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Quantum superposition and Vector Space

  1. Feb 13, 2012 #1
    This is an extract from my third year 'Foundations of QM' lecture notes:

    If ψ1 and ψ2 are admissible states,
    then the superposed state [itex]\alpha[/itex]ψ1 + ψ2[itex]\beta[/itex] ( [itex]\alpha[/itex],[itex]\beta[/itex] [itex]\in[/itex] C ) is also an admissible state.

    [itex]\rightarrow[/itex] complex vector space.



    I understand that a linear superposition of allowed state is also an allowed state. It follows from the linearity of the Schrodinger's equation.

    What I fail to understand is how this leads to the concept of a complex vector space in QM.
    Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 13, 2012 #2

    tom.stoer

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    Linear superposition directly leads to the concept of vector spaces; b/c the coefficients alpha and beta can be complex you get a complex vector space
     
  4. Feb 13, 2012 #3

    bhobba

    Staff: Mentor

    Good question. Here is why you need complex numbers. Lets first consider a possibly loaded dice. How it behaves is given by what is called a probability (or state) vector which is 6 positive numbers that add up to one and gives the probability of the number that comes up when tossed. The vectors with a 1 in it means it will give the number at the position with dead certainty and also gives the possible outcomes of throwing the dice. Such states are called pure states.

    Now an interesting question is this - can you via pure states continuously go from one pure state to another? Well obviously you can't since there is only a finite number of them. But what if you insist. Then by examining the matrix that transforms between pure states and finding its eigenvalues you can actually find such a transformation - but it requires complex numbers.

    So basically you have two possibilities - no continuous transformations between pure states or you allow them. The first is standard probability theory. The second requires complex numbers and is what Quantum Mechanics use. When you think about it it's a better alternative than standard probability theory for physics because the pure states are the possible outcomes and you expect to be able to continuously go from one to the other and not jump between them. But how do you define probabilities on complex numbers? That's where a nifty theorem comes in called Gleasons theorem which basically says there is only one way to do it - the standard way its dome in Quantum Mechanics:
    http://kof.physto.se/theses/helena-master.pdf [Broken]

    So basically that's why you need complex numbers - so you can continuously go between the possible outcomes. It explains the superposition principle and a lot more. Combine it with Gleasons Theroem and you have Quantum Mechanics - pretty neat hey.

    For some more detail on this check out:
    http://www.scottaaronson.com/democritus/lec9.html
    http://arxiv.org/pdf/quant-ph/0111068v1.pdf

    Thanks
    Bill
     
    Last edited by a moderator: May 5, 2017
  5. Feb 13, 2012 #4

    Fredrik

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    The question of why we need a complex vector space is a difficult one, but if we take the things you say that you already understand as the starting point, it's completely trivial. The set of complex valued functions on ##\mathbb R^3## with addition and multiplication by a complex number defined the usual way (i.e. (af)(x)=a(f(x)) and (f+g)(x)=f(x)+g(x)), is a complex vector space. The assumption that you say that you understand says that we're considering a non-empty subset of that vector space that's closed under complex linear combinations. That makes it a subspace of that larger vector space.
     
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