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Measurement question

  1. Aug 26, 2007 #1
    Hi all,

    I have some conceptual questions about measurement which I was wondering if you could help me with.

    Firstly, if the state of a system is expressed in terms of orthonormal (possibly degnerate) eigenfunctions [itex]\Psi = \sum_n c_n\psi_n[/itex] of an operator [itex]A[/itex], does this mean that any measurement of the system will find it to be in one the states [itex]\psi_n[/itex] with probability [itex]|c_n|^2[/itex]? What I'm trying to come to grips with is what stops us from measuring the system to be in both [itex]\psi_n[/itex] and [itex]\psi_{n+1}[/itex], say simulataneously. In quantum mechanics we always talk about systems being in superpositions of wavefunctions right? So why can't we observe some superposition of eigenfunctions which make up the totality our system?

    Secondly, suppose we want to measure the physical quantity associated with the operator [itex]A[/itex] for the the wavefunction above. If what I said is true, then to do this we simply apply [itex]A[/itex] to each of the component eigenfunctions to calculate the possible eigenstates, then calculate the corresponding probabilities. But what if I want to measure a property of [itex]\Psi[/itex] whose operator [itex]B[/itex] has a different set of eigenfunctions than those in which [itex]\Psi[/itex] is expressed. Then what I need to do is expand [itex]\Psi[/itex] in the eigenfunctions of [itex]B[/itex] right? Can I do this by expanding [itex]\Psi[/itex] component-wise? e.g. take an eigenfunction [itex]\psi_n[/itex] with nonzero coefficient in the expansion of [itex]\Psi[/itex] and then express the general eigenfunction of [itex]B[/itex] as a superposition [itex]\sum_n b_n \psi_n[/itex]. The question is, how do I know which [itex]\psi_n[/itex] I should keep in the expansion?

    Last edited: Aug 26, 2007
  2. jcsd
  3. Aug 26, 2007 #2
    Even though it is usually said, that measurement projects the state onto some eigenstate, it would be more correct to say, that it projects the state onto some set of eigenstates, depending on the accuracy of the measurement device. For example, if you measure a position of a particle, somebody might think that the wave function collapses to a delta function, but if your measurement accuracy isn't infinite, then it does not need to collapse to a delta function, but instead to some other localized function.

    So suppose that the state is [itex]|\psi\rangle = |0\rangle + |1\rangle + |10\rangle + |11\rangle[/itex], where [itex]|n\rangle[/itex] are some energy eigenstates. If your measurement accuracy isn't good, it could be that after a measurement the state is [itex]|\psi\rangle = |0\rangle + |1\rangle[/itex] or [itex]|\psi\rangle = |10\rangle + |11\rangle[/itex], and it would collapse further only with a more precise measurement.

    However, I don't think it would be correct to say that here the state would have been measured to be in two states simultaneously. That sounds like you had got two different measurement results, which isn't what happens.

    You can transform a representation of a state into another representation. In B-representation the state is

    |\psi\rangle = \sum_{k} \langle B_k|\psi\rangle | B_k\rangle.

    If you previously had representation

    |\psi\rangle = \sum_{k} \langle A_k|\psi\rangle | A_k\rangle,

    you can combine these to get

    |\psi\rangle = \sum_{k}\Big(\langle B_k|\sum_{j}\langle A_j|\psi\rangle |A_j\rangle\Big) |B_k\rangle = \sum_{j,k} \langle B_k|A_j\rangle \langle A_j|\psi\rangle |B_k\rangle.

    From this you see that

    \langle B_k|\psi\rangle = \sum_{j} \langle B_k|A_j\rangle \langle A_j|\psi\rangle.

    If there is a finite number of eigenstates, this is a usual matrix multiplication, where [itex]\langle B_k| A_j\rangle[/itex] are the elements of the matrix, and [itex]\langle A_j|\psi\rangle[/itex] are the components of the vector.

    If you measure observable B so that the state becomes [itex]|B_k\rangle[/itex], then in A-representation the state is

    |B_k\rangle = \sum_{j} \langle A_j|B_k\rangle |A_j\rangle,

    and in general this is not an eigenstate of A.

    I hope that helps. A direct answer to question
    would have been "keep those for which [itex]b_n[/itex] is nonzero!" But I guess that would not have helped. Anyway, the values of [itex]b_n[/itex] come from the matrix multiplication I showed above.

    About notation: [itex]b_n=\langle B_n|\psi\rangle[/itex]. Those are the same coefficients.
    Last edited: Aug 26, 2007
  4. Aug 27, 2007 #3
    Hi jostpuur,

    Thanks for the detailed reply. However, I'm still a little unsure exactly how the eigenfunctions for [itex]B[/itex] are obtained. Suppose the wavefunction is [itex]\psi = aY_{00} + bY_{10}[/itex], for example, where [itex]Y_{lm}[/itex] are the spherical harmonics. These are eigenfunctions of the operator [itex]\mathbf{L}^2[/itex], so applying it we get the measureable values of L^2.

    Now, since the operators [itex]L_x[/itex] and [itex]\mathbf{L}^2[/itex] do not commute (ie there is uncertainty relation between them), the operator [itex]L_x[/itex] has a different set of eigenfunctions. If we want to measure [itex]L_x[/itex], we must express [itex]\psi[/itex] in terms of its eigenfunctions. To do this, we presumably solve the eigenvalue equation

    [itex]L_x |X\rangle = \lambda |X\rangle[/itex]

    where [itex]|X>[/itex] are the yet to be determined eigenfunctions. If we let [itex]|X\rangle = \sum_{l,m} a_{lm} |Y_{lm}\rangle[/itex] then we get an infinite matrix equation and an infinite number of eigenvalues and corresponding eigenvectors. Only a small subset of these will be necessary to generate the originial function [itex]\psi[/itex]. So how do we know what restriction to place on the test function [itex]|X\rangle[/itex] for general operators? Is it just clever thinking or is there a systematic approach?
  5. Aug 28, 2007 #4
    The spherical harmonics represent the eigenfunctions of [itex]L_z[/itex]. The easiest way to obtain eigenfunctions of [itex]L_x[/itex] is to rotate spherical harmonics around y-axis with angle [itex]\pi/2[/itex].

    Yeah, you will get series. But I must admit right I know I don't remember this stuff very well. I'm not sure how to deal with the radial part in this problem. In one way or another, you will have to start computing inner products of the eigenstates of [itex]L_x[/itex], and the given state [itex]|\psi\rangle[/itex]. They are computed with some kind of integral in spatial space, and these inner products give you the probabilities. There will be infinitely these, but they should form converging series, and for approximations it sould be enough to deal with some of the first terms only, but that of course depends on how rapidly the series converge.

    But how to compute the inner products, and what to do with the radial part in the integral? I'm not sure. If the state is of the kind where the radial part and angle part are separated already, as it seemed to be in your example, then you can probably ignore the radial part, and compute the inner products as two dimensional integrals. Otherwise, you might have to choose some basis for the radial part too.
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