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General solution for the time-dependent Schrödinger equation

  1. Jun 12, 2015 #1
    Hello! I have two uncertainties (hehe) about two concepts from a QM time-dependent Schrödinger equation video.
    The video is
    I cannot move on further if I don't fully grasp everything he explains in the video. My two issues are:
    1) The general solution for the time-dependent Schrödinger equation starts at 55:20.He uses αi for a set of coefficients of states. What exactly are these? the probability densities? probability amplitudes? the absolute probabilities? He mentions any state vector can be written as a superposition of eigenvectors of the energy, so α should be a probability amplitude, so you can calculate it's squared modulus.
    2) At 1:03:04 he comes up with the exponential solution e^-iEjt. Where exactly from does he gets α(0) from?
    Appreciating any kind of answer which would help me shed some light on these two things. Cheers!
     
  2. jcsd
  3. Jun 12, 2015 #2

    blue_leaf77

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    1) Usually one calls the square modulus of ##\alpha## as the probability amplitude.
    2) Didn't he say that he assumes he knows the wavefunction at t=0, and hence the corresponding expansion coefficients?
     
    Last edited: Jun 12, 2015
  4. Jun 12, 2015 #3
    Thanks for the reply.
    1) Fair enough but my question was what does α stand for (he calls it coefficient of states).
     
  5. Jun 12, 2015 #4

    blue_leaf77

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    If your problem is simply being confused as what to call ##\alpha##, you can also adopt what he said in the video. I personally would call it expansion coefficient, the way to obtain it is through the equation ##\alpha_i(t) = \langle i | e^{-iHt/\hbar} | \psi(0) \rangle ##. I guess you know about it already.
     
  6. Jun 13, 2015 #5

    samalkhaiat

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    When we write [tex]|\Psi(t) \rangle = \sum_{n} a_{n}(t) |n\rangle , \ \ \ \ a_{n}(t) = \langle n|\Psi(t)\rangle[/tex] we are simply representing the abstract state vector [itex]|\Psi\rangle[/itex] by a set of numbers [itex]a_{n}[/itex] which have all the information content in the state Psi. You should know that from vector algebra. Recall the similarity with the vector relations [tex]\vec{V}=\sum_{i} v_{i} \hat{e}_{i}, \ \ \ \ v_{i} = \hat{e}_{j} \cdot \vec{V} .[/tex] So, you say that [itex]a_{n}(t)[/itex] are the components of the “vector” [itex]|\Psi\rangle[/itex] in the [itex]|n\rangle[/itex] “direction”, projection of [itex]|\Psi\rangle[/itex] on [itex]|n\rangle[/itex], how much of the state [itex]|n\rangle[/itex] one can find in the state [itex]|\Psi\rangle[/itex], transition (or probability) amplitude from [itex]|\Psi\rangle[/itex] to [itex]|n\rangle[/itex], or (even better) matrix representation of the state vector [itex]|\Psi\rangle[/itex]. Notice that in [itex]a_{n}[/itex]-representation, the Schrodinger equation becomes [tex]i \frac{d}{d t} \langle m | \Psi \rangle = \sum_{n} \langle m |H| n \rangle \ a_{n}(t) ,[/tex] or [tex]i \frac{d a_{m}}{d t} = \sum_{n} H_{m n} \ a_{n}(t) .[/tex] This was the starting equation of the so-called Matrix Mechanics of Heisenberg. It is just an equivalent, matrix form of the differential equation of Schrodinger. Now, suppose that [itex]|n\rangle[/itex] form eigen-states of the Hamiltonian, [itex]H|n\rangle = E_{n}|n\rangle[/itex], then [tex]H_{m n} = \langle m | H | n \rangle = E_{n} \delta_{n m} ,[/tex] and the above matrix equation becomes (no sum on m) [tex]i \frac{d a_{m}(t)}{d t} = E_{m} a_{m}(t) .[/tex] Integrating this from [itex]t=0[/itex] to [itex]t[/itex], we find [tex]\int_{t=0}^{t} \frac{d a_{m}(t)}{a_{m}(t)} = - i E_{m} \int_{0}^{t} dt ,[/tex][tex]\ln |\frac{a_{m}(t)}{a_{m}(0)}| = - i E_{m} t ,[/tex] which we normally write as [tex]a_{m}(t) = a_{m}(0) e^{ - i E_{m} t}.[/tex]
     
  7. Jun 13, 2015 #6
    Thanks a lot, that helped.
     
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