# Is every observable the generator of a symmetry?

1. Jan 6, 2014

### EventHorizon91

We know that observables correspond to hermitian operators on the Hilbert space of physical states of the system. We also know, via Wigner theorem, that for each symmetry there is a linear unitary operator (or anti-linear and anti-unitary). In the case of a continuous symmetry, that is in the case of a Lie group of symmetries, all elements of the group (the connected part of the group) can be written as the complex exponential of a hermitian operator. This hermitian operator is the generator of the corresponding symmetry. So it seems natural to think that all observables may correspond to the generators of some symmetry, given that it is well known that momentum is the generator of spatial translation and angular momentum is the generator of rotations. Is this true? And if so what is the symmetry that correspond to the position observable in quantum mechanics?

2. Jan 6, 2014

### Staff: Mentor

I would say not - eg the number operator in QFT.

I am pretty sure in an abstract way the momentum operator generates the position operator by the symmetry of the situation with each having the representation that's the 'Fourier' transform of the other.

Thanks
Bill

3. Jan 6, 2014

### dextercioby

There's no number operator in finite particle QM. I would say that you should revise the definition of a symmetry generator and then you will find the answer to your question in the title of the thread.

4. Jan 6, 2014

### Bill_K

Of course there is. Simply define an operator that has the eigenvalue n on n-particle states.

5. Jan 6, 2014

### BruceW

This is surely not true. Say for example, you had a Lagrangian that depends on absolute position (not just relative position). Then spatial translation is not a symmetry of the system, and momentum is not conserved. So, for a general observable, it has the possibility of being the generator of some symmetry (like momentum). But it is not always going to be the generator of a symmetry. It depends on what the Lagrangian is like. (in other words, what the system is like).

6. Jan 6, 2014

### WannabeNewton

What is your definition of a symmetry?

7. Jan 6, 2014

### EventHorizon91

For me, a symmetry is a coordinate transformation of the space-time labels or a redefinition of the fields such that the action doesn't change. I think the term symmetry generator is restricted however only to continuous symmetries (a Lie group of symmetries). So, the number operator is hermitian and an observable, but it is not the generator of any symmetry, because its spectrum is discrete. Maybe all observables with continuous spectra are possible generators of a symmetry? As I said, the hamiltonian is the symmetry generator of time translation, momentum of space translations, angular momentum of rotations. So, if it is true in three cases, it must be true in all cases, right? Of course, not all generators of a symmetry are observables, for example the generators of Lorentz boosts are anti-hermitian. Another observable I can think of is charge. Is charge the generator of a symmetry?

8. Jan 6, 2014

### BruceW

yeah, charge is the symmetry related to multiplying a complex field by an unimportant phase factor. But again, this would depend on the Lagrangian. If the Lagrangian has the same form after you do this operation, then there is a symmetry. (although to be honest, I don't know of any Lagrangian which is not invariant under a global phase transform). p.s. In case you didn't notice, I like to talk about Lagrange equations :) that's my definition of symmetry. any operation that leaves the form of the Lagrangian unchanged.

Also, I would guess that you are right that all observables with continuous spectra are possible generators of a symmetry. But I don't know for sure. Maybe there are some extra smoothness conditions that need to be satisfied... But anyway, physically, it seems reasonable.

9. Jan 6, 2014

### Jilang

Well, charge conservation is due to the gauge invariance of the EM field. Does that make it a generator of a shift in the scalar and vector potential by an arbitrary amount?

10. Jan 6, 2014

### BruceW

a global phase shift of the free Dirac field corresponds to charge conservation. the EM gauge invariance is a local phase shift in the Dirac field and a gauge transformation in the scalar and vector potential. So charge conservation is not really related to the EM field. Charge conservation comes about because the free Dirac field is complex, and the Lagrangian is invariant under a global phase shift of this field. But yeah, gauge invariance is another example of a symmetry operation.

OK, I've just picked up my favourite book (Mandl & Shaw QFT), and apparently gauge invariance does not introduce a new conservation law, but it only reproduces the conservation of charge. So I was sort of correct. you can choose either EM gauge transform, or the simpler global phase transform of the Dirac field, and in either case, you get conservation of charge.

11. Jan 6, 2014

### samalkhaiat

I would say ''yes''. However, the definition of observables in QFT is quite subtle.

That will be translation invariance in momentum space.

12. Jan 6, 2014

### WannabeNewton

Yes and they correspond to symmetries of space-time. Similarly position corresponds to the symmetry generator of momentum-translations in momentum space.

13. Jan 6, 2014

### samalkhaiat

In QFT the number operators generate U(1) symmetries, ex: the electric charge, the Baryon number, etc.
The generators of a Lie group can have continuous as well as discreate spectrum.

14. Jan 6, 2014

### EventHorizon91

Ok, I'm fine with the interpretation of the position operator as the generator of translations in momentum space (in the context of QM). But what would be its physical meaning? For the momentum operator, invariance under translations is just a statement of the homogeneity of space. Is invariance under translations in momentum space associated with invariance under Lorentz boosts?

I don't see how the number operator is related to a U(1) symmetry. Also, can generators of a Lie group really have a discrete spectrum? An example would be much interesting.

15. Jan 6, 2014

### Staff: Mentor

Neither do I - I would like a bit more detail on that one.

Thanks
Bill

16. Jan 6, 2014

### strangerep

It would mean that the physics is not fundamentally different at various momentum scales.

But beware -- this is only for the nonrelativistic case, so be careful jumping over to the Lorentz boosts from special relativity. (There is a connection: Gerry Kaiser has shown how the commutator between momentum generators and Lorentz generators contracts (for fixed mass) to the familiar canonical commutators between momentum and position in the non-rel case.)

The ordinary angular momentum generators(!). Have you studied the detail of how the well-known half-integral values arise in the quantum theory of angular momentum?

Edit: Ha! I see below that Sam was composing something similar in parallel with me.

Last edited: Jan 6, 2014
17. Jan 6, 2014

### samalkhaiat

Noether charge of U(1) is the number operator

$$Q = \int \ d^{ 3 } x \ \psi^{ \dagger } ( x ) \psi ( x ) .$$

What is the spectrum of the generators of SU(2) or SO(3)? Did you study the theory of angular momentum in QM?

18. Jan 6, 2014

### strangerep

EventHorizon91,

Let's back up a bit, and straighten out some terminology....

That is usually called a "geometric symmetry", or sometimes a "point symmetry".

Those are usually called a "variational symmetry", usually containing the geometric symmetries as a subgroup.

There are also "contact symmetries" (lousy name) or "simple dynamical symmetries" which map solutions of the equations of motion into other solutions. They typically contain the variational symmetries.

Then there are "generalized dynamical symmetries" which (e.g.) mix the canonical position and momentum variables amongst themselves (i.e., mix the basic variables and their derivatives). Some examples of this can be seen in the symplectic transformations on phase space, often used in classical Hamiltonian dynamics to find a new set of variables in which the equations of motion are easier to solve.

(Caveat: not all authors use precisely the same definitions of these terms. Some use "dynamical symmetry" to mean a symmetry of the Hamiltonian.)

Edit: The point of the above is to help explain that in quantizing a given system, one attempts to construct a Hilbert space on which the symmetries (variational or dynamical) are all unitarily represented (i.e., such that their generators are well-defined self-adjoint operators on that Hilbert space). In general, we don't start with a standard Hilbert space, and then try to find operators. Strictly speaking, it's the other way around. However, it turns out that the usual space of square-integrable wavefunctions is suitable for a large set of cases.

Yes, that's pretty much what "generator" means in this context. A generator is obtained by differentiating a general group element with respect to a parameter, and then setting the parameter to 0 (i.e., differentiating at the identity).

That's for finite-dimensional Hilbert space. If one looks more carefully, it turns out that the only unitary representations of the Lorentz group are infinite-dimensional. This leads to the inevitable necessity for quantum field theory in relativistic cases.

Last edited: Jan 6, 2014
19. Jan 6, 2014

### EventHorizon91

Ops, sorry, you're right! I was thinking for some reason about classical fields... I better know well the ordinary theory of angular momentum quantization in QM!

20. Jan 6, 2014

### EventHorizon91

Oh, I see now there is a new answer, I will read it carefully, but tomorrow morning