# Is this device Hamiltonian Hermitian?

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.

• That term is equal to $ip$ where $p$ is the momentum. To fix it, just multiply by $i$. – knzhou Jan 17 '18 at 19:37
• 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. – DJames Jan 17 '18 at 19:54
• In case A, B, C depend of x the question about hermiticity is not reduced to multiplication by $i$. – 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.