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Observables in Quanum Mechanics

  1. Jun 26, 2005 #1
    In the context of observables in QM the rate of change of the In the contest of observables in QM the rate of change of the expectation value of an observable A is defined by:

    d<A>/dt= d< Y | A Y>/dt = < d Y/dt | A Y > + < Y | dA/dt . Y > + < Y | A dY/dt>

    My question is about the second term (< Y | dA/dt . Y >), is it mathematically correct to differentiate an operator!? Can anyone explain to me what's going on? After all d/dt is an operator that works on a function or on another operator as well? I am confused. :confused:
     
    Last edited by a moderator: Jun 26, 2005
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  3. Jun 26, 2005 #2
    Yes, there's no reason you can't differentiate an operator. For example, in position space the position operator is just [tex]\hat{X}=x[/tex]. So then you can find the time derivative of it as:

    [tex]\frac{d\hat{X}}{dt} = \frac{dx}{dt} = v.[/tex]
     
  4. Jun 26, 2005 #3

    dextercioby

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    There's no problem with "conceiving" operatorial functions.The trouble appears with boundedness,losing selfadjointness...

    Daniel.
     
  5. Jun 27, 2005 #4
    Are you sure? really sure?

    Seratend.
     
  6. Jun 27, 2005 #5

    dextercioby

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    Mikeu,think about Heisenberg picture.:wink:

    Daniel.
     
  7. Jun 27, 2005 #6
    I was pretty sure... I know that in the Schroedinger picture using wave mechanics the position operator is equal to x. I know that the time derivative of x is equal to dx/dt. Why then would the time derivative of the position operator not also be dx/dt? Which part should I reconsider?

    Daniel said to think about the Heisenberg picture, but I haven't learned that yet... My understanding of it is that states are stationary and operators evolve in time I think, which presumably affects their time derivatives somehow, but that wouldn't make what I said incorrect in the Schroedinger picture would it?

    Also, if my answer is incorrect then does anybody have an answer to the original question?

    Mike
     
  8. Jun 27, 2005 #7
    In QM, we cannot associate a path to a particle (HUP allows one to accept this property in a unformal way). Hence you cannot have v=dx/dt. However, you may have the operator equation dX/dt=V, but you should pay attention that V is an operator and is not equal to dx/dt, where x is the eigenvalue of operator X at time t (otherwise you can define a path to a particle), i.e. what you have done on your previous post.

    Seratend.
     
  9. Jun 27, 2005 #8

    George Jones

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    Differentiation is expressed using limits.

    In what follows, I'll proceed formally with very little regard to rigour. Knowing how much rigour to use is a difficult issue - if high standards of rigour had been insisted on at all times, then the amazing progress that has been made in applying quantum theory to physical systems would not have been made. However, this does not mean that rigour is something that can be completely ignored. While traversing Roger Penrose's Road to Reality, I came across the following relevant passage

    "Quantum mechanics is full of irritating issues of this kind. As the state of the art stands, one can either be decidedly sloppy about such mathematical niceties, or else spend the whole time insisting on getting the mathematics right, in which case there is a contrasting danger of getting trapped in a 'rigour mortis'. I am doing my best to steer a middle path, but I am not at all sure what the correct answer is for making progress in the subject!"

    Interesting words from someone who, controversial views notwithstanding, has deep understanding of both pure mathematics and theoretical physics.

    First, a look at what [itex] A(t) [/itex] means. Let [itex] S [/itex] be the state space for the system. For each [itex] t [/itex], [tex] A(t): \mathbb{R} \rightarrow S [/itex].

    Following your lead, I'm not going to use bras for elements of [itex] S [/itex]. The inner product is also a mapping, i.e., [itex] <|>: S \times S \rightarrow \mathbb{C} [/itex]. Let [itex] \psi [/itex] be an [itex] S [/itex]-valued function of [itex] t [/itex], and let [itex] \phi (t) = A(t) \psi (t) [/itex]. Thus, for [itex] A(t) [/itex] self-adjoint, [itex] <\psi | \phi> [/itex] is a real-valued function of a real variable (time), and standard intro calculus applies. So,

    [tex]
    \begin{equation*}
    \begin{split}
    \frac {d} {dt} <\psi|\phi>(t) &= \lim_{h\rightarrow 0} \frac {<\psi|phi>(t+h) - <\psi|phi>(t)} {h} \\
    &= \lim_{h\rightarrow 0} \frac {[<\psi(t+h)|phi(t+h)> - <\psi(t+h)|phi(t)>] + [<\psi(t+h)|phi(t)> - <\psi(t)|phi(t)>]} {h}.
    \end{split}
    \end{equation*} [/tex]

    The [itex] 1/h [/itex] can certainly be "pulled inside" the inner products, but, can the limits also be pulled inside, as you have implicitly done? I think so, but I'm not sure. In intro calculus, [tex]f[/itex] continuous guarantees that [itex] \lim_{x\rightarrow a} f(g(x)) =f(\lim_{x\rightarrow a} g(x)) [/itex]. Without further ado, pull the limits inside, which results in

    [tex]
    \frac {d} {dt} <\psi|\phi> = <\frac {d\psi} {dt}|\phi> + <\psi|\frac {d\phi} {dt}>. [/tex]

    Now, to your question: What does [itex] \frac {d\phi} {dt} = \frac {d} {dt} (A\psi) [/itex] mean?

    [tex]
    \begin{equation*}
    \begin{split}
    \frac {d\phi} {dt}(t) &= \lim_{h\rightarrow 0} \frac {A(t+h)\psi(t+h) - A(t)\psi(t)} {h}\\
    &= \lim_{h\rightarrow 0} \frac {[A(t+h)\psi(t+h) - A(t+h)\psi(t)] + [A(t+h)\psi(t) - A(t)\psi(t)]} {h}\\
    &= A(t) \frac {d\psi} {dt} (t) + \frac {dA} {dt} (t) \psi(t).
    \end{split}
    \end{equation*} [/tex]

    Here, I have written

    [tex] \frac {dA} {dt} (t) = \lim_{h\rightarrow 0} \frac {A(t+h) - A(h)} {h}, [/tex]

    so, I have only succeeded in shifting attention away from your question about differentiation of operators to the question: What does this limit of operators mean? If the operators are bounded, then the limit is with respect to the standard norm on the space of bounded operators. If the operators are unbounded, ...

    Unfortunately, I have had to put all but a handful of my physics and math books into storage, so I can't say more. Again, I've done formal manipulation and not much "real" functional analysis. A rigourous treatment needs to take into account stuff like dextercioby mentioned. I'm not sure you want to see the real thing, but if you do, Reed and Simon is a good place to look.

    Regards,
    George
     
  10. Jun 28, 2005 #9
    What if I selected A = d/dt, what does d(d/dt)/dt means???? :confused:
     
  11. Jun 28, 2005 #10

    dextercioby

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    A=d/dt is not an observable.It needs to be a classical Lagrangian or Hamiltonian obervable (depends on the quantization scheme).

    Daniel.
     
  12. Jun 29, 2005 #11

    reilly

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    Examples of operators that depend on parameters are easily found in many applications of Lie Groups in QM, as well as in dynamics. Perhaps it's a stretch to call a unitary transformation an operator, but that's what we will do. So the operator for a finite rotation is exp(i(Theta)J(i)) for a rotation of theta about the i axis, and theta is your basic parameter. Etc., Etc. (Lie groups are, by definition, composed of continuous group operators, which depend on one or more continuous parameters and which are differentiable as well. (Think of a power series in t with operator coefficients. Definitely differentiable.)

    In quantum optics, the interaction Hamiltonian, often involves an external radiation field, which is time dependent. Thus dH/dt = d(-j*A(x,t))/dt =-j*dA/dt in the Schrodinger picture. In the Interaction picture, often used in QFT, the Hamiltonian and momenta are functions of time.

    What is really tricky is the differentiation of one operator by another, which covers the type of functional differentiation used in QFT. That's another topic.

    Regards,
    Reilly Atkinson
     
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