# Discrete time evolution in a non-Euclidean space?

The time independent schrödinger equation can be written as

$$i\frac{\partial \psi}{ \partial t}=H\psi$$

if we consider the case of a 1D particle we can evolve it in time by discretising the Hamiltonian into

$$H=T+V$$

with $$V$$ some function of choice and the $$T$$ component as

$$T=-\frac{1}{2m}\frac{\partial^2}{\partial x^2}$$

By considering the action on $$\psi$$ we can approximate the matrix of $$T$$ by taylor expanding the wavefunction:

$$\psi(x+\Delta x)=\psi(x)+\Delta x\psi'(x)+\frac 1 2 \Delta x^2 \psi''(x)+\mathcal O(\Delta x^3)$$ $$\psi(x-\Delta x)=\psi(x)-\Delta x\psi'(x)+\frac 1 2 \Delta x^2 \psi''(x)+\mathcal O(\Delta x^3)$$ to get an approximation for the second derivative of the wavefucntion

$$\psi''(x)\approx\frac{1}{\Delta x^2}\left(\psi(x+\Delta x)+\psi(x-\Delta x )-2\psi(x)\right)$$

which allows us to write $$T$$ in matrix form as $$T=\frac{-1}{2m\Delta x^2}\left(\matrix{-2 &1&0&0&0\dots\\1&-2&1&0&0\dots\\0&1&-2&1&0\dots\\\dots}\right)$$

Now consider the case that the space is no longer Eucliedean, with different $$\Delta x_i$$ depending on the position $$i$$. What would the $$T$$ matrix look like in this case?

My thoughts are that when we taylor expand we need to write $$\Delta x'=\Delta x + \epsilon$$ such that

$$\psi(x+\Delta x)=\psi(x)+\Delta x\psi'(x)+\frac 1 2 \Delta x^2 \psi''(x)+\mathcal O(\Delta x^3)$$ $$\psi(x-\Delta x')=\psi(x)-\Delta x'\psi'(x)+\frac 1 2 \Delta x'^2 \psi''(x)+\mathcal O(\Delta x'^3)$$ but it seems this would only allow for constant changes in the $$\Delta x$$ values.

It also seems like we are bound to constant spacing by the fact that the Hamiltonian needs to be Hermitian, is there a way around this?

• In a non-Euclidean space, you would replace $\partial^2/\partial^2x$ in $T$ by the curved-space Laplacian, which involves the components of the metric for the space. The expression is given in en.wikipedia.org/wiki/Laplace_operator in the section “N dimensions”. – G. Smith Apr 7 at 19:01
• @G.Smith In the 1D case it would just recover $g(x)\partial^2/\partial x^2$ but then the Hamiltonian would not be Hermitian right? – James Apr 7 at 19:21