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Deriving <v> from <x> - Introductory Quantum Mechanics

  1. Jun 12, 2013 #1
    (First post, hi everyone.)
    I'm reading Griffith's textbook on intro to quantum mech right now. After establishing the following equation

    $$\left\langle f(x) \right\rangle = \int f(x)p(x)\,dx$$

    And thus

    $$\left\langle x \right\rangle = \int x|\Psi|^2\,dx$$

    He goes on to find ##\left\langle v \right\rangle## by taking the derivative with respect to time, and he writes

    $$\frac{d\left\langle x \right\rangle}{dt} = \int x\frac{\delta}{\delta t}|\Phi|^2\,dx$$

    which I don't completely understand. Why isn't the x inside the partial derivative? I'm not sure if it's a physical or mathematical concept I'm not grasping here.

    I first thought x is the position, so it could be a function of time and written as x(t), but then I figure maybe x(t) doesn't really have any meaning from the statistical perspective. If it can be pulled out of that derivative, then it's just a...constant? Is there a physical meaning to that x or is it just an operator tool that makes the math work out?
     
  2. jcsd
  3. Jun 13, 2013 #2
    Each state of definite x has an associated value of x, which is the x position that the state represents. It is this value of x which appears in the integral to find the average (weighted mean) of the observable x. It's something that is fixed - each definite x position state has an associated x value, which is the value of x that this state represents. The changes in state are changes in [itex]\psi[/itex] - but the correspondence between eigenvectors and eigenvalues remains fixed.

    Here are some examples

    [itex]\mid x_1\rangle[/itex] represents the state of being at [itex]x_1[/itex]. Suppose our wave function is 100% at [itex]x_1[/itex]. Then [itex]\mid \psi \rangle = \mid x_1 \rangle[/itex]

    [tex]\langle \hat{x}\rangle =x_1 \langle x_1 \mid x_1 \rangle = x_1[/tex]

    if instead we had [itex]\mid \psi \rangle = a_1(t)\mid x_1 \rangle + a_2(t)\mid x_2 \rangle[/itex]

    [tex]\langle \hat{x}\rangle = x_1|a_1(t)|^2 + x_2|a_2(t)|^2 [/tex]

    So we have a weighted average of positions [itex]x_1[/itex] and [itex]x_2[/itex]. I wouldn't make sense to have [itex]x_1[/itex] and [itex]x_2[/itex] change with time.

    The integral [itex]\left\langle x \right\rangle = \int x|\Psi(x)|^2\,dx[/itex] is like this only instead of two positions there are infinitely many.

    The eigenvalue (which is the value of position) [itex]x_1[/itex] will always correspond to the state [itex]\mid x_1 \rangle [/itex]. Hopefully this will make sense and be intuitive.
     
    Last edited: Jun 13, 2013
  4. Jun 13, 2013 #3
    That helped a lot. Thank you :]
     
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