Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Canonically conjugate quantities

  1. Apr 22, 2017 #1
    In a lecture on introductory quantum mechanics the teacher said that Heisenberg uncertainty priciple is applicable only to canonically conjugate physical quantities. What are these quantities?
     
  2. jcsd
  3. Apr 22, 2017 #2

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    Quantities A and B are canonically conjugate if their classical mechanical Poisson bracket is equal to unity. The most simple example is the x-coordinate of a particle and the corresponding momentum ##p_x##.
     
  4. Apr 22, 2017 #3
    I do not know what poisson bracket is
     
  5. Apr 22, 2017 #4

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    Saying that the Poisson bracket of ##x## and ##p_x## is unity is equivalent to saying that the quantum commutator ##[x,p_x ] = xp_x - p_x x## has value ##i\hbar##. That's actually how the position and momentum are defined in QM.

    Using that basic commutation relation, you can also show that ##\frac{1}{\sqrt{2}}(x + y)## and ##\frac{1}{\sqrt{2}}(p_x + p_y)## are canonically conjugate with each other, as are ##2x## and ##\frac{p_x}{2}##.
     
  6. Apr 22, 2017 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The uncertainty is very general and does not only apply to canonically conjugate pairs of observables (although in this case it becomes more simple). For any two observables ##A## and ##B##, represented by self-adjoint operators ##\hat{A}## and ##\hat{B}## one can derive the uncertainty relation
    $$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
    where the standard deviations and the average on the right-hand side are evaluated with an arbitrary pure or mixed state.
     
  7. Apr 22, 2017 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Have you skipped classical mechanics classes? Then bad idea to attend QM.
     
  8. Apr 23, 2017 #7
    I have completed 3 courses on classical mechanics but never came across terms like poisson bracket, Hamiltonian etc.
     
  9. Apr 23, 2017 #8

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    How can one do QT without knowing about Poisson brackets? I'm puzzled!
     
  10. Apr 23, 2017 #9
    How can one take 3 courses on classical mechanics without learning about Poisson brackets and Hamiltonians, that's a questiono0)
     
  11. Apr 23, 2017 #10

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Well, I know some professors who don't teach Poisson brackets nor the symplectic structure of phase space. What's also not taught any more is the Hamilton-Jacobi partial differential equation, which could provide Schrödinger's heuristics towards wave mechanics (which for me is the 2nd-best choice compared to the really simple approach via canonical quantization).
     
  12. Apr 23, 2017 #11
    Classical mechanics I have learnt till now -
    Kinematics in plane and space, kinetics, Rotational dynamics, waves and oscillations, Newtonian gravity, properties of materials (fluid dynamics and mechanical properties of materials )
    Throughout the courses we were never introduced to Hamiltonian formalism. Can someone suggest a good refrance which i can use to study Hamiltonian formalism ? As I'm a novice it should start from the very basics.
     
  13. Apr 23, 2017 #12
    I only attended the introductory lecture I'm not taking the course right now. But I'm willing to take the course in future. Please tell me what are all prerequisites for such course in general.
     
  14. Apr 23, 2017 #13

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    F. Scheck, Mechanics - From Newton's Laws to Deterministic Chaos, Springer Verlag (2010)
     
  15. Apr 23, 2017 #14

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

  16. Apr 23, 2017 #15
    Thank you people
     
  17. Apr 23, 2017 #16
  18. Apr 23, 2017 #17

    bhobba

    Staff: Mentor

    Thats understandable.

    First read Susskind for a gentle introduction:
    https://www.amazon.com/Theoretical-Minimum-Start-Doing-Physics/dp/0465075681

    Then read Landau:
    https://www.amazon.com/Mechanics-Third-Course-Theoretical-Physics/dp/0750628960

    Although not about PB's I do recommend going on and studying:
    https://www.amazon.com/Physics-Symmetry-Undergraduate-Lecture-Notes/dp/3319192000

    Its optional but puts it all in context using perhaps the most profound discovery of physics - at rock bottom its about symmetry. Intrigued - Landau uses it but the above book puts it in context with the rest of physics.

    Thanks
    Bill
     
    Last edited by a moderator: May 8, 2017
  19. Apr 23, 2017 #18

    bhobba

    Staff: Mentor

    I would seem hard. But knowing some physics programs its quite possible - sad really - but it happens.

    Thanks
    Bill
     
  20. Apr 23, 2017 #19

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    In some ways, it seems to me that the classical Poisson brackets are more mysterious than the quantum commutators. It's clear that various operators on wave functions don't commute, but the fact that the Poisson brackets are anti-symmetric is a little subtle.

    In (one-dimensional) Hamiltonian dynamics, if you write the Hamiltonian as a function [itex]H(p,x)[/itex] of momentum [itex]p[/itex] and position [itex]x[/itex], then this gives rise to the equations of motion:

    [itex]\frac{dx}{dt} = \frac{\partial H}{\partial p}[/itex]
    [itex]\frac{dp}{dt} = - \frac{\partial H}{\partial x}[/itex]

    That minus sign in the second equation is the source of the antisymmetry of the Poisson brackets. When you write [itex]H = K + V[/itex] where [itex]K[/itex] is the kinetic energy and [itex]V[/itex] is the potential energy, then the minus sign is reflected in the fact that [itex]\frac{dp}{dt} = - \frac{\partial V}{\partial x}[/itex]. Force is the negative of the derivative of the potential energy.

    Anyway, with the minus sign in the equations of motion, you can write for any function [itex]f(p,x)[/itex] of position and momentum:

    [itex]\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial p} \frac{dp}{dt}[/itex]
    [itex] = \frac{\partial f}{\partial x} \frac{\partial H}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial H}{\partial x}[/itex]
    [itex] \equiv \{ f , H \}[/itex] (the definition of the poisson bracket of [itex]f[/itex] and [itex]H[/itex])

    It's a little mysterious as to why that's an important concept in classical mechanics. But the most commonly used examples are:

    [itex]\frac{d}{dt} f(x,p) = \{f, H \}[/itex]
    [itex]\{x, p \} = 1[/itex]

    Then the generalization to multiple dimensions is [itex]\{ A, B \} = \sum_j \frac{\partial A}{\partial x^j} \frac{\partial B}{\partial p^j} - \frac{\partial A}{\partial p^j} \frac{\partial B}{\partial x^j}[/itex], which leads to another famous example:

    [itex]\{L_x, L_y\} = L_z[/itex] (as well as cyclic permutations)

    where [itex]L_x, L_y, L_z[/itex] are components of the angular momentum.
     
  21. Apr 23, 2017 #20

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    The Poisson brackets are important, because they admit a very elegant formulation of symmetry principles in classical mechanics (and also classical field theory, but that's another story). It directly makes the natural structure of the study of Lie groups in terms of their "infinitesimal version", i.e., its Lie algebra available for classical mechanics. The Poisson bracket is a Lie bracket and at the same time a derivation. It provides a symplectic structure to phase space and it allows to define canonical transformations (aka symplectomorphisms) in terms of generating functions and thus admits the definition of symmetries in a very easy way. It turns out that Noether's theorem then is most easily formulated as: "Each generator of a symmetry transformation is conserved along the trajectories of the system and any conserved quantity is the generator of a symmetry of the system."
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted