Why doesn't the phase operator exist? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-17T15:29:59Z https://physics.stackexchange.com/feeds/question/292633 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/292633 19 Why doesn't the phase operator exist? Veteran https://physics.stackexchange.com/users/133801 2016-11-14T12:23:07Z 2019-07-20T08:34:42Z <p>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. </p> <p>Why isn't it possible to define the <a href="http://www.google.com/search?as_epq=phase+operator" rel="noreferrer">phase operator</a>? 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?</p> https://physics.stackexchange.com/questions/292633/-/292650#292650 6 Answer by flippiefanus for Why doesn't the phase operator exist? flippiefanus https://physics.stackexchange.com/users/55261 2016-11-14T14:24:07Z 2016-11-14T14:24:07Z <p>Well there is a <a href="https://en.wikipedia.org/wiki/Susskind%E2%80%93Glogower_operator" rel="noreferrer">Susskind-Glogower</a> 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. </p> https://physics.stackexchange.com/questions/292633/-/329748#329748 0 Answer by D. A. Dubin for Why doesn't the phase operator exist? D. A. Dubin https://physics.stackexchange.com/users/154465 2017-04-29T09:43:54Z 2017-04-29T09:43:54Z <p>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.</p> https://physics.stackexchange.com/questions/292633/-/337669#337669 0 Answer by tparker for Why doesn't the phase operator exist? tparker https://physics.stackexchange.com/users/92058 2017-06-06T00:50:50Z 2017-06-06T00:50:50Z <p>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 <em>is</em> 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.</p> <p>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 <a href="https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms" rel="nofollow noreferrer">exact but not closed differential form</a> with a nontrivial de Rahm cohomology, which is possible because it is defined on a non-simply-connected real-space manifold.</p> <p>There's a detailed discussion of these subtleties <a href="https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.40.411" rel="nofollow noreferrer">here</a>. 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. <a href="http://aapt.scitation.org/doi/abs/10.1119/1.1328351" rel="nofollow noreferrer">Here</a> 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.</p> https://physics.stackexchange.com/questions/292633/-/337795#337795 1 Answer by Arnold Neumaier for Why doesn't the phase operator exist? Arnold Neumaier https://physics.stackexchange.com/users/7924 2017-06-06T16:00:00Z 2017-06-07T16:41:39Z <p>A definite treatment of angle quantization, including a discussion of the extended literature on this problem, is given in</p> <p>H.A. Kastrup, Quantization of the canonically conjugate pair angle and orbital angular momentum, Physical Review A 73.5 (2006): 052104. <a href="https://arxiv.org/abs/quant-ph/0510234" rel="nofollow noreferrer">https://arxiv.org/abs/quant-ph/0510234</a></p> <p>In the main text it is mentioned that</p> <blockquote> <p>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 $p_φ &gt; 0$, i.e. $p_φ ∈ R_+$, or a real number, i.e. $p_φ ∈ R$.</p> </blockquote> <p>The first case corresponds to the question posed in the OP and is treated in fuller detail in another paper of the author (<a href="https://arxiv.org/abs/quant-ph/0307069" rel="nofollow noreferrer">https://arxiv.org/abs/quant-ph/0307069</a>), the second case corresponds to the situation explicitly mentioned in the title of the paper.</p> <p>For both cases, the problem and its resolution is well captured in the first part of the abstract of the paper:</p> <blockquote> <p>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.</p> </blockquote>