Am I right in saying, that one cannot generally assign an order to the Schrödinger-Equation (SE)?

E.g. if one considers a particle in a potential, the hamiltonian contains the second derivative of a spacial coordinate. If one considers the spin of an electron, the hamiltonian is a product of spin-operator and gyromagnetic ratio, in which I cannot find any derivatives.

  • $\begingroup$ In short: the SE is first order in time and second-order in space (when you consider the kinetic term). The associated "eigenvalue problem" is what it is: technically, it's not the "SE initial-value problem" (see the correct comment of @Dvij D.C. below). The discussion here can be of interest: physics.stackexchange.com/q/75363/226902 $\endgroup$
    – Quillo
    Apr 21 at 10:29

2 Answers 2


No. You are not completely right. The Schrodinger equation is a linear partial differential equation that governs the wave function of a quantum-mechanical system. It is first order in the derivative with respect to time.

  • $\begingroup$ This implicitly assumes an abstraction from any implementation of the Hamiltonian operator, which in practice often contains spatial derivatives of second order. When solving TISE, one usually deals with a second-order DE, while solving a TDSE for a matrix Hamiltonian (e.g. for a two-level system) one works with a first-order DE. $\endgroup$
    – Ruslan
    Apr 21 at 0:20
  • $\begingroup$ "no you are not completely right". Does this mean that, apart from the time, I am? $\endgroup$
    – manuel459
    Apr 21 at 9:58
  • 1
    $\begingroup$ @Ruslan Schrodinger equation is simply TDSE, it's a differential equation that describes the time-evolution of the system and it is first-order because it is first order in time-derivatives. TISE is just the eigenvalue problem of the Hamiltonian which will obviously depend on the specifics of a given Hamiltonian. $\endgroup$
    – ACat
    Apr 21 at 10:25
  • $\begingroup$ @DvijD.C. consider a free-particle Hamiltonian in position representation. Construct a TDSE from it, and you get a second-order PDE. The fact that we usually set up an IBVP with initial conditions at a time point doesn't make the PDE first-order. $\endgroup$
    – Ruslan
    Apr 21 at 10:35
  • $\begingroup$ @Ruslan I don't think it's appropriate to say that we get a PDE when we write out the Hamiltonian in position basis, yes, there are second-order differential operators w.r.t. position but it's not an equation to find out the $x$ dependence, i.e., the $x$ dependence is known (i.e., it is an IBVP) because that's precisely the context in which one employs a time-evolution equation in any theory. $\endgroup$
    – ACat
    Apr 21 at 10:52

one cannot generally assign an order to the Schrödinger-Equation

This statement is completely incorrect. If you mean a "Schrodinger equation that describes all systems" and so the order cannot be defined for all systems, your statement is still not true since even though there are other "renderings" of it that are for specific systems, you can't really call them the Schrodinger equation in such cases.

For simple non-relativistic systems, the standard time independent Schrodinger equation is second-order in space derivatives and the time dependent Schrodinger equation is the same but with also a first order time derivative. These equations are applicable to many quantum systems. For other systems:

If one considers the spin of an electron

When we consider (relativistic) spin $\frac 12$ particles, we can use the Dirac equation, which is of the form $$\left(i\hbar\gamma^\mu\partial_\mu-mc\right)\psi=0$$ We see that this equation is first order in space and time (there is of course the Pauli equation for non-relativistic electrons in magnetic fields which is second order in space and first order in time).

In relativistic but spinless systems, we can use the Klein-Gordon equation $$(\Box^2-m^2)\psi=0$$ (in units where $c=\hbar=1$), which also is second order in space derivatives and also second order in time derivatives.

  • $\begingroup$ "we can use the Dirac equation [...] this equation is first order in space and time" — in the non-relativistic context we would use the Pauli equation, which isn't first order in space. $\endgroup$
    – Ruslan
    Apr 21 at 10:37
  • $\begingroup$ Yeah, guess so. I've included the Pauli equation. Thanks for the input @Ruslan. $\endgroup$
    – joseph h
    Apr 21 at 20:18

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