The QH QM QA thread.

Discussion in 'Physics & Math' started by QuarkHead, Mar 22, 2008.

  1. QuarkHead Remedial Math Student Valued Senior Member

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    Mmm.. I think the above may be ever so slightly circular. Let's see.

    Meantime, a pal linked me to this set of lectures, which, on first glance look like a lot of fun.
     
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  3. temur man of no words Registered Senior Member

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    Smoking in class?

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  5. QuarkHead Remedial Math Student Valued Senior Member

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    It seems like the videos were shot in the mid 1970's, when I suppose that smoking was more acceptable? Dunno. Nevertheless, he lived till his 70's, was a much loved and revered teacher at Harvard, and a respectable contributor to QFT.

    Anyway, fukkit - be a rebel, even if it endangers your life, say I.
     
    Last edited: Jun 3, 2008
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  7. temur man of no words Registered Senior Member

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    So life was good back then.
     
  8. QuarkHead Remedial Math Student Valued Senior Member

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    Anyway, it only remains for me figure out why momentum is an operator. Pfft, easy!

    It is easily shown that the set of operators acting on a vector space is itself a vector space. This implies that if an operator, here the Hamiltonian, can be written as the sum of two or more entities, then these entities must themselves be operators. Hence
    is an operator, the kinetic energy operator.

    Vector space theory also says that if any vector, here an operator, can be written as the scalar multiple of another entity, then that entity must itself be an operator. So noting that \(\frac{1}{2m}\) is a scalar multiplier, then \(mv = \frac{\hbar}{i} \nabla ^2 \equiv \vec{p} \) must be an operator, the momentum operator.

    Am I bold enough to say that, since potential energy depends only on position, then I may call it a "position operator"? Dunno. Anybody?
     
  9. D H Some other guy Valued Senior Member

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    QH, what you wrote is not correct. First, look at the units. The reduced Plank's constant, \(\hbar\), has units of length[sup]2[/sup]*mass/time and the Laplacian, \(\nabla^2\), has units of length[sup]-2[/sup]. Thus you expression has units of mass/time.

    In post 180, you wrote

    You cannot deduce the latter expression from the former expression. Eliminating the common denominator from the left and right hand sides of the first expression yields \(||{\mathbf p}||^2 = - \hbar ^2 \nabla ^2 \). This is of course a scalar expression. The desired expression is vectorial, not scalar.

    Taking the square root of a scalar expression to yield a vector expression requires pulling a rabbit out of the hat. Some physicists like to pull rabbits out of the hat. Even though I live in Texas and the physicists here do wear ten gallon hats, taking the square root of the Laplacian operator \(\nabla^2\) to yield the gradient operator is IMHO, an extremely big rabbit to pull out of a hat.

    Anyhow, nothing up my sleeve, taking the square root of both sides of \(||{\mathbf p}||^2 = - \hbar ^2 \nabla ^2\) yields \(\mathbf p = \pm i \hbar \nabla\).

    A better approach is to use a vectorial approach throughout. This will remove the need for a ten gallon hat and will resolve the sign ambiguity problem.
     
  10. QuarkHead Remedial Math Student Valued Senior Member

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    D H: I thank you for that; the fact that I was wrong comes as no great surprise to me, and, no doubt, to the world at large.

    However, lemme say a coupla things in semi-mitigation.

    First in asserting that \(mv = -i\hbar \nabla ^2\) I was copy/pasting from an earlier equality, and neglected to remove the exponent. This was wrong of me.

    Now, the notation for the Laplacian \(\nabla ^2\) I am accustomed to is pretty nasty; operators, differential or otherwise, don't multiply, so, as you say, taking square roots is wrong. {Edit: this is not quite right; operators do multiply, but it's matrix multiplication. So the square root of self multiplication by an operator is not defined.}

    I was aware of this, and wasn't intending to imply that I could do this.

    Let's try again with an alternative notation. Define the Laplacian \(\Delta\) to be \(\Delta\phi=\nabla \cdot \nabla \phi\) for scalar \(\phi\).

    Now recognizing this is scalar (since grad (scalar) = vector, div (vector) = scalar), I reasoned earlier that, since \(p=mv\) is vector, and \(p^2\) is scalar, then by taking the square root of \(p^2\), which is legitimate, I recover the vector \(p = mv\).

    Then with a flourishing hand-wave, I thought I could just decompose the Laplacian in such a way that I was left with a vector on the other side, which is surely \(-i \hbar \nabla\), by the above?

    Let me remind you, and any one else reading, that I have absolutely no pride invested in this subject; I am here to learn. No amount of ridicule or correction can offend me
     
    Last edited: Jun 4, 2008
  11. temur man of no words Registered Senior Member

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    I think the things should go the other way: first find what is the momentum operator and then define the Hamiltonian.
     
  12. QuarkHead Remedial Math Student Valued Senior Member

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    OK, then. How do I do that? Recall, I am trying to derive the momentum operator! I had thought, given various hints, that I could get there via the Hamiltonian. It seems you tough guys think this is a wrong approach

    I am cool with that, but I don't really see another entry point, given my level of knowledge
     
  13. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    I've been trying to do some reading so I can help you out with this problem, QH. Oh man, am I ever feeling swamped right now. I'm reading through Goldstein's "Classical Mechanics" and Dirac's "Principles of Quantum Mechanics", some of the most popular books on this topic, and even then the material feels sorely lacking. I'm still working on it and trying to do my utmost to avoid reading these books from start to finish, but I might just have to do that if I can't find what I'm looking for. Sucks to be away from university at a time like this when I would have so many people I could ask.

    Anyhow, I do believe there is a method of deriving momentum as an operator from Hamiltonian mechanics, with the additional assumption of the correspondence principle connecting classical to quantum mechanics. In the formalism of canonical transformations, it's easy to show that using momentum as the generating function of an inifinitesimal canonical transformation produces an infinitesimal translation in coordinates, so they call it the generator of translations. Classically (I'm working in 1 dimension for simplicity), using the Poisson bracket formalism, you get \(f(x+dx)=f(x)+dx\{f,p\}=f(x)+dx\frac{\partial f}{\partial x}\), which works out as expected. As you probably know, Poisson brackets correspond to quantum commutators with the relation \([u,v]=i\hbar\{u,v\}\). However, I can't yet see how to go from working with functions of x to working with positional bra and ket vectors. Dirac's doing some funny stuff taking derivatives of state kets with respect to x, which doesn't make any sense to me whatsoever, so I have to look into it in more detail.

    Man, now this is getting me all worked up too, because I've decided recently that now is the time when I finally want to learn all the step-by-step historical foundations of QM. In undergrad we were spoon-fed the Dirac formalism without going over the means of its derivation from classical mechanics, and even the graduate course I took did little to improve on this. I've learned a lot of stuff already, like the Sommerfeld model and how they originally discovered quantum numbers, where Bohr's magic \(\frac{h}{2\pi}\) comes from, where Heisenberg got his transitional probabilities from, and much more, but I still have such an incredibly long way to go!

    I really hope someone here knows how to go from the classical generator of translations to the quantum generator, and I'm quite certain there's a direct connection, but I just can't see it at the moment. If we could just get to that step, and derive \(\mathcal{T}(\vec{dx})=1-\frac{i}{\hbar}\hat{\vec{p}}\cdot\vec{dx}\), then I could easily show you from that point where the relationship \(\hat{\vec{p}}\mapsto -i\hbar\vec{\nabla}\) comes from.
     
    Last edited: Jun 5, 2008
  14. BenTheMan Dr. of Physics, Prof. of Love Valued Senior Member

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    Dammit you stole my punchline!

    Serves me right, though, for letting this sit for so long.
     
  15. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    Ben, thank God you've arrived! Help! I'm really stressed out over Dirac's crazy text, how do you go from classical to quantum generators? Save meeeeeee!!!!!!!!!!!!
     
  16. temur man of no words Registered Senior Member

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    Is the following correct?

    In general p and x are elements of some noncommutative algebra satisfying

    \([x_i,p_k]=ih\delta_{ik}I,\qquad[x_i,x_k]=[p_i,p_k]=0,\)

    where \(I\) is the unit element of the algebra.

    The operators \(x_i\) (as multiplier) and \(\frac\partial{\partial x_i}\) acting on, say, the function space \(L^2(\mathbb{R}^n)\) satisfy the above conditions, so this is a representation of the quantum variables x and p as linear operators. These are not the only choice, for example, one can use \(\frac\partial{\partial p_i}\) and \(p_i\) (as multiplier) to represent the coordinate and momentum, which is called I think the momentum representation.
     
    Last edited: Jun 5, 2008
  17. QuarkHead Remedial Math Student Valued Senior Member

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    Let me apologize in advance - I have to leave town for 5 days, so I won't be able to comment on your posts for a while. Don't interpret this as ingratitude!
     
  18. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    Well, if you choose in advance to represent \(\hat{p}\) as \(-i\hbar \frac{\partial}{\partial x}\) in the \(|x>\) basis, then you do get the correct commutation relations and you do get that \(\hat{p}\) generates translations, and similarly you get the right commutations if you do what you suggest to represent \(\hat{x}\) in the \(|p>\) basis. However, I don't see how this guarantees that \(\hat{x}\) and \(\hat{p}\) are the canonical position and momentum operators carrying over from Hamiltonian mechanics, and this is important if you want to write down the correct quantum Hamiltonian.

    I'm still pretty sure there's a unique way to apply the correspondence principle to show that canonical momentum is the generator of translations in QM. I've heard some say otherwise, but there seem to be an awful lot of clues that indeed there is a way to do it, and that Dirac employed it himself. I'm slowly working my way through his text and trying very hard to resist the urge to skim pages, so maybe I'll have a better answer in a bit. If a method does indeed exist to carry the classical generator of translations over to QM, I'd definitely appreciate any help you guys can offer, but I'll do my best to find out in the meantime.

    It's cool, hope you have a safe trip!
     
  19. Vkothii Banned Banned

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    If you guys are still in Reimannian hyperspace, that link I posted uses something called a Dixon-Souriau momentum map (don't ask me what that is), to derive the classic D.P. electron.

    P.S. what does "gauge field of the 2nd kind" mean?
    P.P.S.
    Heh, dig and ye shall find: this might answer some questions about where the operators and Schrodinger meet Hamilton (section 2.4)
    http://philsci-archive.pitt.edu/archive/00002328/01/handbook.pdf
     
    Last edited: Jun 7, 2008
  20. Vkothii Banned Banned

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    The postulates of non-relativistic quantum mechanics are formulated in global inertial reference frames, connected by the transformations of the kinematical (extended) Galilei group, which, due to the Galilei relativity principle, relate the observations of an inertial observer to those of another one. The self-adjoint operators on the Hilbert space, in particular the Hamiltonian operator (governing the time-evolution in the Schroedinger equation and identified with the energy operator in the projective representation of the quantum Galilei group associated with the system), correspond to the quantization of classical quantities defined in these frames. The resulting quantum theory is extremely successful both for isolated and open systems (viewed as sub-systems of isolated systems). At the relativistic level conceptually nothing changes: we have the relativity principle stating the impossibility to distinguish special relativistic inertial frames and the kinematical Poincare' group replacing the Galilei one. Again the energy is one of the generators of the kinematical group and is identical with the canonical Hamiltonian governing the evolution of a relativistic Schroedinger equation.

    --Donato Bini and Luca Lusanna
    Oct. 2007

    Isn't that nice?
     
    Last edited: Jun 7, 2008
  21. AlphaNumeric Fully ionized Moderator

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    Is that the thin one which is basically a short collection of Dirac's lectures from the 50s on the principles of QFT?

    I've read it but, while it's great to read Dirac's own words, it's not the best for a QFT beginner. Constraints 'of the first/second type' is a concept you don't find a great deal in modern texts, unless you read Weinberg, which I do not recommend until you're competant at QFT (his 3 volume text is excellent but not for beginners IMO). Dirac's book is like Weinberg (but MASSIVELY shorter), great to read once you know what it's going to say, if you see what I mean.

    I was once reading that book in Paddington train station, waiting for the inevitably late connection to my home town, and some guy, about 50, walked up to where I was sitting and told me that book was too complicated for me and I should start by learning stuff like Newtonian physics. I didn't have the heart to tell him I'd done my 3rd year exams in quantum mechanics 3 days before....

    Temer, if you're working in 'position space' then sometimes it's useful to change all your descriptions into momentum space. All it involves doing is just a Fourier transform of everything. And to go back, the inverse! Nifty

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  22. CptBork Robbing the Shalebridge Cradle Valued Senior Member

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    I don't think so, because it starts from the very beginnings of building up a bra and ket description of QM. I was under the impression this was how he originally introduced the whole formalism, but maybe I'm wrong and he waited a couple of decades to publish it as a textbook, because the edition I'm using was originally printed in the 50's. It starts with the basic theory of operators, Hilbert spaces, bras and kets, then develops the various commutation relations and generators, and then at the end there's only a couple of chapters on relativistic QM.

    Reminds me of a time about 10 years ago when I was sitting at Heathrow airport waiting for my flight back to Canada, and I was reading some high school math that was a couple years ahead of where we were supposed to be at the time. This old English gentleman sitting across from me smiles and tells me "haw haw, wait 'til you get to calculus!" Except that comment was wholly appropriate, and I told him that's precisely what I was working towards. However, when I did get around to calculus about a year later, it really wasn't nearly as bad as I had expected it to be, so joke's on him!

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    So Alpha, you seem to understand a shedload about this stuff, do you know of any direct connection between infinitesimal translation generators in the Poisson bracket formalism and infinitesimal translation generators in the bra and ket formalism? I'm still sorting through this stuff and it's a complete mess. Feels like I'm being led on a dozen different directions at the same time- maybe that's why they don't bother to teach us this stuff.
     
  23. Vkothii Banned Banned

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    To answer my own question: I think they mean EM is a gauge field of the "second kind" because of second quantisation...?

    The field "operators" are really just a QM way of describing a (classical) field aren't they?

    I know wikipedia isn't considered the ultimate authority, but here's a quite readable entry that discusses field operators for a QFT, and how to transform a classical (scalar) field into operators, and how vector and tensor fields are an extension of the procedure - which essentially is finding the commutating relationship between the operators. The operators themselves are (expansions of coefficients of) Fourier transforms, apparently.

    (for the non-experts, of course)
     
    Last edited: Jun 10, 2008

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