Consider

(Capitals for Operators)

(skipped the delta function integ. step)

does this imply that position operator generates momentum?

**Posted By :** Venkateshan

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Quantum Mechanics 9:24 am

Consider

(Capitals for Operators)

(skipped the delta function integ. step)

does this imply that position operator generates momentum?

**Posted By :** Venkateshan

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I guess so, because anyway the position operator in momentum representation is identical to the momentum operator in position representation and there’s no difference at all. Beyond that it’s a matter of taxonomy isn’t it?

Ok, these are some initial ideas that occured to me :

Kinematical momentum and canonical momentum would be related only by a mass term unless there is coupling to some filed or other complications in the system. Then isnt it reasonable to expect that the generators for boots and the postion operator are related by a mere constant or some simple relation. And of-course, that isnt true.

PS: Typing this, I realize how absurd it sounds but anyway those my the first thoughts.

Anyway, if this true, then why dosent it ever get mentioned anywhere.

Is it just me or the statement that positon operator generates momentum sounds wierd?

Venky asked

Well, the answer is yes. The position and momentum form the generators of what is called the Heisenberg Group. The Algebra obeyed by these generators is precisely the Heisenberg Commutation relations and is called the

Heisenberg Lie Algebra. I’m sure you are familiar with the representation theory of this group which is used almost everywhere in quantum mechanics. The most interesting theorem in this regard is Stone Von-Neumann Theorem which is implicitly used in proving the equivalence of Heisenberg Picture with the Schroedinger Picture in Quantum mechanics.Venky wrote

Hmm, well it would not sound so weird, if you remember the conjugate roles the position and momentum play in Classical Mechanics. For example, there are canonical transformations which interchange position and momentum. And Shanth has already commented on the symmetry between position and momentum representation in QM…

Venky wrote

Hey ! What are you trying to say here ? How does the relation between Kinematical/Canonical momentum or the relation between Boost Generators/Position operator relevant to whatever is being discussed here ? These are completely different issues and hence have to be kept separate.

Canonical Momentum=Kinematical Momentum+ “Potential” Momentumwhich is just the spatial counterpart of the equationEnergy=Kinematical(Rest+Kinetic) Energy + Potential Energy. Both the Kinematical energy/momentum as we know are functions only of the rest mass and the velocity ( viz. p_kinematic=mv , E_kinematic=mc^2 where m is the “relativistic” mass ). And they are called Kinematic because they are independent of dynamics(which is decided by potential momentum/energy). As far as I know, these facts bear no direct relation to the Heisenberg Group.Now, coming to the boost generator – for a free non-relativistic spinless system the boost generator is given by

\vec{N}=m\vec{x} -\vec{p}t. The Noether theorem applied to boost symmetry says that this vector is conserved- which is basically the statement of Centre of Mass Theorem – the centre of mass of an isolated system moves with a uniform velocity.For a relativistic particle in QFT, the very concept of position operator doesn’ t make much sense in general. We have to work with some ad hoc schemes like

Newton-Wigner position operator, which are slightly dirty so I don’t want to go into them unless you are interested . 😉The basic idea of these schemes is that for a free system with non-zero rest mass you can somehow define a “position” using its boost generators and by assuming that a relation similar to the above relation between the boost generator and position operator holds. So, essentially by definition the boost generators are related to the position operators . This is not a simple relation because of

a) the presence of spin – the boost generators should also boost the internal degrees of freedom of the system. So, an additional term involving the spin should appear in the expression for boost generator which boosts the spin.

and

b) the “m” in the relation above is actually the “relativistic” mass or the Hamiltonian(divided by c^2). (Update : Note that, this implies an operator ordering problem in the term m\vec{x} – the m and \vec{x} don’t commute ! In fact, the Heisenberg eqns. of motion require that the commutator between x and m(aka the Hamiltonian) be proportional to the rate of change of x(aka the velocity operator).)

None of which is surprising if you think of it. And all this only becomes dirtier once the interactions are included…… Again, I don’ t see how all these facts are germane to your post !

Don’t confuse boost with translations in the momentum space. Galilean Boost should do more than just translate the momentum, it should also add a term to position for example(x’=x-vt).(I’ll leave it as an exercise for you to show that the boost generator I’ve written above actually does that in non-relativistic QM 🙂 ) And once you come to Einsteinian relativity, the usual velocity addition breaks down and action of boost is altogether different from momentum translation.

I hope this clarifies your confusion. If on the other hand, I’ve confused you more, well then we probably have to sit down and work it out 🙂

Yeah, that's what I completely forgot about when first considering the tarnsformations generated by the position operator. And yes, I verified that the action of boost operator on the state produces x-> x- vt.

Venkateshan

Moin Moin.

Just in order to be able to read that formulae (containing mostly square-signs): Does anyone know, which font I do have to install, in order to view the weblog.

_Tschuess,

__Michael.

@Soliman

Previously, the formulae were just written in HTML which we hoped would render correctly in most browsers…

Anyway, I have converted the equations in the original post so as to render them using the LaTeX plugin in wordpress. Let us know if you still face problems seeing the formulae.

If u keep position fixed momentum is completely arbitrary…

Can anybody discuss the non uniqueness of the momentum operator if x is chosen as the diagonal?

Hi gojib,

The momentum operator does not become ambiguous when the position operator is diagonal !

What happens is that when the position operator is diagonal, the momentum operator becomes non-diagonal which in turn leads to an uncertainty in momentum.

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