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Physical reason for diagonal matrix

  1. Jan 24, 2009 #1
    Whats the physical reason for a diagonalized matrix with the eigenvalues of the system as the values?

    Reading from wiki, it seems that its something to do with the Schrodinger Eqn, but I don't follow that. If someone could explain the point of it to me in plain English (or as close as it gets!) then I'd be very grateful to them.
     
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  3. Jan 24, 2009 #2

    CompuChip

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    Can you be a little more specific? Are you talking about quantum mechanics, where a measurement of a Hermitian operator will always give an eigenvalue of the operator?
     
  4. Jan 24, 2009 #3
    Sorry, yes I am. I didn't realise there was a difference.
     
  5. Jan 24, 2009 #4

    CompuChip

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    If you have Griffith's book on QM, it is explained quite clearly in chapter 3; to be more precise in section 3.2. The argument is roughly as follows:
    • Recall that the expectation value of an observable A, for a system described by wavefunction [itex]\Psi[/itex] is
      [tex]\langle A \rangle = \int \Psi^* \hat A \Psi \, \mathrm dx[/tex]
    • We want the outcome of a measurement to be real, so in particular we want the expectation value (~ average) to be real:
      [tex]\langle A \rangle = \langle A \rangle^*[/tex]
    • It is now easy to see that this must lead to the property of A being Hermitian.
    • If the system is in a determinate state (a state in which a measurement of A yields a well-defined value; for example stationary states are determinate states of the Hamiltonian) then you want the standard deviation
      [tex]\langle \hat A - \langle A \rangle )^2 \rangle [/tex]
      to vanish. This can be shown to lead to
      [tex]\hat A \Psi = \lambda \Psi[/tex]
      for some number lambda. In other words, determinate states are eigenfunctions of [itex]\hat A[/itex].
    • The eigenfunctions of a Hermitian operator can be shown to be orthogonal (or, in case of multiple eigenfunctions for the same eigenvalue, we can choose them to be such). Moreover, the wave-function can be decomposed into the eigenfunctions [itex]f_n[/itex] of [itex]\hat A[/itex], meaning we can write
      [tex]\Psi = \sum_n c_n f_n[/tex] with [tex]c_n = \int f_n(x)^* \Psi(x, t) \, \mathrm dx[/tex]
      (in the discrete case, replace the sum by an integral in the continuum case)
    • Now using all these properties, you can show that
      [tex]\langle A \rangle = \sum_n \lambda_n |c_n|^2[/tex]
      where [itex]\lambda_n[/itex] are the eigenvalues [itex]\hat A f_n = \lambda_n f_n[/itex]. This is precisely the expectation value for an experiment where the outcome [itex]\lambda_n[/itex] has probability [itex]|c_n|^2[/itex]!

    To summarize then: the possible outcomes of a measurement of an observable are the eigenvalues of that observable. This means that we have to calculate the eigenvalues, which you know how to do in the finite case from linear algebra (you can write [itex]\hat A[/itex] as a matrix). Usually, if you're going to work with A (dropping the hat from now on) a lot in your particular problem, it might pay off to express everything in eigenstates of A. From linear algebra you know, that this is equivalent to diagonalizing A such that the eigenvalues are explicit in the way you write it down.

    Hope that clarifies.
     
  6. Jan 24, 2009 #5

    turin

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    You can do the same thing in classical mechanics. The typical example studied is that of a system of coupled oscillators. You can write the Lagrangian in terms of the displacements of the individual oscillators as the generalized coordinates, but you will find that the differential equations can be quite difficult to solve, because they mix the coordinates. You will find that it is much easier to solve the system if you discover some clever generalized coordinates that can be treated independently, so that you obtain a set of differential equations that each involve a single coordinate. This can be done by diagonalizing a matrix, and the characteristic frequencies are the eigenvalues.
     
  7. Jan 25, 2009 #6

    CompuChip

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    In fact, in any part of physics where linear algebra occurs (and since we like to work with linear things because they are so simple, it occurs almost everywhere - either exactly, or in approximation) you can play such tricks. For then you can work with eigenvalues and eigenvectors, which means diagonalizing the matrix, and in the end you can just take linear combinations afterwards to describe the whole system.
     
  8. Jan 25, 2009 #7
    There's a short, generally applicable answer too, but not real physical.

    The diagonalised matrix is a change of basis, where all vector quantities undergo dilation or contraction.

    Edit: I think that was unclear. We cn represent the matrix in a different coordinate system. In the coordinate system where the matrix is in diagonal form all vector quantities operated upon by the diagonalized matix undergo dilation or contraction.
     
    Last edited: Jan 25, 2009
  9. Jan 25, 2009 #8

    turin

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    But Brewer should be aware of the profound physical difference in the interpretation of this mathematics in QM vs. CM. In CM, the interpretation is basically as Phrak suggests, and then one must further interpret what is the meaning of such vector stretching. In QM, the interpretation is completely different, because a stretched version of a state vector is interpretted as the same state, and so the eigenvalues have no effect on the state, but rather the possible results of the measurement. In fact, instead of stretching the vector, in QM the vector is rotated but maintains its length.
     
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