Zero Hamiltonian and its energies

In summary: After a quick look at that it seems that the reparametrisation is just a change from \tau to u, and the fact that the lagrangian is independent of the particular shape of the curve means there is a 'constraint' on the 'possible curves' (which is just a restatement of the fact that the lagrangian is independent of the path taken?).This constraint reduces the degrees of freedom and so we can't find a unique solution from the equations of motion. This is a bit like a potential well in a potential energy graph: different points have different potential energies but the same kinetic energy. This means that we can't recover the path taken from the kinetic energy because a single KE corresponds to many PE. So
  • #1
Karlisbad
131
0
Let's suppose we have a Hamiltonian of the form:

[tex] \bold H =0 [/tex]

then it's obivious that if you want to get its energies you would get [tex] E_{n}=0 [/tex] for every n, this is a non-sense since you must have positive energies (and a ground state) ...then if we "cheat" :rolleyes: and make:

[tex] \bold H \Phi =i\hbar \frac{\partial \Phi}{\partial u}=\lambda_{n} \Phi [/tex] (2)

where u is a parameter that dictates the "evolution" of the system..of course we must have that [tex] \frac{\partial H}{\partial u}=0 [/tex]

then using (2) we get the energies of the system.. :frown: but are they "really" the energies of our initial system with H=0 ? i don't think so..perhaps we should put [tex] \lambda=0=\lambda (n_x ,n_y, n_z,n_t) [/tex] and from it recover the initial energies or consider that the factor..

[tex] \lambda_{n} \Phi [/tex] is an extra term inside our potential.. how you can quantize H=0 and get finite and positive energies?..:cry: :cry:
 
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  • #2
What physical system would possibly have H=0. In what way would this originate
 
  • #3
There's no way the H could be zero, unless you did something wrong along the way.

Daniel.
 
  • #4
You get a zero Hamiltonian for theories that are invariant under reparametrizations, like, for example, a free relativistic particle. H = 0 is a constraint, and at least in principle you can try to quantize it two ways. Either solving the constraints first (obtaining a H different from zero with the new variables) and then applying the quantization procedure, or applying the quantization procedure and imposing the constraints as relations on the Hilbert space.
 
  • #5
H=0 is not a constraint. A constraint is some function of the phase space variables one gets when he cannot invert the Legendre transformation (primary constraint(s)), or when evolution of the primary constraint(s) generates other relation between the momenta and coordinates (secondary constraint(s)).

Daniel.

P.S. For the free relativistic particle one usually works with the einbein formulation specifically to avoid the [itex] \sqrt{\dot{x}^{\mu}\dot{x}_{\mu}} [/itex] ugliness.
 
  • #6
I should have said: The total Hamiltonian is a linear combination of the constraints that vanish on the physical subspace. The Hamiltonian for the free relativistic particle is an example of this. Do you agree with this?
 
  • #7
Surely H = 0 for a particle at rest in empty space with no potentials/fields? I don't have a particularly thorough understanding of Hamiltonian mechanics, but here is my logic:

[tex]H\left(q_j,p_j,t\right) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t)[/tex]

Since all the [itex]q_j[/itex] do not change [itex]\dot{q}_j = 0[/itex] and therefore also [itex]p_j = \frac{\partial L}{\partial\dot{q}} = 0[/itex]. So we have that

[tex]H = - L(q_j,\dot{q}_j,t)[/tex]

Since [itex]L = T - V[/itex] we can get [itex]H = V - T[/itex] from the above, and since [itex]H = T + V[/itex] we can create some equations and subtract them, where T is kinetic and V is potential energy:

[tex]
\begin{equation*}
\begin{split}
H & = T - V\\
& = T + V
\end{split}
\end{equation*}
[/tex]
subtracting
[tex]
0 = V
[/tex]

(Does this imply that potential energy must be zero for a particle to remain at rest?)

Since T is the kinetic energy term it should be zero if the particle is at rest and so [itex]H = L = 0[/itex] too? I don't have a good explanation for this without using some classical mechanics or relativistic mechanics... so I am not sure if this applies in the quantum case. Since there is uncertainty in positions and momentums I would guess that a particle always has the possiblity of having kinetic energy, and the nearest we can get to zero is having a distribution of possible energies centred on zero.

Right? Wrong? Ballpark?
 
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  • #8
hellfire said:
I should have said: The total Hamiltonian is a linear combination of the constraints that vanish on the physical subspace. The Hamiltonian for the free relativistic particle is an example of this. Do you agree with this?

Yes, surely, the Hamiltonian is proportional to the secondary constarint in the einbein formulation.

I'd suggest a path integral quantization for these reparametrization invariant problems. Canonical one is slightly more demanding.

Daniel.
 
  • #9
Surely H = 0 for a particle at rest in empty space with no potentials/fields?

Sorry? Has everybody gone mad?

If you have invariance under reparemtrisations then you gave gauge degress of freedom that have to be fixed. Take the example of the E.M. Field. Integrate the lagrangian by parts, set surface terms to zero to get the lagrangian density L=AG^-1A where G^-1 is the inverse of the Greens function and A is the 4-potential. G^-1 is not invertible, all its eigenvalues are 0, but we have gauge degrees of freedom which need to be fixed (In the path integral formalism, the functional measure contains an infinite term due to U(1) gauge transformations).

In the canonical formalism, this is fixed via the Ghosts fields.
 
  • #10
When I talked about "reparametrization invariance" I actually meant "time reparametrization invariance", which, as far as I know, is the usual meaning of this expression. Lagrangians which are invariant under time reparametrizations lead to a zero Hamiltonian. For this to be true, the Lagrangian must be a homogeneous function of the [itex]\dot q[/itex]. Then, due to Euler's homogeneous function theorem [itex](\partial L / \partial \dot q) \dot q = L[/itex], and it follows that the Hamiltonian is zero. As the E.M. Lagrangian is not homogeneous in [itex]\dot q[/itex], its Hamiltonian is not equal to zero.
 
  • #11
Can you please give me an example of the 'time reparametrisation' you are talking about? I want to make sure that we are actually discussing is the same thing and if you are correct:smile:
 
  • #12
The free relativistic particle:

[tex]\mathcal{L} = \left( g_{\mu \nu} \frac{\partial x^{\mu}}{\partial \tau}\frac{\partial x^{\nu}}{\partial \tau}\right) ^{1/2}[/tex]

You may be interested in sections 1 and 2 of this lecture notes of Edmund Bertschinger:
http://ocw.mit.edu/NR/rdonlyres/Physics/8-962Spring2002/6D03860E-9C15-4EB8-B96A-8B720D68EFE8/0/gr5.pdf
 
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  • #13
That's a nice reference, Hellfire. IIRC from second year analytical mechanics, in proving the Noether theorem for finitely many degrees of freedom, the evolution parameter is switched from time to an arbitrary "\tau" and the newly obtain action leads to a zero Hamiltonian as well.

Daniel.
 
  • #14
Jheriko said:
Surely H = 0 for a particle at rest in empty space with no potentials/fields? I don't have a particularly thorough understanding of Hamiltonian mechanics, but here is my logic:

[tex]H\left(q_j,p_j,t\right) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t)[/tex]

Since all the [itex]q_j[/itex] do not change [itex]\dot{q}_j = 0[/itex] and therefore also [itex]p_j = \frac{\partial L}{\partial\dot{q}} = 0[/itex]. So we have that

[tex]H = - L(q_j,\dot{q}_j,t)[/tex]

Since [itex]L = T - V[/itex] we can get [itex]H = V - T[/itex] from the above, and since [itex]H = T + V[/itex] we can create some equations and subtract them, where T is kinetic and V is potential energy:

[tex]
\begin{equation*}
\begin{split}
H & = T - V\\
& = T + V
\end{split}
\end{equation*}
[/tex]
subtracting
[tex]
0 = V
[/tex]

(Does this imply that potential energy must be zero for a particle to remain at rest?)

Since T is the kinetic energy term it should be zero if the particle is at rest and so [itex]H = L = 0[/itex] too? I don't have a good explanation for this without using some classical mechanics or relativistic mechanics... so I am not sure if this applies in the quantum case. Since there is uncertainty in positions and momentums I would guess that a particle always has the possiblity of having kinetic energy, and the nearest we can get to zero is having a distribution of possible energies centred on zero.

Right? Wrong? Ballpark?
############################

<<Wrong?>>

Quite wrong.

First of all, you are not understanding what he Hamiltonian is. The Hamiltonian is not the value of the energy, it is a relationship between position and momentum for a particular system. If the Hamiltonian is p^2 + q^2, and the value of p^2 + q^2 is zero, then the Hamiltonian is p^2 + q^2, not zero. It is analogous to Bush being the president. Bush is the current VALUE of "president", but the concept of president is not synonymous with "Bush".

As for how you got the obviously wrong result that the the potential energy must be zero for any particle at rest, you need to remember that your initial assumption was that he energy (what you referred to as the Hamiltonian) was zero and that he momentum was zero. So being that you made these assumptions earlier in the derivation, is is not surprising that you reached the result that the potential energy was zero. But you seen to have not realized that you had assumed more than just that the particle was at rest, you also assumed that he energy was zero.
 
  • #15
Zero is zero is zero. 'Nothin from nuthin' is nuthin' (with apologies to the late, great Billy Preston) If H=0 is an operator equation, then "next please". A universally null oR constant operator is of virtually no interest -- what's it good for? Why bother? Every matrix element of a null operator is zero.All eigenvalues of a null operator are zero."Next in line please."

On the other hand, if zero is an eigenvalue among other non-zero ones, all bets are off.

Regards,
Reilly Atkinson
 
  • #16
reilly said:
Zero is zero is zero. 'Nothin from nuthin' is nuthin' (with apologies to the late, great Billy Preston) If H=0 is an operator equation, then "next please". A universally null oR constant operator is of virtually no interest -- what's it good for? Why bother? Every matrix element of a null operator is zero.All eigenvalues of a null operator are zero."Next in line please."

On the other hand, if zero is an eigenvalue among other non-zero ones, all bets are off.

Regards,
Reilly Atkinson

We might be in agreement. What caused you to think otherwise?
 
  • #17
swimmingtoday said:
############################
As for how you got the obviously wrong result that the the potential energy must be zero for any particle at rest, you need to remember that your initial assumption was that he energy (what you referred to as the Hamiltonian) was zero and that he momentum was zero. So being that you made these assumptions earlier in the derivation, is is not surprising that you reached the result that the potential energy was zero. But you seen to have not realized that you had assumed more than just that the particle was at rest, you also assumed that he energy was zero.

I assumed the the kinetic energy, momenta and first derivatives of the coordinates are zero. The only assumptions I made regarding the Hamiltonian are:
[tex]
H\left(q_j,p_j,t\right) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t)
[/tex]
and
[tex]
H = T + V
[/tex]
I don't see how I started out assuming H was zero, unless at least one of the above assumptions is indentical with H=0.

The real error in my maths is that I swapped the order of T and V without noticing in one place, hence all that my working really shows is that the kinetic energy is zero, which was one of my starting assumptions.

[tex]
\begin{equation*}\begin{split}H & = T - V\\ & = T + V\end{split}\end{equation*}
[/tex]

should have been

[tex]
\begin{equation*}\begin{split}H & = V - T\\ & = T + V\end{split}\end{equation*}
[/tex]

Which shows that the kinetic energy is zero and that the Hamiltonian is only made up of the potential energy term, which is just a direct consequence of my starting assumptions anyway...


I still don't see any good reason why H can not be zero. If it is equivalent to the sum of potential and kinetic energies, then a classical particle in free space at rest with no potentials will have H = 0. Perhaps what I am missing is knowing how to go from a classical Hamiltonian to a quantum one... I have been reading some books that cover the topic, but I haven't made much progress yet... the mathematical hurdles are quite high for me.
 
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  • #18
<<I don't see how I started out assuming H was zero, unless at least one of the above assumptions is indentical with H=0.>>

You're right--I was wrong in saying you did that. I went back to your post and found the actual error--it was just a dropping of a minus sign.

<<The real error in my maths is that I swapped the order of T and V without noticing in one place, hence all that my working really shows is that the kinetic energy is zero, which was one of my starting assumptions.
>>

Exactly. I went back to your original post after reading the beginning of your current post, and that is exactly what happened: You argued that when p and q dot were zero, as when the particle is at rest, that the formulala for H gives H = L. You dropped a minus sign. It should have been H = negative L. Proceeding with the minus sign error you set L= T-V to the standard Hamiltonian formula H= T +V, and got T-V = T+V, yielding V=0. However, if the minus sign was not inadvertantly dropped ypou would have argued negative (T-V) = (T+V), which gives the correct result that T (rather than V) equals zero when p and q dot are zero.


<<I still don't see any good reason why H can not be zero. If it is equivalent to the sum of potential and kinetic energies, then a classical particle in free space at rest with no potentials will have H = 0. >>

It's largely a matter of a tricky definition. The Hamiltonian of a particle is a relationship between a particle's position and its momentum, that when calculated out would give its energy. The energy is the actual *number* you get when you do the calculation. The Hamiltonian is a recipe, and the energy is the number you get applying the recipe .

Consider a harmonic oscillator. The Hamiltonian is (p^2)/2m + (k/2)x^2. And suppose it happened to have values of p=0 and x=0. If you are asked "What is the energy of the systen?" you have to answer that the energy is zero. If you are asked "What is the Hamiltonian of the systen?" you have to answer that the Hamiltonian is (p^2)/2m + (k/2)x^2..

As I said earlier, it is like Bush being the president. "President" is like the Hamiltonian, and "Bush" is like the energy. The President is a person with certain powers, and Bush happens to be that particular person under the situation we have before us. But the words "Bush" and "President" do not mean the same thing, just as "Hamiltonian" and "energy" do not mean the same thing.
 
  • #19
swimmingtoday said:
<<I still don't see any good reason why H can not be zero. If it is equivalent to the sum of potential and kinetic energies, then a classical particle in free space at rest with no potentials will have H = 0. >>

It's largely a matter of a tricky definition. The Hamiltonian of a particle is a relationship between a particle's position and its momentum, that when calculated out would give its energy. The energy is the actual *number* you get when you do the calculation. The Hamiltonian is a recipe, and the energy is the number you get applying the recipe .

I understand what you are saying, but I don't think it is particularly relevant or insightful... the Hamiltonian may be identified with an equation which allows you to derive useful equations of motion, but that doesn't reduce the validity of a statement like H=0. Regardless as to what useful things you can do in whatever form any expression takes, it doesn't invalidate the fact that you can evaluate said expression and write an identity just like H=0.

If my notation or terminology is non-standard, then the standard is needlessly confusing imo since there is nothing real to gain by making such a distinction. (Other than confusing me :wink: )


So I am still curious as to why it is that some of the responses have indicated that a Hamiltonian which evaluates to zero is a such a flawed concept, since there hasn't been an actual explanation given. Given that H = T + V it would seem that a particle at rest in empty, potential free space, would always have H evaluate to zero, at least classically. I really can't see anyway that this can be wrong so I am going to need someone to point out the "what probably should be obvious".
 
  • #20
Jheriko said:
I understand what you are saying, but I don't think it is particularly relevant or insightful...

I should point out that it is not me that came up with the way the terms are used!

Here though is why a distinction is made between "the Hamiltonian" and "the energy".

Hamilton's equations are dH/dp = q dot, dH/dq = p dot. (The derivatives are partial derivatives; and I don't care about factors of negative one) If you have a harminic oscillator with an energy of 14, then inserting "14" for H in Hamilton's equations gives q dot= 0, p dot =0, because the derivative of a constant ( "14" is a constant, rather than a function of p and q) is zero. So for H in Hamilton's equations, you cannot use the value of the energy, but ratther the Hamiltonian recipe for energy as a function of p and q.

Likewise in quantum physics if you want to find the averged time derivative of a physical quantity by taking the commutator of the operator for that quantity with the Hamiltonian operator you cannot use "14" as the Hamiltonian operator if 14 is the value of the energy-- a *number* such as 14commutes with anything. You need to write out the Hamiltonian Operator in terms of position and momentum operators.
 
  • #21
With kudos to swimmingtoday, I will reiterate that H=0 describes a totally uninteresting system.That is, the dynamical equations, formed by various derivatives of H classically, or the Schrodinger Eq. are empty; there are no changes in velocity, no force, and any motion stays the same.

There is nothing wrong with H=0, it is just not very interesting.
Regards,
Reilly Atkinson
 
  • #22
reilly said:
There is nothing wrong with H=0, it is just not very interesting.
Why not? As already mentioned, some of the H = 0 systems are invariant under reparametrizations, which is a very interesting property. Of course the task is then to move on to a different Hamiltonian where the physical degrees of freedom are singled out.

Moreover, I came across another kind of systems that have H = 0. These are systems for which the complete energy-momentum tensor and hence the Hamiltonian is identically zero, without being trivial systems. This is the case in condensed matter physics for the Chern-Simons action. This is a topological action in which the space-time metric does not appear, and hence the energy-momentum tensor as the variation of the action wrt the metric vanishes indentically. The task is then to find out what are the states for this ground state degeneracy. I cannot elaborate on this because I just mention what I have read today in chapter VI.1 of Zee's "Quantum Field Theory in a Nutshell". May be someone could elaborate on this.

Anyway, this issue of H = 0 seams to me far more complex and interesting than what you are assuming here.
 
  • #23
hellfire said:
Anyway, this issue of H = 0 seams to me far more complex and interesting than what you are assuming here.
Well, one should be cautious about all of this : H_a = 0 are indeed constraint equations in reparametrization invariant systems. In canonical vacuum gravity, it is well known that the Hamiltonian is ``pure constraint'' (as is the case for all reparametrization invariant systems), the constraints being relations which need to be satisfied by the initial data as well as being the generators of gauge transformations (that is diffeomorphisms) since the time evolution preserves the constraints (they are first class). So, classically, all one has to do is to solve the constraint equations as well as the Hamilton equations (even when the Hamiltonian vanishes on shell, there are nontrivial equations of motion). In order to get classical observables in GR (diffeomorphism invariant quantities), one has to think in the relational way : that is, study properties of physical quantity X with regard to physical quantity Y. However, physical quantities are themselves dirac observables, that is quantities which are transforming well under the (on shell) action of gauge transformations. Therefore, a physical energy (or Hamiltonian) should be a relational quantity and such things are very hard to write down. Quantum mechanically, a similar thing happens : the Schrodinger equation becomes H psi = 0, but the time evolution is hidden in the Heisenberg equations d/dt O = i [H, O] which are the quantized Poisson brackets. Now, to find a physical Hamiltonian in this line of thought, one could use some O as clock and write for another observable K : (d/dt)_O K = -i [O, K]. At least, this is Rovelli's line of thought but it leads to two different notions of time at the quantum level (time as an operator - in either a measurable quantity - and time as a parameter). Personally, I do not wish to consider such things, but in such way one might retrieve the usual positive energies (if one were able to properly quantize the Dirac Algebra). Now, all these problems could be avoided of course by choosing some diffeomorphism gauge at the classical level and quantizing the latter system. However, this leads to problems at the nonperturbative level (breakdown of gauge - this has been studied by Kuchar), but not at the perturbative one (apart from renormalization issues). Anyway, it occurs to me that such problems are better dealt with by modifying the theory at the perturbative level.

Careful
 
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  • #24
- Following to Careful (Rovelli's notion of time showed very interesting to me)..perhaps "time" can be either a "parameter" or an observable for example we make:

[tex] \hat T|\Phi>=t |\Phi> [/tex]

"T" is the time-operator and t (eigenvalue of T) is the parameter, but both are time.

Another form to view the Constraint perhaps..is this if you consider that the metric and its conjugate momenta [tex] \pi _{ab} [/tex] and [tex] g _{ab} [/tex] form a certain Phase-Space of 20 dimension..then we could consider that:

[tex] \bold H(g_ab , \pi _{ab} )=0 [/tex] is a subspace inside the total Phase space..(just as the equation of the sphere [tex] x^{2}+y^{2}+z^{2}-1=0 [/tex] is a subspace of R3 ), i believe that the biggest problem "unifying" QM and GR is that in QM we have always an absolute (all the observers see the same time) concept of time whereas in GR every observer has its own time ...
 
  • #25
Karlisbad said:
- Following to Careful (Rovelli's notion of time showed very interesting to me)..perhaps "time" can be either a "parameter" or an observable for example we make:

[tex] \hat T|\Phi>=t |\Phi> [/tex]

"T" is the time-operator and t (eigenvalue of T) is the parameter, but both are time.

It is not that easy, you have to measure T, but how can we be sure that the order in the T measurements is the same as the order of its eigenvalues (and in ``what'' do we measure ?? usually we measure something like position in time). It is clear that you have to introduce some reference parameter in order for this to work (good old fashioned Newtonian time) as you can see from the math in my previous post. Actually, Arkadiusz Jadcyk has written lots about such ideas and the implication of ``self measuring'' devices ; again not my cup of tea (you cannot avoid the consciousness crap).
 
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  • #26
An apparent paradox??..let be [tex] p_{\mu}p^{\mu}=H=0 [/tex] the Hamiltonian of specia relativity so H=0 then if we make a Wick rotation the energies (if any) of H for a infinite potential well would be:

[tex] \pi ^{2} \hbar ^{2}\frac{1}{L^{2}}(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}+n_{it}^{2})=E_{n} [/tex]

where {0<x<L ,0<y<L ,0<z<L , 0<t<L} are where V=0, elsewhere V=oo (infinite potential well)

the question is..does the expression above for energies of H=0 makes sense?..perhaps the election for the quantum numbers nx,ny,nz,nt is not arbitrary
 
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  • #27
let be [tex] p_{\mu}p^{\mu}=H=0 [/tex]

What? This is not the Hamiltonian for special relativity? A can't even be bothered explaining why it isn't, but just think about how we would go about quantising this.
 
  • #28
Epicurus said:
What? This is not the Hamiltonian for special relativity? A can't even be bothered explaining why it isn't, but just think about how we would go about quantising this.
Correct, one has to apply the Dirac constraint programme correctly. Starting with the action [tex] S(x^{\mu}, \dot{x}^{\mu},t) = m \int_{0}^{t} \sqrt{\dot{x}^{\mu}(s) \dot{x}_{\mu}(s)}ds [/tex] and metric signature (+---), one obtains [tex] p_{\mu}(t) := \frac{\partial L}{\partial \dot{x}^{\mu}(t)} = m \frac{\dot{x}_{\mu}(t)}{\sqrt{\dot{x}^{\nu}(t) \dot{x}_{\nu}(t) } } [/tex]. The momenta satisfy a first class constraint [tex] p^{\mu}(t) p_{\mu}(t) = m^2 [/tex], solving the spatial velocities [tex] v^{\alpha}(t) [/tex] with respect to the spatial momenta [tex] p_{\alpha}(t)[/tex] and the time velocity [tex]v^{0}(t) [/tex] : [tex] v^{\alpha} (t) = \frac{p^{\alpha}(t) v^{0}(t)}{\sqrt{m^2 + p_{\beta}(t) p^{\beta}(t)} }[/tex]. In this way one computes that the Hamiltonian equals :
[tex] H(p^{\mu}, v^{0}) = v^{0}(t) (p_{0} - \sqrt{m^2 + p^{\beta}(t) p_{\beta}(t) } \, ) [/tex]. Now for any observable O defined on the worldline of the particle [tex] O(x^{\mu}(t), p^{\nu}(t)) [/tex], we have as usual that [tex] \frac{d O(x^{\mu}(t), p^{\nu}(t))}{dt} = \{ H, O \} [/tex] the result of which depends upon the Lagrange multiplier [tex] v^{0}(t) [/tex]. Now, [tex] \frac{d\tau(t)}{dt} = \frac{v^{0}(t)}{\sqrt{m^2 + p^{\beta}(t)p_{\beta}(t)}} [/tex] hence [tex] \tau(t) [/tex] really is a nonlocal function of the Lagrange multiplier and the conjugate momenta [tex] p_{\beta}(t)[/tex]. Therefore, [tex] \frac{d O}{d \tau} = \frac{ \{ H , O \} \sqrt{m^2 + p^{\beta}(t) p_{\beta} (t)} }{m v^{0}(t)} = \frac{ \{ p_{0} - \sqrt{m^2 + p^{\beta}(t) p_{\beta} (t)} , O \} \sqrt{m^2 + p^{\alpha}(t) p_{\alpha} (t)}} {m} [/tex] is independent of the Lagrange multiplier and hence a Dirac observable. Now, quantum mechanically [tex] \tau [/tex] becomes an operator, so that's a problem ... how can you measure the derivative of an operator with respect to an operator (which does not commute with the latter) without measuring the latter itself and you measure [tex] \tau [/tex] in what, in t, but t has no real meaning according to the dogma of reparametrization invariance. I guess in real life, one can assume that eigentime and the observable at hand are more or less commuting, but that does not solve the latter issue. Clearly, the parameter t, also plays the role of a kind of ``internal clock'' although the physics does not depend upon it.
 
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  • #29
Someone has pointed here the "Dirac Constraint" method for quantization..could someone provide a good open-access link to understand this to quantizy:

a) zero Hamiltonians [tex] H(q_i , P_i)=0 [/tex]

b) Lagrangians with an acceleration term [tex] L(q,\dot q , \ddot q ) [/tex]

thanks...:redface:
 
  • #30
Karlisbad said:
Someone has pointed here the "Dirac Constraint" method for quantization..could someone provide a good open-access link to understand this to quantizy:

a) zero Hamiltonians [tex] H(q_i , P_i)=0 [/tex]

b) Lagrangians with an acceleration term [tex] L(q,\dot q , \ddot q ) [/tex]

thanks...:redface:

I learned it from Thiemann's notes on ``introduction to canonical quantum gravity'' (look for it on the arxiv, it is a paper written in 2001). He devotes an appendix to this Hamiltonian method for singular legendre transformations.
 
  • #31
By the way if we have [tex] \bold H =0 [/tex] can this H be split into the terms.

[tex] H^{(0)} (g_{00} , \pi _{00} )+H^{(3)}(g_ {ij} , \pi _ {ij} )=0 [/tex]


so we quantizy the term... [tex] H^{(3)}(g_ {ij} , \pi _ {ij})\Phi = E_{n} \Phi [/tex]


and from this we get the energies..:-p :-p
 
  • #32
Karlisbad said:
By the way if we have [tex] \bold H =0 [/tex] can this H be split into the terms.

[tex] H^{(0)} (g_{00} , \pi _{00} )+H^{(3)}(g_ {ij} , \pi _ {ij} )=0 [/tex]


so we quantizy the term... [tex] H^{(3)}(g_ {ij} , \pi _ {ij})\Phi = E_{n} \Phi [/tex]


and from this we get the energies..:-p :-p
First of all in GR, you have four constraints, second the constraints themselves do only contain the [tex] g_{ij}, \pi_{ij} [/tex] :biggrin:. Moreover, these energies as you call them have no physical meaning whatsoever. :-p
 
  • #33
By the way appart from constraint (i finally learned how to apply Dirac's method to them using Wikipedia :-p) Does the Einstein Lagrangian has an acceleration term?..i have some troubles to quantize Lagrangians with an acceleration term so:

[tex] L= A(v)^{2}+BV(x)+C(\dot V)^{2} [/tex]
 
  • #34
Karlisbad said:
By the way appart from constraint (i finally learned how to apply Dirac's method to them using Wikipedia :-p)

Perhaps you should consult a better reference :-p

Karlisbad said:
Does the Einstein Lagrangian has an acceleration term?..

Yes, the Ricci scalar contains second derivatives with respect to time.
 
Last edited:
  • #35
Usually, second order derivative terms in the HE action can be removed by a part integration. See Dirac (1975) for details.

Daniel.
 

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