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chroot

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- Warren

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"It is of course not possible to derive the operators which represent observables in any fundamental way from classical mechanics. However, a simple prescription exists which establishes the bridge between the classical quantities and the operators." After that it goes on to give the operators for the position and momentum.

I guess I should have inserted that classical mechanics part, too.

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chroot

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[tex]\hat p = \frac{\hbar}{i} \frac{\partial}{\partial x}[/tex]

(and the position operator is trivial).

The general "prescription" to go from a classical description to a QM description is thus to replace p with [itex]\hat p[/itex], then evaluate the operator as usual.

- Warren

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That's right. But what I want to know is how does one come up with that particular expression for the momentum operator - that combination of i, h-bar etc. I hope I'm clear.chroot said:Well, all dynamical quantities can be expressed in terms of position and momentum. The momentum operator is

[tex]\hat p = \frac{\hbar}{i} \frac{\partial}{\partial x}[/tex]

(and the position operator is trivial).

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chroot

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Any references on the web that you would recommend? Thanks.

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I just went through my notes. I saw something that goes under the name "the derivation of Schrodinger's equation" (although my teacher stated that it can't be derived in the usual sense) which starts with de Broglie's relation ([tex] E = \hbar k[/tex])and Einstein's relation([tex] E = \hbar\omega[/tex]), and also assumes a one-dimensional wavefunction of the form [tex]\exp(kx - \omega t)[/tex]. Substiute the two equations, differentiate with respect to x...[tex]\hat p[/tex] ([tex]\hat p_x [/tex] actually) falls out.chroot said:

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Is this the one?

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nrqed

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It's true that Schrodinger's equation cannot be "derived". It can be made plausible but it's a question of experimental tests to see if it describes Nature. As for the operator P, again it cannot be formally derived. But it can be made plausible. I know of two approaches and you are mentioning one.neutrino said:I just went through my notes. I saw something that goes under the name "the derivation of Schrodinger's equation" (although my teacher stated that it can't be derived in the usual sense) which starts with de Broglie's relation ([tex] E = \hbar k[/tex])and Einstein's relation([tex] E = \hbar\omega[/tex]), and also assumes a one-dimensional wavefunction of the form [tex]\exp(kx - \omega t)[/tex]. Substiute the two equations, differentiate with respect to x...[tex]\hat p[/tex] ([tex]\hat p_x [/tex] actually) falls out.

Is this the one?

First, a couple of corrections to your equations, it's [itex] p = \hbar k [/itex] (not E) and the exponential contains a factor "i", [itex] e^{i(kx - \omega t)} [/itex].

From the above two, it is clear that the momentum of a plane wave can be obtained by applying the operator [itex] p_{op} = - i \hbar {d \over dx} [/itex].

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One last question (I hope!). Is the other approach as "elementary" as this one?

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The operators operate on the wavefuction which has to satisfy the Schronder equation.

A simplest operator is the position operator which forces the wavefuction to be in a specific position, which is analogous to displacement from classical physics.

We also know from classical physics that momentum p = m * dx/dt so in quatum mechanics you can take the deriviative of the expectetion value of x.

so <p> = m * (d<x>/dt)

hence we get the momentum operator.

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Each physical observable has associated with it an operator, which you know. These operators that correspond to physical entities have a geometric interpretation. In quantum mechanis the relations (i.e. commutation rules) between operators are derived by considering the behaviour and the nature of the generators of infinitesimal symmetry transformations, and then using these generators we can produce unitary representations of finite transformations of the symmetry group that act on wave-functions. From these we can then identify the generators of said objects with dynamical quantities, whose eigenvalues give observables.neutrino said:That's right. But what I want to know is how does one come up with that particular expression for the momentum operator - that combination of i, h-bar etc. I hope I'm clear.

An infinitesimal transformation by a parameter s can be expanded as a taylor series to first order in s, assuming U(0) is the indentity transformation,

[tex]U(s)=1+\left.\frac{dU}{ds}\right|_{s=0}s+\mathcal{O}(s^2)[/tex]

Assuming unitarity [itex]UU^{\dagger}=1[/itex]

[tex]UU^{\dagger}=1+s\left.\left[\frac{dU}{ds}+\frac{dU^{\dagger}}{ds}\right]\right|_{s=0}+\mathcal{O}(s^2)[/tex]

To satisfy unitarity we require [itex]\left.\frac{dU}{ds}\right|_{s=0}=iK[/itex], with K self-adjoint, this K is the generator of infinitesimal transformations. One can also generate finite transformations using this generator.

In non-relativistic quantum mechanics this group of symmetry operations is the Galilei group, consisting of 10 generators of the following symmetry operations on the space-time of classical physics: spatial and temporal translations and non-relativistic rotations and boosts. Respectively, these generators are: momentum, total energy (Hamilontian), angular momentum and some operator that I recall be expressable as a multiple of some other operator (cannot remember precisely). In relativistic quantum mechanics this symmetry group is the Poincare or lorentz group (the lorentz group only consists of rotations and boosts, not translations as well, which Poincare does).

Momentum is responsible for a spatial translations, thus some finite coordinate shift by a vector a of a scalar wave-function is

[tex]\langle\mathbf{x}|\exp\left[\tfrac{i\mathbf{a}\cdot\mathbf{P}}{\hbar }}\right]|\psi \rangle=\psi (\mathbf{x-a})[/tex]

Expanding the exponential on the left gives one a series of ascending powers of the momentum operator. Expanding the right as a taylor series about [itex]\mathbf{x}=\mathbf{a}[/itex] gives one a series with ascending derivatives. Identifying the terms in expansions of the left and right hand side allows one to conclude the form of the momentum operator. This is true in both relativistic and non-relativistic quantum mechanics, as this derivation does not rely on any kind of motion. The same procedure gives one an energy operator.

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Actually, since we know that all infinite-dimensional Hilbert spaces are isomorphic to another, it suffices to choose [itex] L^2 \left(\mathbb{R}^3\right) [/tex] and derive the expression for [itex] \hat{x} [/itex] and [itex] p_{x} [/itex] explicitely.

Daniel.

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Dirac's QM

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Physics Monkey

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Do you happen to know something about the canonical formulation of classical mechanics, Hamiltonians, Poisson brackets, and the like? Here is a tidbit assuming you do. The route to quantum mechanics discovered by Dirac is very simple and is called canonical quantization. Canonical quantization amounts to the following prescription: replace all Poisson brackets in the classical theory with quantum commutators divided by [tex] i \hbar [/tex]. There is where [tex] \hbar [/tex] sneaks in. Now that your observables don't commute, they must be represented by operators, and as you will learn in a good quantum mechanics course, almost everything flows from knowing the commutators. In particular, once you know that [tex] [x,p] = i \hbar [/tex], for example, you can obtain the representation you referred to above for the momentum operator. Of couse, all this is gibberish if you don't know what a Poisson bracket is!

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Haelfix

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Well the first step is to show such an operator 'exists', its fairly trivial and people have described it sufficiently in this post. You then have to show uniqueness. This is nontrivial, for certain operators like say position and momentum its not that hard (again see derivations above), but for other operators this can be *tricky*.

The general rule of thumb is to consider symmetry as tantamount, and pick the most general operator that obeys the symmetries of the system (including unitarity and so forth), this then is *usually* what you want, especially in vanilla QM (in field theory, you will in general *not* have uniqueness, and will have to do some informed guessing). The associated eigenvalues will then describe the observables.

The general rule of thumb is to consider symmetry as tantamount, and pick the most general operator that obeys the symmetries of the system (including unitarity and so forth), this then is *usually* what you want, especially in vanilla QM (in field theory, you will in general *not* have uniqueness, and will have to do some informed guessing). The associated eigenvalues will then describe the observables.

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Umm...nope, nope and nope. Although the syllabus (which is a bunch of ()*$%(*&@$^ ) has "officially been completed" in class, I'm still about QM. The book I mentioned earlier was picked from a set of withdrawn books at a library...and I'm still searching for a good book to follow and to get a grip on the subject. Now, before you start recommending Griffiths and Shankar, I'm looking for a book from a shorter list than usual (basically these are the low-price editions). Some of the popular ones available are Eisberg & Resnick, Liboff, Gasiorowicz(sp?), Merzbacher and Greiner. And some older ones like Scwabl, Schiff and Powell&Crasemann (I have this, but I require a more modern introduction with a balance between mathematics and "wordiness", and I wouldn't mind a little more of the former) . The profesor who taught us QM said that Liboff's appraoch was unconventional, but that book is good for its huge number of problems.Physics Monkey said:Hi neutrino,

Do you happen to know something about the canonical formulation of classical mechanics, Hamiltonians, Poisson brackets, and the like?

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Physics Monkey

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No luck with the canonical approach, eh? Oh well. As for text books, I can't stand Griffiths and would never recommend it to anyone. One really nice book is by Nouredine Zettili. The great thing about this book (besides being a generally clear treatment) is that Zettili provides numerous nontrivial worked example problems at the end of each chapter. There are then lots more problems afterwards for you try your hand at. It's really quite helpful, especially for self study. I don't know if the book is in your price range, but I would recommend taking a look.

Also, you're going to eventually want to read Schwinger's absolutely beautiful quantum textbook. Suprisingly, it's relatively cheap.

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Given a form of the operator in, say, the form of a derivative (e.g. by projecting the momentum operator into position space) it will have one form in terms of derivatives. One can also project the momentum operator into momentum space where it takes on a different form, e.g. a different differential operator. Therefore there must be a rule which maps operators in one space to operators in another space.chroot said:

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This receipt is actually extended for classically constrained systems and it goes that [itex] \mathbb{Z}_{2}[/itex]-graded Dirac bracket goes to [1/ (i*hbar)] * [itex] \mathbb{Z}_{2}[/itex]-graded commutator and a constant (wrt classical canonical variables) goes to the same constant times unit operator in the Hilbert space of states...Physics Monkey said:

Do you happen to know something about the canonical formulation of classical mechanics, Hamiltonians, Poisson brackets, and the like? Here is a tidbit assuming you do. The route to quantum mechanics discovered by Dirac is very simple and is called canonical quantization. Canonical quantization amounts to the following prescription: replace all Poisson brackets in the classical theory with quantum commutators divided by [tex] i \hbar [/tex]. There is where [tex] \hbar [/tex] sneaks in. Now that your observables don't commute, they must be represented by operators, and as you will learn in a good quantum mechanics course, almost everything flows from knowing the commutators. In particular, once you know that [tex] [x,p] = i \hbar [/tex], for example, you can obtain the representation you referred to above for the momentum operator. Of couse, all this is gibberish if you don't know what a Poisson bracket is!

Daniel.

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