The QH QM QA thread.

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

  1. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Ok, so I did a lot of thinking over the last few days since I first joined this thread. I think I might have some clues to the momentum operator puzzle that's got QH and myself so stumped.

    Please let me know what you guys think- I already posted this idea over at in the Quantum Physics section, and I hardly got anything in response. No agreement that the method is good, nor any criticism telling me it's complete trash. All I got was a few comments from a user named lbrits pointing me towards a few notes and theorems when I made my initial inquiry, otherwise I got absolutely nada! The last response I got came before I even posted the following arguments, so I have no way of knowing if it's even a rational approach.

    So here goes, my attempt to construct the momentum operator \(\hat{p}\) from a direct correspondence with classical mechanics:

    In classical mechanics, when working with the Hamiltonian formalism, one can perform what's called an infinitesimal canonical transformation (I believe they're called contact transformations), which is essentially a fancy way of changing variables. Since my understanding of this methodology is limited (been a long time since I learned it, and I learned it in a real hurry at the time), I'm not going to get into the nitty-gritty details. The important thing is that if you use momentum as the "generator" of a contact transformation, you get the following expression applying to any smooth function \(f(x)\):

    (eq.1) \(f(x+dx)=f(x)+dx\{f,p\}\),
    where I'm sure you know which letters correspond to which variables.

    Just to make sure everyone's on the same page, the Poisson bracket is, in general, defined as \(\{a,b\}=\frac{\partial a}{\partial x}\frac{\partial b}{\partial p}-\frac{\partial a}{\partial p}\frac{\partial b}{\partial x}\). Plug this definition into (eq.1) and you get:

    (eq.2) \(f(x+dx)=f(x)+dx\frac{\partial f}{\partial x}\), which essentially says that momentum as the variable in a contact transformation results in an infinitesimal translation, i.e. momentum is the generator of translations. Trivial I know, but perhaps useful nonetheless.

    Now Dirac noticed, I'm assuming from Heisenberg's mechanics, that there's a direct correspondence between the Poisson brackets \(\{a,b\}\), where \(a\) and \(b\) are dynamical variables, and the quantum commutator \(\big[\hat{a},\hat{b}\big]\), given by

    (eq.3) \(\{a,b\}\mapsto\frac{\big[\hat{a},\hat{b}\big]}{i\hbar}\).

    Now I know that up to this point I'm largely just repeating myself from earlier, and I apologize for that, I just want to make sure we're all on the same track. Here comes the new part:

    We define an operator \(\hat{u}(dx)\) such that

    (eq.4) \(\left\langle x'\right|\hat{u}(dx)\left|x \right\rangle=(x+dx)\left\langle x'\right|x\rangle=(x'+dx)\left\langle x'\right|x\rangle\).

    Now the last two expressions may seem incompatible, but in fact they're equivalent because we have

    (eq.5) \(\left\langle x'\right|x\rangle=\delta(x-x')\).

    Now we can plug in our correspondence with the classical Poisson bracket formalism as follows, treating \(u(x+dx)\) as a dynamical variable which is just \(x\) shifted by an amount \(dx\). Classically, we plug the dynamical variable \(u(x)\) into (eq.1) in place of \(f(x)\). The quantum mechanical analogue is then


    Of course, \(\hat{u}(0)\) is just the position operator with zero displacement, so we can substitue \(\hat{u}(0)=\hat{x}\). Then we obtain:


    Now one might argue that this is kind of stupid, because if you immediately substitute \(\big[\hat{x},\hat{p}\big]=i\hbar\), then the relation is obvious. In the \(\left|x\right\rangle\) basis, the operator \(\hat{u}(dx)\) just shifts all the eigenvalues by a value \(dx\), so you'd just get the relation \(\hat{u}(dx)\left|\psi\right\rangle=\hat{x}\left| \psi \right\rangle+dx\left|\psi\right\rangle\). However, let's not do the obvious thing, but instead expand the commutator in (eq.7). Whereas we started by considering active transformations on the position operator in the Heisenberg picture, i.e. allowing the operator itself to be shifted, now we can consider passive transformations acting on the bra and ket vectors instead, which is what's usually referred to as the Schrodinger picture. So then, expanding (eq.7), we get:

    (eq.8) \(\hat{u}(dx)=\hat{x}+dx\frac{\hat{x}\hat{p}-\hat{p}\hat{x}}{i\hbar}\).

    We can then factor this expression, neglecting terms of \(\mathcal{O}(dx^2)\), to obtain:

    (eq.9) \(\hat{u}(dx)=(1+\frac{i}{\hbar}dx\hat{p})\hat{x}(1-\frac{i}{\hbar}dx\hat{p})\)

    Ok wow, that took me a helluva long time to bang out on the keyboard. I'm gonna take a break and finish this in a followup post. Many of you guys are way more advanced than me in these subjects, so please let me know if any of this sounds bogus to you.
    Last edited: Jun 11, 2008
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  3. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Ok, now for part 2:

    If we play with the pieces of (eq.9), and apply the commutation relations in order to complete the calculations, we get:

    (eq.10) \(\hat{x}(1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle=(x+dx)(1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle\),

    So we can infer that

    (eq.11) \((1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle=e^{i\gamma(x)}\left|x+dx\right\rangle\).

    Taking the adjoint of this equation yields

    (eq.12) \(\left\langle x\right|(1+\frac{i}{\hbar}dx\hat{p})=\left\langle x+dx\right|e^{-i\gamma(x)}\).

    Now since the bra and ket vectors are phase shifted in the complex plane by opposite angles, the physics is identical if we ignore this phase shift and set


    It now becomes convenient to define a translation operator \(\hat{\mathcal{T}}(dx)\) which sends \(\left|x\right\rangle\mapsto\left|x+dx\right\rangle\) given by

    (eq.14) \(\hat{\mathcal{T}}(dx)=1-\frac{i}{\hbar}dx\hat{p}\).

    Then the corresponding adjoint of this operator, \(\hat{\mathcal{T}}^\dagger(dx)\) sends \(\left\langle x\right|\mapsto\left\langle x+dx\right|\).

    Now we're almost there, and we can answer (I hope) the original question about why the momentum operator is represented as a derivative in position space.

    Firstly, it should be noted that by considering the commutator \(\big[\hat{x},\hat{\mathcal{T}}(dx)\big]\) acting on the ket \(\left|x\right\rangle\), we can recover the identity \(\big[\hat{x},\hat{p}\big]=i\hbar\). This is the original way I learned about the latter identity- we simply assumed a magic translation operator and assumed it could be magically represented by (eq.13). The only justification they provided to identify the operator \(\hat{p}\) with momentum is that its eigenvalues have the right units. The main book I originally learned this formalism from was Townsend's "A Modern Approach to Quantum Mechanics", which seems to teach the concepts in a very different way from how they were originally derived/discovered. Sakurai doesn't seem to treat it in much more detail, though I haven't read it in a while so I might be forgetting details. I'm hoping that with this whole shpiel I've typed out, that we can now see some justification for how this all works out based on the correspondence between Poisson brackets and quantum commutators.

    Ok, so now consider the following calculations, for an arbitrary wave function \(\psi(x)\):

    (eq.15) \(\psi(x+dx)=\psi(x)+dx\frac{\partial\psi}{\partial x}\)

    The corresponding operator equation is

    (eq.16) \(\psi(x+dx)=\left\langle x\right|\hat{\mathcal{T}}^\dagger(dx)\left| \psi\right\rangle=\left\langle x\right|(1+\frac{i}{\hbar}dx\hat{p}) \left| \psi\right\rangle= \psi(x)+\frac{i}{\hbar}dx\left\langle x\right|\hat{p} \left|\psi\right\rangle\).

    Comparing (eq.15) and (eq.16), we finally obtain what we've been looking for:

    (eq.17) \(\left\langle x\right|\hat{p}\left|\psi\right\rangle=-i\hbar\frac{\partial\psi}{\partial x}\).


    Now if we ignore (eq.13), i.e. \(\gamma(x)\neq 0\), then since the square magnitude of the wave function is all that physically matters, we can multiply the right hand side of (eq.15) by a factor \(e^{i\gamma(x)}\) without cheating, and apply the identity \(\left\langle x+dx\right|=\left\langle x \right|e^{i\gamma(x)}\hat{\mathcal{T}}^\dagger(dx)\), we get

    (eq.18) \(\left\langle x\right|\hat{p}\left|\psi\right\rangle=-i\hbar e^{-i\gamma(x)}\frac{\partial\psi}{\partial x}\).

    So we can say that the representation in (eq.17) is unique up to a unitary scale factor. Out of convenience, it's best to apply (eq.13) and simplify things, since it doesn't change the physics. Now this derivation isn't 100% rigorous, i.e. for instance we neglected all terms of \(\mathcal{O}(dx^2)\), but when I was working on my initial draft of this sequence of arguments at, user lbrits pointed me to the following theorem, which apparently shows that (eq.17) is indeed the unique representation for the momentum operator in position space needed to satisfy the commutation relations, up to a unitary scale factor. Here's the link: Wikipedia: Stone-von Neumann Theorem

    Then from here, it's easy to apply the formalism of vector operators to go from 1-D to 3-D, apply the appropriate commutation relations and get \(\hat{\vec{p}}\mapsto -i\hbar\vec{\nabla}\).

    I hope this helps, QH. To everyone reading this, any criticisms would be gratefully accepted, otherwise I could easily be deluding myself here.
    Last edited: Jun 11, 2008
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  5. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Now I also noticed, QH was trying to do this from the Schrodinger formalism. I myself am not sure how Schrodinger's wave mechanics can give rise to operators, although I know that Schrodinger's equation can be derived from the Dirac operator formalism. Is that how Schrodinger showed that wave mechanics and Heisenberg's matrix mechanics were equivalent? Did he simply take the operators from Heisenberg's mechanics and use them to derive his wave equation? Or is there more to this story?

    Anyhow, when you're just working exclusively with Schrodinger's wave functions \(\psi(x)\) and forgetting about operators and the whole bra and ket formalism, I do have some ideas about how to justify the relation \(\vec{p}\mapsto -i\hbar\vec{\nabla}\), but I'll post that stuff later rather than adding to the clutter I've already created.
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  7. QuarkHead Remedial Math Student Valued Senior Member

    Mon Capitain: Whoa, there pony, you're leaving this carthorse lumbering waaaaay behind! Having given this no thought for a few day let's see if I can catch up a little.....
    which I don't see at all. Given your definition of the Poisson bracket, I don't quite see how you arrive at the last equality.

    Do we have that \(\{f,p\} = \frac{\partial f}{\partial x} \frac{\partial p}{\partial p} - \frac{\partial p}{\partial x} \frac{\partial f}{\partial p}\)? This doesn't quite do it for me. Certainly, the first term is \(\frac{\partial f}{\partial x} \frac{\partial p}{\partial p} = \frac{\partial f}{\partial x} \cdot 1\) But why is the 2nd term zero? (Hint: analysis is not my forte)

    I think I see another way of doing this, by insisting on the isometry of such translations. Lemme give it some thought.

    STOP!! I have this expression as an aside in one of my Lie texts (Azcarra & Izquierdo) but the bastards don't explain, other than saying (I quote) "this is of course the old Dirac quantization prescription"

    I'm going to leave it for now, as I need to get back in gear, so to speak.

    But, Oh.. regarding my attempt to extract the momentum operator from the Schroedinger eqn, D H was completely correct, I'm not sure it can be done, certainly not in the naive way I tried. I think there may be a "parallel" way, though, I have few ideas scribbled down here.....
  8. Vkothii Banned Banned

    He'll get back to you Dirac-ly, I expect.
  9. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Sorry about that! I wanted to bang this stuff out in a hurry before I got distracted by other threads and the usual daily grind. I should have checked first to see how much background you have in the Dirac formalism, how much you know about Poisson brackets and classical mechanics, etc.

    \(x\) and \(p\) are treated as independent variables at this stage in the calculations. The reasoning is that you can pick any initial conditions you want for your system, any initial \(x\) and \(p\), and a functional relationship between the two is only developed after you plug in the equations of motion and solve for \(x(t)\), \(p(t)\), which comes at the very end. I glossed over all this stuff because I was only intending to use it to justify the quantum part, and then one can always go look at the classical equations in more detail as part of another line of inquiry. Still if you want to ask anything about the classical part, I think that's good stuff to know too and I'm planning to re-learn it in full detail. So long story short, you've got the right idea, just set \(\frac{\partial p}{\partial x}=\frac{\partial x}{\partial p}=0\).

    I'd be very interested to see what you come up with.

    Ah yes, the good old "of course", as if we were all born with this knowledge like Plato suggested. It's frustratingly difficult to find good sources on this stuff, isn't it? Even Dirac himself seems to jump the gun here and there without going over the whole foundation (i.e. at points he says "This constant must be set to ___ in order to agree with experiment", where I think he's really talking about Heisenberg's picture when he says "experiment").

    Ok I have to admit my ideas are still vague and naive as well, but here's the heuristics of what Schrodinger originally did. In good 'ole always lovable classical mechanics, Hamilton noticed that the equations of motion for matter could be reformulated as something called the Hamilton-Jacobi equation. In this treatment, the equations of matter are seen as the 0-wavelength analogy to the equations of light, and thus an explanation is given for why both matter and light obey a Least Action principle. So Schrodinger looked to exploit this analogy himself, and indeed his equations can be shown to reduce to the classical Hamilton-Jacobi equation for \(\hbar\mapsto 0\).

    Anyhow, the important thing is that Schrodinger started out by trying to express the wavefunction \(\psi(x)\) for a free particle (no gravity, external forces, walls etc.) as superpositions of plane waves of the form \(cos(kx-\omega t)\). The idea was to basically treat it like a beam of light, using the DeBroglie relations to go from light waves to quantum matter waves. You end up with something like:


    Here, \(\phi(p)\) is just some coefficient representing how much of the wave has momentum \(p\), and \(E=E(p)\) is the energy associated with each momentum component, which we assume for now has the simple classical relation \(E(p)=\frac{p^2}{2m}\). I suppose one of the implicit physical assumptions is that the momentum of a free particle is conserved, since \(\phi(p)\) is assumed in this treatment to be independent of time.

    So, at least according to the textbook where I first learned Schrodinger's approach, the idea is that you want to do some operations on \(\psi(x,t)\) in order to derive a PDE for this function. So a good place to start is by looking at each of the individual pieces in the superposition of momentum waves. I.e. for a wave having only a single momentum component, \(p\), with associated energy \(E\), given by \(\psi(x,t)=cos(\frac{px-Et}{\hbar})\), when we plug in the classical hamiltonian we get

    \(\frac{p^2}{2m}\psi(x,t)=E\psi(x,t)\), or \((\frac{p^2}{2m}-E)\psi(x,t)=0\).

    So you'd hope to find a differential operator \(\box \sim(\ )\frac{\partial^2}{\partial x^2}-(\ )\frac{\partial}{\partial t}\) such that


    The problem with this is that when you take two derivatives of a cosine, you get a cosine back. When you take a single derivative, you get a sine back. So it doesn't quite work out; taking real-valued superpositions of both cosines and sines doesn't work either. However, if you allow the wave function to take on complex values, then we can get something that works out beautifully. So instead of working with cosines, or even a real-valued sum over both cosine and sine waves, we take:


    and for our operator \(\box\) we take \(\box=\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}-i\hbar\frac{\partial}{\partial t}\)

    Then we get \(\box\psi(x,t)=(\frac{p^2}{2m}-E)\psi(x,t)=0\)

    So now we apply that to a general superposition of momentum states,


    and, applying Leibniz' rule for interchanging the order of integrals and partial derivatives, we end up with the famous Schrodinger equation for the free particle,

    \(\left(\frac{-\hbar^2}{2m}\right)\frac{\partial^2\psi(x,t)}{ \partial x^2}=i\hbar\frac{\partial \psi(x,t)}{\partial t}\).

    Now from this point on, in order to, well... transcend the boundaries and ascend to the quantum regime, the Schrodinger equation is taken as fundamental instead of the wave superposition from which it was originally derived. We add a potential energy term to the equations, since the general equation for a wave with one momentum component is \(\left(\frac{p^2}{2m}+V(x,t)\right)\psi(x,t)=E\psi(x,t)\), and this leads to the general (1-dimensional) Schrodinger equation,

    \(\left(\frac{-\hbar^2}{2m}\right)\frac{\partial^2\psi(x,t)}{ \partial x^2}+V(x,t)\psi(x,t)=i\hbar\frac{\partial\psi(x,t)}{\partial t}\).

    So the idea isn't entirely rigorous, and is taken as more of an assumption which we must then check by solving actual problems and seeing if they agree with the Sommerfeld semi-classical models which followed after Bohr. But be that as it may, whenever you're looking to construct an operator, you think of the function \(\psi(x,t)=e^{i(px-Et)/\hbar}\). So if you want to get just the momentum operator \(p\), you take \(p\mapsto -i\hbar\frac{\partial}{\partial x}\).

    The method is then readily extended to 3 dimensions, so the operator for momentum as a vector is then \(\vec{p}\mapsto -i\hbar\vec{\nabla}\). If you want to find the orbital angular momentum operator \(\vec{L}\), you take \(\vec{L}=\vec{r}\times\vec{p}\mapsto -i\hbar \vec{r}\times\vec{\nabla}\), and so on.

    Also, if you want to switch between position and momentum as basis states, in general we assume that the Fourier transform relationship holds true, i.e.

    \(\psi(x,t)=\int_{-\infty}^\infty\phi(p,t)e^{ipx/\hbar}dp\), where the momentum wave coefficients \(\phi(p,t)\) can now be functions of time as well as momentum, and we have included the energy and time dependence \(e^{-iEt/\hbar}\) as part of these coefficients. To go from position to momentum representation, we therefore perform the inverse Fourier transform.

    I apologize for taking so long to get back to you, I wanted to combine this all in one post and it took me a few sessions to type out. If I'm going too fast, I can slow down now since I've pretty much completed my verbal diarhea

    Please Register or Log in to view the hidden image!

    Again, if anyone thinks I'm full of it and can point out some of my mistakes, please speak up.
    Last edited: Jun 12, 2008
  10. QuarkHead Remedial Math Student Valued Senior Member

    Mon Capitain: I thank you for that, it will take me a while to wade through, but I do promise to repay your kindness with my full attention later.

    In the meantime, I think I have found 2 (well 11/2, really) alternative ways of getting at this, which I would like to share.

    I start with the plane wave solution to the general wave equation, which I shamelessly copy from the Wiki;

    \(\psi(\vec{r},t) = Ae^{i(\vec{k} \cdot \vec{r} - \omega t)}\)

    where \(\vec{r} \) the position vector, \( \vec{k} \) the wave propagation vector and \( \omega\) is angular speed.

    For simplicity (and I think it makes sense in context) I will assume that \(||\vec{k}|| = \frac{2\pi}{\lambda}\) where \(\lambda \) is wavelength.

    Looking in my chemistry text, I follow the de Broglie conjecture that \(E = h \nu = mc^2 \Rightarrow \lambda = \frac{h}{\vec{p}}\).

    Then I will have, by the above, that \(\lambda = \frac{2\pi}{||\vec{k}||} = \frac{h}{\vec{p}} \Rightarrow ||\vec{k}|| = \frac {2 \pi \vec{p}}{h} = \frac{\vec{p}}{\hbar}\).

    Plugging this into my plane wave solution above, I get that \(\psi({\vec{r},t}) = Ae^{\frac{i}{\hbar}(\vec{p} \cdot \vec{r}- \omega t)}\).

    I now assume that \(\vec{r}\) and \(\vec{p}\) have components in \(x,y,z\) and partially differentiate w.r.t these coordinate functions. I find that \(\frac{\partial }{\partial x}\psi = (\frac{i}{\hbar})p_x \psi \Rightarrow p_x \psi = -i\hbar \frac{\partial}{\partial x}\) etc.

    Define \(\hat{p}\) to be the vector sum \(\hat{p} = -i \hbar \sum \frac{\partial}{\partial x^{\mu}} =- i \hbar \nabla\), and we're about done.

    Now, noticing that the second derivatives take the form \(\frac{\partial ^2}{\partial x^2}\psi = - (\frac{p_x^2}{\partial \hbar^2})\psi\), after some rearrangements, taking the same sum, and popping in the factor \(\frac{1}{2m}\), I arrive at \( -\frac{\hbar^2}{2m}\nabla^2 \psi\) for the kinetic term in Schroedinger's Hamiltonian.

    Riddled with errors and misconceptions, maybe? You tell me....

    I will, will, will, read my captains post carefully through tomorrow, I promise.

    Oh, I promised another one half derivation. That too will have to have to wait, as I have have a beer that won't!
  11. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    I ain't your captain, you're the one running this ship! Looking at your discussions on topology, there are so many subjects on which you would kick my ass and make me look like an amateur crank, it's not even funny.

    Anyhow, I briefly read through your arguments, but it looks like the gist of it is almost identical to what I wrote above. Check it out, I think you'll be pleasantly surprised. The most important thing to remember, I think, is that there are two main, qualitatively different means of constructing the operators and structure of Quantum Mechanics- wave mechanics, and matrix mechanics. I suppose Dirac's bra and ket formalism is where the two converge.
  12. QuarkHead Remedial Math Student Valued Senior Member

    Right CptBork, I read your stuff. My word, but you are thorough! But you are right, there is a distinct parallel between your approach and mine (though I'm not too happy with all those integrals; but then I hate calculus!)

    You, of course, always have an eye on the physics of the situation, whereas all I do is shuffle symbols around (this confession, no doubt, will be music to the ears of the sort of crank who believes that modern physics does merely that, without taking account of what they call the "real world". My excuse, should I need one, is that at least I am trying to learn, though I can no longer remember why)

    Incidentally, you seem slightly offended by my (teasingly) referring to you as "my captain". I assure you, no disrespect was intended.

    Anyway, last night I grew increasingly unsure of my second approach, but since yours contains elements of it, I'll bang it out nonetheless.

    As always, let \(M\) be a manifold. Define a linear coordinate translation on \(M\) by \(T:M \to M,\; x \to x+\alpha\) for some point \(x \in M\) and \(\alpha\) a scalar.

    Let \(v(x)\) be an element in a vector field defined at the point \(x \in M\).

    Now define the vector translation \(v(x) \to v'(x) = v(x - \alpha)\) (since translating coordinates one way is exactly equivalent to translating vectors the other way).

    Now allow ourselves the liberty of thinking of the expression \(v(x - \alpha)\) as a vector function with it's obvious argument. So this will allow me to take the Taylor series expansion for this "function", whereby

    \(v(x-\alpha) = v(x) - \alpha \frac{d}{dx}(v(x) - \frac{1}{2}(-\alpha\frac{d}{dx})^2v(x) + .... = e^{-\alpha \frac {d}{dx}}v(x)\), unless I made made an error.

    Now for this translation, I insist on unitarity, that is \(UU^{\dag} = I\). In one dimension, a nice choice is of the form \(U = e^{i\beta}\), since \((e^{i\beta})^{\dag} = e^{-i\beta}\). Then I may achieve this by "factoring" \(e^{-\alpha \frac{d}{dx}}\) as \( e^{-i \alpha(-i \frac{d}{dx})}\).

    Since \(\pm i \alpha\) is scalar, this suggests that the infinitesimal generator of one-dimensional linear translations is the operator \(-i\frac{d}{dx}\).

    Ok. This is highly suggestive, but where do I go from here? Do I just wave my hands and say "let's quantize by chucking in an h-bar"?

    Oh look - it's Friday night! The tradition here (UK) is to finish early, go get drunk, go home and beat our wives. I am a traditionalist....... (kidding)
  13. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Ha! No no, I knew it was all in jest. 'Tis all in good fun!


    Ok... no, I'm seething mad; you see, we Bork-kind take great offense to being called "Captain". In our culture, only we may refer to ourselves as "Captain", and for anyone else to do so is a great dishonour to our families and traditions.

    I looked at it briefly, I'll need some more time to read it more carefully; some of it might not be quite so rigorous, a little hand-waving here and there, but I think from an initial readthrough that it contains some legitimate ideas. My bra and ket approach worked somewhat non-rigorously by considering only first-order Taylor expansions. I'm pretty sure the arguments can be made rigorous by considering finite instead of infinitesimal translations, but then the math gets more complicated and if I remember right, you need lots of fancy results such as the Baker-Campbell-Hausdorff formula.

    Happy beating! Just make sure your thumb is bigger than your wife's before you proceed; just to keep everything legal, of course.
  14. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Alrighty, so I looked through your post, and actually spent quite a long time thinking about it and doing some calculations. It's great review, some of this stuff I think I understand far better now than when I originally took the relevant courses. Right now though I don't have access to my most treasured resources, and though I have... unconventional means of accessing them if need be, it's nowhere near as nice as having a hard physical copy to work from. So I'm mostly doing this stuff from memory, and a lil' timely help from Wikipedia. Apologies if I'm screwing up somewhere.

    So... you have some very good ideas in your last post that correspond to a lot of what's done in the Dirac operator formalism, but I think there are a few misconceptions about how to connect it to quantum physics.

    Suppose I define a ket vector \(\left|x\right\rangle\). Then we can define a derivative operator acting on this ket as

    \(\hat{\left(\frac{\partial}{\partial x}\right)}\left|x\right\rangle=\lim_{h\mapsto 0}\frac{\left|x+h\right\rangle-\left|x\right\rangle}{h}=\lim_{h\mapsto 0}\left(\frac{1}{h}\left|x+h\right\rangle-\frac{1}{h}\left|x\right\rangle\right)\)

    Then from this definition, which according to my vague memory is standard, it follows that

    \(\left\langle\psi\right|\hat{\left(\frac{\partial}{\partial x}\right)}\left|x\right\rangle=\frac{ \partial}{\partial x}\left\langle\psi\right|x\rangle=\frac{\partial}{\partial x}\psi^*(x)\)

    Now we hit our first roadblock, which is that we want to talk about the function \(\psi(x)=\left\langle x\right|\psi\rangle\). So we need to define an operator \(\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger\) which is basically identical to the previous operator, only it acts to the left, on bra vectors.

    Now, akin to your reasoning, let's look at the following bra vector:
    \(\left\langle x\right|e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}\)

    Indeed, if we take the inner product of this bra vector with an arbitrary state ket \(\left|\psi\right\rangle\), we end up with the following result:

    \(\left\langle x\right|e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}\left|\psi\right\rangle=e^{a \frac{ \partial}{\partial x}}\left\langle x\right|\psi\rangle=e^{a\frac{\partial}{\partial x}}\psi(x)=\psi(x+a)=\left\langle x+a\right|\psi\rangle\)

    Alternatively, you can show the following:

    \(\left\langle x\right|e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}\hat{x}=(x+a)\left\langle x\right|e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}\)

    To see this, you expand the exponential operator as a Taylor series, and you apply the following commutation rule, which you can prove by induction:

    \(\left[\hat{\left(\frac{\partial^n}{\partial x^n}\right)}^\dagger,\hat{x}\right]=n\hat{\left(\frac{\partial ^{n-1}}{\partial x^{n-1}}\right)}^\dagger\)

    Which then leads to:

    \(\left[e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger},\hat{x}\right]=ae^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}\)

    Either of these methods is sufficient to prove that, up to an arbitrary and physically meaningless unitary constant, we have the following:

    \(\left\langle x\right|e^{a\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger}=\left\langle x+a\right|\)

    So the exponential operator generates finite translations.

    Now we return to working with ket vectors \(\left|x\right\rangle\), and we can see that the equivalent operator, acting to the right, is \(e^{a\hat{\left(\frac{\partial}{\partial x}\right)}}\). i.e.,

    \(e^{a\hat{\left(\frac{\partial}{\partial x}\right)}}\left|x\right\rangle=\left|x+a\right \rangle\)

    The problem is, our previous operator works by operating to the left, whereas this one operates to the right; i.e.

    \(\left\langle x\right|\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger\left|\psi\right\rangle=\frac{ \partial}{\partial x}\left\langle x\right|\psi\rangle=\frac{\partial}{\partial x}\psi(x)\)
    \(\left\langle \psi\right|\hat{\left(\frac{\partial}{\partial x}\right)}\left|x\right\rangle=\frac{ \partial}{\partial x}\left\langle \psi\right|x\rangle=\frac{\partial}{\partial x}\psi^*(x)\)

    So we want to define a derivative operator and its adjoint in such a way that they can both be considered to act to the left, or to the right. In my treatment, derivative operators with a dagger (the adjoint symbol) act to the left, and derivative operators without a dagger act to the right. The relationship we need to compare the two is:

    \(\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger=-\hat{\left(\frac{\partial}{\partial x}\right)}\)

    There are a number of ways to show this, the easiest way to see this is probably to note that translating bra and ket vectors by equal amounts does nothing to change the physics of the system. So we can write, for an arbitrary bra \(\left\langle\psi_1\right|\) and an arbitrary ket \(\left|\psi_2\right\rangle\), and an arbitrary translation \(a\):

    \(\left\langle \psi_1\right|e^{a\hat{\left(\frac{\partial}{ \partial x}\right)}^\dagger}e^{a\hat{\left(\frac{\partial}{\partial x}\right)}}\left|\psi_2\right\rangle=\left\langle \psi_1\right|\psi_2\rangle\ \ \Longrightarrow\ \ e^{a\hat{\left(\frac{\partial}{ \partial x}\right)}^\dagger}=e^{-a\hat{\left(\frac{\partial}{ \partial x}\right)}}\)

    So then we conclude that the operator \(\hat{\left(\frac{\partial}{\partial x}\right)\) is anti-Hermitian, which is no surprise when you look at its relationship to the Hermitian momentum operator. That relationship would be:

    \(\hat{p}=-i\hbar\hat{\left(\frac{\partial}{\partial x}\right)}^\dagger\)

    I'm pretty sure it's around here that we need to invoke some sort of correspondence principle before we can go any further, otherwise we're just taking wild guesses. I have some things to say about the subject of finite translations in Dirac's formalism and I also want to make my earlier arguments a bit more rigorous instead of relying on first order expansions, but that ought to wait for a second post.
    Last edited: Jun 15, 2008
  15. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Alright, so here's one way to complete our derivation of momentum as the generator of translations: From Heisenberg's mechanics, or alternatively by converting Schrodinger's equation into operator form, we know that \(\left[\hat{x},\hat{p}\right]=i\hbar\), and then by applying the Stone-von Neumann theorem, we prove the unique correspondence (up to unitarity)

    \(\hat{p}\leftrightarrow -i\hbar\frac{\partial}{\partial x}\).

    Then we see that \(\hat{p}\) is Hermitian, hence diagonalizable and with real eigenvalues, as we'd hope for any physical operator.

    Alternatively, applying the methods I used in my effort to deduce this relationship, we can then go from infinitesimal translations to finite translations.

    If I define my translation operator as \(\hat{\mathcal{T}}(a)\) which sends \(\left|x\right\rangle\mapsto\left|x+a\right\rangle\), then I may write \(dx=\frac{a}{N}\), which implies:

    \(\hat{\mathcal{T}}(a)=\lim_{N\mapsto\infty}\left( \hat{\mathcal{T}}(dx)\right)^N=\lim_{N\mapsto \infty}\left(1-\frac{i}{\hbar}\frac{a}{N}\hat{p}\right)^N=e^{-ia\hat{p}/\hbar}\)

    I'm not sure how much operator algebra you've learned, so in case you doubt the last equality, you can expand both sides in a Taylor series and prove it rigorously.

    As for the original calculations I did in the Dirac formalism, attempting to use the correspondence principle to show that momentum generates infinitesimal translations, and approximating everything to first order in \(dx\), I believe it can be made rigorous by doing the following:

    In the classical case:
    \(u(a)=f(x+a)\Rightarrow \frac{\partial u(a)}{\partial a}=\{u(a),p\}\)

    The corresponding quantum equation:
    \(\frac{\partial\hat{u}(a)}{\partial a}=\frac{\left[\hat{u}(a),\hat{p}\right]}{i\hbar}\)

    Solving this differential operator equation (can't remember if I've covered this part in my own studies), we can see that shifting the position basis by a length \(a\) gives us the following operator:


    And from here we can deduce everything we need to know about translations

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    Last edited: Jun 14, 2008
  16. AlphaNumeric Fully ionized Moderator

    You 'notice' that the equation is related to the form \(dg = a.g\) and so you make the ansatz \(g = he^{\lambda a p}\). Equating this with the form of \(dg = [A,g]\) gives you that the second term implies \(\lambda = \frac{1}{i\hbar}\) and the first term that \(dh = \frac{p}{i\hbar}h\), which you solve in the same manner, \(h = e^{-\lambda ap}k\), with the restiction becoming that \(\partial_{a}k=0\), which initial conditions give you to be \(k = g(0)\). And so \(g = he^{\lambda a p} = e^{-\lambda a p}ke^{\lambda a p} = e^{-\frac{1}{i\hbar}ap}g(0)e^{\frac{1}{i\hbar} a p}\)
    Last edited: Jun 14, 2008
  17. QuarkHead Remedial Math Student Valued Senior Member

    Lost me, tough guys, completely.

    Talk among yourselves for a while, while I have a beer.
  18. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Sorry, bro! Don't mean to hijack your thread and make it an incomprehensible mess (which is what quantum mechanics is)

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    Tell me any line or phrase or anything that doesn't make sense and I can try to clarify. When I started out at the top of this page, my memory was still very vague on these subjects, but now I think it's been refreshed a great deal and I would write it out a bit differently.

    Anyway as I said, aside from a few formalities I think your translation operator method looks good, I'll post a simpler spiel taking what you wrote and completing the proof without any need to guess at the \(i\)'s, \(p\)'s and \(\hbar\)'s.
  19. CptBork Robbing the Shalebridge Cradle Valued Senior Member

    Ok, just wanted to add that I did a whole bunch of revision to what I wrote above. Some of it was self-contradictory, some of it was unnecessarily complicated, so I tried to clean it up and streamline it a little bit. QH, if it's just meaningless blather to you, I apologize, I'll try to avoid doing anymore of that. Are you comfortable with the bra and ket vectors, or should we stick to Schrodinger's wave approach?
  20. QuarkHead Remedial Math Student Valued Senior Member

    Nah, you're fine. Whatever seems natural to you is OK with me. Lemme say this, though.

    Although my distaste for the bra-ket notation is legendary, I can read it, after a fashion. Any difficulties I have with your or Alpha's posts is entirely due to my own lack of physics knowledge.

    So, for example, when I see terms like "state vector" etc, my knees start to knock. Just take it slow and I'll be cool.

    Anyway, a lovely sunny day, which I resolve to devote to the family. Catch you soon
  21. temur man of no words Registered Senior Member

    Thanks for the links! Now I almost understand the basic idea of gauge theory; but can somebody explain me what is weight of rep and how it is related to a diagram of particles on a plane? How do you directly know from these diagrams that the particle A is in rep 3x3 or something like that? (I can find rigorous, formal, long definitions anywhere but I cannot get these; need some "humanly" explanation that just directly puts the essence into your head.)

    And there is something going on with two kinds of SU(3), one with flavour symmetry and the other with colors. The color SU(3) seems to be not very important since everything must be in a singlet. I am a bit confused since normally people would say the Standard model group is U(1)xSU(2)xSU(3), rather than U(1)xSU(2)xSU(3)xSU(3).
    Last edited: Jun 20, 2008

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