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Eigenvalues in QM

  1. Mar 8, 2007 #1
    If after you apply an operator and hence calculate the expectation value of a measureable entity and if you get an eigenvalue, then does that mean when you do the measurement, you will always get the same value for that operator entity, each time?

    I think yes because otherwise what is so special about eigenvalues. If no eigenvalues exist then it implies you will get any random number for that measurement but after an infinite number of measurements, you should get the average or expected value. However how do you calculate that average if after operating on psi does not give an eigenvalue?
     
    Last edited: Mar 8, 2007
  2. jcsd
  3. Mar 8, 2007 #2

    mjsd

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    calculating expectation value doesn't tell u what you will get in a measurement. it is like saying that calculating the probability of flipping a coin doesn't mean it will show up "head" everytime... but I think you are trying to say that you have already measured the entity and any subsequent measurements will yield the same result... well, in that case, you have "collapsed" the wavefunction to one of its eigen-states.


    ok, let's clear this up a bit. You have a QM state. This state can be written as a linear superposition of an orthonormal set. This set can be made up of the eigenfunctions of say the Hamiltonian operator. If you choose to expand this state in terms of these funtions (convenient if you wish to know anything about the observable associated with the Hamiltonian... energy of course), the eigenvalues in this case is your energy spectrum, the coefficients of the linear expansions will deteremine the probability of measuring each eigenvalues. of course if the state is already in the one of the eigenstates then that simply means that the probability of getting this particular e-values is 100% and repeated measurements will yield the same result.
     
  4. Mar 8, 2007 #3

    nrqed

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    Not necessarily.
    Let's say an operator has eigenvalues 1, -1 and 3 (in some units).
    If the expecattion value is equal to 3, does it mean that every measurement gave 3? Not at all. It could be that 25% of them gave 1, 25% gave -1 and 50% gave 3.

    Or, imagine that the eigenvalues are 2, 5 and 8. You get an expectation value of 5. Maybe it's because 50% of the measurements gave 3 and 50% gave 8, yielding an average of 5, even though *no single measurement gave the value 5*!

    I am losing you here. For any measurement that can be done of an observable, this observable *necessarily* has some eigenvalues! Any single measurement *always* gives one of the eigenvalues of that observable!
    No, you can never get a purely random result. You always necessarily get one of the eigenvalues of the observable you are measuring.
    You calculate the integral [itex] \int \Psi^* {\hat A} \Psi [/itex] of the observable A (if it is written as a differential operator...For spin operators, the expectation value is a row vector times a matrix times a column vector).
     
  5. Mar 8, 2007 #4
    "For any measurement that can be done of an observable, this observable *necessarily* has some eigenvalues! Any single measurement *always* gives one of the eigenvalues of that observable!"

    I didn't know that. If an operator dosen't give an eigenvalue then does that mean no measurable result can be obtained? However with the particle in a box, the momentum operator does not give an eigenvalue but the expectation of the momentum equals 0 in that case. One can still measure the momentum and the average of a large number of measurements will be 0. So each measurement of the momentum of that particle can still be made, even though that operator does not involve an eigenfunction expansion.

    Also how can one get a range of eigenvalues in general? I am only used to there being a single eigenvalue for a particular state. i.e the eigenvalues depend on n, the energy state level.
     
    Last edited: Mar 8, 2007
  6. Mar 8, 2007 #5

    nrqed

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    This is one of the fundamental aspects of the measurement process in QM.
    I am not sure I follow th emeaning of that sentence. If you measure a certain observable, the result of the measurement is necessarily one of the eigenvalues of the operator corresponding to that observable.
    I am not sure what "the momentum operator does not give an eigenvalue" means. You mean the measurement of the momentum operator?
    The measurement of the momentum operator will necessarily give one of its eigenvalues.


    The fact that the average is zero just means that for each eigenvalue of the momentum that is measured, the corresponding negative value is also measured with an equal probability.

    This is a special case. You are thinking of measuring the energy of a state which si an eigenstate of the Hamiltonian, so yes, the measurement will necessarily give the corresponding eigenstate. But what if the atom is in the superposition [itex] (\psi(n=3) - 2 \psi(n=2))/\sqrt{5} [/itex]?
    Then measuring the energy will either give [itex] E_2 [/itex] (with a probability of 4/5 = 80% OR [itex] E_3 [/itex] with a probability of 20%.
     
  7. Mar 8, 2007 #6
    Consider a particle in a box. The wave function is a sine function. The momentum operator is a first derivative in x. So when you apply the momentum operator on this wavefunction of a particle in a box, you don't get back a constant * the original wave function. Hence no eigenvalue or eigenfunction applies in this case. However, it is still possible to measure the momentum of this particle in a box? It is still possible to measure fuzzy variables only that the value you get is an element of the real numbers whereas if the operator gives eigenvalue(s) then the measurement must be an element in this finite set of eigenvalues. Although as an aide, could you have an infinitely long wave function in which case the eigenvalue is infinitely many. However, this set of infinite eigenvalues must still be a proper subset of the real numbers.

    When you apply an operator to the wavefunction and it spits out an eigenvalue than it implies this observable is sharp. However, as you say, this eigenvalue may take on a few different values depending on the size of the superposition of the wavefunction. When the observable is measured however, only one of the eigenvalues which is sharp will eventuate.
     
  8. Mar 8, 2007 #7

    nrqed

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    Right. So the sine wavefunctions are not eigenstates of the momentum operator. But one can always rewrite any wavefunction as a linear combination of the eigenstates of the observable one is measuring!
    In that case, one may write

    [tex] \frac{1}{\sqrt{2}} sin(\frac{n \pi x }{a}) = \frac{1}{2 \sqrt{2}} (e^{i n \pi x/a} - e^{-i n \pi x/a}) [/tex] with the complex exponentials being eigenstates of the momentum operator.
    Measuring the momentum operator in the above state yields either [itex] n \pi \hbar /a [/itex] or [itex]-n \pi \hbar /a [/itex] with equal probabilities.
    It's possible to have continuous eigenvalues (for example a free particle has momenta that can take in principle any value between - infinity and + infinity) so the spectrum fo an observable may be continuous. The position of a particle in an infinite square well may take any value between 0 and a.
    Right.
     
  9. Mar 9, 2007 #8
    d/dx(e(x)-e(-x))=e(x)+e(-x) which is not equal to a constant multiple of e(x)-e(-x)

    In the particle in a box case, the complex numbers cancel as the i is canceled by division by i from the momentum operator constant h(bar)/i

    So no eigenfunction is apparent in this case. How did you get the states you calculated?
     
  10. Mar 9, 2007 #9
    I was applying the operator to the 'whole' wavefunction which like the first time will not give an eigenvalue. I think your point was to break psi into two psi(s) and only one come into existence when a measurement is made on the system (according to the Copehagan Intepretation). The momentum operator is applied to each smaller psi individually and one gets the two eigenvalues you quoted. They both exist in the system until a measurement is made and one come into existence with a 50% chance.

    I can see that if you apply an operator to a complete wavefunction and don't immediately get an eigenvalue then it implies more than one eigenvalue is present in the system so one needs to break the whole wavefunction into linear multiples so that each little psi will give a definite eigenvalue. In that way all measurements give a definite eigenvalue in QM but sometimes there is more than one in a system which makes the system fuzzy. But the measured values are still limited to be an element in the set of eigenvalues.
     
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