with rigid bodies if you instead have a mass density where you take $m_i\rightarrow\rho(\bf{x})$ and integrate over the mass density $\rho(\bf{x})$ for the entire volume. Then, you would just have to replace $m_i$ with $\rho(\bf{x})$ and $\bf{v_i}$ with $\bf{v}$, where $\bf{v}$ is the velocity of the rigid body, and then then take the sum $\sum_i \rightarrow \int_V d\bf{x}$ in the equation in my second post and you would still get the work done on the system is equivalent to the change in KE of the system

 

[cmb]

Gosh! How complicated it all is!

And there I was thinking that energy is, simply, the essential attribute that is required to change a system from one state to another.

 

[Andrew Mason]

Yes, it is still defined as the force applied through a distance, but what I had showed you is that this is equivalent to the change in Kinetic Energy of the system

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing. Energy is a property that a body has - the ability to do work and work is force applied through a distance. Saying that work is defined as the change in kinetic energy confuses the fundamental difference between Work and Energy.

The difference is that Work is a transfer of energy to/from a body and Energy is a property of the state of a body.

This distinction is important. For example, in thermodynamics W=Work and Q = Heat flow are transfers of energy not properties of a state of a thermodynamic system. Energy=U, is a property of the state of a system. Saying ΔU = Q+W does not mean that ΔU is defined as Q+W. It is merely a statement that the magnitude of the change in U is equal to the sum of the heat flow to and work done on the system.

AM

 

[cmb]

Energy is a property that a body has

or a field...

...or a propagating wave...

...or...?

 

[Andrew Mason]

or a field...

...or a propagating wave...

...or...?

Fields and propagating waves are all associated with some body ie. some kind of structure that has inertia or mass. Is the energy in the field or in the body? Does a photon represent energy or the transfer of energy from one body to another?

AM

 

[kinetico]

An Alternative Definition of Energy $E$For a simple particle

$${\vphantom{\int}} \vec{a} = \vec{a}$$$${\vphantom{\int}} m \, \vec{a} = m \, \vec{a}$$$$\int m \: \vec{a} \cdot d\vec{r} = \int m \: \vec{a} \cdot d\vec{r}$$$${\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} + \; constant = \int m \: \vec{a} \cdot d\vec{r}$$$$constant = {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} - \int m \: \vec{a} \cdot d\vec{r}$$$$E = {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} - \int m \: \vec{a} \cdot d\vec{r}$$

For a system of n particles

$$E = \sum_{i=1}^n {\textstyle \frac{1}{2}} \: m_i \, \vec{v}_i^{\: 2} - \sum_{i=1}^n \int m_i \; \vec{a}_i \cdot d\vec{r}_i$$

 

[cbetanco]

Just because the magnitude of the work is the same as the magnitude of the change in kinetic energy does not mean they are the same thing.

Yes it does. This is the work-energy theorem. How can you say the magnitude of work and the magnitude of change in kinetic energy are the same, but then in the same sentence argue they are not the same?

Energy is a property that a body has - the ability to do work and work is force applied through a distance. Saying that work is defined as the change in kinetic energy confuses the fundamental difference between Work and Energy.

The difference is that Work is a transfer of energy to/from a body and Energy is a property of the state of a body.

I never argued otherwise. What I am saying is that the work done on the system is equal to the change in kinetic energy of that system. I never argued that the work is a property of the system. But the work done on or by the system is the same as the change in kinetic energy of that system. Please see pg 9 of the graduate level text on Classical Mechanics by Goldstein, Poole and Safko. In between Eq. 1.29 and 1.30 it is written "Hence, the work done can still be written as the difference of the final and initial kinetic energies." That is ALL I was saying. Its a pretty standard definition that is hard to argue with.

Saying ΔU = Q+W does not mean that ΔU is defined as Q+W. It is merely a statement that the magnitude of the change in U is equal to the sum of the heat flow to and work done on the system.
AM

Yes, it does mean by the definition of being equal, that the change of energy in a thermal system is defined to be the heat flowing in or out plus the work done by or on the system. And the ΔU does not have to be the magnitude in the change, it can also be negative depending if the heat is flowing in or out and the work is done by the system, or on the system.

 

[cmb]

Fields and propagating waves are all associated with some body ie. some kind of structure that has inertia or mass. Is the energy in the field or in the body? Does a photon represent energy or the transfer of energy from one body to another?

AM

I don't see how you come to make that distinction. I might equally argue that bodies are associated with some kind of field structures rather than vice versa. How many 'bodies' would exist without electrostatics, gravitational, and nuclear forces?

I do not see how the energy of a photon in free space is associated with 'a body'.

Simply, we are looking at 'configuration' here. 'Energy' is an attribute of the configuration of 'stuff'. Matter, fields, particles, relative position, relative motion.

Your argument is demonstrated incomplete because energy such as 'kinetic' is relative to other bodies, so cannot be described as 'associated with some kind of body'. A body traveling at the same speed as me has no kinetic energy in my frame, but may have KE in someone else's. Similarly, you cannot have electrostatic or gravitational energy without multiple contiguous bodies. Therefore, we must talk about the 'system' as containing the energy, not by association with a body.

 

[juanrga]

You are talking about corrections and power series expansion. To me, this sounds like there are some underlying assumptions. Can you please give a reference?

As far as I can see, your references are talking about the dynamics of systems approaching equilibrium. From the viewpoint of fundamental dynamics, this means there is again an environment involved. So this just adds a layer of complexity. Again the equations of motion can in principle be derived from the Hamiltonian of a larger system. Namely the combined system "whole" system + relevant part of it's environment. This system evolves accordingly to the Liouville / von Neumann equation.

But I can see where this is going. We're going to end up in another discussion about the question, if the time-evolution of closed systems is unitarian. ;-)

Once again, the expression for the dissipator is exact. It is only for computational reasons that it is often expanded in series expansions. Evidently, specific series expansion depends on assumptions about convergence about the expansion center, but this is a computational problem.

Contrary to what you say, the references study the general evolution of isolated systems. Some of those isolated systems approach equilibrium and others do not. Isolated systems approaching equilibrium cannot be studied with the ordinary Liouville /von Neuman equation (unitary and time reversible). That is the reason for their extension of quantum theory.

 

[juanrga]

I haven't touched particle physics since the 1960s so I am really an (obsolete) interested bystander here.

One thing puzzles me, juanrga.

What exactly is a 'pure state' - it is not a term I am familiar with.

Further since you make the distinction what alternatives are there ie what might non pure state be and what are they called?

Thanks.

This is not about particle physics but about quantum theory.

The purity of a system is given by p=Tr{ρ²}. When p=1 the system is in a pure state, otherwise it is in a mixed state.