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Proper form of schrodinger's equation?

  1. May 9, 2013 #1
    I feel a bit silly asking this, but I've been working through some QM lately and there's one aspect of Schrodinger's equation that's puzzling me. I've typically understood the equation as [itex]i\hbar \frac{d|\psi\rangle}{dt}=\hat H |\psi\rangle[/itex], but I've also seen it written as [itex]i\hbar \frac{\partial |\psi\rangle}{\partial t}=\hat H|\psi\rangle[/itex]. The use of a partial derivative instead of a total derivative is what's got me. In most cases, I know it doesn't matter but I can imagine some where it would. The total derivative indicates that [itex]\psi[/itex] may not explicitly depend on time, but implicitly could through some other variable which is dependent on time. This evolution is generated by the Hamiltonian. In the other case, the partial derivative only concerns explicit time dependence, and hence this form only makes sense to me when there is an explicit [itex]t[/itex] showing up in the equations. Which form is generally correct?
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  3. May 9, 2013 #2

    Simon Bridge

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  4. May 10, 2013 #3


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    Since you've written the SE in the Dirac bra/ket notation, and in the standard formulation time is a mere parameter, you can consider using the total derivative symbol d as correct, sind the SE is about time evolution of kets and nothing more.
  5. May 10, 2013 #4


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    Yes, one should really clearly say that in the bra-ket notation [itex]|\psi(t) \rangle[/itex] does not depend on position, momentum, or any other values of observables. It's (in the here used Schrödinger picture of the time evolution) a function of time only, and thus you should use the total differential.

    The wave function is always the (generalized) scalar product of a (generalized) common eigenvector of a complete set of compatible observables (e.g., position). In the Schrödinger picture observables are represented by time-independent self-adjoint operators, and thus also the (generalized) eigenvectors are time-independent. Thus you have
    [tex]\psi(t,\vec{x})=\langle \vec{x}|\psi(t).[/tex]
    Then you have to use the partial derivative to write
    [tex]\mathrm{i} \hbar \partial_t \psi(t,\vec{x}) = \langle \vec{x}|\mathrm{i} \hbar \mathrm{d}_t \psi(t) \rangle = \langle \vec{x} |\mathbf{H} \psi \rangle = \hat{H} \langle \vec{x}|\psi \rangle=\hat{H} \psi(t,\vec{x}).[/tex]
    Here, [itex]\mathbf{H}[/itex] is the Hamilton operator in abstract Hilbert space and [itex]\hat{H}[/itex] in the position representation.

    It is quite important to distinguish the abtract Hilbert-space objects from their representation wrt. to a given (generalized) basis!
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