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From what I understand, the Schrödinger equation describes how the wave function of a quantum system evolves in space over a given time (I am referring to a relativistic version of the Schrödinger equation). My understanding is that the equation essentially describes the evolution of the probability of a quantum measurement as a classical system. So does this mean that the probabilities determined by the Schrödinger equation depend on the reference frame of the observer (i.e. do time dilation and length contraction affect the probabilities given by the equation)?

EDIT: What I'm ultimately wondering is if the probabilities calculated from the wave function whose evolution is described by the Schrödinger equation depend on the reference frame of the observer (i.e. if two identical systems measured by (1) someone at rest relative to the system and (2) someone in motion relative to the system, are the measurement probabilities different? Does it even make sense so say a measurement is made by someone in motion relative to the system?)

wythagoras
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Chris Laforet
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  • The Schroedinger equation describes the evolution of the wave function. It does not describe the evolution of probabilities. Those only come into play by using the Born rule, which is an independent assumption about the outcomes of classical measurements on a quantum system. Since the Schroedinger equation is non-relativistic, it won't give you the correct results for relativistic quantum systems. For that you need quantum field theory. – CuriousOne Dec 17 '14 at 03:02
  • In short, the Schrodinger Equation is built on top of Classical Hamiltonian Mechanics which means that it is not correct in the relativistic limit. You should read this Wikipedia article. I think that it answers your question. – Geoffrey Dec 17 '14 at 03:19
  • Clarification on question (v1): Are you referring to the Schrodinger equation for time-evolution for "time independent Shrodinger equation" which is unfortunately named since it's really just the equation obeyed by energy eigenvectors? The former is fully general and relativistic. – joshphysics Dec 17 '14 at 05:23
  • @CuriousOne The Schrodinger equation for time evolution, namely $i\hbar d|\psi\rangle/dt = H|\psi\rangle$ is not non-relativistic. In fact, it is the basis of time evolution in any quantum theory of anything, including QFT. Sorry to belabor the point, but I feel that this is one of those misconceptions that needs to be squashed rather assertively. – joshphysics Dec 17 '14 at 05:34
  • @joshphysics: Within the context of the OP's question the reference seems to be to the non-relativistic single particle Schroedinger equation, which is a different thing from a generalized linear evolution equation of QFT (in the context of which the interpretation of the wave function is fundamentally different, not to mention the significant problems of even defining qft problems in a mathematically meaningful way in that notation). – CuriousOne Dec 17 '14 at 05:42
  • @CuriousOne That may be so, and if it is so, then I agree with you, but I'm not convinced that's the context the OP has in mind. Maybe the OP will grace us with clarification. – joshphysics Dec 17 '14 at 05:51
  • @joshphysics: That's fair. Let's ask the OP what he meant. If he had a more general question than I understood, then I gladly withdraw my comment. – CuriousOne Dec 17 '14 at 05:53

3 Answers3

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From what I understand, the Schrödinger equation describes how the wave function of a quantum system evolves in space over a given time (I am referring to a relativistic version of the Schrödinger equation).

First, there is no relativistic Schrodinger equation. The correct relativitic generalisation is Diracs equation, but even that is a kind of approximation to the true theory and one must work with the whole apparatus of QFT = QM + SR.

My understanding is that the equation essentially describes the evolution of the probability of a quantum measurement as a classical system.

The equation directly decribes the evolution of the probability amplitude and not probability per se; the evolution of probabilities is derived. The amplitude in some sense, is the 'square root' of probability, and is one of the basic concepts that distinguishes Quantum Mechanics from Classical Mechanics.

What I'm ultimately wondering is if the probabilities calculated from the wave function whose evolution is described by the Schrödinger equation depend on the reference frame of the observer

Yes it does. The basic problem in canonical quantisation is that we cannot make a covariant choice of creation and annihilation operators.

Mozibur Ullah
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The Schrödinger equation is a non-relativistic approximation to the Klein-Gordon equation. The properties (momentum, energy, ...) described by solutions of Schrödinger equation should depend in the proper way of the Galilei reference frame. In reality they don't. The properties (momentum, energy, ...) described by solutions of Klein-Gordon equation do behave properly under Lorentz transformations, as do the solutions of the Dirac equation, which can be considered the relativistic extension of the Pauli equation.

my2cts
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In what follows I am discussing the time independent Schrodinger equation, the one we all learned in in the first quantum mechanics course, at least during my studies.

shcrodinger

The time independent Schrodinger equation is non relativistic, and yes, it will give different solutions in different frameworks. As the wavefunctions will be different, their square, which will give the probabilities of finding the state at a given (x,y,z) in time t, will be different.

The relevant equations for relativistic situations are the Klein Gordon for bosons and the Dirac for fermions.

In particle physics, the Dirac equation is a relativistic wave equation derived by British physicist Paul Dirac in 1928. In its free form, or including electromagnetic interactions, it describes all spin-½ massive particles, for which parity is a symmetry, such as electrons and quarks, and is consistent with both the principles of quantum mechanics and the theory of special relativity and was the first theory to account fully for special relativity in the context of quantum mechanics.

As matter we observe is mainly composed of fermions the Dirac equations is the most relevant.

Any solution to the Dirac equation is automatically a solution to the Klein–Gordon equation, but the converse is not true.

With the formalism of quantum field theory the building blocks , solutions of the Dirac equation, are not being much discussed.

It was pointed out to me that there exist time dependent equations which are discussed in this question here, and may be of interest for people wanting to pursue this, after reading the discussion in the comments.

anna v
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    -1 (at least for now on answer (v1)): It's unclear to me which Schrodinger equation the OP is referring to. If the OP is referring to the Shrodinger equation for time evolution, then there is nothing non-relativistic about it; it is, in fact, completely fundamental and generally applies to all quantum systems. I think it's very important to make that clear. – joshphysics Dec 17 '14 at 05:25
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    @joshphysics I am assuming the garden variety equation. – anna v Dec 17 '14 at 05:27
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    That doesn't narrow it down I'm afraid. There are two garden-variety equations that go by that name: $i\hbar d|\psi\rangle/dt = H|\psi\rangle$ and $-(\hbar^2/2m)\nabla^2\psi + V\psi = E\psi$. The former is relativistic, but the latter is not. – joshphysics Dec 17 '14 at 05:31
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    @joshphysics the closer I got by searching that the former is relativistic was " it is hard to show that it is" . Do you have a link . I was always of the impression that one needed the Klein Gordon for this , even without spins. ? – anna v Dec 17 '14 at 05:35
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    Any book on relativistic QFT will suffice since the starting point for time evolution is to use some "picture" (e.g. Schrodinger, Heisenberg, Interaction) all of which are equivalent to Schrodinger evolution. But for skeptics, see David Tong's QFT lecture notes equation 2.3 http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf, and the surrounding quote: "All time dependence sits in the states which evolve by the usual Schrodinger equation. We aren’t doing anything different from usual quantum mechanics; we’re merely applying the old formalism to fields." – joshphysics Dec 17 '14 at 05:49
  • @joshphysics: With all due respect to David Tong... a motivated smart student can learn the essentials of non-relativistic QM in a week, but even that student will need a couple of years to even get halfway up to speed in QFT. If we were doing "exactly the same thing", then life would be a heck of a lot easier than it actually is. And if Yangian symmetries happen to play as big a role in QFT as some folks have come to believe, then we may have been looking at this in a completely wrong way for the last 60+ years, anyway. – CuriousOne Dec 17 '14 at 06:05
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    @CuriousOne I think that might be a bit unfair to Tong in the sense that taken out of context, one might interpret the quote in the way you seem to be. Tong is referring specifically to time evolution, and if you ask me, it's quite conceptually important to emphasize that there's nothing new in QFT when it comes to time-evolution. I say this partly because more generally, it's important to realize that QFT is a model within the framework of quantum mechanics, and noting that quantum time evolution is adopted in exactly the same way is an important component of emphasizing that fact. – joshphysics Dec 17 '14 at 06:23
  • @joshphysics I am not disputing QFT. I want a simple demonstration that the S equation is relativistic; I think that the Klein Gordon is the relativistic version of the S. at least the article in wikipedia must also be wrong then. – anna v Dec 17 '14 at 06:39
  • @joshphysics: I do understand what David Tong means, but in practice the "...all we are doing..." piece has taken some of the brightest human beings over half a century to accomplish, and the confusion is probably greater now than the first generation of field theorists could have even imagined. I agree that one can take a somewhat wider view on quantum mechanics than QFT, that, however, does not reduce the complexity of its actual application, neither does it reduce relativistic field theories to a simple first order differential equation as that formal equation suggests. – CuriousOne Dec 17 '14 at 06:43
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    @annav The equation $i\hbar d|\psi\rangle /dt = H|\psi\rangle$ is relativistic in the sense that it determines time evolution of every quantum system, including any relativistic systems such as those described by the standard model of particle physics. I'm rather confused by what exactly you desire. What sort of "demonstration" do you want? I agree that the equations $-(\hbar^2/2m)\nabla^2\psi + V\psi = E\psi$ and $-(\hbar^2/2m)\nabla^2\psi + V\psi = i\hbar \partial\psi/\partial t$ describe non-relativistic systems. (a non-rel massive particle) if that's what you're getting at. – joshphysics Dec 17 '14 at 06:48
  • @joshphysics yes, the mass is non relativistic and it is hard to see what would happen at relativistic energies . So I guess my "no" in the answer above is centered on that, as well as the statement in the wiki article. in the KG you know you are dealing with the rest mass. – anna v Dec 17 '14 at 07:48
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    @joshphysics I guess it is the first order derivative in time that does not allow to see the relativistic form while the KG has it explicitly. – anna v Dec 17 '14 at 07:51
  • I have the impression that you are confusing relativistic with manifestly covariant equation. The Schrödinger equation in its general form $$\tag{1} i \hbar , \frac{d}{d t} , | \psi \rangle = \hat{H} , | \psi \rangle$$ is fully relativistic, provided we give a relativistic hamiltonian $\hat{H}$, but it is not manifestly covariant (or invariant in form). However, it is still valid in any reference frame. For exemple, the Dirac equation (interpreted as a one particle equation) can be expressed as (1) above very easily, with $\hat{H} = \alpha_i , \hat{p}_i , c + m c^2 \mathbb{I}$. – Cham Jan 16 '19 at 14:01
  • @Cham The OP's request about probability in space clearly is asking about covariance, hence this answer addresses his concern. In point of fact, the covariant generalization of the Schr eqn is the arcane, nonlocal Salpeter equation, with a covariant dispersion relation, at the expense of a pseudo differential operator. But an atomic clock governed by the Schr eqn clearly ticks at frame-dependent rates! The question is not an abstract discourse on Hilbert space, but a question on "how do we boost answers"? – Cosmas Zachos Jan 16 '19 at 14:20
  • Just for the sake of clarity (i'm far from an expert), does "relativistic" mean Poincare invariant here? – Daddy Kropotkin Jan 16 '19 at 16:36
  • @N._Steinle : It is not clear! I have an odd feeling they are arguing about Hegerfeldt's and Ruijsennaar's recondite theorems without spelling it out. A space density is not Poincare invariant, but if you look at the simplest explicit answers --or for the kernel of Salpeter's Poincare-invariant operator, they appear to display relativistic kinematics--with a suitable twist. – Cosmas Zachos Jan 16 '19 at 16:48
  • @N._Steinle small clarification: Salpeter's operator is not manifestly Lorentz covariant (which its parent, the K-G is), but possesses the light-cone structure "conducive" to covariance... Its solution space is invariant under Lorentz transformations, Foldy 1956. – Cosmas Zachos Jan 18 '19 at 15:20