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Quantum Mechanics Question

  1. Sep 29, 2007 #1
    Hey all-
    I am new to quantum mechanics so these questions will be elementary and don't be afraid to go over simple concepts. There are two questions on the homework that I barely even know how to approach (I don't know how to type p sub k properly so that is how I will be writing it):

    1. Let f(p,x) be an integer function of operators p sub k,x sub k.
    Derive general functions (df/dx sub k) = -[f,p sub k] and (df/dp sub k) = [f,x sub k]
    With abbreviation [f,g] = i/h bar(fg - gf) from the commutation rules for p sub k and x sub k.
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    Firstly, I thought that [A,B] = i h bar (A,B) Poisson Bracket where (A,B) Poisson Bracket has partial derivatives and a summation. Are the two representations the same?
    Anyway, I absolutely cannot see how to begin proving a relationship between a completely arbitrary function f and its position and momenta. Since I cannot take the partial derivative of an arbitrary function I tried to work backwards and compute [f, p] but by the definition given above (fp - pf) = 0. In fact, it seems like fp-pf will be zero for any values of p and f. However, p and f are not variables, right? It is difficult to see but should they be operators expressed in the bra-ket notation?
    Anyway, I can prove the above relationships using [x,B] = i h bar dB/dp but that seems inadequate because that formula is essentially the same as the problem itself and the book does not really derive it.

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    2. Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi. How does <A> change with time? What follows for <x sub k> and <p sub k>?
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    At first glance it seems that if an operator does not depend on time neither will its expectation value nor will x and p. The only relevant equation I know is <A>=<Psi|A|Psi>. This may just be a ridiculously easy question and what I said before is the answer so please tell me if it's more involved than that.

    Thank you for your help and attention.
     
  2. jcsd
  3. Sep 29, 2007 #2

    Gokul43201

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    Yes, they are not scalars. Do you know how to write the p operator in the x-basis? [Hint: you have probably seen this in the time independent Schrodinger equation.]

    Yes, it is more involved. Notice the part I bolded. [itex]| \psi (t) \rangle \neq | \psi (0) \rangle [/itex]

    How does a general state (of a system, with a time-independent Hamitonian H) evolve in time?
     
  4. Sep 29, 2007 #3
    Yes! p = -i h bar d/dq. That's awesome! So, for the first one, -[f,p] = (-1/i h bar) (f p - p f) = (-1/i h bar)(f -i h bar d/dx - (-i h bar df/dx)) = f d/dx + df/dx! The second part of that is the answer, but what is f d/dx and how does it go away? Should I perhaps make all this stuff act on x? If so, should the x be taken out later (I've only seen this done once)?

    Good grief... I have no idea - this is my first time seeing quantum mechanics mathematically and the class is quantum optics so the prof. assumes I know all this (and there is no textbook, just his notes). Come to think of it, I completely understand next to none of this so let's see if I know anything at all; A is an operator. <A> = <Psi|A|Psi> means the expectation value of A on Psi: so if A were to operate on state Psi, <A> would be the probabilistic distribution of the result. What is a state Psi anyway? Is it a snapshot of a system? So it contains all information about a system at a certain time?
    I cannot think of a time dependent (t.d.) operator but an example of a time independent (t.i.d.) one might be Q^ - the translation operator (because it has no t in it?). So does the question ask what will happen when t.i.d. operator acts upon t.d. system? If so I have no idea.

    To re-iterate, I haven't the foggiest how the general state evolves in time.

    Also, I would appreciate a reference to an introductory QM text that uses bra-ket notation. I'm not sure if the Dirac notation is only useful for QM with systems with infinite degrees of freedom but that was the only book I could find that uses that notation.

    Thank you.
     
  5. Sep 30, 2007 #4

    Gokul43201

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    Start with the commutator and operate it on some arbitrary position-space wavefunction, [itex] \psi (x) [/itex]:

    [tex][f,p] \psi = f \left( -i\hbar \frac{\partial \psi}{\partial x} \right) - (-i \hbar) \frac {\partial}{\partial x} (f \psi) [/tex]

    You know how to find the derivative of a product. Use it to expand the second term on the RHS and see what you are left with.

    Similarly, try the other commutator.
     
  6. Sep 30, 2007 #5
    Brilliant... that's very helpful... thank you very much.

    Now for the other question; I think the state of a system evolves with
    [tex]i\hbar \frac{\partial }{\partial t} \right)|\psi(t)> = H|\psi(t)>[/tex]
    That LaTeX code stuff is helpful, so I'm learning it as I go along... Anyway, I still don't know what is being asked. The original question was "Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi." What is the thing that is "in a time dependent state Psi?" The time independent operator? The expectation value of an operator is in a state? What does that mean?

    I am not pretending; I have never seen any of this before. Should I perhaps somehow use the above time evolution equation with the expectation value equation [tex]<A>=<\psi|A|\psi>[/tex]? Maybe I should then take the derivative of something to find the time variance.

    Perhaps I should solve the above differential equation to get [tex]\psi=e^{H/i\hbar}[/tex]. This doesn't appear helpful and I'm not sure it even makes sense for an operator (H) to be used that way in a solution.

    Last guess... and this one may actually be partially coherent: From the very first equation we know that [tex]\frac{\partial }{\partial t} \right)|\psi(t)> = \frac{H}{i\hbar}|\psi(t)>[/tex]. This is the time derivative of [tex]\psi[/tex] and may be what we want in the first place. So is that the answer? Except we want to know how the expectation change with time so... ummm... [tex]<\frac{\partial A(t)}{\partial t} \right)> = <\psi|\frac{H}{i\hbar}\psi(t)|\psi>[/tex]?

    That looks totally made up and I have no idea how to bring x and p into this. Please help soon...
     
  7. Oct 1, 2007 #6
    I think this might have to do with perturbation theory. This homework is due in a few hours but even of you don't get around to answering by then I'd still like to know. Thanks.
     
  8. Oct 1, 2007 #7
    Check out wikipedia at http://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics). Under Time-dependent perturbation theory there is some helpful looking stuff. I think I might know how to put A in to all this but I totally don't understand the step that get you to d cn/d t. It says something about a partial differential equation but I don't know how to get there and I don't see where the V(t) comes from in the first place.
     
  9. Oct 1, 2007 #8

    malawi_glenn

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    no, perturbation theory is when your system is perturbed..

    Now you have a state that evolves with time, You should instead find info about "Quantum dynamics". Sakurai "Modern Quantum mechanics" has a good chapter about it.

    In shrödinger picture you have time indepentend operators, and time depending states (ket's). And to evolve a state in time, you apply the time evolution operator to it.

    And you could have an operator like this: [tex] e^{-iHt/\hbar} [/tex] Which is the time-evol-operator. The exponential of an operator is defined as the power series expansion of e^x; i.e you have 1 + x + X^2/2 + .. etc; but with your operator instead of x here.
     
  10. Oct 1, 2007 #9

    nrqed

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    never mind my other post, given the way the question is phrased, I thik what they want i sthe following. ( and you were on the right track).

    write [tex] \partial_t <A>A = (\partial_t <\Psi|) A |\Psi> + <\Psi| A ~(\partial_t |\Psi>) [/tex]

    Then use Schrodinger equation for the bra and the ket. You will end up with the expectation value of a simple expression containing A and H (if you use the notation of commutators)
     
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