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A One-dimensional systems

  1. Dec 19, 2017 #1

    SeM

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    Hi, I have heard (or imagined) that a wavefunction, where Psi is on the y-axis and the positions x is naturally on the x-axis, is really a one-dimensional system in Physics (not in mathematics), because the signal or the oscillation of the wavefunction is not really a dimension, and only the position x makes any physical sense in a cartesian system . Is this correct?

    If so, how can a wavefunction Psi(x) generate a one-dimensional subspace as such :

    \begin{equation}
    \mathcal{Y} = [ \phi | \phi = \beta \psi , \beta \in \mathbb{C} ]
    \end{equation}



    where \phi is a function of Y of norm 1 and beta is an arbitrary constant, thus defining the probability density of \psi?
     
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  3. Dec 19, 2017 #2

    George Jones

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    No. "one-dimensional" refers to the real (as opposed to complex) dimension of domain of ##\psi##. As I asked you in another thread, what does "domain" mean?

    Also, if ##\psi## is complex-valued, how can it be on the y-axis?
     
  4. Dec 19, 2017 #3

    PeterDonis

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    Where? Do you have a reference?
     
  5. Dec 20, 2017 #4

    SeM

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    Hi George, and thanks for your comment. The reason I ask this is because of a part in Kreyszig's Functional Analysis, in the chapter 11, "Unbounded linear Operators in Quantum mechanics.", p 574, writes (quote):

    "Psi defines a state of our physical system at some instant to be an element :

    \begin{equation}
    \psi \in L^2(-\infty, \infty) \ \ \ \ \ \ \ \ \ (3)
    \end{equation}

    ..., where \psi in (3) generates a one-dimensional subspace:

    \begin{equation}
    \mathcal{Y} = \phi | \phi = \beta \psi, \beta \in \mathbb{C}
    \end{equation}

    of $L^2(-\infty, +\infty)$. Hence we could equally well say that a state of our system (Psi) is a ONE-DIMENSIONAL subspace $Y \subset L^2(-\infty,+\infty) and then use a $\phi \in Y$ of norm 1 in defining a probability according to the probability density intergral.

    end quote


    Having "heard" or imagined I can't remember which, that Psi on a REAL axis is not really a dimension in physics, and only the x-axis gives some dimensional concept (i.e. distance from nucleus for a bound electron), I thought this was what Kreyszig was talking about here. However, taking Peters word for it, I disregard from this misconception, and want to then ask:

    What does Kreyszig really mean here with saying that Psi is a one-dimensional subspace of Y which si defined by the two other wavefunctions \phi ? Does he mean that the integral of some other wavefunction \phi|\phi in a one dimensional space equals to the full space of the wavefunction (L^2) multiplied by some constant \beta?

    Thanks!
     
  6. Dec 20, 2017 #5

    blue_leaf77

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    That's the dimension of this subspace, this subspace has only one basis which is ##\psi##. Any vector in it must be a linear combination of basis vectors in the subspace but since there is only one such basis in the space under consideration, an arbitrary vector ##\phi## there is written as ##\beta\phi##. I have not read the quoted book but I guess the author brings up this discussion in order to introduce the so-called normalized vectors contained in the square-integrable vector space.
     
  7. Dec 20, 2017 #6

    SeM

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    That is correct.

    Thanks for the extensive explanation. I see one recurring point, that a wavefunction can be presented as vectors, so in this case, those are the ones given in ##\phi## and ##\beta## is some arbitrary constant (such as a coefficient) and makes them a linear combination of ##\psi##
     
    Last edited by a moderator: Dec 20, 2017
  8. Dec 20, 2017 #7

    George Jones

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    The title you chose for the thread is "One-dimensional systems".

    It is not

    that indicates that the system is one-dimensional, it is

    Do you understand why?
     
  9. Dec 20, 2017 #8

    SeM

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    Thanks for asking George. I see that ##\psi## is element of Hilbert space L^2, after Perok explained the nature of Hilbert space and its elements (functions), However, the elements of the functions can be represented as vectors, and are thus in ##\mathbb{R}## or ##\mathbb{C}##. In this case, as Kreyszig says, I am not sure! But part of this suggests that a part of ##\psi## (COMPLEX part?) can be considered as one-dimensional and is in##\mathbb{C}##? Considering the post by Blue leaf, it must, in this assumption, be the complex vectors of ##\psi## (and the vectors are called ##\phi##) which are in ##\mathbb{C}## and these vectors are one-dimensional, but NOT the function.
     
    Last edited by a moderator: Dec 20, 2017
  10. Dec 20, 2017 #9
    Aren't you just mixing up configuration space and Hilbert space?

    A one-dimensional configuration space means that that there is only one spatial dimension/degree of freedom in play, meaning a system that would also be classically considered one-dimensional. This is the case if ##\psi=\psi(x)## without e.g. y or z coordinates.

    Think for example of the standard particle in an infinite well. Even if we look at the standard case which is one-dimensional in the configuration space mentioned above, the wavevector-solutions form an infinite-dimensional Hilbert space.

    And if you want to draw the wavefunction for such a system as a graph on a sheet of paper(which is 2D), both the real and the imaginary part can indeed be visualised where horizontally comes ##x## and vertically ##Re[\psi(x)]## or ##Im[\psi(x)]##. But that is just the drawing, what you draw on the 'y-axis' has nothing to do with dimensionality of the system in any sense!
     
  11. Dec 20, 2017 #10

    SeM

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    This is what I "had heard" or "imagined" I heard, but I could not reproduce it accurately. The y-axis is not a physical dimension, as is x?
     
  12. Dec 20, 2017 #11
    It is a dimension on your drawing, but it doesn't describe a dimension. You even don't need QM to see this actually. Eg. if you would plot a graph of the temperature of the earth's crust as function of depth, there is only one dimension (depth), temperature is not a dimension, but on a graph you would draw this temperature along a y-axis anyway.
     
  13. Dec 21, 2017 #12

    SeM

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    Thanks, this is what I ment when I said that the oscillation of a wavefunction is not a spatial dimension, such as is the position.
     
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