What is the role of Relativistic QM?

[throw]

In the non-relativistic case, time-energy uncertainty involves energy measurements at two different times, it's hard to believe this is from some operator relation since the whole reason for NRQM in the first place is the inability of simultaneously (i.e. at the same time) prescribing values of complementary variables; successive measurements of the same quantity (thus at different times, note we're also talking about exact values for the energy at these times as L&L set it up) is just completely different.

The RQM case means either generalizing this to include time-energy as part of complementary space-time momentum-energy measurements... It simply starts to look like nothing makes any sense until we give up something else - in this case it is the very idea of space-time energy-momentum complementarity that is challenged in the RQM context (see the position vs momentum space wave function comment earlier, or the discussion around the earlier-cited passage, for example).

This has interesting comments on 'Pauli's theorem' even in a non-relativistic context, it seems very hard to believe something can be an operator some of the time but not all of the time in a NRQM context depending on the spectrum which seems to be the argument, and this seems to be the road one has to go down to try to force a time operator.

But regarding the relativistic case - the particle action is gauge invariant, so the Hamiltonian is not even fixed until one fixes a gauge (parametrization), and it possesses primary constraints so it should be treated as a constrained system which also incorporates the massless case. The Schrodinger equation associated to this properly formulated version (in which the gauge choice does not ruin space-time covariance) is not even the naive 3D Schrodinger equation either, for example section 3 here: this appears to be the proper context in which Pauli's theorem should be applied regarding the choice of Hamiltonian and a potentially 'proper time' operator being associated to it, note it regards not 'time' but now a 'proper time' operator.

Thus, in this covariant case, (in which it should again be stressed that the Hamiltonian is determined by the choice of gauge usually taken to keep things covariant) operator commutation relations like $[X^{\mu},P_{\nu}] = i \delta^{\mu}_{\nu}$ are simply unaffected by Pauli's argument about discrete spectra of the Hamiltonian, which would really relate to whether it is possible to establish a proper time operator associated to a covariant-gauge-chosen Hamiltonian.

In a specific gauge in which proper time is taken to be the usual time (section two of the above notes for example), non-locality immediately arises - we now get a non-local Schrodinger equation, the square of which as usual leads to 'negative energy' eigenvalues which then require a direct argument - at best one has to artificially throw away the negative energy eigenvalues and replace them with positive energy eigenvalues for a separate species of plane wave associated to what we call anti-particles, or else talk about going backward in time, to keep those energy eigenvalues positive, and so the meaning of position space is just destroyed as usual making time operators even less relevant, it's unavoidable.

 

[vanhees71]

The counterexample is of course invalid, because the there discussed Hamiltonian
$$\hat{H}=\frac{1}{2m} \hat{P}^2 -q E \hat{Q},$$
where I made deliberately the dimension of the potential right ($q$ electric charge of the particle, $E=\text{const}$ electric field). Of course here $[\hat{H},\hat{P}]=\text{const}$, but this Hamiltonian is indeed not bounded from below, i.e., there's no stable ground state. Of course a homogeneous electric field everywhere is fictitious and thus this is a idealized approximate Hamiltonian not describing a real-world situation.

The energy-time uncertainty relation is of a different kind than the uncertainty relations between observables. To measure time you have to measure an observable, e.g., the hyperfine transition of Cs which defines the SI unit of time, the second. To get this transition, i.e., the difference of the two energy levels, accurately you need to measure for a long time.

Another argument is that for a measurement of time you have to measure some observable $\hat{A}$. Then the uncertainty relation of this observable with the Hamiltonian/energy is
$$\Delta A \Delta E \geq \frac{1}{2} |\langle [\hat{A},\hat{H}] \rangle|=\frac{\hbar}{2}|\langle \dot{A} \rangle|,$$
where $\dot{A}$ is the operator representing the time derivative of the observable $A$. Then
$$\Delta E \geq \hbar \frac{|\langle \dot{A} \rangle|}{2 \Delta A} = \frac{\hbar}{2 \Delta t},$$
where $\Delta t$ is an estimator for the time it takes for the observable $A$ to change by an amount of its standard deviation $\Delta A$, i.e., where a change can be expected to be significantly detected. Again it tells you that to determine the energy of a system you need a sufficiently long time for measuring it.

 

[Demystifier]

In these notes they promote time (not proper time) to an operator and arrive e.g. at the Klein-Gordon equation which tells us point particle wave functions are independent of proper time.

There is no problem of making time an operator in that sense, in a Lorentz covariant way. But this operator is not an operator on a Hilbert space. It is an operator on the space of functions of spacetime position $x=(x^0,x^1,x^2,x^3)$, but those functions are not square integrable. That's because solutions of the Klein-Gordon equation cannot be localized in spacetime; they can be localized in 3-space, but not in time.

 

[throw]

Are you implying that the fact that continuous spectrum eigenfunction solutions of a Schrodinger equation are not square integrable, and so technically not in a Hilbert space (an issue that does not stop books like L&L), somehow let's us rationalize the existence of a time operator in a relativistic context, but not in a non-relativistic context? I'm not sure I could agree with anything like that.

 

[Demystifier]

Are you implying that the fact that continuous spectrum eigenfunction solutions of a Schrodinger equation are not square integrable, and so technically not in a Hilbert space (an issue that does not stop books like L&L), somehow let's us rationalize the existence of a time operator in a relativistic context, but not in a non-relativistic context? I'm not sure I could agree with anything like that.

No. I am saying that the scalar product (both relativistic and non-relativistic) between two wave functions involves a 3-dimensional integration over space, and not a 4-dimensional integral over spacetime. This means that the Hilbert space is a Hilbert space of functions of space only, not of space and time. For reasonable wave packets, this 3-dimensional integral is finite. But there are no wave packets for which the 4-dimensional integral would be finite. Just the opposite, the scalar product, namely the 3-dimensional integral, is conserved in time, it is a constant. If one would then integrate the constant over the infinite time interval, one would get infinity.

 

[vanhees71]

One should, however, also pose the caveat that a "wave function" (or the so-called "first-quantization formalism") has a very different and pretty limited sense in the relativistic case compared to the non-relativistic case. The only successful fully consistent description of relativistic QT for (interacting) particles is local relativistic QFT, providing the only consistent framework to formulate the Standard Model of high-energy particles physics, which still as one of the most successful theories ever discovered.