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I Another energy eigenvalue question

  1. May 6, 2016 #1
    Take the wavefunction $$e^{-i\omega t}$$ which all time dependent functions can be superposed of (right?). You can then get $$ih\frac{\partial}{\partial t}\psi=\hbar \omega \psi$$ and thus if ##\hat{E}=ih\frac{\partial}{\partial t}## then $$\hat{E}\psi=E\psi$$ What did I do wrong?
     
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  3. May 6, 2016 #2

    blue_leaf77

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    What's your question actually?
    Moreover, this doesn't seem like a quantum mechanics problem. You don't even need ##\hbar## as it can be canceled out and the ##\nabla^2## term is also missing, it's just a math eigenvalue problem.
     
  4. May 6, 2016 #3
    I know from previous threads that ##\hat{E} \psi =E\psi## is not a correct QM equation, yet I just "derived" it. What did I do wrong?
     
  5. May 6, 2016 #4

    blue_leaf77

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    I don't know which thread you refer to but the general form of the Schroedinger equation is
    $$
    i\hbar \frac{\partial}{\partial t} \psi(\mathbf{r},t) = H(\mathbf{r},t) \psi(\mathbf{r},t)
    $$
    It may reduce to the time independent Schroedinger equation when the Hamiltonian is time-independent and ##\psi## is an eigenfunction of the Hamiltonian.
    However, the equation you have there is not a Schroedinger equation since it misses the kinetic energy term.
     
  6. May 6, 2016 #5
    Would my equation work?
     
  7. May 6, 2016 #6

    blue_leaf77

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    Your equation is mathematically correct: you come up with a complex exponential function and construct an eigenvalue equation that it satisfies, but this eigenvalue equation is not a Schroedinger equation.
     
  8. May 7, 2016 #7
    So this equation is correct for time dependent quantum systems, right? And because ##\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})##, for stationary states we get ##\hat{H}\psi=E\psi## and for time dependent systems we get ##\hat{H}\psi=\hat{E}\psi##. Is that correct?
     
  9. May 7, 2016 #8

    blue_leaf77

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    There is nothing which can be associated with a quantum system since your equation is not a Schroedinger equation.
    Time-dependent system is known to be difficult to solve analytically. Most time-dependent cases are dealt with the help of some perturbation theory. I can't be certain myself that a concept of eigenstates also applies in TDSE. There is probably not even a stationary state, take for example the time evolution of states in a time-dependent Hamiltonian in which the Hamiltonian in any given time commutes with the Hamiltonian at other times. The time evolution reads
    $$
    \psi(x,t) = \exp\left( -\frac{i}{\hbar} \int_0^t H(t') dt' \right) \psi(x,0)
    $$
    I don't see how states that evolve that way can yield time-independent expectation value for any observable for this state to deserve the calling "stationary".
     
  10. May 7, 2016 #9

    vanhees71

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    Well, the Schrödinger equation indeed reads
    $$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}).$$
    The equation can be separated in the form
    $$\psi(t,\vec{x})=T(t) \Psi(\vec{x}),$$
    if the Hamiltonian is not time dependent, because then
    $$\mathrm{i} \hbar \partial_t \psi = \mathrm{i} \hbar \dot{T}(t) \Psi(\vec{x})= T(t) \hat{H} \Psi(\vec{x}).$$
    Now devide this equation by ##T(t) \Psi(\vec{x})##. Then you get
    $$\frac{1}{T}(t) \mathrm{i} \hbar \dot{T}(t) =\frac{1}{\Psi(\vec{x})} \hat{H} \Psi(\vec{x}).$$
    Now, since the left-hand side is independent of ##\vec{x}## and the right-hand-side is independent of ##t##, it must be a constant, which we call ##E##. This implies the equation for ##T##
    $$\mathrm{i} \hbar \dot{T}=E T \; \Rightarrow \; T(t)=T_0 \exp(-\mathrm{i} \omega t), \quad \omega=\frac{E}{\hbar},$$
    and for ##\Psi##:
    $$\hat{H} \Psi(\vec{x})=E \Psi(\vec{x}),$$
    i.e., ##\Psi## must be an eigenvector of ##\hat{H}## which of course is NOT ##\mathrm{i} \hbar \partial_t## but some differential operator in ##\vec{x}## like the standard case for the motion of a particle in some external potential,
    $$\hat{H}=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$
    So the general solution of the time-dependent Schrödinger equation can be written in terms of the energy-eigensolutions (i.e., the solutions of the time-independent Schrödinger equation)
    $$\psi(t,\vec{x})=\sum_{E} a_E \exp(-\mathrm{i} \omega t) \Psi_E(\vec{x}), \quad \hat{H} \Psi_E(\vec{x})=E \Psi_E(\vec{x}).$$
    The sum goes over all energy eigenvalues. If there are continuous energy eigenvalues ("scattering solutions"), there's also an integral over the corresponding values. Of course, there are cases, where the Hamiltonian has only a continuous spectrum, as in the case of the free particle.
     
    Last edited: May 7, 2016
  11. May 7, 2016 #10

    blue_leaf77

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    @vanhees71 , I think you missed the expansion coefficient (which might as well be dependent on time) in the last equation containing the sum.
     
  12. May 7, 2016 #11

    vanhees71

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    True. I've corrected it, but it must of course be time independent. It's given by the initial wave function at ##t=t_0##.
     
  13. May 7, 2016 #12

    blue_leaf77

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    I thought you were deriving the general form of the wavefunction arising from a time dependent Hamiltonian. If the expansion coefficient is not a function of time, wouldn't ##\psi(t,\vec{x})## as you wrote above just the solution of a Schroedinger equation with a time-independent Hamiltonian. It's just the time evolution operator ##U(t,t_0)## applied to the initial state ##\psi(t_0,\vec{x}) = \sum_E a_E \psi_E(t_0,\vec{x})##. That ##a_E## for time dependent Hamiltonian should be a function of time, you can look at equation 5.5.4 in Modern Quantum Mechanics by Sakurai.
     
  14. May 8, 2016 #13

    vanhees71

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    Yes, sure, for a time-dependent Hamiltonian (in the Schrödinger picture of time evolution!) you cannot separate the Schrödinger equation anymore, and the energy eigenvectors become time-dependent and thus also the expansion coefficients.
     
  15. May 8, 2016 #14

    blue_leaf77

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    Ah sorry I misread
    as "time-dependent Hamiltonian".
     
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