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Featured I Hamiltonian in Quantum vs Classical

  1. Jul 8, 2017 #1
    The Hamiltonian in classical mechanics is not always equal to the total energy of the system. I believe this is only true if there is only a potential field and no vector potential. However, in quantum mechanics for a particle in an EM field, even if a vector potential is used the energy operator E -> H. Is this an inconsistency?

    While H is not always equal to E in classical mechanics, H is still always regarded as the operator for E when there are not exclusively scalar potentials in QM?

    Thanks.
     
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  3. Jul 8, 2017 #2
    Bumping it with a little addition. If there isn't a scalar potential energy term at all, (so no conservative field) how is total energy even defined in quantum mechanics (at least non-relativistic QM) ?

    Is it simply the Eigenvalue of the Hamiltonian? (if it isn't K + U? ) Is it simply just the relative state levels between which photons can be absorbed? Obviously energy is usually a relative concept defined up to a constant, but here it seems that energy is simply defined as being the eigenvalue of the Hamiltonian if it isn't equal to K+U. Just like in classical mechanics where the Hamiltonian isn't always equal to the total energy K+U.
     
  4. Jul 10, 2017 #3

    PeterDonis

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    Can you give a reference? You might be confusing energy with momentum.

    Energy is an observable, so in QM it's a self-adjoint operator, like every observable. The particular operator for the energy is the Hamiltonian operator.
     
  5. Jul 10, 2017 #4

    vanhees71

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    Do you have a concrete example, where the Hamiltonian is not the total energy of the system?

    In #1 you claimed that the Hamiltonian for a particle in an electromagnetic field is not the total energy, but that's obviously not true. For simplicity let's consider the non-relativistic approximation for a single particle in an electromagnetic field. We start from the Lagrangian
    $$L=\frac{m}{2} \dot{\vec{x}^2}-q \phi(t,\vec{x}) + \frac{q}{c} \dot{\vec{x}} \cdot \vec{A}(t,\vec{x}),$$
    where ##\phi## is the scalar and ##\vec{A}## the vector potential of the electromagnetic field.

    The canonical momenta are
    $$\vec{p}_{\text{can}}=\frac{\partial L}{\partial \dot{\vec{x}}}=m \dot{\vec{x}}+\frac{q}{c} \vec{A}=\vec{p}_{\text{mech}} + \frac{q}{c} \vec{A}.$$
    So in this case the canonical momentum is not the mechanical momentum. Maybe that's the cause of your confusion.

    The Hamiltonian is
    $$H=\dot{\vec{x}} \cdot \vec{p}_{\text{can}}-L=\frac{m}{2} \dot{\vec{x}}^2+q \phi(t,\vec{x}),$$
    which is the total energy of the system.

    Of course for the Hamilton formalism you have to express ##\dot{\vec{x}}## in terms of the canonical (!) momentum, leading to
    $$H=\frac{1}{2m} \left [\vec{p}_{\text{can}}-\frac{q}{c} \vec{A}(t,\vec{x}) \right]^2+q \phi(t,\vec{x}).$$
    Since in the square brackets you have ##\vec{p}_{\text{mech}}##, it's still the standard expression for the total energy.

    That's also clear from the equations of motion, following from the Lagrange or Hamilton equations:
    $$m \ddot{\vec{x}}=q \left (\vec{E} + \frac{1}{c} \dot{\vec{x}} \times \vec{B} \right ),$$
    where
    $$\vec{E}=-\vec{\nabla} \phi-\frac{1}{c} \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{A}.$$
    When using the heuristics of canonical quantization in the wave-mechanics (position-space) representation, it's ##p_{\text{can}}## which is substituted with $$\hat{\vec{p}}_{\text{can}}=-\mathrm{i} \hbar \vec{\nabla},$$
    and thus you get the Schrödinger equation in the form
    $$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=-\frac{1}{2m} \left [\vec{\nabla}-\frac{q}{c} \vec{A}(t,x) \right ]^2 \psi + q \phi(t,\vec{x}) \psi(t,\vec{x}).$$
    The Hamiltonian is still the operator representing the total energy of the particle.
     
  6. Jul 10, 2017 #5

    I will look for an example and update when I get the chance to do so. I just have read this statement on other parts of the Internet, but I attempt to provide a source to answer this question.


    However, in terms of what you said. I am not very familiar with canonical momentum, but does that classically capture the total energy of the system? In a classical syste the total energy of a charge in an EM field should be the energy density of the E and B field integrated over all space + kinetic energy of the particle. (And also rest mass if you want to go relativistic)

    Since the canonical momentum isn't simply the momentum of the particle, does it somehow bring in the energy in the magnetic and electric field?
     
    Last edited: Jul 10, 2017
  7. Jul 10, 2017 #6

    hilbert2

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    I guess it's possible to add a term like ##H'(t) = Ae^{Bt}##, where ##A## and ##B## are constants and ##t## is the time variable, in the Hamiltonian function to get an explicit time dependence that doesn't affect any actual dynamics. Then it wouldn't really make sense to say that the Hamiltonian is equal to any total energy that is defined in a practical manner.
     
  8. Jul 10, 2017 #7
    Here is a thread that talks about this issue. https://physics.stackexchange.com/q...ian-of-a-system-not-equal-to-its-total-energy

    I do understand your derivation now, thought would seem to work for a static, conservative electric field & static magnetic field. If the EM field included some non-conservative electric field then it would break down and the Hamiltonian wouldn't equal to the total energy.

    So - it seems that while there are times in classical mechanics where the Hamiltonian isn't equal to the total energy, we still always take the eigenvalue of the quantum Hamiltonian to be defined as the energy simply as a postulate. Obviously there is no requirement for the QM to satisfy this classical analogy, but it would have been nice :)

    In this case, I guess it doesn't really make sense to talk about energy as having any individual components in a classical sense like the sum of kinetic and potential energy, but the energies of the quantum Hamiltonian are simply "some quantity" DEFINED as the eigenvalue of this problem, in which the differences between photon frequency jumps can be calculated from. And in cases where the classical hamiltonian IS equal to the total classical energy, it just so happens that these two "energies" match.

    At least that's what I seem to have gathered.
     
    Last edited: Jul 10, 2017
  9. Jul 10, 2017 #8

    PeterDonis

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    @Electric to be please do not quote the entire post you're responding to. Just quote the specific parts you're responding to. If you highlight a particular part of a post, and use the "reply" popup, it will appear quoted in the edit box.
     
  10. Jul 11, 2017 #9

    vanhees71

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    No, in the static case, i.e., for time-indpendent ##\vec{E}## and ##\vec{B}## the total energy of the particle in the external field is ##\mathcal{E}=T+V=m \vec{v}^2/2+q \phi##.

    This you get unambiguously from the equations of motion
    $$m \ddot{\vec{x}}=q \left (\vec{E}+\frac{\dot{\vec{x}}}{c} \times \vec{B} \right)$$
    Multiplying with ##\dot{\vec{x}}## gives
    $$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \dot{\vec{x}}^2 \right)=-q \frac{\mathrm{d} \phi}{\mathrm{d} t},$$
    if we have static fields, i.e., ##\vec{E}=-\vec{\nabla} \phi##, and then the Hamiltonian is the total energy.

    For time dependent fields, you are right that ##H## is not the total energy of the system, and because then ##\vec{E}=-\vec{\nabla} \phi-\partial_t \vec{A}## is no longer a potential field anymore, you cannot derive the force from a potential, and it's not clear how to define total energy. You can only define the work along the trajectory of the particle, as usual for non-conservative forces! So I was inaccurate in claiming that also here ##H=E##! In the most general case it's even difficult to define what total energy means.
     
  11. Jul 11, 2017 #10
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    You can define the total energy as ##K+V## if you include the field energy term ##\textbf{A}\cdot\textbf{v}## in your definition of the potential energy. In terms of the work function ##U(q_i,\dot{q}_i)##, the Lorentz force emerges as a result of
    $$F_k=\frac{\partial U}{\partial q_k}-\frac{d}{dt}\frac{\partial U}{\partial\dot{q}_k}$$.
    If you define $$V=\sum_i\frac{\partial U}{\partial\dot{q}_i}\dot{q}_i -U$$ then the total energy is potential plus kinetic.
     
  12. Jul 11, 2017 #11

    atyy

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    Could one define the energy as the conserved quantity due to time reversal symmetry in Noether's theorem?

    Then the question is whether H = E even in the presence of a vector potential. I found an interesting discussion in http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf. It seems that there is indeed a problem with potentials. However it turns out the problem goes away, unless there is explicit time dependence in the Hamiltonian.
     
  13. Jul 11, 2017 #12
    When you make time a dynamical variable in phase space and treat it as a variational variable, the canonical momentum of the time coordinate is the Hamiltonian. Does this answer your question?

    EDIT: See Variational principles by "Lanczos"
     
  14. Jul 11, 2017 #13
    For a classical system - a charge interacting with an electric field will always have the total energy conserved. However, instead of attributing the potential energy to saw energy associated with the charge, the energy is store in the field. (Energy density is proportional to B^2 and E^2)

    I believe there are lagrangian formalisms that cover the evolution of the electromagnetic field with a certain amount of charges. However, you're right when a single particle Hamiltonian is introduced with the fields essentially as a set parameter that aren't influenced by the motions of the particle it isn't exacty possible to define total energy. As I stated earlier, this case was my source of confusion in quantizing the Hamiltonian and calling it the operator for energy. Like I said above, it isn't necessary for this to hold true though. Energy is simply defined as the eigenvalues of the Hamiltonian, not as K+U.
     
  15. Jul 11, 2017 #14

    vanhees71

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    Energy is the conserved quantity due to time-translation symmetry. If, however, the external fields are explicitly time-dependent that's no longer a symmet since for this the condition is ##(\partial_t L)_{\text{expl}}=0##.
     
  16. Jul 11, 2017 #15

    vanhees71

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    That's a very tricky issue. A fully self-consistent classical Hamiltonian for classical charged point particles + field, including radiation reaction, doesn't exist. That's the limit, where the classically finally breaks down, and the correct description is QED, which is at least well defined order by order in perturbation theory.
     
  17. Jul 11, 2017 #16
    Well I think there is some form of "Hamiltonian field theory" for classical fields that can describe evolution of fields and particles within it. Though if I had to guess this would probably only work for relatively simple configurations, and would otherwise have to be numerically solvable? I'm not an expert in Lagrangian mechanics. http://www.phys.lsu.edu/~jarrell/COURSES/ELECTRODYNAMICS/Chap12/chap12.pdf

    This wouldn't be used for Schrodinger equation like QM like you said, but would be used (and is) in QED.
     
    Last edited: Jul 11, 2017
  18. Jul 11, 2017 #17

    atyy

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    Have you seen http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture18.pdf ? It discusses that H = E even if there is a vector potential.
     
  19. Jul 11, 2017 #18
    Notice however, they do assume that a scalar electric potential can at least be written (even if it happens to be zero). However, if you have a non-conservative e-field (so time varying EM field) then this concept breaks down. That link also talks about that (must be time independent).
     
  20. Jul 12, 2017 #19

    vanhees71

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    It depends on, how you define "Energy". For a particle in an explicity time-dependent electromagnetic field ##H## is not the expression occuring in the work-energy theorem. Take my calculation from #9 and apply it to this more general case of non-static fields. Then the work-energy theorem reads, using ##T=m \dot{\vec{x}}^2/2##
    $$\dot{T}=q \dot{\vec{x}} \cdot \vec{E}=q \dot{\vec{x}} (-\vec{\nabla} \phi -\partial_t \vec{A})= -q (\dot{\phi}-\partial_t \phi) -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
    Or
    $$\frac{\mathrm{d}}{\mathrm{d} t} (T+q \phi)=q (\partial_t \phi - \dot{\vec{x}} \cdot \partial_t \vec{A}).$$
    Of course you can derive this also from
    $$\frac{\mathrm{d} H}{\mathrm{d} t}=\frac{\partial H}{\partial t}=-\frac{\partial L}{\partial t} = q \partial_t \phi -q \dot{\vec{x}} \cdot \partial_t \vec{A}.$$
    Note that all this holds only for the trajectory of the particle, i.e., the solution of the equations of motion!
     
  21. Jul 15, 2017 #20

    atyy

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    In the case of explicit time-dependence, does the Noether's theorem energy still exist? My intuition is that explicit time-dependence breaks time translation, so the Noether theorem would fail, and energy would not be defined.
     
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