I have a Hamiltonian with the form

$$\hat{H} = A(x)\frac{\partial^2}{\partial x^2} + B(x)\frac{\partial}{\partial x} + C(x)$$

It can be found here (see equation 7). I want to use a finite-difference scheme (central difference), but it looks like my Hamiltonian will no longer be Hermitian. Specifically, if I expand the term $B(x)\frac{\partial}{\partial x}$ using a finite-difference scheme, I get matrix elements

$$h_{ij} = \delta_{i,j+1}B(i\Delta x)\frac{1}{\Delta x} - \delta_{i,j-1}B(i\Delta x)\frac{1}{\Delta x}$$

It looks like $h_{ij} = -h_{ji}$

If the middle term was not there, this wouldn't be an issue.

  • $\begingroup$ That term is equal to $ip$ where $p$ is the momentum. To fix it, just multiply by $i$. $\endgroup$ – knzhou Jan 17 '18 at 19:37
  • $\begingroup$ Can I simply multiply an individual term by $i$? According to the Hamiltonian in the above paper ( arxiv.org/abs/cond-mat/0403739 ), the function B(x) is real. $\endgroup$ – DJames Jan 17 '18 at 19:54
  • $\begingroup$ In case A, B, C depend of x the question about hermiticity is not reduced to multiplication by $i$. $\endgroup$ – Gec Jan 17 '18 at 20:00

For this Hamiltonian been hermitian some conditions should be fulfilled: $$ A^* = A $$ $$ 2A'^* - B^* = B $$ $$ A''^* - B'^* + C^* = C $$ Here $'$ means derivative and $^*$ means complex conjugate. In case when $A, B, C$ are actually independent of $x$, conditions are reduced to the $A$ and $C$ should be real and $B$ should be imaginary.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.