# Cayley's expansion

Is Cayley's expansion $$\exp(-iH\delta t) \psi(x,t)=\frac{1-\frac{i\delta t}{2}H}{1+\frac{i\delta t}{2}H}\psi(x,t)$$ valid for any operator $H$? What conditions should $H$ fulfill?

• What does dividing by an operator even mean? Also, this is not well-known as "Cayley's expansion", and googling returns no unambiguous hints where you might have gotten this from. Additionally, this is currently a pure math question. – ACuriousMind Feb 19 '15 at 15:38
• It appears in the Crank-Nicholson method and I do not now why sometimes this is called "Cayley's expansion". – J.J. Feb 19 '15 at 15:58
• Not seeing anything like this on the Wiki page. Also, even if it appears there, dividing by an operator is still undefined. – ACuriousMind Feb 19 '15 at 16:00
• @ACuriousMind: This post might elucidate the topic. – Kyle Kanos Feb 19 '15 at 16:10
• It seems to me that OP is missing the LHS that says, $\psi(x,t+\delta t)=\exp(-iH\delta t)\psi(x,t)$. You can then clear up the "division by operator" issue. – Kyle Kanos Feb 19 '15 at 16:12

The expansion formally works for any operator. It's breaking the exponential as $$\exp(-x)=\frac{\exp(-x/2)}{\exp(x/2)}$$ and then expanding the numerator and denominator as $e^x\approx1+x$.
However, since the exponential term in the Cayley expansion comes from the time-evolution of a wave-function: $$\psi(x,t+\delta t)=e^{-iH\delta t}\psi(x,t)$$ which itself comes from integrating the Schroedinger equation (and assuming that $H$ is independent of time): $$H\psi=i\frac{\partial\psi}{\partial t}$$ I suspect that the expansion is only going to be found with the Hamiltonian operator, rather than any generic operator.