Symmetry, Lagrangian, Qm, and diff eqs.

[friend]

I'm looking for a summary of what invariance or symmetry of the Action in Feynman's path integral has on the equations of motion and on measurement. Do different symmetry groups of the Action integral result in different equations of motion for different particles? Is the least action principle required in order to construct a valid measurement calculation?

 

[physengineer]

I am not sure if I understood your question exactly and my reply may sound trivial. Anyway,
when you have a symmetry there are invariants associated with each of those symmetries (Noether's Theorem). In other words some quantities will be conserved during, let's say, a collision experiment like charge. Conserved quantities can be helpful in predicting or analysing a measurement.

 

[friend]

I am not sure if I understood your question exactly and my reply may sound trivial. Anyway,
when you have a symmetry there are invariants associated with each of those symmetries (Noether's Theorem). In other words some quantities will be conserved during, let's say, a collision experiment like charge. Conserved quantities can be helpful in predicting or analysing a measurement.

Yes, I'm aware that Noether's Theorem proves a conserved quantity for every continuous symmetry of the Lagrangian. So my question is: Are all quantum mechanical observables necessarily conserved quantities? For example, momentum is both a conserved quantity and a QM observable. And energy is a conserved quantity and a QM observable. So specifically I'm wondering if the quantum mechanical formula for the probability of an event - the integral of the wavefunction times its complex conjugate - only works for observables which are conserved quantities?

I'm thinking that this may be a tautology. For you may not be able to observe things that are not conserved quantities. But then there is the question of actual proof. Thanks.

 

[tom.stoer]

Are all quantum mechanical observables necessarily conserved quantities?

Not necessarily. Observables must be hermitean operators. Think about a two-particle system with conserved c.o.m. momentum P. The individual momenta of the two particles will not be conserved in general, nevertheless they are qm orbservables (e.g. the momentum of an electron in a hydrogen atom)

 

[The_Duck]

Position is observable, but not conserved.

 

[friend]

Not necessarily. Observables must be hermitean operators. Think about a two-particle system with conserved c.o.m. momentum P. The individual momenta of the two particles will not be conserved in general, nevertheless they are qm orbservables (e.g. the momentum of an electron in a hydrogen atom)

I'm thinking in terms that the wavefunction is derived from solving the Schrodinger eq which is a differential equation derived as the equations of motion from the invariance of the Action integral. So maybe there is a connection between symmetry and calculating the probabilities of QM observables.

About your examples, I'm not sure you can isolate the wavefunction of a single particle from the wavefunction of a multi-particle system such that you can even calculate the probabilities for the single particle. And if you cannot calculate the probabilities of a single particle in a multi-particle system, then you cannot say that it does not come from a conserved quantity. Right?

 

[tom.stoer]

Of course you can calculate and measure the probability of observing one particle at position x and with momentum p, even if the particle has to be described by a multi-particle wave function.

 

[friend]

Of course you can calculate and measure the probability of observing one particle at position x and with momentum p, even if the particle has to be described by a multi-particle wave function.

An example of what you are saying escapes me at the moment. But it occurs to me that if you could in general separate single particle wavefunctions from the wavefunction of a multi-particle system, then I would think that the multi-particle system would be easily solved by summation of each of the single particle wave function. I understand that multi-particle systems are very much harder than that due to the cross terms of potential in the Lagrangian/Hamiltonian.

So again the question is whether the measurement process requires an underlying conserved quantity.

tom.stoer mentioned the position. I understand there may be some controversy as to whether position can technically be called an observable. I forget why at the moment.

 

[tom.stoer]

Let's look at the anti-symmetrized two-particle wave-function of a two-fermion system with fermions 1, 2 in state |ab> in momentum space:

\psi_{ab}(p_1, p_2) = \psi_a(p_1)\,\psi_b(p_2) - \psi_b(p_1)\,\psi_a(p_2)

Now let's look at a one-particle observable A_1 like the momentum of one particle.

You have to calculate

\langle A_1 \rangle_{ab} = \int dp_1 \int dp_2 \, \psi_{ab}(p_1, p_2)^\ast \, A_1 \, \psi_{ab}(p_1, p_2)