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Is it possible for two Schrödinger equations describing different systems to have the same wavefunction? And if that is the case, why or why not?

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  • $\begingroup$ It will depend on how broad your definition of a "Schordinger equation" is, but for systems of electrons the answer is essentially no. The 1st Hohenberg-Kohn Theorem states that given the ground state electron density the potential is fixed up to a constant. $\endgroup$ – By Symmetry Nov 16 '16 at 11:57
  • $\begingroup$ When you say "different" equations. Do you simply mean that the potential is different? $\endgroup$ – Mikael Fremling Nov 16 '16 at 11:57
  • $\begingroup$ @MikaelFremling The potential is different. $\endgroup$ – CoffeeIsLife Nov 16 '16 at 12:08
  • $\begingroup$ @BySymmetry I thought that the 1st Hohenberg-Kohn Theorem said that the density is as good as the wavefunction in describing the system. Also, isn't it just for ground states? $\endgroup$ – CoffeeIsLife Nov 16 '16 at 12:17
  • $\begingroup$ @QuantumCAPUCCINO The Hohenberg-Kohn Theorem states thats we can deduce the Hamiltonian from the ground state elctron density. Since from the Hamiltonian we can, in principle, find the energy eigenstate wavefunctions, so the ground state electron density contains as much information as the wavefunctions. I had, however, forgotten about the restriction to the ground state. I don't know whether you can have two Hamiltonian with the same wavefunction for "different" eigenstates. A hand wavy node counting argument suggests that you might be able to generalise the result, at least in a special case $\endgroup$ – By Symmetry Nov 16 '16 at 13:10
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The question is very broadly posed, so there's a bunch of different ways to interpret it, and each of them can give a different answer.

  • Can two different Hamiltonians share an eigenfunction?

    Yes. The answer by ComptonScattering gives an example in finite dimensions; if you want something closer to the usual three-dimensional quantum mechanics of a massive particle, there's even more examples (because the Hilbert space is much bigger).

    For something concrete, you can try the hydrogenic Hamiltonian $\hat H_0$ and $$\hat H = \hat H_0 + f(\hat r) \, \hat L{}^2,$$ where $f(r)\geq0$ is a non-negative function of the radius. Here the ground state will be the same for both (as will all the $\ell=0$ states) but $\ell\neq 0$ states will differ.

  • Can two different Hamiltonians share all their eigenfunctions?

    Yes. For a simple example, take any Hamiltonian $\hat H_1 = \hat H$ that is not the identity operator, and compare it with $\hat H_2 = \hat H{}^2$. Then every eigenfunction of $\hat H_1$ will be an eigenfunction of $\hat H_2$ (though the converse of that is not necessarily true), but the eigenvalues will in general differ.

  • Can two different Hamiltonians share all their eigenfunctions and the eigenvalues?

    No. This is because you can express the Hamiltonian as a function of the eigenvalues and eigenfunctions. In Dirac notation, that reads $$\hat H = \sum_n E_n |n⟩⟨n|.$$

  • Can two different Hamiltonians share at least one solution of the time-dependent Schrödinger equation, for all times?

    Yes. The eigenfunction examples of the first point above are a suitable example.

  • Can two different Hamiltonians share an arbitrary wavefunction in their solutions of the Schrödinger equation, at least for a single time $t_0$?

    Yes. Easy: set up some arbitrary wavefunction $\psi(t_0)$, and let it run under your two arbitrary Hamiltonians $\hat H_1$ and $\hat H_2$. They'll typically take $\psi(t_0)$ in arbitrary, different directions, but hey, the solutions matched at time $t_0$.

  • Can two different Hamiltonians share solutions of the time-dependent Schrödinger equation for all times and for arbitrary initial conditions?

    No. Take an arbitrary wavefunction $\psi_0$, and use that as the initial condition $\psi(t_0) = \psi_0$ for the Schrödinger equation under arbitrary $\hat H_1$ and $\hat H_2$. By assumption, both solutions are equal, and in particular their time derivatives at $t_0$ are also equal. (In fact, that is all you need: agreement of the wavefunction and its time derivative, for arbitrary initial conditions.) This then implies that $$\hat H_1 \psi_0 = i\hbar \partial_t \psi(t_0) = \hat H_2 \psi_0,$$ i.e. that $\hat H_1$ and $\hat H_2$ agree on your arbitrary initial condition $\psi_0$. This then means that they must agree as operators.

As a final note, please note the huge number of questions that folded into your original query, because of its imprecise phrasing, and use it to learn the importance of providing sharp, well-defined questions.

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  • $\begingroup$ @ComptonScattering What does that even mean? You know better than the OP in terms of asking ill-defined questions. $\endgroup$ – Emilio Pisanty Nov 16 '16 at 17:31
  • $\begingroup$ For completeness it is worth mentioning that if there is gauge freedom in your system, there are families of different Hamiltonians related by gauge transformations which lead to the the same evolution of all observable quantities, even though the wavefunctions are different. $\endgroup$ – ComptonScattering Nov 16 '16 at 18:20
  • $\begingroup$ @ComptonScattering Yes and no. The experimental quantities (matrix elements of operators) do stay constant, but the operators that correspond to physical quantities also change. (It's also worth noting that nothing there is new to quantum mechanics, either - just about everything there is already present in the hamiltonian classical mechanics to begin with.) At present that's too nebulous to include in this answer, and it would only serve to further confuse the OP. $\endgroup$ – Emilio Pisanty Nov 16 '16 at 18:47
  • $\begingroup$ @EmilioPisanty How come the first section in your answer (Hydrogenic atom) does not contradict the Hohenberg-kohn Theorem? $\endgroup$ – proton May 19 at 16:07
  • $\begingroup$ @proton Why would it? The added operator is not a function of position (it includes position derivatives via the angular momentum operator). This answer is three years old and its claims can be directly verified without recourse to external theorems. If you have direct concerns about the claims here, then by all means express them, but if you have questions about the relationship to external features of the theory, ask a separate question. $\endgroup$ – Emilio Pisanty May 19 at 16:34
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Interpreting the question as "Can two different time independent Schrodinger equations have the same solution", or "... same ground state solution".

Then yes by construction. Let $| \psi_i \rangle$, $i = 1,2,3$ be three orthogonal wavefunctions.

We can construct two Hamiltonians

$$H_a = E_1 | \psi_1 \rangle \langle \psi_1 | + E_2 | \psi_2 \rangle \langle \psi_2 | + E_3 | \psi_3 \rangle \langle \psi_3 |$$

and

$$H_b = E_1 | \psi_1 \rangle \langle \psi_1 | + \frac{E_2}{2} \left( | \psi_2 \rangle \langle \psi_2 | + | \psi_3 \rangle \langle \psi_3 | + | \psi_3 \rangle \langle \psi_2 | + | \psi_2 \rangle \langle \psi_3 | \right) + \frac{E_3}{2} \left( | \psi_2 \rangle \langle \psi_2 | + | \psi_3 \rangle \langle \psi_3 | - | \psi_3 \rangle \langle \psi_2 | - | \psi_2 \rangle \langle \psi_3 | \right)$$

have the same ground state for $E_1<E_2,E_3,0$ (and the same spectra).

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  • $\begingroup$ If we don't care about locality of the Hamiltonian and treat it simply as a matrix on some abstract Hilbert space then your answer is correct. If we care about locality in some predetermined coordinate representation then your construction will be unsuccessful. $\endgroup$ – OON Nov 16 '16 at 13:59
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    $\begingroup$ Agreed. Though we can make other constructions if we have more stringent requirements. Generally any perturbation to the Hamiltonian that commutes with the ground state projector will produce a new Hamiltonian with the same ground state. $\endgroup$ – ComptonScattering Nov 16 '16 at 14:30
  • $\begingroup$ @OON No, it's pretty easy to cook similar examples in the position representation. Consider e.g. the hydrogenic hamiltonian $\hat H_0$, and its perturbation to $$\hat H = \hat H_0 +f(\hat r)\hat L^2,$$ where $f(r)\geq 0$ throughout, which can be quite different but which share the same ground state. $\endgroup$ – Emilio Pisanty Nov 16 '16 at 15:31
  • $\begingroup$ @EmilioPisanty I don't say that it's impossible, of course not. I say that if you want to preserve locality it's much less simple than what this answer suggests. $\endgroup$ – OON Nov 16 '16 at 15:47
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    $\begingroup$ @OON That's nowhere near what the question asked. $\endgroup$ – Emilio Pisanty Nov 16 '16 at 16:49

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