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I Difference between measurement and passing through a gate

  1. Nov 30, 2016 #1
    My question is as follows, given a hermitian operator, what's the physical difference between performing a measurement on a particle to get an eigenvalue of said operator and acting on the state of the particle with said operator?

    What prompted my question is the swap gate used in quantum information theory, it is hermitian so we can associate it with some measurement an get the particle(s) into an eigenstate of the operator through the measurement process, instead we can insert a swap gate into a circuit to act on the particles and instead of a measurement we get a swap operation, so how are the two processes physically different?
     
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  3. Nov 30, 2016 #2

    PeterDonis

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    You're confusing physics with math. A hermitian operator is a mathematical object, not a physical thing. It doesn't act on the physical state of the particle; it acts on the mathematical representation of that state. That is how the physical action of making a measurement on the particle is represented in the mathematical theory; if the model is correct then the two things will have similar properties. But you still need to keep the mathematical model distinct from the physical process that is being modeled.
     
  4. Nov 30, 2016 #3

    OCR

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    That's true... in fact, it lead to the formation of an axiom...

    It states: " Iff three ranchers are riding in the front seat of a pickup, the rancher in the middle is the most intelligent. "

    Analysis:.. The rancher in the middle will never have to drive, nor, will he ever have to open a gate, or, close a gate, thus... self-evidently true... Q.E.D.

    This is the axiom of TMIR ...


    Thanks for letting me point this out... :approve:
     
    Last edited: Dec 1, 2016
  5. Nov 30, 2016 #4

    atyy

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    The SWAP gate is unitary.

    Although a hermitian operator is associated with measurement, the operation that results is state reduction or collapse, which is not unitary. For example, a simple (not totally general) possibility for what happens after measurement is projection onto some eigenvector of the hermitian operator. The projection is not unitary.
     
  6. Dec 1, 2016 #5
    Maye I didn't state my question clearly enough, what I meant was that the operator is associated with two different processes and I want to know the difference in the physical setup to achieve said processes , i.e. to make a measurement you need to place such and such instrument (or whatever have you), while for acting on particle you need the particle to pass through such and such object.
     
  7. Dec 1, 2016 #6

    PeterDonis

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    No; the two different processes are associated with two different operators. A given operator can only be associated with one process.

    As atyy said, the usual "swap gate" operator in quantum information theory is unitary, not hermitian. It is not associated with a measurement; it is associated with the "swap qubits" process.

    I'm not sure what measurement you were thinking of in the OP; just swapping two qubits isn't a measurement.
     
  8. Dec 1, 2016 #7

    vanhees71

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    Why shouldn't a unitary operator represent an observable? E.g., it's wise to use ##\exp(\mathrm{i} \phi)## instead of ##\phi## to characterize an angle since in this case you have some trouble to define a self-adjoint operator to represent the angle but no problem to define the corresponding unitary operator by exponentiating an Hermitean operator and continuate it to the entire Hilbert space.
     
  9. Dec 1, 2016 #8

    PeterDonis

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    If by "represent" you mean that given a Hermitian operator ##H## that represents an observable, we can always define a unitary operator ##U = \exp(iH)##, then yes, we could always "represent" an observable by a unitary operator. But the eigenvalues of the observable, i.e., the possible measurement results, are the eigenvalues of ##H##, not ##U##.
     
  10. Dec 2, 2016 #9

    vanhees71

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    Fine, but the really important feature of self-adjoint (and I mean really self-adjoing; Hermitean is not sufficient!) is that there is a spectral decomposition, and unitary operators also have a spectral decomposition, and if you know ##\exp(\mathrm{i} \phi)## of course you know ##\phi## modulo ##2\pi##, which is precisely the definition of an angle. There's of course one subtlety in QT. If you define the angle operator by ##\exp(\mathrm{i} \hat{\phi})##, e.g., in the space of ##2 \pi## periodic square-integrable functions in the sense of the standard scalar product,
    $$\langle \psi_1|\psi_2 \rangle=\int_0^{2 \pi} \mathrm{d} \phi \psi_1^*(\phi) \psi_2(\phi),$$
    as
    $$\exp(\mathrm{i} \hat{\phi} \psi(\phi)=\exp(\mathrm{i} \phi) \psi(\phi),$$
    which obviously defines a unitary operator, the expection value of ##\phi## in an arbitrary pure state is defined as ##\langle \exp(\mathrm{i} \phi) \rangle=\exp(\mathrm{i} \bar{\phi})##.

    The point here is that you cannot define ##\hat{\phi}## itself as an operator, because it is obviously not self-adjoint, because with ##\psi(\phi)## in the Hilbertspace of ##2 \pi## periodic functions, that's no longer the case for ##\hat{\phi} \psi(\phi)=\phi \psi(\phi)##. So the operator is Hermitean, but of course, ##\psi(\phi) \mapsto \exp(\mathrm{i} \phi) \psi(\phi)## provides a perfectly fine unitary operator!
     
  11. Dec 2, 2016 #10

    PeterDonis

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    That, plus the fact that the eigenvalues are real. Unitary operators don't have real eigenvalues.

    I understand that there are issues involved with some observables; but the fact remains that the measurement results have to be real numbers, so, for example, in the case of angles, the measurement result is ##\phi##, not ##\exp(i \phi)##, even if the mathematical representation of how that result is obtained is somewhat roundabout.
     
  12. Dec 2, 2016 #11

    vanhees71

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    Since there is no self-adjoint operator representing ##\phi## you are not even able to define an angle as observable. That's why one uses ##\exp(\mathrm{i} \hat{\phi})## as explained above. If you insist on a real description (which I don't understand at all since exponentials are easy to deal with) you can as well use ##\cos \hat{\phi}## and ##\sin \hat{\phi}##. For details about the problem of angle/phase variables, see

    Phase and Angle Variables in Quantum Mechanics
    P. CARRUTHERS and MICHAEL MARTIN NIETO
    Rev. Mod. Phys. 40, 411 – Published 1 April 1968

    https://doi.org/10.1103/RevModPhys.40.411
     
  13. Dec 2, 2016 #12

    PeterDonis

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    Are those operators self-adjoint? The abstract of the paper you linked to (I can't read the full paper since it's behind a paywall) appears to indicate that they are. If so, that would address the issue; yes, ##\phi## itself is not an observable, but ##\cos \phi## and ##\sin \phi## are, and given both of those you can always compute a value for ##\phi## that serves all required purposes.
     
  14. Dec 2, 2016 #13

    PeterDonis

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    Sure, mathematically they are. But the OP was asking about the physical difference between a Hermitian (or more properly self-adjoint, as you point out) operator and a unitary operator. What does ##\exp(i \hat{\phi})##--that operator itself, regardless of its mathematical relationship to other operators--represent physically?
     
  15. Dec 3, 2016 #14
    Sorry it took me so long to reply to the thread, anyway I still think I'm not explaining my self properly, my question is as follows:

    given some operator U which represents an observable, one can define a measurement as a projection to one of the operator's eigenstates, alternatively one can use the operator to act on a physical state and change it, these two operations do not yield the same result, yet they are derived from the same operator, what I'm asking is when the time comes and I need to realize these two procedures physically ( let's say in a lab ), how are they related, if at all, do I use the same apparatus with some extra modification to perform a projection?

    additionally, is there any connection between the apparatus and the mathematical description of the operator ( other than the obvious if you pass the appropriate physical object through the apparatus you get the state predicted by acting on the original state by the operator)?
     
  16. Dec 3, 2016 #15

    vanhees71

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    It depends on what you want to describe. First of all, if you have a self-adjoint operator, its eigenvectors to different eigenvalues are orthogonal to each other and you can always choose a complete set of orthonormalized eigenvectors (I assume that there is no continuous part in the spectrum, where things become slightly more complicated, and you have to deal with generalized functions, but all that happens on a non-rigorous level is that the sums I write down in the follogin become integrals). So let ##|a_n \rangle## be a complete orthonormal set of eigenvectors of the self-adjoint operator ##\hat{A}## you have
    $$\sum_{n} |a_n \rangle \langle a_n|=1, \quad \langle a_{j}|a_k \rangle=\delta_{jk}.$$
    If the system is prepared in a pure state, represented by a normalized vector ##|\psi \rangle## then
    $$|\psi_j|^2=|\langle a_j|\psi \rangle|^2$$
    is the probability for measuring the value ##a_j## of the observable represented by the self-adjoint operator ##\hat{A}## (provided the eigenvalues are non-degenerate). This is Born's Rule.

    Now suppose you want to describe an ideal von Neumann filter measurement (which almost never can be performed in practice, by the way). Then you use your measurement device to just select systems for which you find ##a_j##, and through away all other. Then you prepare a new pure state that is represented by ##|\psi' \rangle=|a_j \rangle##.

    If you want to evaluate the expectation value of the operator, given the system is prepared in a pure state ##|\psi \rangle##, then you have to evaluate
    $$\langle A \rangle=\langle \psi|\hat{A} \psi \rangle.$$
    That this is compatible with Born's rule, you easily see by inserting a unit operator in terms of the eigenvectors of ##\hat{A}##:
    $$\langle A \rangle = \sum_j \langle \psi|\hat{A} a_j \rangle \langle a_j|\psi = \sum_j a_j \langle \psi | a_j \rangle \langle a_j|\psi \rangle = \sum_j a_j |\langle a_j|\psi \rangle|^2.$$
    This is the usual rule to evaluate the expectation value of a quantity given that the probability that it takes the value ##a_j## is ##|\langle a_j|\psi \rangle|^2##, i.e., that's what Born's Rule postulates, and that shows that the formalism is consistent.
     
  17. Dec 3, 2016 #16

    atyy

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    For a unitary operation, one can use the known interaction terms in eg. QED to design the desired operation.

    For a measurement, one uses the known quantum physics to describe the interaction of the measurement apparatus, treated as a quantum system, with the system of interest. Then one makes a position measurement of the pointer on the measurement apparatus. The position of the pointer, if the measurement apparatus has been correctly designed, will have the same distribution as the observable you wish to measure, eg. http://www-bcf.usc.edu/~tbrun/Course/lecture08.pdf
     
  18. Dec 3, 2016 #17

    PeterDonis

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    Yes, they do, because from a strictly "apply the operator" point of view, a measurement is not a projection to one of the eigenstates; it's just applying the operator to the state. For example, suppose I have a particle in some spin state ##|s>##, and I want to measure its spin in the ##z## direction. I apply the operator ##S_z## to the state ##|s>##, and, assuming ##|s>## is not itself an eigenstate of ##S_z##, I get a superposition of ##S_z## eigenstates--not a single eigenstate. The single eigenstate only comes into play if I adopt a collapse interpretation, but I don't need to do that to get the correct experimental result; I could use the MWI or another no collapse interpretation and still make correct predictions.

    So by applying the ##S_z## operator (more precisely, by using a physical mechanism that is mathematically modeled by that operator), I acted on a physical state and changed it, but I also made a measurement. So the two operations are not different if I'm taking this point of view; the only difference is which operator I use. Operators that are not Hermitian (or more precisely self-adjoint, as vanhees71 pointed out) cannot represent observables, because they don't have real eigenvalues and measurement results must be real numbers. In other words, the operator ##S_z## above must be self-adjoint; but there are other operators I could apply to the same state that are not self-adjoint, but which still change the state. (AFAIK the "swap qubits" operator is such an operator.)
     
  19. Dec 3, 2016 #18

    vanhees71

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    I don't understand what you mean that applying ##\hat{s}_z## to the state vector represents measuring ##s_z##. That's nowhere in the quantum-mechanical formalism.

    As I tried to say before, there's nothing that prevents using complex numbers as the results of measurements, e.g., I can define an angle ##\phi## uniquely by giving the corresponding phase factor ##\exp(\mathrm{i} \phi)##, and that's much more convenient than (but equivalent to represent it by the two real quantities ##\cos \phi## and ##\sin \phi##.

    The reason that self-adjoint operators (or equivalently unitary operators) are well suited to represent observables is that they have a complete orthonormal set of eigenvectors (modulo the refinements necessary due to the possibility of continuous spectra) and thus allow for a consistent interpretation of the states in the sense of Born's rule.

    In short, the formalism goes by the following postulates: Associated with any quantum system is a Hilbert space and an algebra of observables (usually represented by self-adjoint operators on the Hilbert space). A state of a quantum system is uniquely determined by a statistical operator, i.e., a self-adjoint positive semidefinite operator with trace 1, ##\hat{\rho}##. A state is by definition called pure state if it is described by a projection operator, i.e., if ##\hat{\rho}^2=\hat{\rho}##, which implies that there exists a vector of norm 1 such that ##\hat{\rho}=|\psi \rangle \langle \psi|##. The possible values (if you restrict yourself to the standard case of self-adjoint operators to represent observables indeed necessarily real ones) of an observable are the eigenvalues of the corresponding operator, and there always exists a complete orthonormal set of such eigenvectors. Take ##|a_j,\beta_k \rangle## as the complete set of orthonormal eigenvectors of the self-adjoing operator ##\hat{A}## (with ##\beta_k## labeling possible degeneracies of eigenvectors), then the probability to find the value ##a_j## when measuring ##A##, provided the system is prepared in the state described by ##\hat{\rho}## is given by (Born's rule)
    $$P(a_j|\hat{\rho})=\sum_k \langle a_j,\beta_k|\hat{\rho}|a_j,\beta_k \rangle.$$
    This implies that the value of an observable is determined if and only if
    $$\hat{\rho}=\sum_k p_k |a_j,\beta_k \rangle \langle a_j,\beta_k|.$$
    Nowhere in this standard formulation represents the application of an operator on a state vector (or the statistical operator for the more general formulation provided here) a measurement of the corresponding observable!
     
  20. Dec 3, 2016 #19

    PeterDonis

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    Ordinary language is vague, and I am evidently not using it in a way that is useful for this discussion. You have given the mathematical formalism, which is a much better way of doing it.
     
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