0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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”. $\endgroup$ – G. Smith Apr 7 at 19:01
  • $\begingroup$ @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? $\endgroup$ – James Apr 7 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.