# Is Hamiltonian a scalar or tensor in Quantum Mechanics?

According to Wikipeida, a scalar operator is invariant under rotations, and the Hamiltonian satisfies this definition. But at the same time, a Hamiltonian can be written as a matrix, which means it is a rank-2 tensor. Does it mean that a "scalar operator" may also be a tensor?

• It depends: rotations on what? The Hamiltonian is a scalar in physical space, a tensor in Hilbert space, and a component of a vector in spacetime. – Javier Jul 1 '19 at 1:32
• @Javier, I feel you should write that up as an answer not a comment. – KF Gauss Jul 1 '19 at 6:52

Here, we have (at least) three vector spaces at play: a Hilbert space $$\mathcal{H}$$, spacetime $$\mathbb{R}^4$$, and, given an observer, a subspace isomorphic to $$\mathbb{R}^3$$ representing space for that observer. Now, the Hamiltonian $$H$$ is first of all a linear operator $$\mathcal{H} \to \mathcal{H}$$, which makes it a (1,1) tensor. That is, given a basis $$|n\rangle$$ the Hamiltonian has components $$H_{mn}$$ (though we don't usually write it like this), with two indices.
Under rotations of the coordinates we use in 3D space, operators are changed as $$A \mapsto U^\dagger A U$$, with $$U$$ a unitary operator representing the rotation. And it so happens that $$H$$ doesn't change: $$U^\dagger H U = H$$. We say that $$H$$ is a scalar with respect to rotations. Some operators, like the components of momentum $$P_i$$, do change and mix among themselves, so it makes sense to pack them together as $$(P_x, P_y, P_z)$$ and say that they transform as a vector. And if we do Lorentz transformations, we find that $$H$$ does change, mixing with the $$P_i$$, so we pack them all together in a four-vector $$(H, P_x, P_y, P_z)$$, whose first component doesn't change if we only do rotations.
The TL;DR is that it depends on what vector space you're looking at, and what basis changes on that vector space. As another example, consider the wavefunction for a spin-1/2 particle, $$\psi = (\psi_1, \psi_2)^T$$. It has two components, each of which is an element of $$\mathcal{H} = L^2(\mathbb{R}^3)$$. So, is it a spinor or a state-vector? How does it transform under rotations: with the 2x2 spinor matrix, or with the operator $$U$$? The answer, obviously, is both. Or look at this answer of mine, where I discuss this same issue but regarding the adjoint of a vector.