Normally, a free particle Hamiltonian is written

$$ \hat{H} = - \frac{\hbar^2}{2m} \Delta $$

which is a differential operator because Laplacian $\Delta$ is.

On the other hand, in second quantization notation a Hamiltonian for free particle system is

$$ \hat{H} = \sum_j \varepsilon_j a^\dagger_j a_j $$

where $j$ are states with energies $\varepsilon_j$ and creation operator $a^\dagger_j$. It is not obvious if this is a differential operator.

Foundation: For example, the result known as Lippmann-Schwinger equation assumes in its derivation that the Hamiltonian is a differential operator so Schrödinger equation can be solved as a differential equation. Is it true in second quantization notation, as well?

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    $\begingroup$ What do you mean by differential operator? In the usual mathematical sense of the term, the second quantization operators are not differential operators. They are linear operators acting on a suitable Hilbert space, but not partial or pseudo differential operators. $\endgroup$
    – yuggib
    Commented Jul 27, 2015 at 12:36
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    $\begingroup$ Note that the derivation in the linked answer does not need to make any assumptions about whether $H$ is a differential operator. Exactly the same manipulations could be performed for a finite-dimensional matrix: it is just matrix inversion, after all. In the case of an infinite-dimensional matrix the "inverse" is more frequently called the Green's function, but the principle is the same. $\endgroup$ Commented Jul 27, 2015 at 14:36
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    $\begingroup$ Formally, $\hat{H}$ is not a differential operator but a linear operator on the (Hilbert) space of states. Technically one should always write (using the one-particle case as an example): $$ \langle x | \hat{H} | \psi \rangle \equiv -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \langle x | \psi \rangle \equiv -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) $$ Where $|x\rangle$ is a position eigenstate, $|\psi \rangle$ is the state the particle is in, and $\psi(x)$ is the corresponding wavefunction. That is, $\hat{H}$ is only the Laplacian in the position representation. This is so even in standard QM. $\endgroup$
    – gj255
    Commented Jul 27, 2015 at 14:56
  • $\begingroup$ @MarkMitchison I don't think so. The crucial thing there is expressing the solution as a sum of particular and homogenous solutions. If $H_0$ was simply a linear operator then "particular" solution would suffice. $\endgroup$
    – Minethlos
    Commented Jul 27, 2015 at 15:00
  • $\begingroup$ @Minethlos The homogeneous solution is just a zero eigenvector of the linear operator $(E - H_0)$. It is clear that one can always add such a zero eigenvector to the solution of $(E - H_0)\lvert \psi\rangle = V\lvert\psi\rangle$. So I don't see any need to restrict the conclusions to a statement about differential/infinite-dimensional operators. $\endgroup$ Commented Jul 27, 2015 at 19:05

3 Answers 3


The expression $$ \hat{H}=\sum_j \varepsilon_j\,a_j^\dagger a_j $$ is not the most general expression for free particles hamiltonian because it implies that you already found the eigenvalues $\varepsilon_j$ and diagonalized $\hat{H}$, i.e. already solved the Schrödinger equation.

Maybe you should look at the problem in a different basis. Let say $\{\vert j\rangle\}_j$ is a single particle state basis. Then you can chose to express $\hat{H}$ in another one particle basis, the position state basis $\{\vert x\rangle\}_x$ for instance.

The associated change of basis on the $a$ and $a^\dagger$ operators is perfomed with : $$ a(x)=\sum_j \langle x\vert j\rangle\,a_j\,. $$ Conversly, one can change back to the previous basis with : $$ a_j=\int\mathrm{d}x\,\langle x\vert j\rangle\,a(x) $$ Then you get the general expression of $\hat{H}$ in such basis : $$ \hat{H}=\int\mathrm{d}x\,a^\dagger(x)\left[\frac{\hat{p}^2}{2m}\right]a(x) $$ Such expression gives you back the fact that $\hat{H}$ is kind of a sum on one particle differential operators , provided that : $$ \hat{p}=-\mathrm{i}\hbar\partial_x $$


A 1-particle Hilbert space (neglecting spin for simplicity) is usually modelled as $L^2(\mathbb R^3)$ which is a function space, and the Hamiltonian is a differential operator. N-particle Hilbert spaces are usually constructed as tensor products of this 1-particle Hilbert space, but there exists an isomorphism such that you can again interpret them as function spaces. E.g. for 2 particles you have

$L^2(\mathbb R^3) \otimes L^2(\mathbb R^3) \simeq L^2(\mathbb R^3 \times \mathbb R^3)$.

So you can again interpret the Hamiltonian as a differential operator on these spaces.

The Fock space on which your second-quantized operators act is modelled as a direct sum of all N-particle Hilbert spaces. But unfortunately I don't know a simple way how this Fock space could be modelled as a function space. So my guess is: No, the Hamiltonian in second quantization is not a differential operator.

  • $\begingroup$ Does this result give the required interpretation? $\endgroup$
    – Minethlos
    Commented Jul 27, 2015 at 14:53

The second-quantization hamiltonian you wrote above is the hamiltonian for a collection of independent harmonic oscillators, where the quantities oscillating are now the amplitudes of the normal modes of the field being quantized. So, yes, the hamiltonian is a differential operator on the space complex functions of $N$ variables, one for each normal mode. Explicitly: $$ \hat{H}f(\phi_1,\ldots,\phi_i,\ldots,\phi_N) = \sum_j^N \varepsilon_j a^{\dagger}_j a_j f(\phi_1,\ldots,\phi_i,\ldots,\phi_N) = \sum_j^N \varepsilon_j(-\frac{d^2}{d\phi_j^2} + \phi_j^2 - \frac{1}{2}) f(\phi_1,\ldots,\phi_i,\ldots,\phi_N) $$ (in suitably chosen units) where $\phi_i$ is the amplitude of the $i^{th}$ normal mode.


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