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One foundational postulate of QM is that a closed physical system at one instant of time, say $t$, is completely described by a wavefunction $\psi \in S^1\subset H$ (where $H$ is a Hilbert space and $S^1$ its unit sphere). Another foundational postulate is that the wavefunction of a closed system should evolve deterministically along some orbit $t \mapsto \psi(t)$. It is therefore possible to define a time-evolution operator $U(t,t'):S^1\subset H \to S^1 \subset H:\psi(t') \mapsto \psi(t)=U(t,t')\psi(t')$. From the considerations so far we still need the linearity property for $U$ (after extending its (co)domain to $H$) to arrive at the conclusion that $U(t,t')$ be unitary (for all $t$).

Some would suggest that the linearity of $U$ is simply foundational itself and an experimentally falsifiable part of QM which does not rely on some deeper philosophical underpinning.

Many texts (see also Weinberg's "lectures in QM") however adopt the point of view that time-translation is a symmetry à la Wigner and that all transition probabilities $t \mapsto |\langle \psi(t),\phi(t)\rangle|^2$ should therefore be constant in time. Wigner's theorem then tells us that $U(t,t')$ should be either unitary or anti-unitary. A very plausible continuity argument rules out the anti-unitary option.

So which way carries the greater truth? Or is the distinction between the two viewpoints only apparent?

Personally, I have difficulty understanding the second viewpoint. Are wavefunctions not supposed to describe the system at one instant of time and not be a spacetime description? e.g. the inner product in usual QM Hilbert spaces asks you to integrate or perform certain sums related to the wave-function at one such instant of time. Therefore, it seems not evident to me that time-translation should be a symmetry in the sense of Wigner. Does Lorentz-invariance somehow force the constancy of $t\mapsto |\langle\psi(t),\phi(t)\rangle|^2$?

5th decile
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  • There is a very nice explanation of the 2nd viewpoint in Fonda & Ghirardy, "Symmetry Principles in Quantum Physics", Sec.1.4, starting pg.23 bottom, https://www.scribd.com/doc/30539019/Symmetry-Principles-in-Quantum-Physics-L-Fonda-G-C-Ghirardi-Marcel-Dekker-s. See if this answers your question. – udrv Jan 17 '17 at 10:05
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    @udrv unfortunately not: the constancy of $t \mapsto |\langle \psi(t),\phi(t)\rangle|$ is, as usual, forced down our throat in that text. – 5th decile Jan 17 '17 at 11:00
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    I see what you mean, and in this case the problem is indeed non-trivial. The more refined argument I am aware of is as follows: 1) Given the usual QM representation of states and observables but without any assumptions as to their time evolution, 2) Relativity imposes linear evolution through the no-signaling theorem, which invokes the speed limit but not explicitly Lorentz transformations, and 3) Time-reversal (not time-translation) imposes unitarity. – udrv Jan 17 '17 at 16:26
  • That's an interesting comment, thx. 2) is new to me and very interesting (some googling seems to yield papers/expositions confirming that view). I don't understand what you mean by 3) however: it's one of the basic operating assumptions to have states represented by wavefunctions of unit norm. So time-evolution should better leave the unit-sphere invariant and the only non-trivial part about the unitarity is (by definition then) the linearity, no? No physics must be brought in to justify norm-preservation. – 5th decile Jan 17 '17 at 20:35
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    Point (3) is needed because QM states may also be mixed states given by density matrices, of which the pure states are a particular case. The most general linear evolutions on the convex set of mixed states that are i) not only positive definite, but also completely positive (a.k.a. positive, local and separable), ii) homogeneous (insensitive to state normalization), and iii) preserve probability (the density matrix trace), are the generally irreversible evolutions with Lindblad-type generators. Requiring time-reversal restricts the set of acceptable evolutions to the unitary ones. – udrv Jan 17 '17 at 23:32
  • One point that is often confused: once QM is defined on the Hilbert space of states, with the associated convex set of mixed states and density matrices and the algebra of observables, the superposition principle is firmly in place and completely independent of any particular time evolution, linear or nonlinear. In other words, there exist nonlinear evolutions perfectly compatible with the superposition principle, although overlaps like $|\langle \phi(t)| \psi(t) \rangle|$ are no longer conserved in time, and that is no big deal. – udrv Jan 17 '17 at 23:33
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    The major problem with nonlinear evolutions is that they are expected to distinguish between mixed states as statistical ensembles of pure states and mixed states as local states arising from, say, some entangled state shared with other non-interacting systems. This is so far untenable under the no-signaling theorem. – udrv Jan 17 '17 at 23:38
  • Aren't "superposition principle" and "linearity" (or perhaps "invariance under linear combinations") not synonyms? To be clear, I mean this in a dynamical sense: the coëfficients in the relevant linear combination are not allowed to depend on time. Do you have advice on where to have a good read on these subjects (in particular the relation between the no-signaling theorem and linearity) – 5th decile Jan 18 '17 at 01:04
  • A Lindblad-type evolution $e^{\mathcal{L}t}$ is linear on density matrices, in the regular sense that $e^{\mathcal{L}t} \left(\lambda_1 \rho_1(0) + \lambda_2 \rho_2(0)\right) \rightarrow \lambda_1 e^{\mathcal{L}t}\rho_1(0) + \lambda_2 e^{\mathcal{L}t}\rho_2(0)$ (generally one takes $\lambda_i > 0$ to keep positivity in check, and $\lambda_1 + \lambda_2 = 1$ for normalization). – udrv Jan 18 '17 at 01:52
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    But since it generally takes pure states to mixed states, it doesn't take $|\Psi(0)\rangle = c_1|\psi_1(0)\rangle + c_2 |\psi_2(0)\rangle$ into $c_1|\psi_1(t)\rangle + c_2 |\psi_2(t)\rangle$, but $|\Psi(0)\rangle\langle \Psi(0)| \rightarrow e^{\mathcal{L}t}|\Psi(0)\rangle\langle \Psi(0)| \neq |\Psi(t)\rangle\langle \Psi(t)|$, although superpositions work as usual in the Hilbert space. – udrv Jan 18 '17 at 01:53
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    As for refs., the original paper is https://arxiv.org/abs/quant-ph/0102125, and there is an old list of papers on linearity/non-linearity in QM, https://arxiv.org/abs/quant-ph/0410036, but I'll have to come back with something more recent. – udrv Jan 18 '17 at 02:06
  • Ok, a newer starting point may be http://lanl.arxiv.org/pdf/1411.1768v2, and see also http://lanl.arxiv.org/pdf/quant-ph/0508092v4. For the statistical mixture problem that appears in linearity discussions see the nice intro in http://lanl.arxiv.org/pdf/quant-ph/0402094v2. – udrv Jan 18 '17 at 05:45
  • @udrv what do you mean by "requiring time-reversal" exactly (3)? That the time-Evolution must be a reversible process? I read the paper on linearity and the no signalling condition, and it was very insightful. Do you know similar papers on the topic of unitarity? – Quantumwhisp Oct 23 '20 at 11:54

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First of all, pure states in quantum mechanics don't correspond to elements of the "unit sphere" of the Hilbert space; they correspond to elements of the projective Hilbert space (which for a finite-dimensional Hilbert space is a complex projective space), which is not the same thing. State vectors that differ by a global phase (which represent the same physical pure state) correspond to distinct elements of the Hilbert space's "unit sphere" but the same element of its projective Hilbert space. (Also, using the notation $S^1$ to refer to an arbitrary hypersphere is extremely confusing, because that notation almost always refers specifically to a circle.)

To answer your question: different physicists will answer differently, but I personally am very much in the first camp, that the unitarity of quantum time-evolution is simply an experimentally verified postulate that cannot be derived from more fundamental principles (other than the Schrodinger equation itself). I think that basing the unitarity assumption on the time-translational invariance of the Hamiltonian reflects an unfortunate bias (mostly by high-energy theorists) toward the idea that quantum mechanics always describes "fundamental" laws of physics. In virtually every field of physics, it's often very useful to consider Hamiltonians that depend explicitly on time. In these cases the translational symmetry argument immediately falls apart, but the time translation is still perfectly unitary. Therefore, I think that unitarity is more fundamental than time-translational invariance.

And if I could editorialize a bit, anyone who objects that these time-dependent Hamiltonians are just "special limiting cases" of the Standard Model is full of themself - after all, the Standard Model itself is probably just a limiting case of an even more fundamental theory, and we have no idea whether or not that more limiting theory is time-translationally invariant, or if that idea even makes sense.

Pages 4-6 of this paper discuss some nasty implications of non-unitary time evolution - e.g. it would allow faster-than-light signalling and efficiently solving PP-complete problems. But the frequent claim that the probabilistic interpretation of QM requires unitary time evolution is incorrect; non-unitary time evolution has consequences that seem very difficult to reconcile with our world, but it is a perfectly logically consistent theory.

tparker
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  • "non-unitary time evolution has consequences that seem very difficult to recoincile with our world, but it's a perfectly logically consistent theory" - To make this consistent, would that mean one has to alter the way probabillities are calculated in QM? – Quantumwhisp Oct 03 '19 at 05:43
  • @Quantumwhisp No. One could imagine a non-unitary time-evolution map which is nonlinear but norm-preserving, in which case you wouldn't need to change anything at all. Even if the non-unitary time-evolution map changes the norm, then you just need to use the fact that the expectation value of a quantum-mechanical observable $\hat{O}$ is $$\frac{\langle \psi(t) | \hat{O} | \psi(t) \rangle}{|\psi(t)|^2}$$ (in the Schrodinger picture). If the time-evolution map is not norm-preserving, then the denominator becomes a non-constant normalization function. This is sometimes called "manual ... – tparker Oct 03 '19 at 16:30
  • ... renormalization." – tparker Oct 03 '19 at 16:30
  • I see the point about the linearity (which is needed for the norm preserving map to be unitary). About the option (manual renormalization), I would imagine that for any linear non-unitary time-evolution operator, I can find an unitary one that will yield the same expectation values for every operator, which makes non-unitary time-evolution with manual normalization equal to unitary time evolution with standard normalization. – Quantumwhisp Oct 03 '19 at 19:54
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    @Quantumwhisp See https://physics.stackexchange.com/q/506267/92058. – tparker Oct 04 '19 at 01:28
  • @ChiralAnomaly I don't understand your claim that the Heisenberg picture treats space and time more symmetrically. I would say that in non-relativistic QM, neither states nor operators are parameterized by space in either picture. Are you thinking about quantum field theory? – tparker Oct 04 '19 at 04:08
  • @ChiralAnomaly Also, I disagree that proposals for non-linear QM have only been formulated in the Schrodinger picture. Non-linear QM simply postulates a nonlinear form for the time-translation map, and as such is independent of your choice of picture - as usual, you're free to put the parentheses wherever you want. – tparker Oct 04 '19 at 04:09
  • @tparker (1) Yes, I'm thinking of quantum field theory. Considering modifications of QM that can't be extended to modifications of QFT is like considering justifications of QM that can't be extended to cases with time-dependent background fields. I think what your answer rightly critiqued about the latter applies also to the former. (2) Maybe I misunderstood what is meant by nonlinear. I was thinking of modifications of the Schr. equation that include higher powers of the wavefunction, and it's not clear to me how a Heisenberg-picture version of that would be defined at all. – Chiral Anomaly Oct 04 '19 at 13:06
  • @tparker I'm agreeing with your statement that the assumption of unitary time-evolution shouldn't be based on time-translation symmetry. My first comment was meant to give another reason, one that takes into account an even larger set of quantum theory's empirically-successful applications. – Chiral Anomaly Oct 04 '19 at 13:17
  • @ChiralAnomaly If we just assume that the pre-measurement time evolution is deterministic, then we can write $|\psi(t)\rangle = U(t) |\psi_0\rangle$ for some (not necessarily linear) map $U$. Then all time differential operators on $|\psi(t)\rangle$ can be trivially rewritten as acting on $U$ instead. This leads to the "picture-independent" version of the Schrodinger equation $i \hbar, dU/dt = H U$, which holds equally well in the Schrodinger and Heisenberg pictures. A nonlinear version of QM simply means that this equation is modified to be nonlinear in $U$. It's just as well-defined in ... – tparker Oct 05 '19 at 01:55
  • ... the Heisenberg as in the Schrodinger picture. – tparker Oct 05 '19 at 01:56
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    You're right: nonlinearity doesn't necessarily prevent defining a Heisenberg picture. If $U(t)$ is invertible, even if nonlinear, then we can define $A(t)$ by $A(t) := U^{-1}(t) A U(t)$. Maybe nonlinear mods of QM are crafted to ensure that $U$ is invertible; I haven't been motivated to study them that carefully. But in any case, my original comment addressed to @Quantumwhisp was flawed, so I deleted it. – Chiral Anomaly Oct 05 '19 at 03:12
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    @ChiralAnomaly While I haven't thought much about this either, I think that $U$ doesn't even have to be invertible; the Heisenberg picture operator is $A(t) = U^\dagger A U$, not $U^{-1} A U$. The adjoint of an arbitrary map on vectors is defined by $\langle \psi | U^\dagger := | U \psi \rangle^\dagger$ as usual (that definition does not assume linearity, only an inner product). – tparker Oct 05 '19 at 03:56
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    LOL, I was just coming back to replace my broken comment when I saw that you had already replied. Of course you're right again: it should be $U^\dagger A U$. But in that case, if $U$ is nonlinear, then I don't see any reason to expect $(U^\dagger A U)(U^\dagger B U) = U^\dagger AB U$, because I don't see why we should expect $U U^\dagger = 1$. ... so I'm back to not seeing a way to make sense out of the Heisenberg picture in the nonlinear case. Seems weird if the time-translation of $A(0)B(0)$ were not the same as $A(t)B(t)$. Anyway, good answer (+1), and sorry for my mixed-up comments. – Chiral Anomaly Oct 05 '19 at 04:28