The WDW equation is: $\hat{H}(x)|\psi \rangle=0.$ Schrödinger’s time dependent wave function equation says: $$i\hbar \frac{\mathrm d}{\mathrm dt} | \Psi(t)\rangle=\hat{H}|\Psi(t)\rangle.$$

Does it make any sense at all to equate these two wave functions? If Wheeler-DeWitt is true, does that imply $$i\hbar \frac{\mathrm d}{\mathrm dt}|\Psi(t)\rangle=0?$$

I doubt this is important, but the question does intrigue me because I am curious about how the wave function works mathematically and I only recently learned about the Wheeler-DeWitt equation.

Ultimately, what would this mean physically?

  • $\begingroup$ Wikipedia: “Although the symbols $\hat H$ and $|\psi \rangle$ may appear familiar, their interpretation in the Wheeler–DeWitt equation is substantially different from non-relativistic quantum mechanics. $|\psi \rangle$ is no longer a spatial wave function in the traditional sense of a complex-valued function that is defined on a 3-dimensional space-like surface and normalized to unity. Instead it is a functional of field configurations on all of spacetime. … $\endgroup$
    – Ghoster
    Commented Sep 25, 2022 at 3:55
  • $\begingroup$ … This wave function contains all of the information about the geometry and matter content of the universe. $\hat H$ is still an operator that acts on the Hilbert space of wave functions, but it is not the same Hilbert space as in the nonrelativistic case, and the Hamiltonian no longer determines evolution of the system [emphasis mine], so the Schrödinger equation $\hat{H} |\psi\rangle = i \hbar \partial / \partial t |\psi\rangle$ no longer applies. This property is known as timelessness.” $\endgroup$
    – Ghoster
    Commented Sep 25, 2022 at 3:55
  • $\begingroup$ I am sure that the theorists can make sense of Wheeler-DeWitt for you, but from an experimental standpoint it is very questionable. A wave function is the description of a quantum mechanical ensemble. Where does one get a quantum mechanical ensemble of universes from? And what would the time variable be? There is no notion of one global time in a relativistic universe. In such a universe all physics is necessarily local and observer dependent. And if we get past that, where do we get quanta from? Quanta are irreversible energy exchanges. What does the universe exchange energy with??? $\endgroup$ Commented Sep 25, 2022 at 4:02
  • $\begingroup$ @FlatterMann You are saying experimental physicists do not have a consensus on this equation because it does not necessarily pan out experimentally? Ie, little to no experiments exist which support this? $\endgroup$
    – user
    Commented Sep 25, 2022 at 4:45
  • 1
    $\begingroup$ Is there a time independent Schrödinger equation? Yes. $\endgroup$
    – Ghoster
    Commented Sep 25, 2022 at 4:51

1 Answer 1


The Wheeler deWitt equation (WDE) describes a quantum system that doesn't evolve with respect to a parameter $t$. In quantum theory real physical systems are described by quantum observables, not by parameters, so the parameter time is unobservable. So the fact that the WDE state $|\Psi\rangle$ doesn't evolve with respect to parameter time highlights a problem that already existed. The WDE state is a state in which a clock system $C$ with a time observable $\hat{T}$ is entangled with the rest of the universe: $$|\Psi\rangle=\sum_t|t\rangle|\psi(t)\rangle,$$ where the $|t\rangle$ states are eigenstates of the time observable, see



The relative state of the rest of the universe $|\psi(t)\rangle$ evolves over time according to some relevant equation of motion such as the Schrodinger equation. There is a corresponding treatment for the Heisenberg picture:


This understanding of time has been experimentally tested too



  • $\begingroup$ It sounds like a magic trick because the clock system emulates an irreversible process either way, doesn't it? How is this different from an irreversible classical clock? Is the argument that one can make the correlation at the level of the wave function without having to reduce to classical observables and still observe dynamics? By the same argument every other classical observable can also be emulated by a suitable internal quantum system, which is really just a complicated way of saying that "a ruler consists of atoms that are made of quantum fields". I believe we already knew that. $\endgroup$ Commented Sep 25, 2022 at 9:46
  • $\begingroup$ In science, people propose theories, work out their consequences and then conduct critical discussion of those consequences, including experimental tests. When we apply quantum physics to clocks we may learn something new about clocks, or about quantum theory and we may make progress toward a replacement for quantum theory by critical discussion and experimental testing of the theory of quantum clocks. $\endgroup$
    – alanf
    Commented Sep 25, 2022 at 10:59
  • $\begingroup$ @alanf Thank you. I am still not yet ready to understand this. However, I've accepted it as the answer. $\endgroup$
    – user
    Commented Oct 12, 2022 at 19:59

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