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Are stationery states eigenstates?

  1. May 24, 2006 #1

    I have a couple of newbie questions. Are all stationery solutions to a SE eigenstates of observables? Is the converse true (that is, are all eigenstates stationery)? If the answer is no, can a measurement collapse a stationery wave-function onto a non-stationery one? And finally, is the probability of a superposition collapsing into a component state given by the squared modulus of the co-efficient of the respective state? Thanks.

    Molu, a clueless high-school boy who thinks QM is weird
  2. jcsd
  3. May 24, 2006 #2
    There are many observables. The relevant one here is the Hamiltonian of the system, and its eigenstates are precisely the stationary states. This is clear from the Schrodinger equation;

    [tex]\hat{H}\psi = i\hbar \frac{\partial \psi}{\partial t}[/tex]

    In eigenstates of the Hamiltonain, the left side is a real multiple of the wavefunction [tex]\hat{H}\psi=E\psi[/tex] (E being the [real] eigenvalue). The solution of this ODE is of course [itex]\psi(t)=e^{-\frac{i}{\hbar}E t}\psi(0)[/itex], a stationary solution. The converse is also true.

    A measurement collapses a wavefunction into an eigenstate of the observable corresponding to the measurement (i.e., position measurement~X operator, momentum measurement~P operator, etc.). If that observable is the Hamiltonian, of course this eigenstate is a stationary state. For other observables, this is not generally true. However, it is true when your observable A commutes with the Hamiltonian observable - i.e. [tex]\hat{A}\hat{H}=\hat{H}\hat{A}[/tex].

    "And finally, is the probability of a superposition collapsing into a component state given by the squared modulus of the co-efficient of the respective state?"

    Yes, that coefficent being in the expansion of the state in the eigenstate basis of the observable. The expansion must be a basis, and it must contain only eigenstates of your observable.
    Last edited: May 24, 2006
  4. May 25, 2006 #3
    Thanks a lot, that's very helpful.

    So if a state has a definite total energy, measuring its position may cause it to no longer have a definite total energy?
  5. May 25, 2006 #4
  6. May 25, 2006 #5
    what crap is this???:surprised

    How can a state change its energy when measured?
    One knows what energy a state has only be measuring it. This very process of measurement, collapses the wavefunction to the eigenstate, whose corresponding eigenvalue ( here energy ), our measurement yeilded.

    Remember, that in a "no-hidden-variable theory", like our present QM, there is no pre-existing value, the measurement process itself ascribes a value to the system.
    Last edited: May 25, 2006
  7. May 25, 2006 #6
    If you make a measurement you simply change the state of the system. If it was in a state described as a linear combination of eigenstates then it shifts to one of the possible eigenstates. Measurement does not affect any eigenstate's energy. It simply causes a wavefunction to 'collapse'. So if your system was described by a linear combination of eigenstates, i.e.

    [tex]\Psi(x,t) = \sum_{n=1}^{\infty}c_{n}\psi_{n}(x)exp(-i\frac{E_{n}t}{\hbar})[/tex]

    (where some of the [itex]c_{n}[/itex]'s could be zero beyond some maximum value of N)

    then upon a measurement the wavefunction collapses and is not described by the wave equation. The wavefunction evolves according to the Schrodinger equation under no measurement condition and when a measurement is taken, it collapses. These are two different processes.

    [itex]\Psi_{n}(x,t)[/itex] is the wavefunction corresponding to the n-th eigenstate. So,

    [tex]\Psi_{n}(x,t) = \psi_{n}(x)exp(-i\frac{E_{n}t}{\hbar})[/tex]

    After measurement which eigenstate is occupied by the system cannot of course be determined. (Separation of variables gives rise to an exponential time varying factor that is a function of the energy of the n-th eigenstate multiplied by the spatial part of course. I hope you understand my notation)
    Last edited: May 25, 2006
  8. May 25, 2006 #7
    An eigenstate, |u> of an observable B is a state which satisfies the relation

    B|u> = b|u>

    where b is the eigenvalue. A stationary state is one whose probability density is time independant. These are not the same things. It is possible for a state to be an eigenstate of an operator and the state not be a stationary state. An example is when the Hamiltonian operator is time-dependant, i.e. H = H(t). The Eigenvalues are the different energies the particle may have. The eigenstates will be a function of time. E.g. a charged particle in a time-dependant electromagnetic field yields such energies and non-stationary energy egenstates. I.e.

    H(t)|u,t> E(t)|u,t>

  9. May 25, 2006 #8

    George Jones

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    I assume that you're referring to the exchange

    with which I have no problem.

    If a quantum system is prepared in a state of definite energy, i.e., in an eigenstate of the Hamiltonian, then, in general, after a position measurement, the system will no longer be in an eigenstate of the Hamiltonian.

    Now, more detail. Assume an ensemble of identical physical systems are all prepared in the same energy eigenstate, and that this eigenstate has corresponding eigenvalue E. Immediately after state preparation, an energy measurement is made on each of the systems. With probability 1, i.e., with complete certainty, it can be prdicted that the result of each measurement is E. This is what is meant by "definite total energy."

    Suppose that instead of energy measurements, position measurements are made on each each system immediately after state preparation. Immediately following the position measurements, energy measurements are made on each system. Then, there will be a statistical spread in the results of the energy measurements. This is what is meant by "measuring its position may cause it to no longer have a definite total energy".

    I assume that you refer to the fact that in general, quantum system scannot be said to possess properties, as demonstrated, for example, by the Kochen-Specker theorem. However, exceptions in language are often allowed for eigenstates of observables. In his book Quantum Theory: Mathematical and Structural Foundations, Chris Isham writes "a situation in which it arguable is meaningful to say that a system 'possesses' this value for A."

    His italics.

  10. May 25, 2006 #9
    I'm not sure what you mean, we are talking about changing position, not energy. In any case, what is meant by energy or position here is the expectation value. A simple calculation reveals that for a stationery state (perfectly separable solutions to the TDSE) the standard deviation of the Hamiltonian is 0, so all measurements of energy yield the expectation value. Distinguishing between the measured energy and the expectation value of the enrgy is simply a matter of terminology here.
  11. May 25, 2006 #10
    So separable solutions to the TDSE are energy eigenstates only when the Hamiltonian is time-independent? I never saw anyone mention this condition. Griffiths just says all stationery solutions are eigenstates of the Hamiltonian with the eigenvalue being the energy expectation value and the standard deviation 0. There's no mention of time-dependence of the Hamiltonian.
  12. May 25, 2006 #11
    And one more question, for any given hermetian operator A, do the eigenstates of A constitute an orthonormal basis spanning the complete Hilbert space?
  13. May 25, 2006 #12
    He probably never got into situations with a time-dependant Hamiltonian. Look for such a Hamiltonian in his text book as well as look for a place where he might have said that he will only address time-independant Hamiltonians. Note: If the Hamiltonian of a system is a function of time then the energy of the system is not conserved.

  14. May 25, 2006 #13

    George Jones

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    This is a deep and subtle question to which physicists and mathematicians might give different answers.

    Do you mean hermitian or self-adjoint. If you mean self-adjoint, then a phyicist might answer "Yes, for each self-adjoint A there are eigenstates [itex]\left| a \right>[/itex] such that for every [itex]\left| \psi \right>[/itex] in Hilbert space,

    [tex]\left| \psi \right> = \sum_{i} \left| a_{i} \right> \left< a_{i} | \psi \right> + \int \left| a_{\lambda} \right> \left< a_{\lambda} | \psi \right> d \lambda,[/tex]"

    while a mathematician might answer "No, since, for example, the position operator has no eigenvalues or eigenvectors."

    Other subtleties include: What does span mean in an infinite dimensional space? What is a Hilbert space? etc.

    Last edited: May 25, 2006
  15. May 25, 2006 #14


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    Thats correct George, and the mathematicians will be completely correct. Hermiticity and self adjointness only coincide when the Hilbert space under consideration is finite dimensional (eg on a lattice or somesuch).

    What physicists are really doing is working in a certain subspace of the full thing (often called the physical Hilbert space) where these things now are well defined.
  16. May 25, 2006 #15

    George Jones

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    I don't think the question is one of dimension, I think it is one of boundedness. There exist Hermitian operators on infinite-dimensional Hilbert spaces, but these operators are necessarily (Hellinger-Toeplitz theorem) bounded. It is quite easy to show that for a quantum system, at least one of the conjugate position and momentum operators has to be unbounded, so Hermitian operators are not sufficient for observables in quantum theory. This introduces a lot a of grit (like operator domains that are not all of Hilbert space) into the mathematics that physicists can often ignore.

    I'd like to elaborate on this a bit.

    Given a Hilbert space H (kets), there is a natural bijection between the space of continuous linear functionals H' (bras) on H. If we take the space of kets to be S, a proper subset of H, then the set S' of continuous linear functionals on S is "larger" than H'. This is because a mapping that is continuous on H is automatically continuous on S since S < H, but a mapping that is continuous on S does not have to continuous on elements of H that are outside of S.

    So we have a (Gelfand) triple

    S < H = H' < S'

    Instead of taking ket space to be H and bra space to be H', take ket space to be S and bra space to be S', so there are "more" bras than kets. Another name for this setup is rigged Hilbert space.

    Included in the bra space S' that is dual of of S, the space of physical states, are delta functions and plane waves, so this allows physicists to work with distributions as "bras", and these can be used as weak eigenvectors of, e.g., the position operator.

  17. May 25, 2006 #16
    George, do you know of any good reference books (or articles or websites) on this topic that would be accessible at the early graduate level? I'm interested in learning about Gelfand triples in general from a mathematical point of view, and specifically how they allow us to use delta distributions and plane waves as eigenstates in physics.

  18. May 26, 2006 #17
    That went way over my head:( What I was asking was can all solutions to a Schroedinger equation be represented as linear combinations of the eigenstates of a particular observable? I'm only in high-school so please don't pull Gelfand triples and whatnots on me :D Thanks.
  19. May 26, 2006 #18
    The answer to this question posted here is yes.

  20. May 26, 2006 #19
    May I suggest Eisberg/Resnick and/or Griffiths...take the latter only if you have some background in differential and integral calculus, Fourier Series (and later Fourier Transforms) and complex variables. There are several other good books to chose (sadly some are out of print). Once you have the hang of the basic postulates then you could move on to the so called advanced texts.
  21. May 28, 2006 #20
    I don't know about Eisberg/Resnick, but I've been trying Griffiths as leisure reading, though I don't know anything about Fourier Analysis (they don't teach that in our school :).
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