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Canonical momentum in QM

  1. Nov 18, 2004 #1
    Why is it canonical momentum that is replaced by the momentum operator in quantum mechanics? What is the physical significance of canonical momentum anyway?
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  3. Nov 18, 2004 #2


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    Basically,QM is nothing but a quantized version of Hamiltonian classical mechanics,in which the phase space variables are the generalized coordinates and momenta.If your simple particle system (implied by classical mechanics of point particles) is non degenerate (that is the symplectic structure is still the Poisson paranthesis),then the quantization is simple,classical obervables become densly defined self adjoint operators which must obbey the quantized version of the fundamental Poisson paranthesis on [tex]\Gamma [/tex ],which are the fundamental graduaded commuting relations.
  4. Nov 19, 2004 #3
    But the Schrodinger equation is sometimes written as

    [tex]i\hbar\frac{\partial\Psi}{\partial t}=\frac{\hat{p}^2}{2m}\Psi+V\Psi[/tex]

    but here this can only be correct when the momentum and cononical momentum are the same, right? I mean the expression on the right is recognisable as the calssical hamiltonian with p the (normal) momentum. And yet it is the canonical momentum that is substituted with the momentum operator? Does this mean the above is only correct in cases where the hamiltonian is [itex]p^2/2m+V[/itex] with p the canonical momentum?
  5. Nov 19, 2004 #4


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    The reason is that the commutation relations in QM have to correspond to the Poisson brackets in classical mechanics in order to find back the correspondence principle.

    1/(i hbar) [ A , B]_qm = {A_cl, B_cl}_PB

  6. Nov 19, 2004 #5


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    Normally, the hamiltonian (in classical mechanics) is expressed in the generalized coordinates q and the *canonical* momenta p. So there is a natural transposition in the QM hamiltonian. If you deduced your classical hamiltonian from the classical lagrangian, you automatically get the right expression. If you want to write down the classical hamiltonian directly (total energy) then you have to be careful of course: the total energy will probably be expressed using physical momenta, and then you have to substitute for the expression using the canonical momenta.
  7. Nov 19, 2004 #6
    So the Schrodinger equation should be:

    [tex]i\hbar\frac{\partial\Psi}{\partial t}=\frac{\hat{P}^2}{2m}\Psi+V\Psi[/tex]

    With [itex]P[/itex] the expression for the momentum in terms of the canonical momentum. And [itex]\hat{P}[/itex] this expression with the *canonical* momentum substituted by [itex]-i\hbar\nabla[/itex]?!
  8. Nov 20, 2004 #7
    I read in White (Quantum Mechanics) that the "physical" momentum p equals the "canonical" momentum p_can plus the "electromagnetic radiation" momentum which is the electronic charge times the "magnetic" vector potential qA. I guessed the adjectives from your posts when I noticed that White used the operator p-qA when there is electromagnetic radiation-electron interaction.
    So here are my two questions:
    1- Could someone experienced tell me the difference in the physical meaning (NOT in the use in equations) between the two momenta?
    2- Does it really make any difference in the following results ,e.g. the absorption rate become the gain rate, if the sign of qA were positive?
    Finally a younger fellow instructor asked me a related question that I could not get a conclusive answer so try to give me your best answers to this:
    Are the operators of the quantum mechanics, e.g. the momentum operator, are unique and why? I appreciate if you give link to a formal proof if you can.

    Mahmoud-SifedDin A. Taha
    MSc. in salvoing Schrodinger equation for QW Semiconductor Laser!
  9. Nov 21, 2004 #8
    I guess a good reason for the "canonical momentum" to be the cross-point between CM and QM is invariance.

    Start on the CM side. You may choose any coordinates (q's) you want, determine the corresponding momenta and built the hamilton equations that determine the trajectories. The hamiltonian formulation of CM applies to any system of coordinates (q's): this is invariance. The translation to QM is always obtained from the correspondance principle ([q,p]=ihb), whatever the coordinates you have choosen. Using the canonical momenta is necessary for this invariance to be recovered on the QM side.

    Note that the correspondance principle is the underlying physics you are looking for.
    It is related, as you know, to the uncertainty principle.
    Try to measure cartesian coordinates of a particle, and the uncertainty principle will relate the precision of your measurement to the precision of the linear momentum measurement. ​
    Go now to cylindrical coordinates. Try to measure the angular position of the particle, and the uncertainty principle will relate the precision of your measurement to the precision of the angular momentum measurement, which is the canonical momentum associated to angular position.​
    If you have more courage than myself for calculations, you may prove -for example- that the uncertainty principle in cartesian coordinates implies the uncertainty principle in cylindrical coordinates with the cylindrical canonical momenta. And more generally you could prove that it implies the uncertainty principle for any q's and their correponding canonical momenta.

    Invariance maybe helpful to show the consequences of a symmetry. Indeed, from the hamiltonian formulation it is easy to see that a symmetry implies a conserved quantity (Noether theorem). For example, if a system has a rotational symmetry, then the hamiltonian does not depend on the angular coordinate, and the angular momentum will be a conserved quantity. The correspondance principle ensures that this also applies in QM.

    The example above shows you that a simple 1-particle system can be studied with different kind of coordinates and their associated canonical momenta. For a system which includes electromagnetic radiation, it will not be a surprise to you that a canonical momentum may be more complicated. For example, this additional complication will simply ensure the momentum conservation when the system has a translation symmetry. The momentum will then include particle and field.
  10. Nov 21, 2004 #9


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    "Are the operators of the quantum mechanics, e.g. the momentum operator, are unique and why? I appreciate if you give link to a formal proof if you can."

    This is a good question, and slightly hard to discuss in this forum, theres quite a bit too it (and you have to be more specific in what you want).. For the specific case of the non relativistic momentum operator for a particle in say R^3, responsible for translations, I think you will find it easy to show (try it). The idea is to send classical variables into quantum operators via a correspondance principle, but they naively don't look unique (rather they seem to be unique only up to some commutation relation).

    The Stone Von Neumann theorem guarentees that the commutation relation in quantum mechanics is unique (given a suitable set of reasonable constraints and domain.. unfortunately not constrictive enough). That is, given two operators that satisfy an uncertainty principle, you will not be able to find another 2 that satisfy the same relation, and mantain selfadjointness + the correspondance principle + all the symmetry principles of the operators.

    Where it starts to get really frustrating is when you start moving into systems with infinite degrees of freedom. The Stone Von Neumann theorem breaks down, and you have to consider various algebraic extensions of it (C* algebras) or supersymmetric extensions.

    Ok, maybe im saying this wrong (its late here and im nursing a hangover), but for simple systems satisfying a bunch of good properties, you can show specific operators are unique (at least up to conjugacy)
  11. Nov 29, 2004 #10
    To find the Hamiltonian Operator, we replace canonical momentum by the momentum operator, coordinate by position operator.

    How do we find the operator for a variable like [tex]xp_x[/tex]? coordinate multiplied by momentum.

    We do not qeustion about the physical meaning of the variable at this moment. If we simply do the substitution: [tex]\hat{x}\hat{p_x}[/tex]

    There are two problems:
    (1) How do we define multiplication of operators? In the expression, x-operator is multiplied by p-operator, but only composition of operators is well-defined.

    (2) The operator obtained is not Hermitian.
  12. Nov 29, 2004 #11


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    If you've read carefully the postulate of quantization,you may have seen that,when dealing with observables described by noncommuting selfadjoint operators,you need to symmetrize wrt to these observables all classical expressions containing products of these observables.
    In the simple case of the product between one coordinate and its corresponding momentum (at quantum level we know that their corresponding selfadjoint operators don't commute),you make at classical level the substitution:
    [tex] xp_{x} \rightarrow \frac{1}{2} (xp_{x}+p_{x}x) [/tex] and to this apply the second postulate of QM,to get the correct operator,who needs,above all,to be selfadjoint.
    What do you mean,define the multiplication of operators??The composition and the multiplication are the same thing,but care needs to be taken with unbounded operators.
    Basically the definiton is as follows (hope i can remember it properly):
    Let A and B be two linear operators applying the same separable Hilbert space H onto himself:
    [tex] A:H\rightarrow H;B:H\rightarrow H [/tex].
    Assuming we're dealing with bounded operators,the product of the operators A and B is himself a bounded linear operator applying the same separable Hilbert space onto himself.This definiton is very particular.In general,bounded operators on separable Hilbert spaces are characterized by a domain (a subset of vectors from H) and an image (a subset of vectors from H,not necessarily the same vectors like in the domain).
    Let's pick two arbitrary bounded linear operators A and B.We know that:
    [tex] A:D_{A}\rightarrow Im_{A} ;B:D_{B}\rightarrow Im_{B} [/tex],where the notations used i hope to be obvious.
    In the case that [tex] Im_{B} \subseteq D_{A} [/tex],then we can define the bounded linear operator AB,with the domain the entire domain of B and the image,the entire image of A.
    Note,that this is the most general definition for the product/composition of linear bounded operators in Hilbert spaces.
    QM uses separable QM spaces and imposes the operators describing physically measurable quantities (observables) to be densly-defined selfadjoint linear operators (not necessarily bounded,most operators are not) as stated in the second postulate (the postulate of quantization) .

  13. Nov 29, 2004 #12
    The momentum operator is not unique in QM. For example in the coordinate basis <x|P=-ih d/dx+f(x), where f is arbitrary. So long as the commutation relation [X,P]=ih is satisfied, then you're okay. Your wavefunctions will be different, but the probabilities will be the same for all observables.

    As with regards to the P in the Schrodinger eqn., my book (Shankar) suggests to write the Schrodinger equation in cartesian coordinates, and then change variables. That's the way he got orbital angular momentum about an axis is: -ih d/d@, where @ is the azimuth angle. I'm sure there are plenty of expressions too for angular momentum, as long as they obey [L_x, L_y]=ih L_z, which is just a property of rotations (in this universe at least).

    Shankar does mention that there is a way to directly quantize into different coordinate systems, but says it's complicated and usually not worth it.

    So basically I don't know :grumpy:
  14. Nov 30, 2004 #13

    When you said the 2nd postulate of QM, I assume you meant the postulate: every operator that represents a physical variable must be Hermitian? (which book you referred to? The 2nd postulate on my book says only the eigenvalue equation.)

    That postulate does not give any hint "how to make it Hermitian"
    The way you suggested [tex] xp_{x} \rightarrow \frac{1}{2} (xp_{x}+p_{x}x) [/tex]
    is only one of the countless number of ways to make it Hermitian. How do you ensure that the operator you obtained is the correct one?
  15. Nov 30, 2004 #14
    I appreciate your debate in answering this question but these two questions seem to be untouched would someone give a response on them please

  16. Nov 30, 2004 #15
    You do't need QM to answer these questions...

    1. The canonical momentum is like the mechanical momentum but with a little bit added to take into account the electromagnetic field. This is usually called "A" and is a four vector. It is sometimes referred to as the "4-vector potential"

    This is so that things like momentum conservation can apply - if I take a particle and fire it into an electric field it will accelerate, and the particle's momentum is not constant. But its canonical momentum will be constant (I think).

    Well, also there are some very technical reasons why we talk about canonical momentum. For example, they allow us to derive the Lorentz forces classically using Hamiltonian mechanics. See the excellent text Goldstein, "Classical Mechanics" for more info. I ordered a copy from Amazon but it never arrived, otherwise I would be able to give a better reply :)

    2. Well, magnetic field and electric field are related to "A". If A's sign is reversed, this is like reversing the sign of the electric field and magnetic field.

    Being precise,
    E = grad (A_0) and
    B = curl (A_vector).
    Here A is a four vector with components (A_0, A_vector). Try an advanced book on electromagnetism for more on this stuff, although I don't know a good one to recommend.
  17. Nov 30, 2004 #16


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    I believe that my tendency towards mathematical rigurosity made me less understood. :wink: In my prior and future posts,when i speak about principle
    number "x" from QM of nondegenerate systems (the one usually taught in school) i mean the postulates from my QM course taken home,and splendidly taught by Mr.Assoc.Proff,Ph.D,Lucian Saliu.In that course the axiomatical part of QM in the classical Copenhagen formulation comprises 5 principles,the last one being the famous "state vector collapsing"principle.
    I know that the course taken home (and apparently mastered by myself :approve: )is not perfect,it lacks some chapters that belong to the normal discussion of Q phenomenology,but what has been taught is very rigurous and well organized,especiallly the axiomatical core of QM.The reasons for the deficits is less time for QM courses at my hometown faculty thans it used to be in the past.The reduction from 84->56 hours for the course meant that,in order to keep the essential first part of the course untouched,some chapters (like Relativistic QM,Spin,Scattering)needed to be erased.Actually that was a good thing for my collegues... :tongue2:
    What kind of book do u use?
    My teacher reccomended me some books for further reading(on those 3 chapters actually),but i believe your book is not among them.
    The classical symbol for a Hamiltonian observable of a nondegenerate sysytem is uniquely determined by the method of symmetrizing wrt to classical observables whose corresponding QM densly-defined selfadjoint linear operators do not commute.And that's a fact stated in the comments made by the teacher to the stated version of the second postulate.

    Last edited: Nov 30, 2004
  18. Nov 30, 2004 #17


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    "The momentum operator is not unique in QM. For example in the coordinate basis <x|P=-ih d/dx+f(x), where f is arbitrary. So long as the commutation relation [X,P]=ih is satisfied, then you're okay. Your wavefunctions will be different, but the probabilities will be the same for all observables."

    Which is fine, your just shifting your system around. In a coordinate invariant language (say geometric quantization) the operator is still unique (in a conjugacy class sense).

    So like I said, the limiting factor is the commutation relations, which are unique by Stone Von Neumann, *nearly always*.

    Note, *nearly always* is subtle. If your classical symplectic space is topologically nontrivial (say it has nonvanishing homology classes), the correspondance can and will break down, and you have to do things by hand. Also, there can be dirac constraints imposed on the system.
  19. Dec 1, 2004 #18
    >The classical symbol for a Hamiltonian observable of a nondegenerate sysytem is
    >uniquely determined by the method of symmetrizing wrt to classical observables
    >whose corresponding QM densly-defined selfadjoint linear operators do not commute.
    >And that's a fact stated in the comments made by the teacher to the stated version
    >of the second postulate.


    I will give you a counter-example :) :)
    Now I am reading something about Weyl Ordering.
  20. Dec 1, 2004 #19


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    Well,give a counterexample!!

  21. Dec 1, 2004 #20
    Your confusion seems to rest in what you think Schrodinger's equation is, i.e.

    [tex]i\hbar\frac{\partial\Psi}{\partial t}=\frac{\hat{p}^2}{2m}\Psi+V\Psi[/tex]

    That isn't Schrodinger's equation exactly. That is an example of Schrodinger's equation when the canonical momentum is the linear mechanical momentum. Schrodinger's equation is

    [tex]\hbar \frac{d}{dt}|\Psi(t)> = H(t)|Psi(t)>[/tex]

    The momentum that appears in the Hamiltonian is the canonical momentum. I.e. the source of the Hamiltonian comes from an integral of motion in classical mechanics. If (after any dF(x,y,z,t)/dt is dropped) the Lagrangian is not an explicit function of time then the quantity h defined as

    [tex] h(x,v,t) = \pi_k v^k - L [/tex]

    where vk = dqk/dt are the components of the generalized velocity and qk are generalized coordiinates and

    [tex] \pi_k = \frac{\partial L}{\partial v^k}[/tex]

    are the components of the canonical momentum (aka generalized mometum). It can be shown that

    [tex]\frac{dh}{dt} = -\frac{\partial L}{\partial t} [/tex]


    And, of course, when h is expressed in terms of the canonical momentum and the generalized coordinates then its called the Hamiltonian.

    When the system is a charged particle moving in an EM field then

    [tex]\pi_k = p_k + qA_k[/tex]

    pk are the components of linear mechanical momentum Similarly the canonical 4-momentum is defined as

    [tex]\PI_{\mu} = \frac{\partial L}{\partial U_{mu}}[/tex]

    and the canonical 4-momentum for a charged particle in an EM field is

    [tex]\PI_{\mu} = P_{\mu} + qA_{\mu}[/tex]

    where [itex]P_{\mu} = m_0 U_{\mu}[/itex].

    Last edited: Dec 1, 2004
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