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Consider a $2 \times 2$ Hermitian (or symmetric) matrix-valued function $$g(x) = \{ g_{jk}(x)\}_{j,k=1,2}, \quad x \in \mathbb{R}^{2},$$ such that $0 < m_{-}I \leq g(x) \leq m_{+}I$, for some $m_{-}, m_{+} > 0$. Set $A = (A_{1}, A_{2}) := (-bx_{2}/2, bx_{1}/2)$, $b > 0$ and define the operator $$H = \sum_{j,k=1,2} (-i\partial/\partial x_{j} - A_{j}) g_{jk} (-i\partial/\partial x_{k} - A_{k}),$$ selfadjoint in $L^{2}(\mathbb{R}^{2})$ and essentially selfadjoint in $C_{0}^{\infty}(\mathbb{R}^{2})$.

If $g(x) = I$ for all $x \in \mathbb{R}^{2}$, this operator is called the Landau Hamiltonian and has the quantum mechanical interpretation of a spinless, charged particle under a constant magnetic field $b > 0$. I would like to understand further physical interpretations for the operator $H$ (in its general form).

Also, I understand that this operator can be given an interpretation in geometrical terms. Could you please explicit these interpretations and give some references to better understand them?

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I'm afraid I don't know any references off the top of my head, but let me at least give you a differential-geometric translation of sorts.

  1. If $g$ is real-valued, then it defines a Riemannian metric on $\mathbb{R}^2$.

  2. The vector field $A$ is a vector potential for the constant magnetic field $b$, in the sense that $\operatorname{curl}(A) \cdot \hat{\mathbf{k}} = b$. Even better, the magnetic potential $A$ can be interpreted as the connection $1$-form $$ A := -\frac{1}{2}bx^2\mathrm{d}x^1 + \frac{1}{2}bx^1\mathrm{d}x^2 $$ of the unitary connection $\nabla = d - iA$ on the (trivial) line bundle $\mathbb{R}^2 \times \mathbb{C}$ defined by $$ \nabla_{\frac{\partial}{\partial x^1}} = \frac{\partial}{\partial x^1} + i \frac{1}{2}bx^2, \quad \nabla_{\frac{\partial}{\partial x^2}} = \frac{\partial}{\partial x^2} - i \frac{1}{2}bx^1, $$ with curvature (viz, magnetic field strength) $$ B := \mathrm{d}A + A \wedge A = \mathrm{d}A = b \,\mathrm{d}x^1 \wedge \mathrm{d}x^2; $$ observe that sections of the line bundle $\mathbb{R}^2 \times \mathbb{C}$ are just complex-valued functions on $\mathbb{R}^2$. Parallel transport along this connection, then, is none other than magnetic translation.

  3. Given a complete Riemannian manifold $(X,g)$, a Hermitian vector bundle $E \to X$, and a unitary connection $\nabla$ on $E$, you get a canonical essentially self-adjoint second order elliptic differential operator called the Bochner Laplacian $\nabla^\ast \nabla$, which satisfies, as one would expect, $$ \nabla^\ast \nabla = \sum_{j,k} g^{jk}(x)\nabla_{\frac{\partial}{\partial x^j}}\nabla_{\frac{\partial}{\partial x^k}} + \text{lower order terms}. $$ If I'm not too mistaken, your magnetic Schroedinger operator $H$ should basically be the Bochner Laplacian for the connection $\nabla = d - iA$ on the trivial line bundle $\mathbb{R}^2 \times \mathbb{C}$ over $(\mathbb{R}^2,g)$.

I might have some signs and constants out of place here or there, but I think this should give at least the moral translation into differential-geometric terms. Lurking in the background is Weyl's famous characterisation of electromagnetism as a $U(1)$ gauge theory, i.e., in terms of a unitary connection (viz, electromagnetic potential) on a line bundle over spacetime, whose curvature therefore gives the electromagnetic field strength.

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  • $\begingroup$ Am I right to think that the form associated to the Bochner Laplacian (denoted $L$) is $\langle Lu,u \rangle = -\int_{\mathbb{R}^{2}} g^{jk} (\nabla u - iuA)\cdot(\nabla u - iuA) \mu dx$, in $L^{2}(\mathbb{R}^{2}, \mu dx)$, where $\mu = \sqrt{\det g}$? If that is the case, then the Bochner Laplacian seems to be unitarily equivalent to my operator plus a multiplication operator by $\frac{1}{4}|\nabla \ln \mu|_{g} + \frac{1}{2} \mathrm{div } g \nabla\ln\mu$. Do you know if this quantity has a geometrical interpretation? Does it have a definite sign? $\endgroup$
    – Geno Whirl
    Dec 2, 2014 at 18:35
  • $\begingroup$ Sorry, I mistyped the form, I meant: $\langle Lu,u \rangle = - \sum_{j,k=1,2}\int_{\mathbb{R}^{2}} g^{jk} (\partial u /\partial x_{k} - iA_{k}u) \overline{(\partial u /\partial x_{j} - iA_{j}u)} \mu dx$ $\endgroup$
    – Geno Whirl
    Dec 3, 2014 at 0:20
  • $\begingroup$ That is the correct form. From a purely mathematical perspective, your operator is a sort of truncation of the Bochner Laplacian, so that the difference, from one perspective, is just the price you pay for the truncation. On the other hand, if you really take $g$ seriously as a Riemannian metric, and if that difference really is a scalar function, then I guess, if nothing else, you can think of $H = \nabla^\ast \nabla + V$, where the difference $V$ is treated as a scalar potential? I must admit I'm only culturally semi-literate in these particular matters, so take this with a grain of salt. $\endgroup$ Dec 3, 2014 at 8:46
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your $\mathbf{g}$ is the position-dependent inverse effective mass tensor; if you calculate the classical equations of motion $$ {{\ddot{x}_1}\choose{\ddot{x}_2}}=2b\,\mathbf{g}(\mathbf{x})\cdot{{-\dot{x}_2}\choose{\dot{x}_1}} $$ then the semiclassical quantization condition is that the area inside a closed orbit must equal $(2\pi/b)(n-1/2)$, $n=1,2,3,\ldots$.

For a discussion in a three-dimensional setting, but without the position dependence of $\mathbf{g}$, see these lecture notes.

For a physical application to a semiconductor with position dependent effective mass, see Landau levels and cyclotron resonance in graded mixed semiconductors.

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  • $\begingroup$ Could you give me a precise physical example in which an effective mass tensor (with dependence on the position) could be applied? $\endgroup$
    – Geno Whirl
    Dec 2, 2014 at 16:58
  • $\begingroup$ I can't access the reference, could you help me out please? $\endgroup$
    – Geno Whirl
    Dec 3, 2014 at 0:15

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