Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solving the Schödinger equation - R.Shankar's PQM, mistake?

  1. Apr 15, 2015 #1

    On p.145 of Shankar's Principles of Quantum Mechanics, the author derives the general propagator for the Schrodinger equation in the following manner.

    Shankar's working
    Expanding the state vector in the energy basis,

    [itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]

    then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]

    he gets [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}-Ea_{E})|E\rangle [/itex]

    and so [itex]\,\,\imath\hbar\dot{a_{E}} = Ea_{E} [/itex]

    from this [itex]\,\, a_{E}(t) = a_{E}(0)\cdot e^{\frac{-\imath Et}{\hbar}} [/itex]

    My working

    [itex] |\psi(t)\rangle = \sum_{E} |E\rangle \langle E| \psi(t) \rangle \equiv \sum_{E} a_{E}(t)|E\rangle [/itex]

    then acting on both sides with [itex]\,(\imath\hbar\frac{\partial}{\partial t} - H) [/itex]

    I get [itex]\,\,\,0 = \sum(\imath\hbar\dot{a_{E}}|E\rangle + \imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle). [/itex] Note the extra (middle) term that I get here; this is missing in Shankar's corresponding step.

    But [itex] \,\imath\hbar a_{E}|\dot{E}\rangle - a_{E}E|E\rangle \,\,[/itex] is the schrodinger equation with an eigenket (of H, of course) substituted in and multiplied throughout by [itex] \dot{a_E}[/itex]. So this term equals zero.

    and so [itex]\,\,\,\dot{a_{E}} = 0 [/itex]

    Clearly, this is at odds with Shankar's result, whose first time derivative is not zero. Is there a mistake anyone can spot in either of the workings above?

  2. jcsd
  3. Apr 15, 2015 #2


    User Avatar
    Science Advisor
    2016 Award

    Shankar has silently assumed that ##\hat{H}## is time-independent, and thus ##|E \rangle## is also time-independent. So Shankar's analysis is correct. You can also get to it in the following way. From
    $$\mathrm{i} \hbar \partial_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle$$
    you get the formal solution
    $$|\psi(t) \rangle=\exp \left (-\frac{\mathrm{i}}{\hbar} t \hat{H} \right) |\psi(0) \rangle.$$
    Now inserting a completeness relation with the energy eigen-states
    $$\sum_{n} |E_n \rangle \langle E_n |=1,$$
    you get
    $$|\psi(t) \rangle=\sum_{n} |E_n \rangle \exp \left (-\frac{\mathrm{i}}{\hbar} t E_n \right) a_n(0).$$
  4. Apr 15, 2015 #3
    ##|E>## is time independent.
    Nevertheless, if one introduces the base vectors as ##|E(t)>=exp(iEt/h)|E>## then the time dependent part vanishes when writing the unity operator as ##{\mathbb{I}}=\sum_E |E(t)><E(t)|##
  5. Apr 15, 2015 #4


    User Avatar
    Science Advisor
    2016 Award

    No, you misunderstood it somewhat. You seem to mix the Heisenberg and the Schrödinger picture, because your ##|E(t) \langle## is a Heisenberg-picture energy eigenstate, but then you use the Schrödinger-picture state ket ##|\psi(t) \rangle##. You must never take products of eigenvectors from one picture and state kets from another. This will almost always lead to something wrong!
  6. Apr 15, 2015 #5
    Sorry, I can't recover my previous message.

    When I wrote ##|E(t)>=\exp(iEt/h)|E>## I was thinking of stationary states (eigenstates + phase), but this has nothing to do here. I missed the assumption of time independent hamiltonian from which everything else follows.
  7. Apr 15, 2015 #6
    thanks for your reply, vanhees71.

    I see what you mean. I've used [itex] {|E_{n}\rangle}[/itex] as a basis; the Heisenberg picture rotates the basis with time and the state vector [itex] |\psi\rangle [/itex] remains fixed; the Schrödinger picture rotates the state vector while the basis itself remains the same.

    Writing [itex]\,i\hbar a_{E}\dot{|E\rangle}\,[/itex] (and cancelling it with the other term in the equation where this appears) is erroneous as the ket [itex] |E\rangle [/itex], used as a basis vector here, is time-independent (using the Schödinger picture) and this term is the source of my error. Am I correct in my understanding?
    Last edited by a moderator: Apr 15, 2015
  8. Apr 15, 2015 #7
    The following is the same derivation as done by J.Binney and D.Skinner in the book, The Physics of Quantum Mechanics. The derivation has been reproduced here verbatim from p.31-32.

    """ The TDSE tells us that states of well-defined energy evolve in time as

    [itex] i\hbar\partial_{t}|E_{n}\rangle = H|E_{n}\rangle = E_{n}|E_{n}\rangle [/itex]

    which implies [itex] |E_{n}, t\rangle = |E_{n}, 0\rangle e^{-iE_{n}t/\hbar}\,\,\,\,[/itex] (Equation 2.29)

    That is, the passage of time simply changes the phase of the ket at a rate [itex] E_{n}/\hbar [/itex]. We can use this result to calculate the time evolution of an arbitrary state [itex]|\psi\rangle [/itex]. In the energy representation, the state is

    [itex] |\psi, t\rangle> = \sum_n{a_{n}(t)|E_{n}, t\rangle}\,\,\,\,[/itex] (Equation 2.30)

    Substituting this expansion into the TDSE, we find

    [itex] i\hbar\partial_{t}|\psi\rangle = \sum_n{i\hbar(\dot{a_{n}}|E_{n}\rangle + a_{n}\partial_{t}|E_{n}\rangle)} = \sum_n{a_{n}H|E_{n}\rangle} [/itex]

    where a dot denotes differentiation with respect to time. The right side cancels with the second term in the middle, so we have [itex]\dot{a_{n}} = 0 [/itex]. Since the [itex] a_{n} [/itex] are constant, on eliminating [itex] |E_{n}, t\rangle [/itex] between equations (2.29) and (2.30), we find that the evolution of [itex] |\psi\rangle [/itex] is simply given by

    [itex] |\psi, t\rangle = \sum_n{a_{n} e^{-iE_{n}t/\hbar} |E_{n}, 0\rangle} [/itex] """

    Binney seems to have mixed up the Heisenberg and Schrödinger representations too, hasn't he? His equation (2.30) has a time-evolving state vector on the left, and a time dependent basis on the right. Is my understanding flawed here?

    Last edited by a moderator: Apr 15, 2015
  9. Apr 16, 2015 #8


    User Avatar
    Science Advisor
    2016 Award

    Here you use and energy eigenstate as a state ket in the Schrödinger picture. Then the ##a_n## are of course constant in time. The treatment of the different pictures of time evolution is confused in many textbooks. The best book I know with respect to this issue is unfortunately in German:

    E. Fick, Einführung in die Grundlagen der Quantenmechanik, Aula-Verlag Wiesbaden
  10. Apr 16, 2015 #9
    I don't speak German, but thanks for the reference anyway.

    The first two steps of Binney's derivation, I understand. He has, as you say, used an energy eigenstate as a state ket.

    However, in his third step (Equation 2.30 in my previous post), he has expanded an arbitrary state ket [itex] |\psi, t\rangle [/itex] in a linear combination of time-evolving energy eigenstate state kets [itex] |E, t\rangle [/itex]. Isn't this erroneous?

    He has expanded a time-evolving state ket in terms of a time-evolving basis, thereby mixing the Heisenberg and Schrödinger representations, has he not?
  11. Apr 16, 2015 #10


    User Avatar
    Science Advisor
    2016 Award

    No, that's fine. Then of course, the ##a_n## must be time-independent.

    It's easy to derive with the time-independent energy eigenvectors (the correct ones for the Schroedinger picture):
    $$|\psi(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |\psi(0) \rangle=\sum_n a_n \exp(-\mathrm{i} \hat{H} t) |E_n \rangle =
    \sum_n a_n \exp(-\mathrm{i} E_n t/\hbar) |E_n \rangle.$$
    The ##a_n## are given by
    $$a_n=\langle E_n|\psi(0) \rangle$$
    and thus time-independent as claimed!
    Last edited: Apr 21, 2015
  12. Apr 21, 2015 #11
    Got it now. Thanks for the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Solving the Schödinger equation - R.Shankar's PQM, mistake?
  1. Shankar PQM is wrong? (Replies: 1)