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A Quantum challenge: mathematical paradoxes

  1. Apr 22, 2016 #1

    micromass

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    There are many apparent paradoxes in quantum mechanics. Luckily, a careful application of math reveals that all is well. But can you figure out why the following ##7## challenges are not paradoxes?

    Rules:
    • Do not look at paper [1] before answering. It contains all the answers in detail. Any other use of outside sources is allowed, but make sure to quote them.
    • Give a careful explanation why you think the speciic situation is not a paradox. Use any math/physics you wish.

    Challenge 1
    For a particle in one dimension, the operators of momentum ##P## and position ##Q## satisfy Heisenberg’s canonical commutation relation
    [tex][P, Q] = \frac{\hbar}{i} \boldsymbol{1}[/tex]
    By taking the trace of this relation, one finds a vanishing result for the left-hand side, ##\text{Tr}[P, Q] = 0##, whereas ##\text{Tr}\left(\frac{\hbar}{i}\boldsymbol{1}\right)\neq 0##. What went wrong?

    Challenge 2
    Consider wave functions ##\varphi## and ##\psi## which are square integrable on ##\mathbb{R}## and the momentum ##P = \frac{\hbar}{i}\frac{d}{dx}##. Integration by parts yields
    [tex]\int_{-\infty}^{+\infty}\overline{\varphi(x)}(P\psi)(x)dx = \int_{-\infty}^{+\infty} \overline{(P\varphi)(x)}\psi(x)dx + \frac{\hbar}{i} [ (\overline{\varphi}\psi)(x)]_{-\infty}^{+\infty}[/tex]
    Since ##\varphi## and ##\psi## are square integrable, one usually concludes that these functions vanish for ##x\rightarrow \pm\infty##. Thus, the last term in the previous equation vanishes, which implies that the operator P is Hermitian.

    However, the textbooks of mathematics tell us that square integrable functions do, in general, not admit a limit for ##x\rightarrow \pm \infty## and therefore they do not necessarily vanish at infinity. There are even functions which are continuous and square summable on ##\mathbb{R}## without being bounded at infinity: an example of such a function is given by
    [tex]f(x) = x^2 e^{-x^8\sin^2x}[/tex]
    Can one conclude that the operator P is Hermitian in spite of these facts and if so, why?

    Challenge 3
    Consider the operators ##P = \frac{\hbar}{i} \frac{d}{dx}## and "##Q = ## multiplication by ##x##" acting acting on wave functions depending on ##x ∈ \mathbb{R}##. Since ##P## and ##Q## are Hermitian operators, the operator
    [tex]A = PQ^3 + Q^3P[/tex]
    also has this property, because its adjoint is given by
    [tex]A^\dagger = (PQ^3 + Q^3P)^\dagger = Q^3P + PQ^3 = A[/tex]
    It follows that all eigenvalues of ##A## are real. Nevertheless, one easily verifies that
    [tex]Af = \frac{\hbar}{i}f[/tex]
    with
    [tex]f(x) = \left\{\begin{array}{ll}
    \frac{1}{\sqrt{2}} |x|^{-3/2}\text{exp}\left(-\frac{1}{4x^2}\right) & \text{for}~x\neq 0\\
    0 & \text{for}~x=0
    \end{array}\right.[/tex]
    which means that ##A## admits the complex eigenvalue ##\hbar/i##. Note that the function ##f## is infinitely differentiable on ##\mathbb{R}## and that it is square integrable, since
    [tex]\int_{-\infty}^{+\infty} |f(x)|^2dx = 2\int_0^{+\infty}|f(x)|^2dx = \int_0^{+\infty} x^{-3} \text{exp}\left(-\frac{1}{2x^2}\right)dx= [\text{exp}(-1/(2x^2))]_0^{+\infty} = 1[/tex]
    Where is the error?

    Challenge 4
    Let us consider a particle confined to the interval ##[0, 1]## and described by a wave function ##\psi## satisfying the boundary conditions ##\psi(0) = 0 = \psi(1)##. Then the momentum operator
    [tex]P = \frac{\hbar}{i}\frac{d}{dx}[/tex]
    is Hermitian, since the surface term appearing upon integration by parts vanishes:
    [tex]\int_0^1 \left(\overline{\varphi}(P\psi) - \overline{(P\varphi)}\psi\right)(x)dx = \frac{\hbar}{i}[(\overline{\varphi}\psi)(x)]_0^1 = 0[/tex]
    Since P is Hermitian, its eigenvalues are real. In order to determine the latter, we note that the eigenvalue equation,
    [tex](P\psi_p)(x) = p\psi_p(x)[/tex]
    is solved by ##\psi_p(x) = c_p \text{exp}\left(\frac{i}{\hbar}px\right)## with ##p\in \mathbb{R}## and ##c_\in \mathbb{C}\setminus\{0\}##. The boundary condition ##\psi_p(0) = 0## now implies ##\psi_p = 0##, therefore ##P## does not admit any eigenvalues. Nevertheless, the spectrum of ##P## is the entire complex plane and ##P## does not represent an observable. How can one understand these results which seem astonishing?

    Challenge 5
    If one introduces polar coordinates in the plane or spherical coordinates in space, then the polar angle ##\varphi## and the component ##L_z## of angular momentum are canonically conjugate variables in classical mechanics. In quantum theory, the variable ##\varphi## becomes the operator of "multiplication of the wave function ##\psi(\varphi)## by ##\varphi## " and ##L_z = \frac{\hbar}{i} \frac{\partial}{\partial \varphi}##, which implies the commutation relation
    [tex][L_z,\varphi] = \frac{\hbar}{i}\boldsymbol{1}[/tex]
    These operators acting on periodic wave functions (i.e. ##\psi(0) = \psi(2\pi)##) are Hermitian. Furthermore, ##L_z## admits a complete system of orthonormal eigenfunctions ##\psi_m## for ##m\in \mathbb{Z}## and
    [tex]L_z \psi_m = m\hbar\psi_m~\text{with}~\psi_m(\varphi) = \frac{1}{\sqrt{2\pi}}\text{exp}(im\varphi).[/tex]
    By evaluating the average value of the operator ##[L_z,\varphi]## in the state ##\psi_m## and by taking
    into account the fact that ##L_z## is Hermitian, one finds that
    [tex]\begin{align*}
    \frac{\hbar}{i}
    & = \left\langle \psi_m, \frac{\hbar}{i}\boldsymbol{1} \psi_m\right\rangle\\
    & = \left\langle \psi_m, L_z\varphi \psi_m\right\rangle - \left\langle \psi_m, \varphi L_z\psi_m\right\rangle\\
    & = \left\langle L_z^\dagger\psi_m, \varphi \psi_m\right\rangle - m\hbar\left\langle \psi_m, \varphi \psi_m\right\rangle\\
    & = (m\hbar - m\hbar)\left\langle \psi_m, \varphi \psi_m\right\rangle\\
    & = 0
    \end{align*}[/tex]
    Where is the problem?

    Challenge 6
    Let us add a bit to the confusion of the previous example! In 1927, Pauli noted that the
    canonical commutation relation implies Heisenberg’s uncertainty relation ##\Delta P \Delta Q\geq \frac{\hbar}{2}## by virtue of the Cauchy-Schwarz inequality. One can derive, in the same way, the uncertainty relation
    [tex] \Delta L_z \Delta \varphi\geq \frac{\hbar}{2}[/tex]
    The following physical reasoning shows that this inequality cannot be correct. One can always find a state for which ##\Delta L_z < \frac{\hbar}{4\pi}## and then the uncertainty for the angle ##\varphi## has to be larger than ##2\pi##, which does not have any physical sense, since ##\varphi## takes values in the interval ##[0, 2\pi)##. How is this possible?


    By the way, this example shows that the uncertainty relation ##\Delta A \Delta B\geq \frac{1}{2}|\left\langle [A,B]\right\rangle|## for any two observables ##A## and ##B## (whose derivation can be found in most quantum mechanics texts) is not valid in such a generality.

    Challenge 7
    Let us consider a particle of mass ##m## in the infinite potential well (for ##a>0##)
    [tex]V(x) = \left\{\begin{array}{ll}
    0 & \text{if}~|x|\leq a\\
    \infty & \text{otherwise}
    \end{array}\right.[/tex]
    The Hamiltonian for the particle confined to the inside of the well is simply
    [tex]H = - \frac{\hbar^2}{2m} \frac{d^2}{dx^2}[/tex]
    Let
    [tex]\psi(x) = \left\{\begin{array}{ll}
    \frac{\sqrt{15}}{4a^{5/2}}(a^2 - x^2) & \text{if}~|x|\leq a\\
    0 & \text{otherwise}
    \end{array}\right.[/tex]
    be the normalized wave function of the particle at a given time. Since
    [tex]H^2\psi = \frac{\hbar^4}{4m^2}\frac{d^4}{dx^4}\psi = 0,[/tex]
    the average value of the operator ##H^2## in the state ##ψ## vanishes:
    [tex]\left\langle H^2\right\rangle_\psi = \left\langle\psi,H^2\psi\right\rangle = \int_{-a}^a\overline{\psi(x)}(H^2\psi)(x)dx = 0[/tex]
    This average value can also be determined from the eigenvalues and eigenfunctions of H,
    [tex]H\varphi_n = E_n \varphi_n[/tex]
    with
    [tex]E_n = \frac{\pi^2\hbar^2}{8ma^2} n^2~\text{for}~n=1,2,3,...[/tex]
    by applying the formula
    [tex]\left\langle H^2\right\rangle_\psi = \sum_{n=1}^{+\infty} E_n^2 p_n[/tex]
    with ##p_n = |\left\langle\varphi_n,\psi\right\rangle|^2##.

    Proceeding in this way, one definitely does not find a vanishing result, because ##E_n^2>0## and ##0\leq p_n\leq 1##, ##\sum_{n=1}^{\infty} p_n = 1##. In fact, the calculation yields ##\left\langle H^2\right\rangle_\psi = \frac{15\hbar^4}{8m^2a^4}##. Which one of these two results is correct and where does the inconsistency come from?

    References
    [1] Mathematical surprises and Dirac's formalism in quantum mechanics by François Gieres.
    https://arxiv.org/pdf/quant-ph/9907069
     
    Last edited: Apr 22, 2016
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  3. Apr 22, 2016 #2

    A. Neumaier

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    Shouldn't that be an Insight article rather than a question?
     
  4. Apr 22, 2016 #3

    A. Neumaier

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    In Challenge 6, the final statement is not justified since $\phi$ is not an observable in the traditional sense of being a self-adjoint operator. The textbook formula is correct under the stated assumptions.
     
  5. Apr 22, 2016 #4

    Demystifier

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    Such paradoxes teach us the importance of rigorous functional analysis in QM.
    But does anybody know a mathematical paradox in QM the solution of which is based on some other branch of mathematics?

    I would especially appreciate a physical paradox based on logic or set theory. :smile:
     
  6. Apr 22, 2016 #5

    A. Neumaier

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    Consider a measuring device that measures all measuring devices which don't measure themselves.
     
  7. Apr 22, 2016 #6

    Demystifier

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    Physically not interesting. :confused:
    Perhaps a cat which is dead and not dead?
    But of course, I want something more challenging.
     
  8. Apr 22, 2016 #7

    Strilanc

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    Perhaps the undecidability of the spectral gap in the limit?

    Bell tests, or related results like the one from the 'quantum pigeonhole principle' paper, might also fit the bill. (The pigeonhole thing certainly seemed to do a number on the authors of the paper :oldeyes:.) Classical intuitions suggest an impossibility result, but then quantum mechanics slips by, creating an apparent paradox.

    There's also super-basic stuff like the "unwinnable coin flip", where choosing whether or not to hit a qubit with a NOT gate has no effect on the outcome of a circuit (because NOT rotates around the X axis, and the qubit happened to be pointing along the X axis).
     
  9. Apr 24, 2016 #8
    Nice post, great paper, exciting discussion.

    Would make a very nice Insight, as suggested above.

    Thanks for posting.
     
  10. Apr 24, 2016 #9

    strangerep

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    Isn't the "insights article" already the Gieres paper itself? I'm guessing that Micro intended this for QM students who don't already know the answers. :oldbiggrin:
     
  11. Apr 25, 2016 #10

    Demystifier

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    Yes, that's very interesting. But I already made some comments on it in
    https://www.physicsforums.com/threads/spectral-gap-or-gapless-undecidable.847554/#post-5315726

    As a further insight, let me add that undecidability is really not a big deal for physics and applied mathematics. Many problems which are "simple" from a practical point of view are technically "undecidable".
    For instance, let f : R → R be an arbitrary elementary function. The question
    “Is there a real solution to the equation f (x) = 0?”
    is undecidable [B. Robič -The Foundations of Computability Theory, https://www.amazon.com/Foundations-Computability-Theory-Borut-Robic/dp/3662448076/ref=sr_1_1?s=books&ie=UTF8&qid=1461571546&sr=1-1&keywords=robič+computability+theory ]

    That's interesting too, but to resolve the paradox one does not need a deep understanding of combinatorics (or any other branch of mathematics). In fact it is not a paradox at all once you understand that QM is contextual, meaning nothing more but the fact that physical properties of the system can be changed by the physical process of measurement. Any magician can create an illusion of having 3 pigeons in 2 boxes, with 1 pigeon in each box. Quantum mechanics is not much different, and it's a paradox only for those who believe that Nature doesn't cheat in a way that magicians do.
     
    Last edited: Apr 25, 2016
  12. Apr 26, 2016 #11

    TeethWhitener

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    That's what I'm guessing too, and since I don't know the answers (and in order to get what has the potential to be a very interesting thread back on topic), I'll take the first crack at embarrassing myself. Warning: this isn't going to be very rigorous.

    For challenge 1, assuming that ##P## and ##Q## have continuous eigenspectra, I strongly suspect that the trace of the commutator is not well-defined. If we right- and left-multiply the original expression by some normalized wavefunction on both sides, we get:
    $$\langle \psi | PQ | \psi \rangle - \langle \psi | QP | \psi \rangle = \frac{\hbar}{i}\langle \psi | \psi \rangle$$
    Taking the trace is equivalent to summing up over all eigenfunctions, an infinite sum. So the right side will be
    $$\frac{\hbar}{i}\sum_{k=0}^{\infty} 1$$
    which is infinite. Using completeness on the left side, we get
    $$\sum_p \sum _{p'} \sum_q \sum_{q'} \langle \psi | p\rangle \langle p | P | p' \rangle \langle p' | q \rangle \langle q | Q | q' \rangle \langle q' | \psi \rangle$$
    minus an equally ugly looking expression for the other half of the commutator. Here's where I'm stuck. I'm guessing the left hand side ends up looking like ##\infty - \infty## (maybe because of the sum over the ##\langle p'|q \rangle## term?) and you can't simply cancel pairwise for some mathy reason. But I don't have the rigor to show that.

    Anyway, that's my first stab at it.
     
  13. Apr 26, 2016 #12

    micromass

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    Yes, that was my intention of course. I didn't post this as an insight since the OP is obviously copied directly from the Gieres paper. I know the Gieres paper is pretty famous among some here, this thread is not meant for them. But I hope some newcomers to this kind of QM-mathematics may find this useful.
     
  14. Apr 26, 2016 #13

    micromass

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    That is kind of the answer. The trace is only well-defined for some kind of operators, not for all operators. The trace is well-defined for finite matrices of course, but also for operators which are called trace-class. The ##P## and ##Q## operators cannot be trace-class as challenge 1 shows. In fact, they cannot even be continuous, but that's another story.
     
  15. Apr 26, 2016 #14
    In challenge 1 none of the operators involved have a finite trace.
     
  16. Apr 27, 2016 #15

    Zafa Pi

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    Will this do?
    Mr. ER can only put an even number of Xs in each row of a tic tac toe board. (one X or O per square)
    Mr. OC can only put an odd number of Xs in each column.
    They will face the following task: They will be separated and may not communicate. A judge will tell ER one of the 3 rows and tell OC one of the 3 columns. Neither hear what the other was told. They win if and only if ER fills out his row and OC fills out his column such that they agree at the intersection.
    Before being separated they can strategize and take anything with them.

    Is the following statement valid? It is impossible for them to guarantee a win since it's not possible to fill a tic tac toe board satisfying the requirements and thus no amount of strategizing will suffice.
    The paradox is that the answer is no.
    No mathematician I've given this problem to has found the correct answer.
     
  17. Apr 28, 2016 #16

    Demystifier

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    It will not do because I don't see that as a physical problem the solution of which requires a deeper understanding of mathematics.
     
  18. Apr 28, 2016 #17
    Obviously you must use entangled qubits. The following would work, if it can be done.

    Prepare 2 3x3 grids, for ER and OC, of EPR pairs.

    Entangle each row to have even number of X's (spin up, or |1>)
    Entangle each col to have odd number of X's (spin up, or |1>)

    Suppose even parity is |1>, odd is |0>

    Then, one way to prepare the two grids:

    Entangle the three in each col to make the first one the parity bit for the other two. (That means that each is parity bit for the other two, of course.) For each row, entangle the three in the row to make each NOT the parity bit for the other two.

    Now, after they are assigned one row or col by the Judge, ER and OC both read only the row or col they need to know, and it should work. Of course it would be no good if they read all nine of their half of the EPR pairs. They must read only their three.

    Can they (Quantum Computer workers) actually do that much entangling of qubits these days?

    Unfortunately I don't know how to accomplish this exactly (assuming it's possible at all). Use CNOT, Hadamard gates, no doubt. Ask a physicist.
     
  19. Apr 29, 2016 #18

    Zafa Pi

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    I find this problem amazing. There is no hint of QM lurking anywhere., as opposed to the Bell business. Any mathematician not versed in QM will always get it wrong.
    You're on the right track - google: quantum pseudo telepathy. The wiki choice is direct and clear.
     
  20. Apr 29, 2016 #19
    Thanks Zafa Pi, this is a great "paradox", with real pedagogical value.

    If you give them that keyword "telepathy" and a day to think about it, they'll always get it right! They can google lots of relevant hits.

    No doubt it's clear if you already know the answer; but with only that wiki page, Gauss himself couldn't figure it out. One non-trivial mistake: it says Alice and Bob need only one EPR pair - surely that's wrong, they need two? But there are plenty of papers to study; if I have questions I'll start a new thread.

    ----------------------------

    In this thread "paradox" means paradoxical to a student, not to an expert. With that in mind here's a "paradox".

    Suppose you can prepare electrons in the lab with identical momenta. Measure one electron's speed, get (say) 2 m/s. Measure another more accurately, get 2.3. Another even more accurately, get 2.34.

    The question is, if you keep measuring them more and more accurately, what value will you get for the speed, in the limit?

    There is a definite answer: you don't need any more information. And, it appears to violate at least three well-known laws of physics. That seems paradoxical, to a student.

    Note - with a nod to Demystifier - the resolution requires a deeper understanding of mathematics.

    See, for instance, "Principles of Quantum Mechanics" 4th edition, P.A.M Dirac, Section 69 "The motion of a free electron".
     
    Last edited: Apr 29, 2016
  21. Apr 29, 2016 #20

    Zafa Pi

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    You're correct on both accounts. However it wouldn't take long to bring Gauss up to speed on QM. And they do need two EPR pairs.

    You are also correct about your paradox - how can the accuracy of measurements have no limit?
     
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