
#1
Mar512, 02:15 AM

P: 1,583

In quantum mechanics, the Hamilltonian operator is constructed as the infinitesimal generator of the time translation group, which is a 1parameter group. Yet it can still depend on time. So you have a situation where the generator of a 1parameter 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. 



#2
Mar512, 08:03 PM

P: 1,030

Hello lugita15,
I think the operator " generator of temporal translation " is [itex]g_t = i\hbar \partial/\partial t[/itex], in the same sense as the operator " generator of coordinate translation " is [itex]g_k = i\hbar \partial/\partial q_k [/itex]. Schroedinger's equation says that the actual psifunction [itex]\psi(q,t)[/itex] is such that the operator [itex]i\hbar \partial/\partial t[/itex] has the same effect as the Hamiltonian operator [itex]H(q,p,t)[/itex]. You can write then [itex]e^{\Delta t g_t/i\hbar} \psi = e^{\Delta t H(q,p,t)/i\hbar}\psi [/itex]. 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. 



#3
Mar512, 08:51 PM

P: 1,583





#4
Mar512, 11:23 PM

P: 1,030

Why isn't momentum a function of position?
Forgive me, I have a difficulty understanding your question.
My point was that the operator [itex]g_t = i\hbar \partial/\partial_t[/itex] is the generator of temporal translation. It does not contain [itex]t[/itex]. This is parameterindependent the same way as [itex]i\hbar \partial/\partial x[/itex] and we should use this to shift any arbitrary function [itex]f(q,t)[/itex] to another time. Generators are derivatives hence there are no parameters in them. In quantum theory, in special case of temporal translation of function [itex]\psi(q,t)[/itex] that satisfies Schroedinger's equation, we can use the Hamiltonian operator instead of [itex]g_t[/itex]. 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 wavefunction, 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 [itex]L^2[/itex]. 



#5
Mar612, 12:07 AM

Sci Advisor
P: 5,307

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. 



#6
Mar612, 12:10 AM

P: 1,583

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




#7
Mar612, 12:32 AM

Sci Advisor
P: 5,307

you mean why energy can depend on time?




#8
Mar612, 12:38 AM

P: 1,583

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 timedependent Hamiltonian. But from a more rigorous point of view, what privileges the Hamiltonian in this regard? 



#9
Mar612, 12:50 AM

Sci Advisor
P: 5,307

Introducing an xdependence 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.




#10
Mar612, 02:42 AM

P: 1,583

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




#11
Mar612, 03:40 AM

Sci Advisor
P: 5,307





#12
Mar612, 06:15 AM

P: 563

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? 



#13
Mar612, 07:46 AM

P: 1,583





#14
Mar612, 10:21 AM

P: 1,583





#15
Mar612, 10:50 AM

Sci Advisor
P: 5,307

You can write down something like
[tex]\psi(\Omega) = \langle\Omega\psi\rangle = \sum_{lm}\psi_{lm}\,Y_{lm}(\Omega)[/tex] [tex]1 = \sum_{lm}lm\rangle\langle lm[/tex] Now acting with an operator O on a wave function means [tex]\psi \to \psi^\prime = \mathcal{O}\,\psi = \sum_{lm} \psi^\prime_{lm}\,Y_{lm} [/tex] This is nothing else but the anglerepresentation of the operator O. Of course one can construct the lmrepresentation as well [tex]\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\mathcal{O}\,Y_{lm}[/tex] Of course the function OY can be written as a linear combination of Y’s again [tex]\mathcal{O}\,Y_{lm} = \sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}[/tex] And therefore [tex]\mathcal{O}\psi = \sum_{lm}\psi_{lm}\,\sum_{l^\prime m^\prime}o_{ll^\prime mm^\prime}\,Y_{l^\prime m^\prime}[/tex] Now interchange summations [tex]\mathcal{O}\psi = \sum_{l^\prime m^\prime}\left[\sum_{lm} o_{ll^\prime mm^\prime}\,\psi_{lm}\right]\, Y_{l^\prime m^\prime}[/tex] 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. 



#16
Mar612, 11:24 AM

P: 1,583





#17
Mar612, 12:17 PM

Sci Advisor
P: 5,307





#18
Mar612, 12:19 PM

P: 1,583




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