Why isn't momentum a function of position?

[lugita15]

In quantum mechanics, the Hamilltonian operator is constructed as the infinitesimal generator of the time translation group, which is a 1-parameter group. Yet it can still depend on time. So you have a situation where the generator of a 1-parameter group can depend on the parameter. Yet the momentum operator, the generator of infinitesimal spatial translations, cannot depend on the parameter(s) of the spatial translation group, namely position. And similarly the angular momentum operator cannot depend on angle (or direction), the parameter(s) of the rotation group. Is there a fundamental reason for this, or is it simply that we happen to already know the properties of the dynamical variables from classical mechanics, so we don't bother with it? I'd be disappointed if it was the latter, because that might undermine the elegance of treatments like Sakurai or Townsend (at the graduate and undergrad levels respectively) which try to derive QM from minimal first principles.

Any help would be greatly appreciated.

Thank You in Advance.

 

[Jano L.]

Hello lugita15,

I think the operator " generator of temporal translation " is g_t = i\hbar \partial/\partial t, in the same sense as the operator " generator of coordinate translation " is g_k = i\hbar \partial/\partial q_k.

Schroedinger's equation says that the actual psi-function \psi(q,t) is such that the operator i\hbar \partial/\partial t has the same effect as the Hamiltonian operator H(q,p,t). You can write then e^{\Delta t g_t/i\hbar} \psi = e^{\Delta t H(q,p,t)/i\hbar}\psi. But mathematically, these operators are not the same thing.

Group theory and symmetries are important concepts, but personally I am sceptical about the role of symmetry in " deriving " quantum mechanics. It is better to study historical papers and try to understand how people came to it. Besides, the whole concept of translation generators works only for functions that are equal to their Taylor expansion. Such functions are not sufficient.

 

[lugita15]

Hello lugita15,

I think the operator " generator of temporal translation " is g_t = i\hbar \partial/\partial t, in the same sense as the operator " generator of coordinate translation " is g_k = i\hbar \partial/\partial q_k.

Schroedinger's equation says that the actual psi-function \psi(q,t) is such that the operator i\hbar \partial/\partial t has the same effect as the Hamiltonian operator H(q,p,t). You can write then e^{\Delta t g_t/i\hbar} \psi = e^{\Delta t H(q,p,t)/i\hbar}\psi. But mathematically, these operators are not the same thing.

I agree that there is a difference between abstract Hilbert space operators and their position basis representations, just as there is a difference between quantum states |ψ> and their position basis representations <x|ψ>=ψ(x), i.e. position-space wave functions. But what does that have to do with my question? I was asking about the explicit dependence of the Hamiltonian operator on a parameter, but the lack of the same for the momentum and angular momentum operators.

 

[Jano L.]

Forgive me, I have a difficulty understanding your question.

My point was that the operator g_t = i\hbar \partial/\partial_t is the generator of temporal translation. It does not contain t. This is parameter-independent the same way as i\hbar \partial/\partial x and we should use this to shift any arbitrary function f(q,t) to another time.

Generators are derivatives hence there are no parameters in them.

In quantum theory, in special case of temporal translation of function \psi(q,t) that satisfies Schroedinger's equation, we can use the Hamiltonian operator instead of g_t. The Hamiltonian operator contains time to account for the external forces.

There is no corresponding Schroedinger's equation for spatial or angular derivative of the wave-function, so we cannot replace these derivatives by something else that would contain coordinate or angle of rotation.

However, there are operators that contain both derivatives and also the coordinates, like L^2.

 

[tom.stoer]

the shift operators are constructed from id/dx and id/dt; these are purely 'kinematical' quantities

H is not this shift operator but the energy operator; writing down the Schrödinger equation demanding that the action of id/dt and H on special wace functions goes beyond kinematics.

 

[lugita15]

So Tom, do you have any thoughts on my question?

 

[tom.stoer]

you mean why energy can depend on time?

 

[lugita15]

Yes, why can the Hamiltonian operator depend on time, but the momentum cannot depend on position and the angular momentum operator cannot depend on angle? The time translation group, the spatial translation group, and the rotation group are on equal footing as far as dependence on parameters go, so why shouldn't their generators be on equal footing?

I know, from a physical perspective you can add or remove energy from a system as time progresses, which corresponds to a time-dependent Hamiltonian. But from a more rigorous point of view, what privileges the Hamiltonian in this regard?

 

[tom.stoer]

Introducing an x-dependence in p would spoil the canonical commutation relations or the Poisson brackets {x,p}=1; but t and H are no canonically conjugate variables, t is not an operator in QM, and therefore there is no such restriction.

 

[lugita15]

But angle is not an operator in QM, and yet angular momentum is not allowed to depend on angle. Why is that?

 

[tom.stoer]

But angle is not an operator in QM, and yet angular momentum is not allowed to depend on angle. Why is that?

Of course; the angle acts as an operator on states

 

[PhilDSP]

But angle is not an operator in QM, and yet angular momentum is not allowed to depend on angle. Why is that?

That's really a very interesting question. The classical usage of angular momentum presumes you are measuring or calculating its value around a fixed axis, right? The angular momentum should not change in value if you orient the axis in some other direction, some other fixed angle. You have locked the value onto the axis by definition.

But what is meant by angular momentum in a QM context? Are there actually 3 different orthogonal (independent) axes where the total angular momentum is a composite of the contribution of each? Yet they are free floating as a whole because no assumption is made that any of the axes are locked onto a simply spinning object.

That would mean that whatever fixed angle in a reference frame you initially choose to orient the object which has angular momentum, the value would be the same. The orthogonality of the 3 axes is what makes the value invariant. Nicht wahr?

 

[lugita15]

Of course; the angle acts as an operator on states

Um, I'm not familiar with an angle operator. Do you mean rotation operators? Those are analogous to time evolution operators; that is they each form a 1-parameter (or possibly many parameter) group.

 

[lugita15]

You can write down something like

\psi(\Omega) = \langle\Omega|\psi\rangle

1 = \sum_{lm}|lm\rangle\langle lm|

\psi(\Omega) = \sum_{lm}\psi_{lm}Y_{lm}(\Omega)

Yes, you can write down the angular dependence of the position space wave function as a linear combination of spherical harmonics, which are position-basis representations of the angular momentum eigenstates. But how does that have anything to do with my question about why the angular momentum operator cannot have a parametric dependence on angle?

 

[tom.stoer]

You can write down something like

\psi(\Omega) = \langle\Omega|\psi\rangle = \sum_{lm}\psi_{lm}\,Y_{lm}(\Omega)

1 = \sum_{lm}|lm\rangle\langle lm|

Now acting with an operator O on a wave function means

\psi \to \psi^\prime = \mathcal{O}\,\psi = \sum_{lm} \psi^\prime_{lm}\,Y_{lm}

This is nothing else but the angle-representation of the operator O.

Of course one can construct the lm-representation as well

\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\mathcal{O}\,Y_{lm}

Of course the function OY can be written as a linear combination of Y’s again

\mathcal{O}\,Y_{lm} = \sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}

And therefore

\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}

Now interchange summations

\mathcal{O}\psi = \sum_{l^\prime m^\prime}\left[\sum_{lm} o_{ll^\prime mm^\prime}\,\psi_{lm}\right]\, Y_{l^\prime m^\prime}

The term in […] is nothing else but the rep. of the operator O acting as “matrix” o on the coefficients \psi.

Attention: the functions O cannot be arbitrary but have to respect periodicity in the angles, so there is (e.g.) not operator θ but (e.g.) cosθ only.

 

[lugita15]

You can write down something like

\psi(\Omega) = \langle\Omega|\psi\rangle = \sum_{lm}\psi_{lm}\,Y_{lm}(\Omega)

1 = \sum_{lm}|lm\rangle\langle lm|

Now acting with an operator O on a wave function means

\psi \to \psi^\prime = \mathcal{O}\,\psi = \sum_{lm} \psi^\prime_{lm}\,Y_{lm}

This is nothing else but the angle-representation of the operator O.

Of course one can construct the lm-representation as well

\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\mathcal{O}\,Y_{lm}

Of course the function OY can be written as a linear combination of Y’s again

\mathcal{O}\,Y_{lm} = \sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}

And therefore

\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}

Now interchange summations

\mathcal{O}\psi = \sum_{l^\prime m^\prime}\left[\sum_{lm} o_{ll^\prime mm^\prime}\,\psi_{lm}\right]\, Y_{l^\prime m^\prime}

The term in […] is nothing else but the rep. of the operator O acting as “matrix” o on the coefficients \psi.

Attention: the functions O cannot be arbitrary but have to respect periodicity in the angles, so there is (e.g.) not operator θ but (e.g.) cosθ only.

If by angle operator you mean one of the position operators in spherical coordinates, then I agree such a beast exists (following your suggestion, we could say there's an operator O such that O|r,θ,φ>=cosθ|r,θ,φ>). But how does the existence of this operator answer the question of why, say, the spin angular momentum operator (which isn't defined in terms of position operators) cannot have a parametric dependence on angle?

 

[tom.stoer]

But how does the existence of this operator answer the question of why, say, the spin angular momentum operator ... cannot have a parametric dependence on angle?

Why should it have such a dependence? Then it would be something else, not spin.

 

[lugita15]

Why should it have such a dependence? Then it would be something else, not spin.

Well, what I'm trying to understand is why the Hamiltonian operator can have a dependence on time, the parameter of the time translation group, but the spin angular momentum operator cannot have a dependence on angle, the parameter of the (intrinsic) rotation group.

 

[Ilmrak]

Well, what I'm trying to understand is why the Hamiltonian operator can have a dependence on time, the parameter of the time translation group, but the spin angular momentum operator cannot have a dependence on angle, the parameter of the (intrinsic) rotation group.

In a time translational invariant theory, total energy can't depend on time because it's the Noether charge related to that symmetry.

Ilm

 

[lugita15]

In a time translational invariant theory, total energy can't depend on time because it's the Noether charge related to that symmetry.

Yes, I know that, but you can have a system with a time-dependent external force, and you'll get a time-dependent Haimiltonian operator. Is the reason why you can't have analogous situations for momentum or angular momentum something rigorous in the quantum mechanical formalism, or is it just due to our classical knowledge of how things are supposed to work?

I suppose my question could be phrased in purely mathematical terms: under what conditions do the generators of a Lie algebra inherit the parametric dependence of the associated Lie group?

 

[kith]

Is the reason why you can't have analogous situations for momentum or angular momentum something rigorous in the quantum mechanical formalism, or is it just due to our classical knowledge of how things are supposed to work?

Interesting question.

I don't think the answer is something special to quantum mechanics. Also in the classical Lagrangian/Hamiltonian formalism, conjugate momenta don't depend on their generalized coordinates. Maybe the question is similar to "why do we have second order differential equations?".

Time is special, because it is not a conjugate quantity to anything or a function of such quantities. I.e. it's just a parameter and not an observable in Hamiltonian mechanics.

 

[lugita15]

Interesting question.

I don't think the answer is something special to quantum mechanics. Also in the classical Lagrangian/Hamiltonian formalism, conjugate momenta don't depend on their generalized coordinates. So the answer seems to be related to the fact, that a classical state is characterized by two independent values for every degree of freedom.

Time is special, because it is not a conjugate quantity to anything or a function of such quantities. I.e. it's just a parameter and not an observable in Hamiltonian mechanics.

But as I told Tom, for the spin angular momentum operator there is no "intrinsic angle operator" conjugate to it, so why can't the spin angular momeentum operator depend on an angle parameter?

 

[kith]

But as I told Tom, for the spin angular momentum operator there is no "intrinsic angle operator" conjugate to it, so why can't the spin angular momeentum operator depend on an angle parameter?

Well, spin is a very special case because in NRQM, it is introduced ad hoc. If we want to think about your question in terms of the Hamiltonian formalism, we have to use a theory which describes the relevant observables in this framework. For spin, this would be QFT.

 

[lugita15]

Well, spin is a very special case because in NRQM, it is introduced ad hoc.

That's actually not true. If you construct Hilbert space operators out of the representation theory of the Galilei group, just as you do the analogous thing with the Poincare group in QFT, you will naturally get spin angular momentum; see Ballentine for details. What is genuinely relativistic is the spin-statistics theorem.

 

[kith]

That's actually not true. If you construct Hilbert space operators out of the representation theory of the Galilei group, just as you do the analogous thing with the Poincare group in QFT, you will naturally get spin angular momentum; see Ballentine for details. What is genuinely relativistic is the spin-statistics theorem.

Which section in Ballentine do you have in mind? Substituting QxP by QxP+S (as he does in chapter 3) seems pretty ad hoc to me.

For relativistic QM, there are physical reasons for the existence of spin. I really wonder what such reasons could be in the non-relativistic case.

 

[strangerep]

What is genuinely relativistic is the spin-statistics theorem.

Actually, the boson-fermion superselection rule can be derived from the properties
of the rotation group alone.

Hegerfeldt, Kraus, Wigner,
"Proof of the Fermion Superselection Rule without the
Assumption of Time-Reversal Invariance",
J. Math. Phys., vol 9, no 12, (1968), p2029.

Abstract: The superselection rule which separates states with integer
angular momentum from those with half-integer angular momentum is proved using
only rotational invariance.

Their argument is essentially a more rigorous version of the one in Ballentine's
section 7.6 about rotations by $2\pi$. This sort of thing can be developed to
reveal a spin-statistics theorem for non-relativistic QM.

I suppose my question could be phrased in purely mathematical terms: under
what conditions do the generators of a Lie algebra inherit the parametric
dependence of the associated Lie group?

Never. The Lie algebra is derived from the Lie group by differentiating wrt
the parameters and then setting the parameters to 0.

The confusion about time in dissipative Hamiltonian systems is a different
issue. Let's go back to your original question:

[...] the Hamiltonian operator is constructed as the infinitesimal generator
of the time translation group, which is a 1-parameter group. Yet it can still
depend on time.

Cases where the Hamiltonian depends on time involve (at least) a second
subsystem in some way. E.g., a dissipative system can gain or lose energy
from/to another system, a system under the influence of a time-dependent
external force presumes the existence of another system responsible for that
force, etc.

So in general we have a composite system whose total Hamiltonian is
time-independent. But for the component subsystems, their individual evolution
parameters might not coincide with a global time parameter associated with the
total Hamiltonian. One chooses the component-specific evolution parameters to
make the maths as convenient as possible, and (presumably) to coincide with
some notion of local clock associated with that subsystem.

Herein lies a deep question about the distinction between kinematics and
dynamics. There is a (no-acceleration) theorem of Currie-Jordan-Sudarshan
which shows that assuming a common evolution parameter associated with
interacting particles is not viable in general: their respective worldlines in a
common Minkowski space fail to transform in a way which is compatible with
the interacting versions of the Hamiltonian and Lorentz boost operators.

The usual way to construct dynamics is the so-called "instant form" in which
we add an interaction term to the Hamiltonian (and to the Lorentz boost
generators in the relativistic case). This is motivated by our familiarity
with our everyday picture of Euclidean 3D space and our imagined reference
frame coordinatized implicitly in a way which is compatible with free
dynamics. Ballentine describes this briefly on p83 where he justfies modifying
only the Hamiltonian in the Galilean algebra to accommodate external fields.

But this is not the only possible way that we can try to make a "split" between
kinematics and dynamics. There's also the "point form" and "front form" which
modify other generators, but I won't delve into the details here.

It may even be the case that none of these relatively simple approaches are
truly adequate for all purposes. Sudarshan and collaborators also experimented
with more general alternatives in which the evolution parameter is determined
dynamically rather than via a once-and-for-all split between kinematics and
dynamics (which is what's done in the other forms of dynamics I listed above).

I can probably dig out more references for the above if necessary, but that's
probably enough for now.

 

[dextercioby]

Which section in Ballentine do you have in mind? Substituting QxP by QxP+S (as he does in chapter 3) seems pretty ad hoc to me.

For relativistic QM, there are physical reasons for the existence of spin. I really wonder what such reasons could be in the non-relativistic case.

Spin occurs naturally in the theory of vector representations of a central extension of the universal covering group of the (proper) Galilei group in the same way it emerges from the theory of vector representations of the universal cover of the restricted Poincaré group, i.e. through one of its Casimirs. This is a result known from the work of Bargmann and especially Levy-Leblond way back from the 1960's.

In the same way spin is embedded in the Dirac equation, it is also embeded in a linearized version of the Schrödinger equation as shown by Levy-Leblond and explained neatly in one of Greiner's books (either the one on symmetries, or the one on wave equations, don't remember which).

 

[lugita15]

Spin occurs naturally in the theory of vector representations of a central extension of the universal covering group of the (proper) Galilei group in the same way it emerges from the theory of vector representations of the universal cover of the restricted Poincaré group, i.e. through one of its Casimirs. This is a result known from the work of Bargmann and especially Levy-Leblond way back from the 1960's.

OK, this seems pretty convincing.

In the same way spin is embedded in the Dirac equation, it is also embeded in a linearized version of the Schrödinger equation as shown by Levy-Leblond and explained neatly in one of Greiner's books (either the one on symmetries, or the one on wave equations, don't remember which).

This seems less convincing. Why in the world would you choose to linearize the Schrodinger equation, other than the fact that you know it somehow gives you spin?

 

[dextercioby]

[...] This seems less convincing. Why in the world would you choose to linearize the Schrodinger equation, other than the fact that you know it somehow gives you spin?

Why would you linearize the Klein-Gordon equation (as Dirac did in 1928), if the square rooted equation was enough to cure the <negative norm issue> ?

 

[lugita15]

Actually, the boson-fermion superselection rule can be derived from the properties of the rotation group alone.

Hegerfeldt, Kraus, Wigner,
"Proof of the Fermion Superselection Rule without the
Assumption of Time-Reversal Invariance",
J. Math. Phys., vol 9, no 12, (1968), p2029.

Abstract: The superselection rule which separates states with integer
angular momentum from those with half-integer angular momentum is proved using
only rotational invariance.

Their argument is essentially a more rigorous version of the one in Ballentine's
section 7.6 about rotations by $2\pi$. This sort of thing can be developed to
reveal a spin-statistics theorem for non-relativistic QM.

Are you talking about the Jabs paper discussed in this thread? You said this in that thread:

The thing that still leaves me a little perplexed is this: although demanding a consistent sense for the rotation transformations sounds asthetically pleasing, I have trouble seeing why it's essential (a priori) from a physical perspective. But hey, the double-valuedness of rotations is tricky at the best of times -- needing the "Dirac belt trick" or similar devices to illiustrate it.

Have your qualms been resolved since then?

Never. The Lie algebra is derived from the Lie group by differentiating wrt
the parameters and then setting the parameters to 0.

The confusion about time in dissipative Hamiltonian systems is a different
issue.

So are you saying that for a dissipative Hamiltonian system, the time-dependent operator does not generate the associated time evolution operators of the system? How can that possibly be?

Cases where the Hamiltonian depends on time involve (at least) a second
subsystem in some way. E.g., a dissipative system can gain or lose energy
from/to another system, a system under the influence of a time-dependent
external force presumes the existence of another system responsible for that
force, etc.

So in general we have a composite system whose total Hamiltonian is
time-independent. But for the component subsystems, their individual evolution
parameters might not coincide with a global time parameter associated with the
total Hamiltonian. One chooses the component-specific evolution parameters to
make the maths as convenient as possible, and (presumably) to coincide with
some notion of local clock associated with that subsystem.

OK, but you only deal with different time parameters in relativity, don't you? (And in relativity, isn't time not a parameter anyway, but rather part of the position 4-vector operator?) In nonrelativistic QM all the time parameters are the same.

Herein lies a deep question about the distinction between kinematics and
dynamics. There is a (no-acceleration) theorem of Currie-Jordan-Sudarshan
which shows that assuming a common evolution parameter associated with
interacting particles is not viable in general: their respective worldlines in a
common Minkowski space fail to transform in a way which is compatible with
the interacting versions of the Hamiltonian and Lorentz boost operators.

Does this limitative theorem extend even to nonrelativistic QM? That would be surprising.

The usual way to construct dynamics is the so-called "instant form" in which
we add an interaction term to the Hamiltonian (and to the Lorentz boost
generators in the relativistic case). This is motivated by our familiarity
with our everyday picture of Euclidean 3D space and our imagined reference
frame coordinatized implicitly in a way which is compatible with free
dynamics. Ballentine describes this briefly on p83 where he justfies modifying
only the Hamiltonian in the Galilean algebra to accommodate external fields.

Here is what Ballentine has to say on the subect:

One may ask why only the time displacement generator H should be changed by the interactions, while the space displacement generators P are unchanged. If the system under consideration were a self-propelled machine, we could imagine it displacing itself through space under its own power, consuming fuel, expelling exhaust, and dropping worn-out parts along the way. If P generated that kind of displacement, then the form of the operators P certainly would be altered by the interactions that were responsible for the displacement. But that is not what we mean by the operation of space displacement. Rather, we mean the purely geometric operation of displacing the system self-congruently to another location. This is the reason why P and the other generators of symmetry operations are not changed by dynamical interactions. However, H is redefined to be the generator of dynamic evolution in time, rather than merely a geometric displacement along the time axis.

This seems rather hand-wavy to me. Why is the Hamiltonian operator singled out to redefined as "the generator of dynamic evolution in time, rather than merely a geometric displacement along the time axis"? Why can't you have, akin to a dissipative time-dependent Hamiltonian operator, a position-dependent momentum operator or a angle-dependent angular momentum operator?

 

[lugita15]

Why would you linearize the Klein-Gordon equation (as Dirac did in 1928), if the square rooted equation was enough to cure the <negative norm issue> ?

Probably because Dirac wasn't aware of this solution to the problem, so the only available resolution he saw was linearization. But the problem with the Klein-Gordon equation doesn't arise for the Schrodinger equation, so you wouldn't feel a need to linearize it, unless you knew in advance that it would yield spin.

 

[martinbn]

Klein-Gordon and Schrodinger's equations are linear. Surely you don't mean linearization but finding degree one equation.

 

[dextercioby]

Yes, linearization = linear dependence of first derivatives.

 

[strangerep]

Are you talking about the Jabs paper

No.

So are you saying [...]

No.

[...] but you only deal with different time parameters in relativity, don't you? (And in relativity, isn't time not a parameter anyway, but rather part of the position 4-vector operator?) In nonrelativistic QM all the time parameters are the same.

Consider classical Hamiltonian dynamics. One can apply time-dependent canonical transformations (which include mapping one time parameter to another), obtaining a different description of the same dynamical physical situation.

[...]what Ballentine has to say [...] seems rather hand-wavy to me. Why is the Hamiltonian operator singled out to redefined as "the generator of dynamic evolution in time, rather than merely a geometric displacement along the time axis"? Why can't you have, akin to a dissipative time-dependent Hamiltonian operator, a position-dependent momentum operator or a angle-dependent angular momentum operator?

If you mean "why can't you have an interaction term in a momentum operator?", well you can: it's called the "point form" of dynamics. In principle, one can find canonical transformations between one form of dynamics and another (if they correspond to the same physical situation).

One can sometimes think more clearly about this stuff by reviewing some classical dynamics theory, and concentrating on canonical coordinates and momenta, and canonical transformations which preserve the dynamics (Hamilton's equations). E.g., if we start with a canonical pair (q,p) we can find transformations to a new pair (q'(q,p), p'(q,p)) such that canonical commutations relations (or Poisson brackets) still hold for (q',p'). But p' is a function of the old coordinate and momentum -- this is not important, because we also have a new canonical coordinate q'.

More generally, there are also "extended" canonical transformations which involve the time parameter. For more detail, try Goldstein, or Jose & Saletan.

 

[lugita15]

Are you talking about the Jabs paper

No.

So then where can I find out how to go from the superselection rule derivation presented in Ballentine to the spin-statistics theorem?

So are you saying [...]

No.

I'm a little confused. You said that the generators of a Lie algebra can never inherit the parametric dependence of the associated Lie group, but you're also agreeing with me that for a dissipative system the time-dependent Hamiltonian operator generates the time translation group, which is a group whose parameter is time.

Consider classical Hamiltonian dynamics. One can apply time-dependent canonical transformations (which include mapping one time parameter to another), obtaining a different description of the same dynamical physical situation.

But in the nonrelativistic case the Galilean transformation treats time as trivial, so what's the point of even fiddling with the time parameter?

If you mean "why can't you have an interaction term in a momentum operator?", well you can: it's called the "point form" of dynamics.

So are you saying that you can have the momentum operator have a parametric dependence on position, and the spin angular momentum operator have a parametric dependence on angle?

 

[strangerep]

I[...] you're also agreeing with me that for a dissipative system the time-dependent Hamiltonian operator generates the time translation group

I didn't say that. I'm finding it quite hard to discuss this with you. Please try to phrase your future responses differently, in a way that doesn't put words in my mouth. I don't like trying to un-distort "so you're saying[...]" assertions.

But in the nonrelativistic case the Galilean transformation treats time as trivial, so what's the point of even fiddling with the time parameter?

In classical Hamiltonian dynamics the relevant group becomes a much larger symplectic group. How much classical Hamiltonian dynamics have you studied? Maybe I'm pitching my answers at the wrong level.

So are you saying that you can have the momentum operator have a parametric dependence on position, and the spin angular momentum operator have a parametric dependence on angle?

This is another "are you saying [...]" response that I find tedious to deconstruct. But it's bedtime for me now. Maybe I'll have more energy tomorrow.

 

[lugita15]

I didn't say that. I'm finding it quite hard to discuss this with you. Please try to phrase your future responses differently, in a way that doesn't put words in my mouth. I don't like trying to un-distort "so you're saying[...]" assertions.

Sorry about that strangerep, I'll try to phrase my questions more neutrally. First of all, is it true that for a dissipative system, the time-dependent Hamiltonian operator is the infinitesimal generator of the 1-parameter time translation group? By that I mean the following: is it true that U(t,dt)=1-iH(t)dt? (modulo a factor of h-bar)? If so, is it fair to say that the Lie algebra generators inherit the parametric dependence of the associated Lie group?

In classical Hamiltonian dynamics the relevant group becomes a much larger symplectic group. How much classical Hamiltonian dynamics have you studied? Maybe I'm pitching my answers at the wrong level.

Yes, you may be. I've mainly studied advanced mechanics in the Lagrangian formalism. And I'm not really familiar with terms like "symplectic group". My knowledge of Hamiltonian mechanics mainly comes from allusions to it in quantum mechanics books. And just to have all my cards on the table, I know nonrelativistic QM at least at the level of Sakurai, and what I know of Lie groups and Lie algebra also largely originates from reading about QM.

This is another "are you saying [...]" response that I find tedious to deconstruct. But it's bedtime for me now. Maybe I'll have more energy tomorrow.

OK, again let me make my question less loaded. Can the momentum operator have a parametric dependence on position, and can the spin angular momentum operator (in a given direction) have a parametric dependence on angle? If not, is the reason they can't a mathematical reason or a physical reason?

 

[kith]

In the same way spin is embedded in the Dirac equation, it is also embeded in a linearized version of the Schrödinger equation as shown by Levy-Leblond and explained neatly in one of Greiner's books (either the one on symmetries, or the one on wave equations, don't remember which).

This is really something new for me. Just to get you right: you say that the Schrödinger equation and the Klein-Gordon equation are on equal footing, as well as the Dirac equation and a first-order Schrödinger-like equation? This implies, that the Klein-Gordon equation describes the behaviour of electrons correctly, in cases where spin doesn't matter. So can I really calculate the hydrogen spectrum including relativistic effects (but obviously without spin-orbit coupling) from it?

And thanks for the further references, but they are currently out of my scope.

 

[dextercioby]

This is really something new for me. Just to get you right: you say that the Schrödinger equation and the Klein-Gordon equation are on equal footing, as well as the Dirac equation and a first-order Schrödinger-like equation?[...]

Yes.

[...] This implies, that the Klein-Gordon equation describes the behaviour of electrons correctly, in cases where spin doesn't matter.[...]

Yes.

So can I really calculate the hydrogen spectrum including relativistic effects (but obviously without spin-orbit coupling) from it?[...]

Of course you can. This is done in some books, where an exact formula for the bound states spectrum is obtained, as opposed to applying first order perturbation theory to the relativistic correction (due to energy varying with speed, the p^4 term) which is exposed in almost all books of quantum mechanics/atomic physics .

 

[kith]

This is done in some books, where an exact formula for the bound states spectrum is obtained[...]

Ok, thanks for clarifying (wikipedia says the same).

 

[strangerep]

I've mainly studied advanced mechanics in the Lagrangian formalism. And I'm not really familiar with terms like "symplectic group". My knowledge of Hamiltonian mechanics mainly comes from allusions to it in quantum mechanics books

I'm unlikely to have time to compose a detailed answer in the next few days, but I can say for sure that the fastest way to a better understanding of all this is to get hold of one of the textbooks I mentioned (Goldstein, or Jose & Saletan) to read up on Hamiltonian dynamics, canonical transformations, and how it all fits together. You could try these Wiki articles to get a (very brief) overview:

http://en.wikipedia.org/wiki/Hamiltonian_dynamics
http://en.wikipedia.org/wiki/Canonical_transformation

but they are no substitute for a textbook if you haven't already studied this stuff.

 

[naffin]

In quantum mechanics, the Hamilltonian operator is constructed as the infinitesimal generator of the time translation group, which is a 1-parameter group. Yet it can still depend on time. So you have a situation where the generator of a 1-parameter group can depend on the parameter. Yet the momentum operator, the generator of infinitesimal spatial translations, cannot depend on the parameter(s) of the spatial translation group, namely position. And similarly the angular momentum operator cannot depend on angle (or direction), the parameter(s) of the rotation group. Is there a fundamental reason for this, or is it simply that we happen to already know the properties of the dynamical variables from classical mechanics, so we don't bother with it?

I think it's simply because spatial translations and rotations form always a group, while time translations not. Stone's theorem says that the generator is fixed.

 

[lugita15]

I think it's simply because spatial translations and rotations form always a group, while time translations not. Stone's theorem says that the generator is fixed.

Under what circumstances do the unitary time evolution operators not form a group?

 

[naffin]

(I was obviously referring to one-parameter groups)
If your system is a particle in a time-dependent potential it is clearly not true that U(t_1)U(t_2) = U(t_1 + t_2) in general, so you can't use Stone's theorem.

 

[lugita15]

(I was obviously referring to one-parameter groups)
If your system is a particle in a time-dependent potential it is clearly not true that U(t_1)U(t_2) = U(t_1 + t_2) in general.

That's completely new to me. So you're saying that if H(t1)H(t2)≠H(t2)H(t1), then U(t1)U(t2)≠U(t2)U(t1)? I thought U(t1)U(t2)=U(t2)U(t1) regardless, but you just have to be more careful. In QM books, they often do an informal derivation where they say that the Hamiltonian is constant on each infinitesimal interval (t,t+dt), and so you can define U(t,dt) quite easily. Then just take an infinite product to find the time evolution operator for a finite time interval. Is there anything wrong with that kind of approach?

 

[naffin]

In QM books, they often do an informal derivation where they say that the Hamiltonian is constant on each infinitesimal interval (t,t+dt), and so you can define U(t,dt) quite easily. Then just take an infinite product to find the time evolution operator for a finite time interval. Is there anything wrong with that kind of approach?

Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation i \hbar \frac{dU}{dt} (t) = H(t) U(t).
The solution involves something like the Dyson operator, but I only know that is (rigorously) obtained by iterative method, as you say.

The evolution from time t0 to time t1 U(t_0, t_1) depends in general on t0 and t1 separately, not only on their difference U(t_1 - t_0).
All you can say is that U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1) and not that U(0, t_1 + t_2) = U(0, t_2) U(0, t_1).
If U(t_2, t_1) =U(t_1 - t_2) holds for any t1,t2 you can apply Stone's theorem, so you have a time-independent Hamiltonian.

 

[lugita15]

Hamiltonian is still defined by the Schrodinger equation, so it satisfies the equation i \hbar \frac{dU}{dt} (t) = H(t) U(t).
The solution involves something like the Dyson operator, but I only know that is (rigorously) obtained by iterative method, as you say.

Do you have any details about this, or do you know where I can find more information?

The evolution from time t0 to time t1 U(t_0, t_1) depends in general on t0 and t1 separately, not only on their difference U(t_1 - t_0).

I agree with this.

All you can say is that U(0, t_1 + t_2) = U(t_1, t_1 + t_2) U(0, t_1) and not that U(0, t_1 + t_2) = U(0, t_2) U(0, t_1).
If U(t_2, t_1) =U(t_1 - t_2) holds for any t1,t2 you can apply Stone's theorem, so you have a time-independent Hamiltonian.

So then can you consider the time-evolution operators to form a two-parameter group or something?

 

[naffin]

No, two-parameter group implies one-parameter group, the other parameter fixed.

 

[lugita15]

No, two-parameter group implies one-parameter group, the other parameter fixed.

OK, so then what structure does the set of unitary time evolution operators possess?

 

[naffin]

Do you have any details about this, or do you know where I can find more information?

Ballentine, pag. 89.
For the exact solution I know "Quantum Mechanics I", Galindo-Pascual.

In general they are just unitary operators, I don't know other structures.