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Complex Wavefunctions

  1. Jun 28, 2014 #1
    Why do wavefunctions in quantum mechanics need to be complex? What are the drawbacks of using real valued wavefunctions like: Asin(kx+ωt+ø) etc...or a standing wave equation: Asin(kx)cos(ωt)?
    I'm an undergraduate student and recently passed 12th grade...So any answer of my level would be appreciated...
     
  2. jcsd
  3. Jun 28, 2014 #2
    Its give anther dimension to play
     
  4. Jun 28, 2014 #3
    You can use them but then A ends up being complex.
     
  5. Jun 28, 2014 #4
    This issue has been discussed several times here. See my take at https://www.physicsforums.com/showpost.php?p=3799168&postcount=9
     
  6. Jun 28, 2014 #5
  7. Jun 28, 2014 #6
    At least in some important cases, you can describe the matter field (say, electrons) with one real function. A gauge transform (you can google it, if you don't know what it is) is required for that, but, e.g., in the case of the Dirac equation, some additional tricks are needed.
     
  8. Jun 28, 2014 #7

    PeterDonis

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    The usual answer is, because you can't represent quantum interference phenomena using real-valued wavefunctions. A more precise answer would be that, while you can, in principle, use real-valued functions to represent quantum interference, as other posters in this thread have described, doing so makes the math a lot more complicated; things are a lot simpler when you use complex numbers.
     
  9. Jun 28, 2014 #8
    The reason why they need to be complex overall is to avoid discontinuities in the wavefunction. If you want a function to be reversible it needs to be continuous. Consider something like a nuclear decay or a reflection. In the real world this is all or nothing, there is no halfway house. If you want to model it probabilistically that extra degree of freedom is required. For example for a reflection to be reversible the image needs to pass out of the mirror and into the other side, it requires that extra degree of freedom to be continuous.
     
  10. Jun 29, 2014 #9
    I sense maybe you are not as comfortable with complex numbers as you could be. They usually tend to be explained terribly. They are not as scary and mysterious as they are usually made out to be. So, this gives me another chance to plug my favorite math book, Visual Complex Analysis, which may be expensive to buy, but if you're lucky you can find a copy at the library and there are sample sections available online:

    http://usf.usfca.edu/vca//PDF/vca-toc.pdf
     
  11. Jun 29, 2014 #10

    bhobba

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    It's because they are representations in the position basis of complex vector spaces.

    So your question really is why complex vector spaces.

    Here's why. Suppose we have a system in 2 states represented by the vectors [0,1] and [1,0]. These states are called pure. These can be randomly presented for observation and you get the vector [p1, p2] where p1 and p2 give the probabilities of observing the pure state. Such states are called mixed. Probability theory is basically the theory of such mixed states. Now consider the matrix A that say after 1 second transforms one pure state to another with rows [0, 1] and [1, 0]. But what happens when A is applied for half a second. Well that would be a matrix U^2 = A. You can work this out and low and behold U is complex. Apply it to a pure state and you get a complex vector. This is something new. Its not a mixed state - but you are forced to it if you want continuous transformations between pure states.

    QM is basically the theory that makes sense of such weird complex pure states (which are required to have continuous transformations between pure states) - it does so by means of the so called Born rule.

    Thanks
    Bill
     
    Last edited: Jun 29, 2014
  12. Jun 30, 2014 #11
    My impressions from self study so far:

    As an exercise, I derived a "real valued" second order in time version of the Schrodinger equation before. It didn't look very pretty (del^4 operators and whatnot) but there it was.

    I think the reason why we have chosen to represent this second-order wave behavior as a complex number (and the use of complex numbers, or more complicated mathematical objects than the reals always seems to be a choice of representation, IMO) as opposed to some real-valued (field, field time derivative) or other representation is that
    1. It allows us to reduce the time order of the PDE.
    2. We can only measure relative phase of the wavefunction and have no real reason to privelege any given phase as being the "peak" of a physical wave, like we do for ex voltage in an electric line. (BTW, waves in electrical engineering are also represented by complex numbers, though it is understood that the imaginary parts are thrown away eventually to recover the "real" wave behavior.)
     
  13. Jun 30, 2014 #12

    bhobba

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    No.

    The use of complex numbers (specifically complex vector spaces) is absolutely fundamental to QM.

    One reason is its the only way to generalise the equations of classical physics:
    http://arxiv.org/pdf/1204.0653.pdf

    The other is if you wish to model physical processes in a reasonable way as a theory of observations you basically end up with two choices - bog standard probability theory and QM with complex numbers:
    http://arxiv.org/pdf/quantph/0101012.pdf

    Another reason is its required for entanglement:
    http://arxiv.org/abs/0911.0695

    Thanks
    Bill
     
  14. Jul 2, 2014 #13
    As Bill says it is fundamental to the theory, not just a more practical math formalism. I like the explanation in #10, and would like to see how it would relate to the more common explanation based on the de Broglie dispersion relations between angular frequency and wavenumber.
     
  15. Jul 2, 2014 #14
  16. Jul 6, 2014 #15
    Also it might be of interest to note that if you only allow real solutions to the Schrödinger equation then there is no time evolution. I.e. a universe with only real wavefunctions would be static!

    This is easy to see: the Schrödinger equation is [itex]\boxed{i \frac{\partial \psi}{\partial t} = H \psi }[/itex] and H is real, so if psi is also real, then the left-hand side is imaginary and the right-hand side is real, hence both are zero and in particular [itex]\frac{\partial \psi}{\partial t} = 0[/itex].
     
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