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Why isn't momentum a function of position?

by lugita15
Tags: function, momentum, position
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lugita15
#1
Mar5-12, 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 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.
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Jano L.
#2
Mar5-12, 08:03 PM
PF Gold
P: 1,165
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 psi-function [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.
lugita15
#3
Mar5-12, 08:51 PM
P: 1,583
Quote Quote by Jano L. View Post
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 psi-function [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.
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.
#4
Mar5-12, 11:23 PM
PF Gold
P: 1,165
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 parameter-independent 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 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 [itex]L^2[/itex].
tom.stoer
#5
Mar6-12, 12:07 AM
Sci Advisor
P: 5,441
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
#6
Mar6-12, 12:10 AM
P: 1,583
So Tom, do you have any thoughts on my question?
tom.stoer
#7
Mar6-12, 12:32 AM
Sci Advisor
P: 5,441
you mean why energy can depend on time?
lugita15
#8
Mar6-12, 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 time-dependent Hamiltonian. But from a more rigorous point of view, what privileges the Hamiltonian in this regard?
tom.stoer
#9
Mar6-12, 12:50 AM
Sci Advisor
P: 5,441
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
#10
Mar6-12, 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?
tom.stoer
#11
Mar6-12, 03:40 AM
Sci Advisor
P: 5,441
Quote Quote by lugita15 View Post
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
#12
Mar6-12, 06:15 AM
P: 612
Quote Quote by lugita15 View Post
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
#13
Mar6-12, 07:46 AM
P: 1,583
Quote Quote by tom.stoer View Post
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
#14
Mar6-12, 10:21 AM
P: 1,583
Quote Quote by tom.stoer View Post
You can write down something like

[tex]\psi(\Omega) = \langle\Omega|\psi\rangle[/tex]

[tex]1 = \sum_{lm}|lm\rangle\langle lm|[/tex]

[tex]\psi(\Omega) = \sum_{lm}\psi_{lm}Y_{lm}(\Omega)[/tex]
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
#15
Mar6-12, 10:50 AM
Sci Advisor
P: 5,441
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 angle-representation of the operator O.

Of course one can construct the lm-representation 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.
lugita15
#16
Mar6-12, 11:24 AM
P: 1,583
Quote Quote by tom.stoer View Post
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 angle-representation of the operator O.

Of course one can construct the lm-representation 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.
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
#17
Mar6-12, 12:17 PM
Sci Advisor
P: 5,441
Quote Quote by lugita15 View Post
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
#18
Mar6-12, 12:19 PM
P: 1,583
Quote Quote by tom.stoer View Post
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.


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