# What does “Relativistic” mean in Quantum Mechanical Terms?

I was reading recently how the compatibility of quantum mechanics with special relativity was initially a problem for physicists and then Dirac succeeded in formulating a relativistic, quantum-mechanical theory through the Dirac equation.

From my understanding, the whole idea of relativity is to keep the speed of light or photons the same in all frames. But in in quantum mechanics, particles don't have a definite position except when measured-let alone velocity (which is it's derivative) so how is "speed of a photon" even defined quantum mechanically? How do we generalize relativistic definitions like speed to be compatible with quantum mechanical concepts? How exactly did Dirac incorporate SR into his wave equation?

I read related Wikipedia articles but I couldn't understand much.

I am somewhat familiar with elementary concepts like the Lorentz-Transform, proper-time, the Schrodinger equation, etc. but as you might tell have no formal education in it so please be easy with the technicality while answering.

• Quantum field theory is the marriage of QM with SR. QM does not have a limit on information transfer; it's possible for particles to spontaneously zip to the other side of the universe in QM, which is a problem. Additionally, The energy of a particle from SR is $E^2=m^2c^4+p^2c^2$, but QM doesn't enforce that rule. QFT is made by introducing mechanisms to limit the transfer of information to $c$ and making sure the energy always comes out to the SR equation – Jim May 24 '17 at 12:18
• relativistic effect will destroy some symmetry in QM; and hence destroy some degeneracies. (e.g. a moving super fast election in a H atom) – Shing May 24 '17 at 12:18
• "But in in quantum mechanics, particles don't have a definite position except when measured-let alone velocity" - but there are, in QM, minimum uncertainty wave-packets with a well-defined group velocity. – Alfred Centauri May 24 '17 at 12:57
• @AlfredCentauri so does this mean that it's the group velocity of photon wavepackets that is kept constant to generalize the theory? – alex May 24 '17 at 13:08
• @Draco18s I think the question of whether particles do have definite properties at all times is essentially the same as the hidden variable idea (although it is usually explained in terms of spin rather than position) and that was settled by the Bell test experiments in the favour of them not having certain definite measurable properties at all times. or more precisely, its meaningless to ask whether they do – alex May 24 '17 at 15:36

In quantum mechanics, only measurements have the statistical distributions, the "uncertainties" and all the things that are (validly) bothering you. As you point out, this makes notions of measured spacetime co-ordinates problematic.

But the underlying theory that lets one calculate these statistical distributions can be Lorentz-invariant.

It is not emphasized enough, particularly in many lay expositions, that much, if not most of quantum mechanics is utterly deterministic. This deterministic part is concerned with the description and calculation of the evolution of a system's quantum state. Aside from some more modern mathematical techniques and notations, this part of quantum mechanics probably wouldn't look very alien or physically unreasonable to even Laplace himself (whom we can take as a canonical thinker from the philosophy of determinism). This quantum state evolution takes place on an abstract spacetime manifold just like classical physics. When a quantum theory is said to be relativistic or Lorentz invariant, it is usually the deterministic quantum state space evolution that is being talked about. Note, in particular, that no measurement takes place in this part of the description, so there's no problem with a spacetime manifold parameterized by zero uncertainty spacetime co-ordinates.

We model measurements with special Hermitian operators and recipes for handling them called observables. Given a system's quantum state, these operators let us work out the statistical distributions of outcomes of the measurements we can make on a system with that quantum state. When people talk of quantum uncertainty, Heisenberg's principle and all the rest of it, they are speaking about the statistical distributions that come from these measurements. So, whilst Schrödinger's equation (the deterministic, unitarily evolving description) for the electron in a hydrogen atom is written in terms of zero uncertainty space and time co-ordinates (known as parameters to emphasize that they are not measurements), the outcome of a position measurement is uncertain and the position observable lets us calculate the statistical distribution of that outcome.

• So instead of trying (most probably in vain) of trying to modify the meaning of measurables so that we can apply Lorentz invariance to it we just apply the principle of Lorentz invariance to the deterninistic but not directly measurable "parameters" amd just accept whatever effect this has on the behaviour of measurables? – alex May 24 '17 at 13:40
• @alex That's a pretty good summary, yes! I think you've got it. – WetSavannaAnimal May 24 '17 at 21:51
• Wet, re you last paragraph, do we actually model measurements with Hermitian operators? As I understand it, acting on a state with a Hermitian operator (corresponding to an observable) is not a measurement since this does not 'collapse' the state into an eigenstate of the observable. Would you mind elaborating or setting me straight on this? – Alfred Centauri May 25 '17 at 1:14
• @AlfredCentauri we model them with operators AND special recipes for handling them. So I meant the operator, together with the idea that the state winds up randomly in one of the operator's eigenstates with the measurement yielding that particular eigenstate's eigenvalue as the measurement outcome. Your comment is exactly the reason why I don't like the word "operator" and prefer "observable", because there's a "recipe"as well. Also, "operator" implies a mapping of a quantum state, which is not what an observable is - even though the operator's image must lie in the .... – WetSavannaAnimal May 25 '17 at 1:23
• ...... quantum state space so that one computes the $n^{th}$ moment of the measurement statistics as $\langle\psi | A^n| \psi\rangle$. – WetSavannaAnimal May 25 '17 at 1:24

There is no full consensus about what relativistic means, but as a good start, we can take the criterion given by Tim Maudlin in his paper "Space-Time in the Quantum World":

[A] theory is compatible with Relativity if it can be formulated without ascribing to space-time any more of different intrinsic structure than the [...] relativistic metric.

This directly implies that the laws must be invariant under Lorentz transformations, there is no preferred frame as such and nothing can propagate faster than light. This is a good notion since it does not need things like "speed of a photon" that are, as you pointed out, a bit problematic.

• classically lorentz invariance constitutes the invariamce of proper time i.e. the "distance" through spacetime between two events that occur at specific points in space at specific times from each observers point of view. what does Lorentz invariance mean in qm? what does proper time measure the "distance" between? – alex May 24 '17 at 12:32
• Lorentz invariance is lorentz invariance, if classical or quantum does not matter. This is a property of a physical law - the physical law contains space and time coordinates, and the Lorentz invariance then says: If I transform to another Lorentz frame, the physical law takes the same form. This is true for the Dirac equation, and false for the Schrödinger equation. – Luke May 24 '17 at 12:37
• @alex In QM only measurements have the statistical distributions, the "uncertainties" and all the things that are (validly) bothering you. AS you point out, this makes notions of measured spacetime co-ordinates problematic. But what Luke is saying is that the underlying theory that lets you calculate these statistical distributions can be Lorentz-invariant. Things like the Dirac equation describes the probability amplitude (or a vector-like quantity whence one calculates the PA) as a function of the spacetime co-ordinates. The PA is not measured, and nor are the co-ordinates .... – WetSavannaAnimal May 24 '17 at 12:40
• @alex ..... When you measure an electron's position (that in itself is a little bit of a complicated notion for the Dirac equation, but the problems are not really relevant here - let's just assume that there is some calculation you can do from the equation's solution to come up with a probability amplitude for position), the result will give you a statistical spread of co-ordinates for possible answers. – WetSavannaAnimal May 24 '17 at 12:44
• @Luke But does just making the formula look the same when the x and t terms in it are tranformed in a certain way guarantee that the rest of the significant features of SR (like the constancy of the speed of photon wavepackets which Alfred seems to me to be implying )will also be inherited? – alex May 24 '17 at 13:21

How exactly did Dirac incorporate SR into his wave equation?

Relativity treats space and time on equal footing. Lorentz transformations ("boosts") "mix" space and time more or less analogously to the way a 3-dimensional rotations "mixes" the usual $(x,y,z)$ coordinates of space.

The Schrödinger equation describes non-relativistic quantum mechanical systems well, but is second-order in spatial derivatives and first-order in the temporal derivatives. In other words, space and time are not on equal footing.

Dirac, being a formal theorist, looked for an equation that would reproduce predictions (e.g., the energy levels of Hydrogen in the limit of slowly moving electrons) of the non-relativistic theory while treating space and time on equal footing. The equation which he discovered and now bears his name is a first-order differential equation in space and time. This "equal treatment of space and time" is the sense in which the Dirac equation incorporates special relativity.

To put it concretely: In my course on QED, I can remember two conditions which were imposed due to relativity.

• The fields transform via a Lorentz transformation, and the Lagrangian should not change form under this Lorentz transformation.
• Since measurements cannot affect each other at spacelike separation, the commutator of two fields at equal times but different locations must be 0. This was in particular important for the photon field $A^\mu (t, x)$.

It was by imposing that his equation is Lorentz invariant (among other things) that Dirac was able to find his equation. This follows from imposing that the lagrangian is Lorentz invariant.

## protected by Qmechanic♦May 24 '17 at 20:06

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