Why does the Hamiltonian generate time evolution?

[Chopin]

The first part of QFT seems to be almost entirely mathematical formalism, without really requiring a whole lot of physical insight to proceed. For instance, we can start with the free-field scalar Lagrangian, minimize it using the Euler-Lagrange equation to arrive at the Klein-Gordon equation, and observe that solutions to this equation are complex exponential waves, expressed using creation and annihilation operators. We can also observe that the Lagrangian is symmetric under infinitesimal time translations, and use that fact to produce a conserved quantity called H, which is [tex]\dot{\phi}\pi - L[/tex].

Finally, we can observe that H can be expressed in terms of the creation and annihilation operators, in a way which shows that they produce eigenstates of H, and observe that an eigenstate [tex]a(k)|0\rangle[/tex] with eigenvalue [tex]\omega_k[/tex] has a time derivative of [tex]i\omega_k a(k)|0\rangle[/tex]. From this, we can conclude that we can evolve a free field through time by exponentiating by H, i.e. [tex]\phi(x,t)=e^{iHt}\phi(x,0)[/tex]. This can all be determined entirely through mathematical reasoning, without ever needing to introduce any physical insight into what [tex]\phi[/tex], [tex]a(k)[/tex], etc. actually mean.

The next part of QFT will then build on the idea that [tex]e^{iHt}[/tex] is the time evolution operator, by introducing an interaction Hamiltonian and expanding it out in a power series, which leads to Wick's Theorem and Feynman diagrams and all that fun stuff. My question, though, is how do we know that H will always be the generator of time evolution? We can actually prove it in free theory because we can expand a state into its Fourier basis and watch them all evolve, but that requires explicitly solving the equations of motion, which of course we can't do in an interacting theory. So is there actually a way to show that [tex]e^{iHt}[/tex] will do the same thing in the interacting theory as in the free theory, or do we simply have to assume that it still will?

Sorry that this is a bit wordy, hopefully I managed to adequately communicate my question. I can't quite tell whether this question is really stupid or really profound, so I'd appreciate any insight either way.

 

[meopemuk]

You have presented the traditional logic of introducing QFT, in which one starts from the Lagrangian of a classical field theory, solves Euler-Lagrange equations, then proceeds to quantization, etc. etc. I always found this approach rather confusing. I think that an alternative approach presented in S. Weinberg, "The quantum theory of fields", vol. 1 is more logical and transparent. In this logic, the Hamiltonian is [tex]\textbf{defined} [/tex] as the generator of time translations. This definition is valid in both free and interacting cases. So, the answer to your question is rather simple.

Eugene.

 

[xaos]

isn't in the schroedinger's pde over a hilbertspace, the time evolution part is itself proportional to a hamiltonian (laplacian plus potential)? (sorry, this isn't my field)

 

[Chopin]

isn't in the schroedinger's pde over a hilbertspace, the time evolution part is itself proportional to a hamiltonian (laplacian plus potential)? (sorry, this isn't my field)

Yeah, I think my question basically amounts to "Why is the Schrodinger equation true?" In basic QM you're usually told that it's not possible to derive the Schrodinger equation from any previous physical knowledge, but I was wondering if by the time we get to QFT, we have enough tools to do it. The fact that the Hamiltonian generates time evolution makes sense intuitively, given the fact that it's derived from the Lagrangian by an infinitesimal time translation, so it seems as though there ought to be a way to formally prove the connection between the two.

You have presented the traditional logic of introducing QFT, in which one starts from the Lagrangian of a classical field theory, solves Euler-Lagrange equations, then proceeds to quantization, etc. etc. I always found this approach rather confusing. I think that an alternative approach presented in S. Weinberg, "The quantum theory of fields", vol. 1 is more logical and transparent. In this logic, the Hamiltonian is [tex]\textbf{defined} [/tex] as the generator of time translations. This definition is valid in both free and interacting cases. So, the answer to your question is rather simple.

Eugene.

I've heard of Weinberg a few times, but haven't read a whole lot. I may try to look through it. However, there has to be something more to it than just defining the Hamiltonian as the time translation generator, because you have to be consistent about it. I can't just pick some operator out of the air and say "this will produce d/dt of my states", because I would run into logical inconsistencies. There has to be something about the Hamiltonian in particular that allows it to fill this role without running into any mathematical contradictions.

 

[Ben Niehoff]

The Hamiltonian is the generator of time translation because it is the conserved charge associated with time translation invariance. Conserved charges are always generators of the symmetry related to their conservation. Here are the other main examples:

Momentum is the conserved charge associated with spatial translation. Also, the momentum operator is the generator of translations.

Angular momentum is the conserved charge associated with rotations, and it is also the generator of rotations.

Electric charge is the conserved charge associated with global gauge transformations, and it is also the generator of global gauge transformations.

To see why these are true, try acting on a state with each of these various operators.

 

[tom.stoer]

I think this discussion can be answered already in the classical canonical formalism w/o any reference to quantum mechanics.

 

[Chopin]

Conserved charges are always generators of the symmetry related to their conservation.

This makes sense intuitively. Is there an explicit proof for it somewhere, or does one basically just run the proof for Noether's Theorem backwards to get it?

 

[The_Duck]

Section 3.4 of Ballentine's QM book gives an argument that, in basic QM at least, the generators of translations, boosts, rotations, and time translations are the operators representing momentum, position, angular momentum, and energy respectively.

 

[schieghoven]

Section 3.4 of Ballentine's QM book gives an argument that, in basic QM at least, the generators of translations, boosts, rotations, and time translations are the operators representing momentum, position, angular momentum, and energy respectively.

I've never heard of the boost generator == position operator. I don't think it can be right, because the commutator of boosts gives a rotation operator:
[tex]
[K_i, K_j] = -i \epsilon_{ijk} J_k
[/tex]
(e.g. Weinberg eq 2.4.20 or Foldy Phys Rev 102 568, 1956)

In contrast, we know that position operators commute (commutator vanishes).

 

[The_Duck]

Yes--in the Lorentz group. Ballentine discusses nonrelativistic quantum mechanics and the Galilean group. There the position operator generates boosts. To boost a wave function by momentum p in NRQM do you not multiply by exp(ipx)? So x generates boosts. I don't know to what extent the ideas presented in Ballentine apply to QFT.

 

[Chopin]

I've been watching Sidney Coleman's QFT lectures, and during his discussions on symmetries and conservation laws, I think I remember him saying that the boost symmetries lead to the conservation of the velocity of the center of mass, or something like that. Does that sound right?

 

[meopemuk]

I've never heard of the boost generator == position operator. I don't think it can be right, because the commutator of boosts gives a rotation operator:
[tex]
[K_i, K_j] = -i \epsilon_{ijk} J_k
[/tex]
(e.g. Weinberg eq 2.4.20 or Foldy Phys Rev 102 568, 1956)

In contrast, we know that position operators commute (commutator vanishes).

The center-of-mass position is given by the Newton-Wigner position operator. In the spin-zero case this operator has simple form

[tex] \mathbf{R} = -\frac{c^2}{2}\left(\mathbf{K}H^{-1} + H^{-1}\mathbf{K} \right) [/tex]

The three components of [tex] \mathbf{R} [/tex] do commute with each other.

Eugene.

 

[Chopin]

Cool. That still leads me back to my original question, though--does anybody know a general, relativistic, field-theory-based proof that conserved quantities are generators of their corresponding symmetries? It seems like this is probably fairly straightforward, but I just don't quite know enough about the machinery to make it work. As tom.stoer said, though, it can probably be done in the classical field formalism, without using the commutation relations.

 

[Deric Boyle]

Hey Chopin why don't you refer Sir Dirac's work.
For more basic dobts you may refer M. Jammer's "The conceptual development of quantum mechanics" Tata McGrawHill publication

 

[tom.stoer]

@Chopin: let's start with a basic question (omitting details like domain, boundedness, etc.)

1) you have a conserved quantity Q, that means [H,Q] = 0
2) you can use Q as generator of a unitary operator U[a]=exp(iaQ)
3) of course H' = U*HU = H b/c of (1)

So you have this U[a] is a unitary, one-parameter symmetry.

OK so far?

 

[Chopin]

Maybe. Are we simply defining an operator Q such that [H, Q] = 0, and will later go on to show that it is conserved, or are we making the assertion at this point that any operator which commutes with the Hamiltonian is conserved? I know that it is true that conserved quantities commute with the Hamiltonian, but I wanted to check whether we are taking this as a prerequisite to the argument, or whether we will derive it as part of the argument.

 

[tom.stoer]

As H generates the time evolution (which is a consequence of the classical Hamiltonian formalism) [H,Q]=0 is equivalent to dQ/dt = 0 (Heisenberg equation of motion). So this is nothing else but quantizing well-known classical formulae {H,Q} where {,} is nothign else but the Posson bracket.

Constructing Q may be a difficult task. It can be done using Noether's theorem, educated guessing, ... Anyway. Once you have found an observable Q that commutes with H it is by definition a conserved quantity and you can always construct a unitary operator that represents a symmetry of the system.

So all arguments are already present in the classical formalism - except for applying U to states.

 

[Chopin]

I'm not sure that really helps me, though--the original reason I was looking for the proof about generators was to help show how the Hamiltonian generates time evolution. This proof seems to assume that up front, and go on to show how other generators are created in terms of it.

Assume we have a field [tex]\phi(x)[/tex], which we arrived at by extremizing a Lagrangian density [tex]L[/tex] (this is supposed to be script L, but I can't figure out how to do that in TeX.) We derive a Hamiltonian [tex]H[/tex] from that Lagrangian, by noting that it has a time symmetry, and using Noether's Theorem to arrive at a conserved quantity. We also derive [tex]\pi(x)[/tex] from the Lagrangian in the usual way. Also assume that we have imposed the equal time commutation relations on [tex]\phi(x)[/tex] and [tex]\pi(x)[/tex]. Finally, define a unitary operator [tex]U(t) = e^{iHt}[/tex].

My question is, using only the above facts, and nothing else, is it possible to show that [tex]\phi(\textbf{x},t) = U(t)\phi(\textbf{x},0)[/tex]? This is essentially what all the QFT texts seem to do, but I don't quite understand how they get there.

 

[A. Neumaier]

Assume we have a field [tex]\phi(x)[/tex], [...]
define a unitary operator [tex]U(t) = e^{iHt}[/tex].

My question is, using only the above facts, and nothing else, is it possible to show that [tex]\phi(\textbf{x},t) = U(t)\phi(\textbf{x},0)[/tex]? This is essentially what all the QFT texts seem to do, but I don't quite understand how they get there.

Not quite. The formula in the usual QFT textbooks is

[tex]\phi(x,t) = U(t)\phi(x,0)U(-t)[/tex]

It is a special case of the fact that for any operator A in the Schroedinger picture, the corresponding time-dependent operator in the Heisenberg picture is

[tex]A(t) = U(t)AU(-t)[/tex]

 

[tom.stoer]

And this is nothing else but the formal integration of the time-dependent Heisenberg e.o.m.

 

[Chopin]

Not quite. The formula in the usual QFT textbooks is

[tex]\phi(x,t) = U(t)\phi(x,0)U(-t)[/tex]

Sorry, I was thinking of a state, not an operator.

[tex]\Psi(t) = U(t)\Psi(0)[/tex]

And this is nothing else but the formal integration of the time-dependent Heisenberg e.o.m.

Yes, and my question is, why? Why does this specific operator appear in the equations of motion? Or, alternatively, why does the Schrodinger Equation take the form that it does?

With a free field, you can solve the equations of motion to arrive at the creation/annihilation operators, and you can express the Hamiltonian in terms of them. Then you can actually show mathematically how the Hamiltonian arises in the derivative of the field. So it's clearly not just a convention that we pulled out of the air (i.e., I couldn't just decide to invent the "Steve Picture", where states evolve according to the angular momentum operator or something.) However, since we can't solve interacting theory, that same proof doesn't work there. So there must be some other way to show that the Hamiltonian still serves the same role in the interacting theory that it did in the free theory.

 

[A. Neumaier]

Why does this specific operator appear in the equations of motion? Or, alternatively, why does the Schrodinger Equation take the form that it does?

The Hamiltonian is defined as the infinitesimal generator of the symmetry group of time translations. This produces the Schroedinger equation. Particular Hamiltonians are chosen to match this definition.

 

[Chopin]

I've been groping around online a little more, and ran across a few things that seem to be related to my question here. Most notably, https://www.physicsforums.com/showthread.php?t=187612", the latter of which is way over my head, but gives me some clues on how to proceed. It looks like to really understand the answer to this question completely, I'll need to do some reading up on abstract algebra. Buzzwords like Lie algebras, conformal mappings, representation theory, and symplectic manifolds seem to come up a lot. Does that sound like the right track, or am I digging unnecessarily deep here?

 

[strangerep]

Buzzwords like Lie algebras, conformal mappings, representation theory, and symplectic manifolds seem to come up a lot. Does that sound like the right track, or am I digging unnecessarily deep here?

If you haven't yet studied Lie groups/algebras and some associated representation
theory, that would certainly be a good investment if you're interested in quantum physics.

You could a physics-oriented introduction such as Greiner & Muller's
"Quantum Mechanics -- Symmetries", or possibly some of the early sections
of Ballentine where he shows how to go from physical symmetries (Galilean)
to QM.