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Can someone give me a geometrical description of the math in QM?

  1. Nov 12, 2009 #1
    So I’m trying to visualize what is going on in QM geometrically. More specifically I would like to visualize the time dependent wave equation in 3 dimensions. So let’s start with dimensionality. Normally when I think of a function of some variables, I picture it in a space with the number of dimensions equal to the number of variables plus one for the function. For instance, y(x) is usually mapped in 2 dimensions. So then my first impression of the wave equation is that it is a function of four variables, psi(x,y,z,t), so I should picture it in 5 dimensional space right?


    This seemed obvious to me but when I talked to my professor about it he didn’t like the idea but he didn’t really explain himself. So I thought about it some more and I guess his point was that because you can still describe any point in that function with 4 coordinates then it is only 4 dimensional (the definition of dimensionality). I think that does make sense. If we take y(x)=x^2, you could transform the coordinate system so that it is just a straight line with unique points or you could map it as a scalar field on the x axis. So then it really is only 1 dimensional. We then think of the wave function as a scalar field in 4 dimensions so I guess problem solved right?


    I’m not so sure. Look back at y(x)=x^2 and add y(x)=x to that plot. Now there is no way to describe those two functions in one dimension because you need more than one coordinate to describe any point. Even if you transform the coordinate system or plot as a scalar field you will have overlap. This system is truly 2 dimensional. So what if we consider two electron guns firing together once every second. If you plot the scalar field of the probabilities of the two electrons then you need 5 coordinates to describe any point in that system (the x y z position, the time, and which particle). Trust me this sounds crazy to me too so I’m thinking I’m just missing something simple and obvious. Anyone care to explain?
     
  2. jcsd
  3. Nov 12, 2009 #2
    Correct me if I misunderstood but adding these would produce a single value. The function would be f(x)=x+x^2.
     
  4. Nov 12, 2009 #3
    your wave function psi(x,y,z,t) maps R^4 onto C^1, four dimensions onto a single one in the complex plane (which does require 2 real numbers to specify).

    How would you transform the coordinate system such that y=x^2 is a straight line? No linear transformation exists. Consider a cannon ball shot in the air; its path is parabolic. In whatever coordinate system you chose to apply this, it always reaches an intermediate hight twice during the path. The physical system dictates this, and it is necessary that physical phenomenon are independent of the coordinate system you chose to express them in.
     
  5. Nov 12, 2009 #4
    Ya but that is not what I'm talking about. That is still a function. I'm talking about the two functions combined in such a way that the result is not a function (2 y's for every x).
     
  6. Nov 12, 2009 #5
    I guess coordinate transformation was a bad example. Can you elaborate on the first part you wrote?
     
  7. Nov 13, 2009 #6
    what MikeyW is saying is that your wave function is a mapping from Cartesian 4-space (R4) to the complex numbers (C1). That is each possible state in 4-space mapped to a single complex probability amplitude. Visualizing complex valued functions is not at all like real analysis.
    The square of the wave function, however, is a real number, and is proportional to the probability of finding a particle in 4-space at position (x,y,z,t).
     
    Last edited: Nov 13, 2009
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