Is it possible to "construct" the Hamiltonian of a system if its ground state wave function (or functional) is known? I understand one should not expect this to be generically true since the Hamiltonian contains more information (the full spectrum) than a single state vector. But are there any special cases where it's possible to obtain the Hamiltonian? Some examples would be really helpful.


4 Answers 4


IF you know that your Hamiltonian is of the form $$ \hat H=\frac{-\hbar^2}{2m}\nabla^2+V(\mathbf r) \tag 1 $$ for a single massive, spinless particle, then yes, you can reconstruct the potential and from it the Hamiltonian, up to a few constants, given any eigenstate. To be more specific, the ground state $\Psi_0(\mathbf r)$ obeys $$ \hat H\Psi_0(\mathbf r) =\frac{-\hbar^2}{2m}\nabla^2\Psi_0(\mathbf r)+V(\mathbf r)\Psi_0(\mathbf r) =E_0 \Psi_0(\mathbf r), $$ which means that if you know $\Psi_0(\mathbf r)$ then you can calculate its Laplacian to get $$ \frac{ \nabla^2 \Psi_0(\mathbf r) }{ \Psi_0(\mathbf r) } = \frac{2m}{\hbar^2}\left(V(\mathbf r)-E_0\right). $$ If you know the particle's mass, then you can recover $V(\mathbf r)-E_0$, and this is all you really need (since adding a constant to the Hamiltonian does not change the physics).

However, it's important to note that this procedure guarantees that your initial $\Psi_0$ will be an eigenstate of the resulting hamiltonian, but it does not preclude the possibility that $\hat H$ will admit a separate ground state with lower energy. As a very clear example of that, if $\Psi_0$ is a 1D function with a node, then (because 1D ground states have no nodes) you are guaranteed a unique $V(x)$ such that $\Psi_0$ is an eigenstate, but it will never be the ground state.

If you don't know that your Hamiltonian has that structure, there is (in the general case) no information at all that you can extract about the Hamiltonian from just the ground state.

  • As a simple example, without staying too far from our initial Hamiltonian in $(1)$, consider that Hamiltonian in polar coordinates, $$\hat H=\frac{-\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} + \frac{1}{\hbar ^2r^2}L^2\right)+V(r),$$ where I'm assuming $V(\mathbf r)=V(r)$ is spherically symmetric, and encapsulating the angular dependence into the total angular momentum operator $L^2$.

    Suppose, then, that I give you its ground state, and that it is an eigenstate of $L^2$ with eigenvalue zero (like e.g. the ground state of the hydrogenic Hamiltonian). How do you tell if the Hamiltonian that created it is $H$ or a similar version, $$\hat H{}'=\frac{-\hbar^2}{2m}\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} +V(r),$$ with no angular momentum component? Both versions will have $\Psi_0$ as a ground state (though here $\hat H'$ will have a wild degeneracy on every eigenspace, to be fair). Carrying on with this thought, what about $$\hat H{}''=\frac{-\hbar^2}{2m}\left(\frac{1}{r^2}\frac{\partial}{\partial r} r^2\frac{\partial}{\partial r} + f(r)L^2\right)+V(r),$$ where I've introduced an arbitrary real function $f(r)$ behind the angular momentum? This won't affect the $\ell=0$ states, but it will take the rest of the spectrum to who knows where. (In fact, you can even tack on an arbitrary function of $L_x$, $L_y$ and $L_z$, while you're at it.)

  • A bit more generally, any self-adjoint operator which vanishes on $\left|\Psi_0\right>$ can be added to the Hamiltonian to get you an operator that has $\left|\Psi_0\right>$ as an eigenstate. As a simple construction, given any self-adjoint operator $\hat A$, the combination $$\hat H {}''' = E_0 \left|\Psi_0\right>\left<\Psi_0\right| + \left(\mathbf 1 - \left|\Psi_0\right>\left<\Psi_0\right| \right) \hat A \left(\mathbf 1 - \left|\Psi_0\right>\left<\Psi_0\right| \right) $$ (where the factors in brackets are there to modify $\hat A$ into vanishing at $\left|\Psi_0\right>$ and its conjugate) will always have $\left|\Psi_0\right>$ as an eigenstate.

Even if you know all the eigenstates, it's still not enough information to reconstruct the Hamiltonian, because they do not allow you to distinguish between, say, $\hat H$ and $\hat{H}{}^2$. On the other hand, if you know all the eigenstates and their eigenvalues, then you can simply use the spectral decomposition to reconstruct the Hamiltonian.

In general, if you really insist, there is probably a trade-off between what you know about the Hamiltonian's structure (e.g. "of the form $\nabla^2+V$" versus no information at all) and how many of the eigenstates and eigenvalues you need to fully reconstruct it (a single pair versus the whole thing), particularly if you allow for approximate reconstructions. Depending on where you put one slider, you'll get a different reading on the other one.

However, unless you have a specific problem to solve (like reconstructing a Hamiltonian of vaguely known form from a specific set of finite experimental data) then it's definitely not worth it to explore the details of this continuum of trade-offs beyond the knowledge that it exists and the extremes I noted above.


Assume for simplicity that all the operators are bounded. If you know the wave function $\psi$ associated with the ground state of the unknown Hamiltonian $H$, then $H$ has the form $$H = E_0|\psi\rangle\langle\psi| \oplus K$$ where $K$ is another Hamiltonian defined on a subspace of the original Hilbert space of co-dimension 1, and $E_0$ is the energy of the ground state, which we might as well assume to be zero (in particular, the spectrum of $K$ is bounded below by $E_0$). This shows that in general it is not possible to reconstruct $H$, since there is an infinite family of Hamiltonians $H$ parametrised by strictly positive self-adjoint operators $K$ that will solve the original problem.

  • $\begingroup$ So, if I understand it correctly, you mean a particular wave function can be ground state to many different systems (Hamiltonians). $\endgroup$ Jul 23, 2016 at 17:50
  • $\begingroup$ Correct. A wave function is just a normalised vector in a Hilbert space. There are many operators on such Hilbert space that you can write that have this vector as an eigenvector. $\endgroup$
    – Phoenix87
    Jul 23, 2016 at 18:20
  • $\begingroup$ Do you know of any very special example(s) where one can fix the Hamilton? $\endgroup$ Jul 23, 2016 at 18:24

There's no analytic proof, but numerical evidence suggests that if you know that the Hamiltonian is local, and it satisfies the Eigenstate Thermalization Hypothesis (which most local Hamiltonians do), then you can extract the entire Hamiltonian from a single excited eigenstate, though not from the ground state: https://arxiv.org/abs/1503.00729.

  • $\begingroup$ Can you clarify what "local" means in this context? (Given the theme of my answer and Vladimir, you could take it to be a hamiltonian such that $⟨x_0|H|\psi⟩$ is uniquely determined by the values of $⟨x|\psi⟩$ at $x=x_0$ and possibly a neighbourhood around it, or even restricted to $⟨x_0|\psi⟩$ and a finite number of derivatives. However, if you're talking about the ETH and therefore about many-body hamiltonians, it could mean an entirely different thing.) $\endgroup$ Jul 23, 2016 at 23:51
  • $\begingroup$ @EmilioPisanty In this context a many-body Hamiltonian defined on a graph of discrete sites is defined to be "local" if it can be written as a sum of terms with finite spatial extent. In others words, in the form $\hat{H} = \sum_i \hat{O}_i$, where $i$ runs over the sites and each operator $\hat{O}_i$ is supported only on sites that are within a distance $r$ of site $i$, where the "radius" $r$ is an integer that does not depend on the total system size. $\endgroup$
    – tparker
    Jul 24, 2016 at 1:03
  • $\begingroup$ @EmilioPisanty So for instance the $\hat{O}_i$ might consist of first-, second-, and third-nearest-neighbor couplings. The maximum number of neighbors away that are coupled can be arbitrarily large, but it must be finite, so that in the infinite-system limit each coupling only stretches an infinitesimal way across the entire system. Then if you "zoom out" enough, the couplings can be made to appear arbitrarily short-range. $\endgroup$
    – tparker
    Jul 24, 2016 at 1:07
  • $\begingroup$ Yeah, I figured that would be the case. Given the level of the OP and the broad (HNQ) audience of this post, it's worthwhile including some of that explanation and context in the post. $\endgroup$ Jul 24, 2016 at 1:45

If unknown part of Hamiltonian is potential $V({\bf{r}})$, then you can write a stationary Schrodinger equation and figure out what the potential should be.

  • $\begingroup$ Sorry, I didn't follow your suggestion. Can you expand your answer a bit? $\endgroup$ Jul 23, 2016 at 18:10
  • $\begingroup$ You have $\psi$, you differentiate it $d^2/d{\bf{r}}^2$, you write $E\psi$ next to it, and you must obtain $V\psi$. Are you familiar with the Schrodinger equation? $\endgroup$ Jul 23, 2016 at 18:39
  • $\begingroup$ Yes, I am familiar with it. Actually I thought along the similar line but then got confused because I thought Hamiltonian has more info than a single state. That's why I am looking for particular "class" of Hamiltonians where this reasoning works. $\endgroup$ Jul 23, 2016 at 18:52

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