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In many articles about quantum optics, the phase-number uncertainty relation $$\Delta \phi \Delta n \ge 1$$ has been mentioned and used as a heuristic argument, but they say that the phase-number uncertainty relation does not exist in a strict sense. Also many textbooks say there does not exist the phase operator.

Why isn't it possible to define the phase operator? If I define in such way $$\hat{\phi}\vert \phi \rangle=\phi \vert \phi \rangle$$ does this cause any problem? What is the major obstacle in defining the phase operator? Furthermore, how can I derive the phase-number uncertainty relation?

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    $\begingroup$ For a good review, see The Quantum Phase Operator, by S.M. Barnett and J.A. Vaccaro (eds.). $\endgroup$ – Emilio Pisanty Nov 14 '16 at 14:48
  • $\begingroup$ A very simple heuristic argument for why the uncertainy relation cannot make sense in that form is that the phase is only defined up to multiples of $2\pi$. Suppose you know $n$ exactly. That should mean that $\Delta\phi=\infty$, which makes no sense. $\endgroup$ – Adomas Baliuka Jun 1 '17 at 20:00
  • $\begingroup$ @Adomas Sure, but where does that uncertainty relation come from, what does each term in it mean, and what guarantees are there that you can define them to begin with? The question implicitly accepts that heuristics say that kind of thing, but the point is to move beyond that and into a more solid footing. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 22:04
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Well there is a Susskind-Glogower operator, which is a fairly good candidate for a phase operator. The problem is that the eigenstates for this operator does not have well-defined orthogonality/completeness relations. The reason for this, as I understand it, is because the Fock states (which are in a Fourier relationship to these eigenstates) only extend to positive infinity and not to negative infinity.

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  • $\begingroup$ Many thanks for the hint. Sounds like there are many mathematical conflicts with defining the phase operator. $\endgroup$ – Veteran Nov 18 '16 at 20:00
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A definite treatment of angle quantization, including a discussion of the extended literature on this problem, is given in

H.A. Kastrup, Quantization of the canonically conjugate pair angle and orbital angular momentum, Physical Review A 73.5 (2006): 052104. https://arxiv.org/abs/quant-ph/0510234

In the main text it is mentioned that

There are two typical (generic) examples where the unit circle $S^1$, parametrized by the momentum angle φ ∈ R mod 2π, represents the configuration space, whereas the canonically conjugate momentum variable $p_φ$ represents is either a positive real number $p_φ > 0$, i.e. $p_φ ∈ R_+$, or a real number, i.e. $p_φ ∈ R$.

The first case corresponds to the question posed in the OP and is treated in fuller detail in another paper of the author (https://arxiv.org/abs/quant-ph/0307069), the second case corresponds to the situation explicitly mentioned in the title of the paper.

For both cases, the problem and its resolution is well captured in the first part of the abstract of the paper:

The question how to quantize a classical system where an angle φ is one of the basic canonical variables has been controversial since the early days of quantum mechanics. The problem is that the angle is a multivalued or discontinuous variable on the corresponding phase space. The remedy is to replace φ by the smooth periodic functions cos φ and sin φ.

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    $\begingroup$ I'm not sold that phase and coordinate angle are equivalent, to be honest - you expect (among other objections) the canonical conjugate to phase to be bounded from below, which is not the case for coordinate angle. Coordinate angles are azimuthal angles on a plane whose coordinates commute, but for phase the quadratures are canonical conjugates. And finally there is the whole time-in-QM thing. $\endgroup$ – Emilio Pisanty Jun 6 '17 at 16:09
  • $\begingroup$ @EmilioPisanty: I don't see how ''the whole time-in-QM thing'' could have the slightest relevance for the present thread. Time involves no phase, and the time-energy uncertainty relation while superficially related, is in fact a completely different question with a completely different resolution. $\endgroup$ – Arnold Neumaier Jun 7 '17 at 16:47
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    $\begingroup$ This solution has the same problems as Carruthers and Nieto: why do you need two functions of an observable to describe a single observable. $\endgroup$ – ZeroTheHero Jun 8 '17 at 16:10
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    $\begingroup$ @ArnoldNeumaier There is nothing ridiculous in insisting that one single observable be expressed in terms of one self-adjoint operator, and this has nothing to do with algebraic geometry. You can be happy with the solution in terms of sine and cosine if you want but that problem is still considered open. $\endgroup$ – ZeroTheHero Jun 9 '17 at 16:30
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    $\begingroup$ But the angle is a multivalued variable since it is determiend only mod $2\pi$. Thus one has exactly the same problems as in algebraic geometry for representing multivalued roots of equations. The solution is known for more than 100 years. You just ignore the insight from algebraic geometry to be able to consider the problem as still open. But from the more abstract point of view that algebraic geometry provides it is obvious that there cannot be a solution of the kind you seem to look for. $\endgroup$ – Arnold Neumaier Jun 9 '17 at 16:59
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May I suggest that you start from the Zentralblatt fur Mathematik review Zbl. MATH 981. 81002 by N. P. Landsmann of the book "Mathematical aspects of Weyl quantization and phase". The short answer to your question is that a canonical phase operator does not exist. A number of proposals for non-canonical operators have been made. Most are legitimate operators on the usual Hilbert space but only one (as far as I know) is independent of the amplitude in the classical limit.

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The difficulty arises in trying to impose periodic boundary conditions. We would like to require that $\theta(0) = \theta(L)$, where $L$ is the system length. But since phases are only defined up to multiples of $2 \pi$, we actually only require that $\theta(0) = \theta(L) + 2 \pi n$ for some integer $n$. Therefore, the phase operator can be multivalued, as long as all the different values differ by multiples of $2 \pi$. This makes it difficult to formalize mathematically. Technically, the subtlety with the boundary conditions means that $\theta(x)$ cannot be a "self-adjoint operator." It is Hermitian, but for infinite-dimensional Hilbert spaces, self-adjointness is actually a strictly stronger condition that Hermiticity, and is required for all the nice properties that we'd like (e.g. having only real eigenvalues). The subtle nature of the domain of the $\theta$ operator is the reason why the naive application of the uncertainty relation doesn't work.

For example, we'd like to be able to freely integrate by parts and neglect boundary terms, but the boundary conditions on $\theta$ mean that we need to be careful of the possibility that the boundary term actually contributes a multiple of $2 \pi$. This is because $d\theta$ is an exact but not closed differential form with a nontrivial de Rahm cohomology, which is possible because it is defined on a non-simply-connected real-space manifold.

There's a detailed discussion of these subtleties here. It turns out that while $\theta(x)$ cannot be made into a self-adjoint operator, $\sin(\theta)$ and $\cos(\theta)$ can, so if you only work with those operators than everything more or less works out okay. Here is a shorter and somewhat more elementary discussion of how different self-adjoint operators that extend the same Hermitian operator can lead to the observably different physics.

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  • $\begingroup$ I'm not sold that phase and coordinate angle are equivalent, to be honest - you expect (among other objections) the canonical conjugate to phase to be bounded from below, which is not the case for coordinate angle. Coordinate angles are azimuthal angles on a plane whose coordinates commute, but for phase the quadratures are canonical conjugates. And finally there is the whole time-in-QM thing. $\endgroup$ – Emilio Pisanty Jun 6 '17 at 16:11
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Attempt to construct phase state from number state of a single oscillator has been failing, because state represented, if any, is absolute phase state, which should not exit.

Technically, as numbers cannot be negative, it is impossible to have absolute phase state, which, if exists, is conjugate state of number state, by fourier transform.

I have recently found that, using two oscillators, it is trivially easy to construct relative phase state representing relative phase between the oscillators.

As number difference between the oscillators can be arbitrary large negative or positive integer, it is trivially easy to have relative phase state from number difference by fourier transform.

Similarly, while absolute time state should not exist, it is possible to construct relative time state using two oscillators with different angular velocity.

See http://vixra.org/pdf/1906.0222v2.pdf for discussion with mathematical expressions.

Using such state, it is trivial to construct relative phase or relative time operator acting on two oscillators.

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    $\begingroup$ This is not an answer to the question as posed. $\endgroup$ – Emilio Pisanty 3 hours ago

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