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Time independence of probability amplitudes

  1. Jul 26, 2011 #1
    What's wrong with the following:

    [itex] \frac {d \left< E_{k} | \Psi \right>}{dt} = \frac{\partial \left< E_{k} \right|}{\partial t} \left| \Psi \right> + \left< E_{k} \right| \frac{\partial \left| \Psi \right>}{\partial t} = \frac{i}{\hbar} \left< E_{k} | H | \Psi \right> - \frac{i}{\hbar} \left< E_{k} | H | \Psi \right> = 0[/itex]

    and thus probability amplitude for energy [itex]E_{k}[/itex] should be time independent in any state [itex]\left| \Psi \right>[/itex].
  2. jcsd
  3. Jul 26, 2011 #2


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    First, to avoid confusion you must write the time-dependence of states explicit.
    Second, by your procedure you can show for ANY two states that

    [itex] \displaystyle\frac {d \left< \Phi(t) | \Psi(t) \right>}{dt} = 0[/itex]

    But that's not what you want. The probability amplitude to have energy E at time t is

    [itex] \left< E | \Psi(t) \right>[/itex]

    Here <E| is a time-independent state (i.e., it is a state in the Heisenberg picture). The rest should be easy.
  4. Jul 26, 2011 #3
    Eigenkets of observables are time-independent in the Schroedinger picture. Since you used the equation of motion for an arbitrary state ket, you are working in this picture.
  5. Jul 26, 2011 #4
    @Demystifier: Yes thanks, but this,

    [itex]\displaystyle\frac {d \left< \Phi(t) | \Psi(t) \right>}{dt} = 0[/itex]

    is exactly my point. I do understand it shouldn't come out to be this but I'm using proper rules for taking Hermitian adjoints. What exactly is wrong with my algebra?

    Do you mean that since [itex] \left< E_{k} \right| [/itex] is time-independent [itex] \frac{\partial \left< E_{k} \right|}{\partial t} = 0 [/itex] and thus,

    [itex]\frac {d \left< E_{k} | \Psi \right>}{dt} = \left< E_{k} \right| \frac{\partial \left| \Psi \right>}{\partial t} = - \frac{i}{\hbar} \left< E_{k} | H | \Psi \right>[/itex]

    ; but the first equation in my first post should still hold, shouldn't it?
  6. Jul 26, 2011 #5


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    Nothing is wrong with your algebra. That is the correct result. In words, the scalar product of time-dependent states is time independent. Why do you think that "it shouldn't come out to be this"?


    Yes, if they are time-dependent states. But read again my first sentence in #2.

    Or perhaps you wonder what is the difference between |E> and |E(t)> ?
    Last edited: Jul 26, 2011
  7. Jul 26, 2011 #6


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    I think, one has to be a bit more careful with which quantity you are considering. In addition the time dependence of the state vectors and the operators representing observables is dependent on the picture of the time dependence, but the physical outcome of quantum mechanics is independent of the choice of the picture. For simplicity's sake, I'll work in the Schrödinger picture, where the state kets (or equivalently and more general the statistical operator) evolve with the time-evolution operator generated from the full Hamiltonian of the system, which I assume to be not explicitly time dependent. The equation of motion for that operator reads

    [tex]\frac{\partial}{\partial t} \hat{C}(t,t_0)=-\mathrm{i} \hat{H} \hat{C}(t,t_0), \quad \hat{C}(t,t_0)=\hat{1}.[/tex]

    I've set [itex]\hbar=1[/itex]. The solution of this equation reads

    [tex]\hat{C}(t,t_0)=\exp[-\mathrm{i} H (t-t_0)].[/tex]

    This is obviously a unitary operator.

    The operators representing observables are constant in time (if not explicitly time dependent, which case I exclude here).

    Now let [itex]|o_1,\ldots,o_n \rangle[/itex] denote the simultaneous (generalized) eigenvector of a complete set of operators. Further let [itex]|\psi_0 \rangle[/itex] denote the state ket at time, [itex]t_0[/itex]. The state ket at any later time is then given by

    [tex]|\psi,t \rangle=\hat{C}(t,t_0) |\psi_0 \rangle.[/tex]

    The wave function wrt. to the complete set of (generalized) eigenvectors is given by

    [tex]\psi(o_1,\ldots,o_n,t)=\langle o_1,\ldots,o_n|\psi,t \rangle[/tex]

    and generally time dependent. It's modulus squared gives the probability (densitity) for simultaneously getting the values [itex]o_j[/itex] when measuring the complete set of compatible observables. This probability is of course also time dependent.

    The wave function obeys the equation of motion,

    [tex]\mathrm{i} \partial_t \psi(o_1,\ldots,o_n,t) = \hat{H}(o_1,\ldots o_n) \psi(o_1,\ldots,o_n,t),[/tex]

    where [itex]\hat{H}(o_1,\ldots,o_n)[/itex] denotes the Hamiltonian in the representation of the complete set of operators. In general that's a differential operator (as, e.g., in the position representation) or an integral kernel (as, e.g., in the momentum representation). This is of course nothing else than the Schrödinger equation for the wave function and, btw., independent on the picture of time evolution as it should be since the wave function (or better said its modulus squared) is a physically observable probability.

    Now, if you have a complete set of energy eigenvectors, [itex]|E,\alpha_j[/itex], where the [itex]\alpha_j[/itex] count a possible degeneracy of the energy eigenvectors, then the time evolution is particularly simple since

    [tex]\psi(E,\alpha_j,t)=\langle E,\alpha_j|\exp(-\mathrm{i} \hat{H} t)|\psi_0 \rangle
    =\langle \exp(+\mathrm{i} \hat{H} t E,\alpha_j|\psi_0 \rangle = \langle \exp(+\mathrm{i} E t E,\alpha_j|\psi_0 \rangle = \exp(-\mathrm{i} E t) \psi(E,\alpha_j,t_0)[/tex].

    The probability to measure an energy, [itex]E[/itex] (eventually simultaneously with some other observables if there exist observables which are compatible with the Hamiltonian, i.e., that are conserved quantities that are not pure functions of the Hamiltonian itself) is therefore time independent.
  8. Jul 26, 2011 #7
    Because if this is correct, using equation analogous to one in my first post, we can then say that for an arbitrary observable [itex]Q[/itex] with spectrum [itex]\{ q_{n} \}[/itex], the probability amplitude [itex]\left< q_{k} | \Psi \right>[/itex] for some value [itex]q_{k}[/itex] in spectrum [itex]\{ q_{n} \}[/itex]

    [itex]\frac {d \left< q_{k} | \Psi \right>}{dt} = 0[/itex]

    ; which means that probability amplitude for any value, of any observable, never changes in any state. The property of constant probability amplitudes is what characterizes stationary states; not just any state in general.
  9. Jul 27, 2011 #8


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    As I've shown in my posting, this is of course wrong. The wave function wrt. to a basis of eigenfunctions of non-conserved (compatible) quantities is time dependent!
  10. Jul 27, 2011 #9
    @vanhees71: Thanks, but, I can't follow your line of reasoning completely; I'm new to QM and couldn't understand a lot of stuff in your previous post :D

    So someone please clear this up; I haven't exactly derived everything from the basic underlying mathematical structure of the vectorspace, but, I think I'm using the correct rules of manipulating derivatives and adjoints and bras and kets but the answer seems physically unreasonable.
  11. Jul 27, 2011 #10


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    Of course, I only gave a very short summary of quantum dynamics in one special picture of time evolution (the Schrödinger picture). You find this in nearly any quantum-theory textbook. My favorite as an introduction is

    Sakurai, Modern Quantum Mechanics, Addison-Wesley

    and for the interpretative part

    L. Ballentine, Quantum Mechanics, A modern development
  12. Aug 5, 2011 #11


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    Saim, read again my post #2 because it seems that you haven´t digested what I have written there.
  13. Aug 5, 2011 #12
    OK, from the first line of that post I understand that the result is correct for time dependent states; their scalar product must be time independent. So this "derivation" works for time dependent states. My problem is why wouldn't it work for a time-independent bra [itex]\left< q_{k} \right|[/itex] and a time-dependent [itex]\left| \Psi ,t \right>[/itex].

    Say we have a time dependent state [itex]\left| \Psi ,t \right>[/itex] and we want to find the probability amplitude for some value [itex]q_{k}[/itex] of some observable [itex]Q[/itex] such that the eigenstate [itex]\left| q_{k} \right>[/itex] is time-independent; this amplitude will be [itex]\left< q_{k} | \Psi ,t \right>[/itex]. Now lets consider the time derivative of this amplitude

    [itex]\frac {d \left< q_{k} | \Psi ,t \right>}{dt} = \frac{\partial \left< q_{k} \right|}{\partial t} \left| \Psi, t \right> + \left< q_{k} \right| \frac{\partial \left| \Psi , t \right>}{\partial t} = \frac{i}{\hbar} \left< q_{k} | H | \Psi , t \right> - \frac{i}{\hbar} \left< q_{k} | H | \Psi , t \right> = 0[/itex]

    Is this true as well? I understand that you are emphasizing that the result is true for two time-dependent but what about this situation? If its true then it means probability amplitudes for observables cannot depend on time for any state [itex]\left| \Psi ,t \right>[/itex], which seems physically absurd. If this is untrue and not correct, where is the mistake?
  14. Aug 5, 2011 #13


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    On your assumptions

    [tex] \frac{\partial \left\langle q_{k}\right|}{\partial t} = 0 [/tex]

    so the 0 at the end of your multiple equality is wrong, because the 2 terms before the last equality sign are not equal.
  15. Aug 5, 2011 #14
    Regarding your original post:

    \frac{\partial \left\langle E_{k} \right|}{\partial t} = 0


    \frac{\partial \left\langle E_{k} \right|}{\partial t} \neq \frac{i}{\hbar} \left\langle E_{k} \right| H
  16. Aug 5, 2011 #15
    I see the point. Thank you for pointing out the silly overlook.
  17. Apr 14, 2012 #16
    Why exactly is [itex]\frac{\partial \left\langle E_{k} \right|}{\partial t} \neq \frac{i}{\hbar} \left\langle E_{k} \right| H[/itex] ? [itex]\left| E_{k} \right>[/itex] is a valid quantum state and should thus solve the Schrodinger equation, namely, [itex]\frac{\partial \left| E_{k} \right\rangle}{\partial t} = \frac{i}{\hbar} H \left| E_{k} \right\rangle[/itex], taking the Hermitian adjoint of which we get [itex]\frac{\partial \left\langle E_{k} \right|}{\partial t} = \frac{i}{\hbar} \left\langle E_{k} \right| H[/itex]. Why is this not true?

    P.S: I'm sorry for beating a dead horse but my last post in which I thought I understood my overlook was an overlook itself since I never really understood this point.
  18. Apr 14, 2012 #17
    What is [itex]\langle E_k \vert[/itex]?
  19. Apr 14, 2012 #18
    Bra vector corresponding to [itex]\left| E_{k} \right>[/itex]
  20. Apr 14, 2012 #19
    ok, what is [itex]\vert E_k \rangle[/itex]? How is it fundamentally defined?
  21. Apr 14, 2012 #20
    The energy eigenstate for our system; a state with definite value of energy [itex]E_{k}[/itex]
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